The Solar System Design Disproves Newton Theory (Revised No.2)

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
I do this research
Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
The Solar System Design Disproves Newton Theory (Revised No.2)
The Author Authorized To Be Used By
Mr. Gerges Francis Tawdrous
A Student–Physics Department- Physics
& Mathematics Faculty –
Peoples' Friendship University of Russia
(RUDN University) – Moscow – Russia
Dr. Budochkina, Svetlana Aleksandrovna
Associate Professor (Mathematical Analysis
and Theory of Functions Department)
Peoples' Friendship University of Russia
(RUDN University) – Moscow – Russia
Phone +201022532292
E-Mail: mrwaheid@gmail.com
Curriculum Vitae http://vixra.org/abs/1902.0044
Phone +7 (495) 952-35-83
E-Mail: budochkina-sa@rudn.ru, sbudotchkina@yandex.ru
Website
http://web-local.rudn.ru/web-local/prep/rj/index.php?id=2944&p=19024
The Assumption Of S. Virgin Mary -Written in Cairo –Egypt –28th
September 2021
Abstract
Paper hypothesis
The Solar Planets Revolutions Around The Sun Be Done In Equal Periods of Time
Paper hypothesis Explanation
(I)
- The Solar Planets move as gears in one machine- each planet motion depends on
the other planets motions.
- The solar planets motions be integrated with one another to create one unified
integrated motion.
- The unified integrated motion causes the planets revolutions around the sun to be
done in equal periods of time, to save the timing adjustments of the planets
motions to be integrated with one another.
- The equal periods causes the planets motions to use different rates of time to adjust
the planets orbital distances increasing and their velocities decreasing
- The equal periods of time and the using of the different rates of time causes the
solar system distances to be created in a network form
- The distances network means, the distances be created based on one geometrical
design and be distributed as a network.
- This vision attributes 3 new features to the solar planets motions which are:
- (1) The Planets Move An Integrated Unified Motion
- (2) The Planets Motions Use Different Rates Of Time

I do this research
2nd
2
- (3) The Solar System Distances Be Created In A Network Form
- By that, the solar group be similar to a great mechanical clock, each planet works
as a gear moves depending on the other planets motions and defines a part of the
solar system general rate of time (the general rate of time is the solar day 86400s)
(II)
- Mars was the next planet after Mercury with original orbital distance 84 mkm
- Pluto was the Mercury moon revolving around it
- The Earth Moon and Saturn weren't created in that time
- Also the sun wasn't created in that time. The Sun is the last created piece in the
solar system
- Then
- Mars And Pluto had migrated from their original orbital distances to their current
positions.
- The solar system distances network was effected greatly by the planets migration.
- The solar system network design tried to repair the negative effects of the planets
migration.
- As a result, the solar system network forced the outer and inner planets motions
periods of time to use Mars and Pluto rate of time as their basic rate.
- That put Mars and Pluto as 2 heads for the solar system motion.
(III)
- The Equal periods theory disproves Newton Theory Of The Sun Mass Gravity, for
many basic reasons (1st
The sun is the last created piece in the solar system) and
(2nd
the planets motions be integrated disproving planet motion independency
concept) and (3rd
the solar planets motion be transported or caused in a series from
one planet to another in sequence series but Newton vision told all planets connect
with the sun by the mass gravity)

I do this research
2nd
3
Paper Objectives
- (1st
Objective) The paper proves, The Solar Planets Revolve Around The Sun In
Equal Periods Of Time
- (2nd
Objective) The Paper Provides The Solar System Creation Theory.
Please scan the figure (ORCID)

I do this research
2nd
4
Paper Contents
Subject Page No.
1- Introduction 6
2- Methodology 10
Paper Part No. (I) The Equal Periods Proof 13
3- Can The Solar Planets Motions Use Different Rates Of Time? 14
3-1 Preface 15
3-2 The Rates Of Time Using Proof Idea 16
3-3 The Planets Rates of time Definition 22
4- The Solar Planets Rates Of Time Analysis 26
4-1 Preface 27
4-2 Venus Motion Rate of time 28
4-3 Earth Motion Rate of time 30
4-4 Mars Motion Rate of time 32
4-5 Jupiter Motion Rate of time 34
4-6 Saturn Motion Rate of time 36
4-7 Uranus Motion Rate of time 38
4-8 Neptune Motion Rate of time 40
4-9 Pluto Motion Rate of time 42
4-10 The General Discussion 44
5- The Planets Orbital Distances Depend On Their Rates Of Time 46
5-1 Preface 47
5-2 The Planets Orbital Distances Test 48
5-3 Mars, Jupiter and Saturn Motions Analysis 54
5-4 One Equation controls The planets orbital periods and distances 60
5-5 Why Saturn And The Moon Use Equal Rates Of Time? 72
5-6 Why Mercury Use A Double Of Its Orbital Distance? 78
5-7 Pluto And Saturn Motions Rate of Time 80
5-8 The Rate (4.61) be used between Pluto and the moon motion 86
5-9 The Moon Orbital Motion Equation 90
6- Mars Migration Theory 106
6-1 Mars Migration Theory 107
6-2 Pluto Migration Theory 110
6-3 Planets Migration Theories Proves 112
7- Planets Unified General Motion 115
7-1 The Planets Unified General Motion Description 116
7-2 The Solar Planets Motions Are Complementary One Another 118
7-3 The Solar Planets Move An Unified Motion (A Team Motion) 122
8- Mercury Motion Distance During Its Day Period 127
8-1 Preface 128
8-2 Why Mercury Jupiter Distance =720.7 mkm? 129

I do this research
2nd
5
8-3 Jupiter Motion In A Comparison With Light Motion 133
9- Jupiter orbital Circumference Analysis 138
9-1 Preface 139
9-2 Jupiter orbital Circumference Analysis 140
10- The Sun Age Description 154
10-1 The Sun Circles The Earth 155
10-2 The Rate (1.0725) 156
10-3 The Sun Diameter Analysis 160
10-4 The Sun Rays Analysis 163
Paper Part No. (II) The Solar System Creation Theory 168
11- The Solar System Creation Theory 169
11-1 Preface 170
11-2 The Solar System Creation Theory 172
11-3 The Inner Planets Motions Reason 177
11-4 Light Beam (1.16 mkm/sec) Proves Discussion 181
11-5 Saturn Motion More Analysis 218
11-6 Equal Distances Method 225
12-The Solar System Distances Be Created In A Network Form 231
12-1 Preface 232
12-2 The Continuum effect Through the Solar System Distances 234
12-3 The Solar System Distances Distribution 239
12-4 The Solar System Distances Dependency On One Another 243
Appendix No.1 The Solar System Equal Distances List 245
References and Biography 247

I do this research
2nd
6
1-Introdcution
- Many discovered new features be attributed to the planet motion…
- The question is, why all these features weren't discovered before? Because in fact,
the features give us a different vision completely from Newton Vision about the
solar system motion?
- The question is valid, it asks "Why We Are So Far From The Truth"?
(I)
- The solar system is designed by some accurate geometrical design, the secret of its
study is to use the planets motions data.
- If someone studied any ancient building will find this fact. the ancient building
geometrical design doesn't depend on its outer shape, to be as a pyramid or a cube.
But depends on its dimensions. For that reason when we just look at some ancient
building we learn nothing. The geometrical design be found in the dimensions.
- The basic mistake was the removing of the planets motions data from the theory
analysis and to accept the theories regardless the planets data.
- By that, while Newton gave us the gravitation equation, Jupiter be in order no. (5)
from the sun, when we asked why Jupiter isn't the most near planet to the sun
because of it's greatest mass, the answer be because of (the initial conditions!),
regardless the (initial conditions) claim be a fact or not. in all cases, the planets
order don't support Newton theory and his gravitation equation. By that the theory
be built without support from the planets motions data.
- Titus bode law is an example for this problem, the law be refused because it
couldn't predict Neptune orbital distance but it predicted all other planets. The
point is that, the people who refused the law didn't know why the law be used by 8
planets and why Neptune only didn’t follow it, they refused and removed the law
without understanding for its error. This example is useful because it shows the
unfairness of treatment with the planets motions data. no one search behind the
data to know how this data be created

I do this research
2nd
7
(II)
- The moon orbital motion can explain this idea as clear as possible, let's summarize
it in following
- The moon displacement per a solar = 88000 km
- The moon day period =29.53 solar day
- By that, the distances total which be passed by the moon in its day period should
be =(88000 km x 29.53 days = 2598640 km = 2π x 413600 km)
- Based on this data, The Moon Apogee Radius Should Be = 413600 km
- But it's not a fact
- The moon apogee radius = 406000 km
- Means,
- The moon can't be far from the Earth at a distance more than 406000 km, and the
moon never visit the point 413600 km
- This data shows clearly we can't understand the moon motion trajectory by the
observation only but we need to analyze the motion data to know what's going on.
- Again
- How the moon apogee radius be (406000 km)? it should be (413600 km?)
- Notice
- There's one more feature should be found as a result because the distance
(2598640 km) is so long the moon would have to revolve around Earth through its
(new) apogee orbit
- Shortly
- Based on this data, the moon would revolve around Earth along month through the
apogee orbit whose radius (r=413600 km) and would be prevented to revolve
around Earth through any near orbit!
- The Fact Is That,
- The moon creates an angle (θ) between the moon displacement direction and the
moon orbit horizontal level, by that, while the moon moves daily 88000 km but

I do this research
2nd
8
this distance doesn't be accounted for the moon orbit. The orbit considers another
displacement depend on the angle, let's call it (the real displacement L)
- (The Real Displacement) L = 88000 km cos (θ)
- As a result
- The real displacement be shorter than (88000 km) and the total displacements
during (29.53 days) be less than (2598640 km) and be (2550973 km) (-2%)
- By this intelligent technique, the genius moon can revolve around Earth through
more near orbit and avoid to visit the point 413600 km forever
(III)
- The moon orbital motion is a good example to explain the solar system motion,
because the observation doesn’t give all details, we need to analyze the data to
discover in more clear the planets motions …
- For example, One new feature tells, The Planets Move Their Revolutions
Around The Sun In Equal Periods!
- How can that be possible? If the planets orbital distances be greater and their
velocities be lower, how the periods be equal?
- The planets use different rates of time,
- By that, 1 hour of Mercury Motion = 24.6 hours of Jupiter motion. the different
rates of time compensates the 2 other features (increasing distances and decreasing
velocities)
- I want to say,
- We can't understand the solar system motion by the observation only, we need to
analyze the planets motions data to know how this data be created. this is the
method by which the solar system geometrical design can be discovered and by
that we can protect ourselves from the imaginary ideas- as Newton's!-
- The point is that, the sun is the last created piece in the solar system and all planets
were created and moving before the sun creation. Based on that, Newton theory of
the sun mass gravity is absolute imaginary one.

I do this research
2nd
9
- Let's take a look on the paper contents
- Paper Part no. (I)
- The Planets Revolution Equal Periods be proved in 3 points (no.3, 4 and 5), in
these 3 points, we proves strongly that the solar planets use different rates of time
and these rates of time define their orbital periods and distances
- Mars Migration Theory be proved in point no. (6)
- The Planets Unified Motion be proved in point no. (7)
- Jupiter And Mercury Motions Be Analyzed in points no. (8 and 9) to be detailed
example of proof for the planets unified motion.
- The sun is created after all planets, this theory be proved in point (no.10)
- Paper Part no. (II)
- The solar system creation theory be proved in point (no.11)
- The solar system distances network theory be proved in point (no.12)

I do this research
2nd
10
2-Methodology
- I use the planets motions data analysis in comparison with the theories which try to
explain the planets motions- let's use one example to explain how this method
works?
- Example No. 1
- Newton told us– The planet motion be independent from all other planets motions
- This is the theory – let's see some planets motions data for a comparison
- Data
(1)
Saturn Orbital Distance = Saturn Uranus Distance
= Mars Orbital Circumference
= Pluto Neptune Distance (error 1.5%)
= Pluto eccentricity Distance (error 1.5%)
= Neptune Orbital Distance/π
= Uranus Orbital Distance /2
(2)
Mercury Neptune Distance = Saturn Pluto Distance
Jupiter Pluto Distance = Uranus Neptune Circumference
Earth Neptune Distance = Mercury Saturn Circumference (0.5%)
(3)
Jupiter Mercury Distance = 2 Mercury Orbital Circumference
Jupiter Venus Distance = Venus Orbital Circumference (1.5%)
Jupiter Earth Distance = Earth Orbital Circumference (1.2%)
(Earth and Jupiter at 2 different sides from the sun)
(4)
Jupiter Mercury Distance = Mars Orbital Distance x π (0.6%)
Jupiter Uranus Distance = Venus Jupiter Circumference (0.8%)
Pluto Orbital Distance = Earth Orbital Circumference x 2π
Why These Distances Are Equal?

I do this research
2nd
11
- The solar system has around (45 distances) and these distance are around (25
distances) why they are equal one another? Can Planets Independent Motions
Cause These Distances Equality?
- In some desert someone sees a mirage and thought it's a water! How can we prove
to him that this is a mirage and Not a water?
- Newton told that, Planets Motions be Independent From One Another, and as a
result - The solar system distances be created based on each other in a Network
form where more than 50% of all distances are equal one another!!
- I try to show that –Newton Theory mistake is some how clear.
- The planets creations and motions data should be similar to a creature genes which
can tell us how this planet be created and moving – so whatsoever the theory be
the planets motions data be more important than it – and by that – even if Newton
was correct, there's no reason to neglect the planets motions.
- Let's use one more example
- Example No. 2
- 778.6 mkm = 1.0725 x 720.7 mkm (Mercury Jupiter Distance)
- 720.7 mkm = 1.0725 x 671 mkm (Venus Jupiter Distance)
- 671 mkm = 1.0725 x 629 mkm (Earth Jupiter Distance)
- 629 mkm = (1.0725)2
x 550.7 mkm (Mars Jupiter Distance)
- 550.7 mkm = (1.0725) x 511 mkm
- 4900 mkm = (1.0725) x 4495.1 mkm (Neptune Orbital Distance) (1.6%)
- 1375 mkm = (1.0725) x 1284 mkm (Earth Saturn Distance)
- 4900 mkm = Jupiter orbital circumference
- 1375mkm = Mercury Saturn Circumference
- The previous distances are 13 distances (30%) of all distances- they are example of
many others- the question is why they are rated with the same one rate (1.0725)?

I do this research
2nd
12
- I try to show, some geometrical design effects through the planets motions data.
This effect is stronger than any planet motion and force all of them to use the same
one rate (1.0725)– No Independent Motion Be Found Here – the data simply
disproves Newton Theory Of The Sun Mass Gravity and Planet Independency
Motion.
- The previous data is example for wide range of data – I add some more of them
here to show the strong force range of using this rate (1.0725)
- (28.3 deg / 26.7 deg) =(26.7 deg /25.2 deg) = (25.2 deg /23.4 deg) = (122.5 /115.2)
= (1.0725)
- Where
- 28.3 degrees = Neptune Axial Tilt
- 26.7 degrees = Saturn Axial Tilt
- 25.2 degrees = Mars Axial Tilt
- 23.4 degrees = Earth Axial Tilt
- 115.2 degrees = 90 degrees + 25.2 degrees (Mars Axial Tilt)
- 122.5 degrees = Pluto Axial Tilt.
- Shortly
- The planets data analysis put the data in comparison with the theories to test the
theory sufficiency.

I do this research
2nd
13
Paper Part No. (I)
The Solar Planets Revolutions Around The Sun Be
Done In Equal Periods of Time

I do this research
2nd
14
3- Can The Solar Planets Motions Use Different Rates Of Time?
3-1 Preface
3-2 The Rates Of Time Using Proof Idea
3-3 The Planets Rates of time Definition

I do this research
2nd
15
3-1 Preface
- The paper supposes that, The Solar Planets Use Different Rates Of Time
- The Rates of time is a meaning we have learnt from special theory of relativity,
and it means, for example, (1 h of Mercury motion =24.6 h of Jupiter motion), that
means, (While the clock moves 1 hour in Mercury the other clock moves 24.6
hours in Jupiter).
- We accept this meaning, and based on it, the paper supposes, Planets Motions Use
Different Rates Of Time.
- Now
- Can That Be A Fact?
- I have a proof for this idea, and the proof is useful to improve our understanding
for the solar planets motions data and its meaning.
- But
- The proof can prove that, the solar planets motions use different rates of time, but
can't explain why or how to do that. that means, I have a proof for a discovery but
how to explain this discovery I can't answer. The proof which I have provides the
fact but no a theoretical explanation be provided neither geometrical mechanism to
explain how these rates of time be created.
- Let's write the proof idea in the next point as clear as possible before to discuss it.

I do this research
2nd
16
3-2 The Proof Of The Rates Of Time Using.
- Let's summarize the proof idea in following…
(I)
- The solar planets rates of time be defined for any planet based on the period this
planet needs to move a distance = this planet orbital distance.
- Let's provide these periods for better explanation:
- Mercury needs a period =14.13 solar days to move a distance =57.9 mkm =
Mercury Orbital Distance (So the required period is 14.13 Solar Days)
- Venus needs a period =35.8 solar days to move a distance =108.2 mkm = Venus
Orbital Distance (So the required period is 35.8 Solar Days)
- Earth needs a period =58.1 solar days to move a distance =149.6 mkm = Earth
- The Earth moon a period =1700 solar days to move a distance =149.6 mkm =
Earth Orbital Distance by using its daily displacement 88000 km (So the required
period is 1700 Solar Days)
- Mars needs a period =109.4 solar days to move a distance =227.9 mkm =Mars
- Jupiter needs a period =687 solar days to move a distance =778.6 mkm = Jupiter
Orbital Distance (So the required period is 687 Solar Days)
- Saturn needs a period =1710 solar days to move a distance =1433.5 mkm = Saturn
- Uranus needs a period =4890 solar days to move a distance =2872.5 mkm =
Uranus Orbital Distance (So the required period is 4890 Solar Days)
- Neptune needs a period =9635 solar days to move a distance =4495.1 mkm =
Neptune Orbital Distance (So the required period is 9635 Solar Days)
- Pluto needs a period =14547 solar days to move a distance =5906 mkm =Pluto

I do this research
2nd
17
(II)
- The Planet Motion Rate of Time be defined depends on these periods.
- For example
- 58.1 days (Earth period) = 1.622 x 35.8 days (Venus Period)
- Means
- 1 hour of Venus Motion = 1.622 hours of Earth Motion
- This is the rate of time we search for. And by this method we define it clearly.
- Then
- This rate of time does 2 basic Job which are
- (1st
Job)
- The rate of time defines the planets orbital periods, because of that
- 365.25 days (Earth Orbital Period) = 1.622 x 224.7 days (Venus Orbital Period)
- This rule is used for all planets orbital periods, as we will prove that later in our
discussion.
- (2nd
Job)
- The rate of time defines the planets orbital distances, let's do that in following…
- 1 day of Venus Motion = 1.622 days of Earth Motion (this is the rate of time)
- But
- Venus during one solar day moves a distance =3.024 mkm
- Earth during 1.622 solar days moves a distance =4.176 mkm
- And
- (4.176 mkm /3.024 mkm) = (940 mkm /680 mkm)
- Where
- 940 mkm = Earth Orbital Circumference
- 680 mkm = Venus Orbital Circumference
- As the data shows, the rate of time defines the planets orbital circumferences.

I do this research
2nd
18
(III)
- Now the idea be clear
- The periods of planets motions for their orbital distances define the rates of time be
used by these planets motions..
- And
- These rates of time define all planets orbital periods and distances
- This is a good understanding for the planets data
- But
- What do we need?
- We need to know if the planets motions required periods be defined independently
or dependently on one another…
- Now
- We have 2 different visions as result for these 2 options. Let's explain that in
following as clear as possible
- (1st
Option)
- If the planets required periods be defined independently for each planet, that
means, each planet motion can be independent from one another. because the
periods are independent periods from one another
- (2nd
Option)
- If the planets periods be defined dependently on one another, that will make the
solar group as one machine of gears.
- The solar planets in this case will be similar to carriages in one train which move
together one unified motion. or the planets be similar to gears in one machine of
gears. They move one unified motion.
- Or Better Example,
- the solar planets be similar to a mechanical clock, each planet is a gear in the
mechanical clock and causes to create a rate of time as a part of the machine
general rate of time.

I do this research
2nd
19
- I want to say that,
- The second option of the periods dependency on one another isn’t so simple option
but it's a very effective option and even its effect be understood by any student of
astrophysics, because
- The planets orbital periods and distances be defined based on the motions rates of
time which are defined based on these required periods the planets need to pass
their orbital distances- So If these periods be created based on one another – that
makes all planets orbital periods and distances be defined based on one another.
- Shortly, that means,
- We don't need to observe to know any planet orbital period or distance, if any data
of it be absent, we can simply conclude it by using the rates of time – the second
option simply changes the solar planets data into data be defined by one Equation
only and by that we can remove all planets data (periods and distances) and can
conclude them correctly by using the rates of time
- Shortly
- If the planets required periods be defined based on one law and not independently
from one another the whole vision of the solar system will be changed by a very
strong proof
- To make this discussion more interesting, I want to say that,
- There's one law controls all planets required periods of time to move their orbital
periods. And this one law disproves and refuted completely Newton Theory of the
sun mass gravity and Planet Motion Independency Concept which be inherited
form Newton theory.
- We simply have a chance to create a progress in the solar system motion
understanding.
- So, Let's Summarize Our Discussion Steps In Following

I do this research
2nd
20
(IV)
- We Should Follow The Following Steps In Our Discussion
- (1st
Step)
- The Planets Motions Rates Of Time Definition
- And
- The Planets Orbital Periods Definition Based On The Rates Of Time
- (2nd
Step)
- The Planets Orbital Distances Definition Based On The Rates Of Time
- (3rd
Step)
- We discuss one law controls the planets required periods for the solar planets
orbital distances motions.
(V)
- We should notice that
- The previous steps are steps for data providing, that means, these steps are
methods to arrange the data
- Because
- I prove that, the planets motions use different rates of time. In this proof discussion
I provide the data in some arrangement to show that this data be understandable
only based on the hypothesis of the planets motions different rates of time.
- By that, these steps are a method of Data arrangement.
- Means
- We don't discuss the question (how the planets motion use different rate of time?)
neither the question (why the planets motions use different rates of time?) because,
we try to prove the discovery which is (the planets motions use different rates of
time), and to prove that, I put the planets data in the best arrangement form by
using these 3 steps to show that, this data can't be understandable except by using
the hypothesis (the planets motions use different rates of time).
- So, our process aims basically to prove this fact

I do this research
2nd
21
- Then
- We should analyze the planets required periods for their orbital distances motions
to prove that one law controls all periods and by that we can change the solar
system current description. for that reason the periods should be analyzed in a
separated dedicated point.
- I want to say
- We have to provide the data which prove the hypothesis (the planets motions use
different rates of time)
- After the proof discussion be completed we have to start the solar system
description discussion because one great effect be done as a result for the
hypothesis proof and the one law proof.
- Let's explain how we arrange the planets data in following:
- In next point (3-3) we provide Mercury Rates of time as an example and use it to
define all planets orbital periods based on Mercury Day Period
- Mercury is an example, and we should do the same job with each planet, for that
we dedicate the next point (NO. 4) for all planets rates of time test. Based on that,
- Mercury is our example for test and
- All planets periods be tested as Mercury rate in (Point No. 4)
- Then
- In point no. (5) we prove that, all planets orbital distances be defined based on the
planets motions rates of time.
- And in this point we prove that one law controls all periods

I do this research
2nd
22
3-3 The Planets Rates Of Time Definition
(I)
The Required Periods For The Solar Planets Orbital Distances Motions Are
- Mercury needs 14.13 Solar Days (27.9 Solar Days)
- Venus needs 35.8 Solar Days
- Earth needs 58.1 Solar Days
- The moon needs 1700 Solar Days
- Mars needs 109.4 Solar Days
- Jupiter needs 687 Solar Days
- Saturn needs 1710 Solar Days
- Uranus needs 4890 Solar Days
- Neptune needs 9635 Solar Days
- Pluto needs 14547 Solar Days
- The rates between these period define the rates of time used by each planet in
comparison with other – as we will explain in following
- But
- For Mercury we use the period (27.9 solar days) = (2 x 14.13 days) (error 1%) ,
because Mercury uses its Day Period and not its orbital period and also uses the
distance 720.7 mkm and not Mercury orbital circumference 360 mkm. This Data
Mercury uses because of some geometrical necessity we should explain in a point
no (8-2) in which we analyze the period of time Mercury needs to pass a distance =
its orbital distance.
- For now we use for Mercury the period (27.9 Solar Days) and we explain the
reason in this point (8-2).

I do this research
2nd
23
(II)
- Mercury Motions Rates Of Time
- (35.8 /27.9) = 1.279, means
- 1 hour of Mercury Motion = 1.28 hours of Venus Motion
- (58.1 /27.9) = 2.078, means
- 1 hour of Mercury Motion = 2.078 hours of Earth Motion
- (1700 /27.9) = 60.8, means
- 1 hour of Mercury Motion = 60.8 hours of The Moon Motion
- (109.4 /27.9) = 3.91, means
- 1 hour of Mercury Motion = 3.91 hours of Mars Motion
- (687 /27.9) = 24.6, means
- 1 hour of Mercury Motion = 24.6 hours of Jupiter Motion
- (1710 /27.9) = 61.1, means
- 1 hour of Mercury Motion = 3.91 hours of Saturn Motion
- (4890 /27.9) = 174.8, means
- 1 hour of Mercury Motion = 174.8 hours of Uranus Motion
- (9635 /27.9) = 344.8, means
- 1 hour of Mercury Motion = 344.8 hours of Neptune Motion
- (14547 /27.9) = 520.2, means
- 1 hour of Mercury Motion = 520.2 hours of Pluto Motion

I do this research
2nd
24
MERCURY MOTION RATES OF TIME LIST
1 hour of Mercury motion = 1.279 h of Venus Motion
1 hour of Mercury motion = 2.078 h of Earth Motion
1 hour of Mercury motion = 3.91 h of Mars Motion
1 hour of Mercury motion = 24.6 h of Jupiter Motion
1 hour of Mercury motion = 61.1 h of Saturn Motion
1 hour of Mercury motion = 174.8 h of Uranus Motion
1 hour of Mercury motion = 344.6 h of Neptune Motion
1 hour of Mercury motion = 520.2 h of Pluto Motion
1 hour of Mercury motion = 60.8 h of The Earth Moon Motion
- Notice
- The list tells, Saturn and the Moon use equal rate of time!! Why? we should
answer that in point no. (5-5) of this paper
- Now
- Can these rates of time define the planets orbital periods? Let's do that in
following:

I do this research
2nd
25
Mercury Rates Of Time Test
224.7 days (Venus orbital period) = 1.279 x 175.94 days (Mercury Day Period)
365.25 days (Earth orbital period) = 2.078 x 175.94 days (Mercury Day Period)
687 days (Mars orbital period) = 3.91 x 175.94 days (Mercury Day Period)
4331 days (Jupiter orbital period) = 24.6 x 175.94 days (Mercury Day Period)
10747 days (Saturn orbital period) = 61.1 x 175.94 days (Mercury Day Period)
30589 days (Uranus orbital period) = 174.8 x 175.94 days (Mercury Day Period)
59800 days (Neptune orbital period) = 344.6 x 175.94 days (Mercury Day Period)
90560 days (Pluto orbital period) = 520.2 x 175.94 days (Mercury Day Period)
- The errors are 1.4% and 1% with Neptune and Pluto orbital periods respectively,
no other error more than 1%
- The data tells, Mercury rate of time defines all planets orbital periods based on
Mercury Day Period.
- The data tells that, the planets orbital period be created in proportionality with one
another.
- We have to test all planets rates of time as Mercury to know if all periods be
definition correctly similar to Mercury test. For that reason we will test all planets
periods of time in the next point (No.4)

I do this research
2nd
26
4- The Solar Planets Rates Of Time Analysis
4-1 Preface
4-2 Venus Motion Rate of time
4-3 Earth Motion Rate of time
4-4 Mars Motion Rate of time
4-5 Jupiter Motion Rate of time
4-6 Saturn Motion Rate of time
4-7 Uranus Motion Rate of time
4-8 Neptune Motion Rate of time
4-9 Pluto Motion Rate of time
4-10 The General Discussion

I do this research
2nd
27
4-1 Preface
- In this point we test the planets motions rates of time definition as we have done
with mercury in the previous point. We bring Mercury test here in this preface to
be a reference for the other planets test.
MERCURY LIST
224.7 days (Venus orbital period) = 1.279 x 175.94 days (Mercury Day Period)
365.25 days (Earth orbital period) = 2.078 x 175.94 days (Mercury Day Period)
687 days (Mars orbital period) = 3.91 x 175.94 days (Mercury Day Period)
4331 days (Jupiter orbital period) = 24.6 x 175.94 days (Mercury Day Period)
10747 days (Saturn orbital period) = 61.1 x 175.94 days (Mercury Day Period)
30589 days (Uranus orbital period) = 174.8 x 175.94 days (Mercury Day Period)
59800 days (Neptune orbital period) = 344.6 x 175.94 days (Mercury Day Period)
90560 days (Pluto orbital period) = 520.2 x 175.94 days (Mercury Day Period)
Errors are 1.4% 1% with Neptune and Pluto respectively
- Each point uses a similar test with a planet of the solar group in order
- In point no. (4-10) we provide the general discussion of all data.

I do this research
2nd
28
4-2 Venus Motion Rate Of Time
I – Data
VENUS MOTION LIST
1 hour of Venus motion = (0.78) h of Mercury Motion
1 hour of Venus motion = 1.62 h of Earth Motion
1 hour of Venus motion = 3.057 h of Mars Motion
1 hour of Venus motion = 19.233 h of Jupiter Motion
1 hour of Venus motion = 47.8 h of Saturn Motion
1 hour of Venus motion = 136.7 h of Uranus Motion
1 hour of Venus motion = 269.4 h of Neptune Motion
1 hour of Venus motion = 406.7 h of Pluto Motion
1 hour of Venus motion = 47.54 h of The Moon Motion
- In Following We Provide The Data Analysis

I do this research
2nd
29
II–Data Analysis
175.94 days (Mercury day period) = 0.78 x 224.7 days (Venus orbital period)
365.25 days (Earth orbital period) = 1.62 x 224.7 days (Venus orbital period)
687 days (Mars orbital period) = 3.057 x 224.7 days (Venus orbital period)
4331 days (Jupiter orbital period) = 19.2 x 224.7 days (Venus orbital period)
10747 days (Saturn orbital period) = 47.8 x 224.7 days (Venus orbital period)
30589 days (Uranus orbital period) = 136.7 x 224.7 days (Venus orbital period)
59800 days (Neptune orbital period) = 266.2 x 224.7 days (Venus orbital period)
90560 days (Pluto orbital period) = 403 x 224.7 days (Venus orbital period)
- There Are Errors (1%) Only With Neptune And Pluto Orbital Periods
- Shortly
- All planets orbital periods be created based on Venus orbital period (224.7 days)
by using the rates of time between Venus and these planets –
- No error in all data more than 1% -
- The Data Proves That – The Rate of Time Controls The Solar System motion

I do this research
2nd
30
4-3 Earth Motion Rate Of Time
I – Data
EARTH MOTION LIST
1 hour of Earth motion = (0.48) h of Mercury Motion
1 hour of Earth motion = (0.617) h of Venus Motion
1 hour of Earth motion = 1.88 h of Mars Motion
1 hour of Earth motion = 11.8 h of Jupiter Motion
1 hour of Earth motion = 29.4 h of Saturn Motion
1 hour of Earth motion = 84.1 h of Uranus Motion
1 hour of Earth motion = 165.8 h of Neptune Motion
1 hour of Earth motion = 250.4 h of Pluto Motion
1 hour of Earth motion = 29.3 h of The Moon Motion

I do this research
2nd
31
II–Data Analysis
175.94 days (Mercury day period) = 0.48 x 365.25 days (Earth orbital period)
224.7 days (Venus orbital period) = 0.617 x 365.25 days (Earth orbital period)
687 days (Mars orbital period ) = 1.88 x 365.25 days (Earth orbital period)
4331 days (Jupiter orbital period) = 11.8 x 365.25 days (Earth orbital period)
10747 days (Saturn orbital period) = 29.4 x 365.25 days (Earth orbital period)
30589 days (Uranus orbital period) = 84.1 x 365.25 days (Earth orbital period)
59800 days (Neptune orbital period) = 163.8 x 365.25 days (Earth orbital period)
90560 days (Pluto orbital period) = 248 x 365.25 days (Earth orbital period)
- Errors (1.2% and 1%) With Neptune And Pluto Orbital Periods Respectively
- Shortly
- All planets orbital periods be created based on Earth orbital period (365.25 days)
by using the rates of time between Earth and these planets –
- No other error in all data more than 1% -

I do this research
2nd
32
4-4 Mars Motion Rate Of Time
I – Data
MARS MOTION LIST
1 hour of Mars motion = (0.255) h of Mercury Motion
1 hour of Mars motion = (0.327) h of Venus Motion
1 hour of Mars motion = (0.531) h of Earth Motion
1 hour of Mars motion = 6.3 h of Jupiter Motion
1 hour of Mars motion = 15.6 h of Saturn Motion
1 hour of Mars motion = 44.71 h of Uranus Motion
1 hour of Mars motion = 88.1 h of Neptune Motion
1 hour of Mars motion = 133 h of Pluto Motion
1 hour of Mars motion = 15.55 h of The Moon Motion

I do this research
2nd
33
II–Data Analysis
175.94 days (Mercury day period) = 0.255 x 687 days (Mars Orbital Period)
224.7 days (Venus orbital period) = 0.327 x 687 days (Mars Orbital Period)
365.25 days (Earth orbital period = 0.531 x 687 days (Mars Orbital Period)
4331 days (Jupiter orbital period) = 6.3 x 687 days (Mars Orbital Period)
10747 days (Saturn orbital period) = 15.6 x 687 days (Mars Orbital Period)
30589 days (Uranus orbital period) = 44.7 x 687 days (Mars Orbital Period)
59800 days (Neptune orbital period) = 88.1 x 687 days (Mars Orbital Period)
90560 days (Pluto orbital period) = 133 x 687 days (Mars Orbital Period)
- Errors (1.2% and 1%) With Neptune And Pluto Orbital Periods Respectively
- Shortly
- All planets orbital periods be created based on Mars orbital period (687 days) by
using the rates of time between Mars and these planets

I do this research
2nd
34
4-5 Jupiter Motion Rate Of Time
I – Data
JUPITER MOTION LIST
1 hour of Jupiter motion = (0.040) h of Mercury Motion
1 hour of Jupiter motion = (0.0419) h of Venus Motion
1 hour of Jupiter motion = (0.084) h of Earth Motion
1 hour of Jupiter motion = (0.1589) h of Mars Motion
1 hour of Jupiter motion = 2.48 h of Saturn Motion
1 hour of Jupiter motion = 7.1 h of Uranus Motion
1 hour of Jupiter motion = 14 h of Neptune Motion
1 hour of Jupiter motion = 21.1 h of Pluto Motion
1 hour of Jupiter motion = 2.47 h of The Moon Motion

I do this research
2nd
35
II–Data Analysis
175.94 days (Mercury day period) = 0.040 x 4331 days (Jupiter Orbital Period)
224.7 days (Venus orbital period) = 0.0419 x 4331 days (Jupiter Orbital Period)
365.25 days (Earth orbital period = 0.084 x 4331 days (Jupiter Orbital Period)
687 days (Mars orbital period) = 0.1589 x 4331 days (Jupiter Orbital Period)
10747 days (Saturn orbital period) = 2.48 x 4331 days (Jupiter Orbital Period)
30589 days (Uranus orbital period) = 7.1 x 4331 days (Jupiter Orbital Period)
59800 days (Neptune orbital period) = 14 x 4331 days (Jupiter Orbital Period)
90560 days (Pluto orbital period) = 21.1 x 4331 days (Jupiter Orbital Period)
- Errors (1.4% and 1%) With Neptune And Pluto Periods respectively
- All planets orbital periods be created based on Jupiter orbital period (4331days)
by using the rates of time between Jupiter and these planets

I do this research
2nd
36
4-6 Saturn Motion Rate Of Time
I – Data
SATURN MOTION LIST
1 hour of Saturn motion = (0.016) h of Mercury Motion
1 hour of Saturn motion = (0.0209) h of Venus Motion
1 hour of Saturn motion = (0.0339) h of Earth Motion
1 hour of Saturn motion = (0.0639) h of Mars Motion
1 hour of Saturn motion = (0.403) h of Jupiter Motion
1 hour of Saturn motion = 2.857 h of Uranus Motion
1 hour of Saturn motion = 5.63 h of Neptune Motion
1 hour of Saturn motion = 8.5 h of Pluto Motion
1 hour of Saturn motion = 1 h of The Moon Motion

I do this research
2nd
37
II–Data Analysis
175.94 days (Mercury day period) = 0.040 x 10747 days (Saturn Orbital Period)
224.7 days (Venus orbital period) = 0.0209 x 10747 days (Saturn Orbital Period)
365.25 days (Earth orbital period = 0.0339 x 10747 days (Saturn Orbital Period)
687 days (Mars orbital period) = 0.0639 x 10747 days (Saturn Orbital Period)
4331 days (Jupiter orbital period) = 0.403 x 10747 days (Saturn Orbital Period)
30589 days (Uranus orbital period) = 2.857 x 10747 days (Saturn Orbital Period)
59800 days (Neptune orbital period) = 5.63 x 10747 days (Saturn Orbital Period)
90560 days (Pluto orbital period) = 8.5 x 10747 days (Saturn Orbital Period)
- Errors (1%) Only With Neptune And Pluto Orbital Periods
- Shortly
- All planets orbital periods be created based on Saturn orbital period (10747 days)
by using the rates of time between Saturn and these planets

I do this research
2nd
38
4-7 Uranus Motion Rate Of Time
I – Data
URANUS MOTION LIST
1 hour of Uranus motion = (0.0057) h of Mercury Motion
1 hour of Uranus motion = (0.0073) h of Venus Motion
1 hour of Uranus motion = (0.011) h of Earth Motion
1 hour of Uranus motion = (0.022) h of Mars Motion
1 hour of Uranus motion = (0.14) h of Jupiter Motion
1 hour of Uranus motion = (0.3499) h of Saturn Motion
1 hour of Uranus motion = 1.97 h of Neptune Motion
1 hour of Uranus motion = 2.97 h of Pluto Motion
1 hour of Uranus motion = (0.349) h of The Moon Motion

I do this research
2nd
39
II–Data Analysis
175.94 days (Mercury day period) = 0.0057 x 30589 days ( Uranus Orbital Period)
224.7 days (Venus orbital period) = 0.0073 x 30589 days ( Uranus Orbital Period)
365.25 days (Earth orbital period = 0.011 x 30589 days ( Uranus Orbital Period)
687 days (Mars orbital period) = 0.022 x 30589 days ( Uranus Orbital Period)
4331 days (Jupiter orbital period) = 0.14 x 30589 days ( Uranus Orbital Period)
10747 days (Saturn orbital period) = 0.3499 x 30589 days (Uranus Orbital Period)
59800 days (Neptune orbital period) = 1.97 x 30589 days ( Uranus Orbital Period)
90560 days (Pluto orbital period) = 2.97 x 30589 days ( Uranus Orbital Period)
- No Error In All Data More Than 1% -
- All planets orbital periods be created based on Uranus orbital period (30589 days)
by using the rates of time between Uranus and these planets

I do this research
2nd
40
4-8 Neptune Motion Rate Of Time
I – Data
NEPTUNE MOTION LIST
1 hour of Neptune motion = (0.0029) h of Mercury Motion
1 hour of Neptune motion = (0.0037) h of Venus Motion
1 hour of Neptune motion = (0.006) h of Earth Motion
1 hour of Neptune motion = (0.011) h of Mars Motion
1 hour of Neptune motion = (0.0713) h of Jupiter Motion
1 hour of Neptune motion = (0.177) h of Saturn Motion
1 hour of Neptune motion = (0.507) h of Uranus Motion
1 hour of Neptune motion = 1.509 h of Pluto Motion
1 hour of Neptune motion = (0.1773) h of The Moon Motion

I do this research
2nd
41
II–Data Analysis
175.94 days (Mercury day period) = 0.0029 x 59800 days (Neptune Orbital Period)
224.7 days (Venus orbital period) = 0.0037 x 59800 days (Neptune Orbital Period)
365.25 days (Earth orbital period = 0.006 x 59800 days (Neptune Orbital Period)
687 days (Mars orbital period) = 0.011 x 59800 days (Neptune Orbital Period)
4331 days (Jupiter orbital period) = 0.0713 x 59800 days (Neptune Orbital Period)
10747 days (Saturn orbital period) = 0.177 x 59800 days (Neptune Orbital Period)
30589 days (Uranus orbital period) = 0.507 x 59800 days (Neptune Orbital Period)
90560 days (Pluto orbital period) = 1.509 x 59800 days (Neptune Orbital Period)
- Neptune orbital period creates an error 1.3% with all data except Pluto and
Uranus
- All planets orbital periods be created based on Neptune orbital period (59800 days)
by using the rates of time between Neptune and these planets

I do this research
2nd
42
4-9 Pluto Motion Rate of time
I – Data
PLUTO MOTION LIST
1 hour of Pluto motion = (0.00192) h of Mercury Motion
1 hour of Pluto motion = (0.00245) h of Venus Motion
1 hour of Pluto motion = (0.00399) h of Earth Motion
1 hour of Pluto motion = (0.0075) h of Mars Motion
1 hour of Pluto motion = (0.0472) h of Jupiter Motion
1 hour of Pluto motion = (0.1174) h of Saturn Motion
1 hour of Pluto motion = (0.336) h of Uranus Motion
1 hour of Pluto motion = (0.662) h of Neptune Motion
1 hour of Pluto motion = (0.11743) h of The Moon Motion

I do this research
2nd
43
II–Data Analysis
175.94 days (Mercury day period) = 0.00192 x 90560 days (Pluto Orbital Period)
224.7 days (Venus orbital period) = 0.00245 x 90560 days (Pluto Orbital Period)
365.25 days (Earth orbital period = 0.00399 x 90560 days (Pluto Orbital Period)
687 days (Mars orbital period) = 0.0075 x 90560 days (Pluto Orbital Period)
4331 days (Jupiter orbital period) = 0.0472 x 90560 days (Pluto Orbital Period)
10747 days (Saturn orbital period) = 0.1174 x 90560 days (Pluto Orbital Period)
30589 days (Uranus orbital period) = 0.336 x 90560 days (Pluto Orbital Period)
59800 days (Neptune orbital period) = 0.662 x 90560 days (Pluto Orbital Period)
- Pluto orbital period creates error 1% with all data.
- No error In All Data More Than 1% -
- All planets orbital periods be created based on Pluto orbital period (90560 days)
by using the rates of time between Pluto and these planets –
The Data Proves That – The Rate of Time Controls The Solar System motion

I do this research
2nd
44
4-10 The General Discussion
- Shortly
- The data shows that
- All Planets Orbital Periods are rated with one another by the rates of time we
have calculated based the required periods for the solar planets orbital distances
motions.
- Why?
- If the solar planets motions are independent from one another, how all planets
orbital periods be defined based on these rates?
- In the next point (No.5) we prove that, the solar planets orbital distances be created
depending on these rates of time also. How that can be possible?
- Why all planets orbital periods and distances be defined based on these rates of
time?
- BUT
- We should keep an eye on the required periods because they are the source based
on which the rates of time be created.
- And
- The question should be,
- Are these required periods be defined independently from one another? or all these
periods be defined based on one another by using one law only?
- I want to say,
- We have to know perfectly the answer of this question, because if there's one law
or one Equation by which the planets required periods can be defined that
summarizes all data of the solar planets orbital periods and distances into one
Equation only
- In this case we can use the one equation in place of all planets orbital periods and
distances, and as a result no need for any observation for this data because we can
conclude them theoretically

I do this research
2nd
45
- That's why we need to move step by step in our discussion, and proves each step
strongly because it will result one equation is place of ten planets orbital periods
and distances.
- After we do that, we have to discuss the reason behind.
- Now let's move to the next point to prove,
- The Planets Orbital Distances Be Defined Based On These Rates Of Time.
- NOTICE
- If we prove that, the solar planets distances be based on these rates of time, the
solar group description should be changed into the following one…
- (Solar Group Is)
- The Solar Group is similar to a machine of gears each planet is a gear in it, or
- The Solar Group is similar to one building and each planet is a part of this same
building, or
- The Solar Group is similar to one body and each planet is a member of it, or
- The Solar Group is similar to puppets theater, all puppets are connected together
by one force and causes their motions, or
- The Solar Group is similar a canal, the water moves through it and causes the
rotation of 9 waterwheels. The Motion Direction of The waterwheels is different
from the water motion direction but no waterwheel can rotate without the water
motion, or
- The solar planets are similar to train carriages, they move together one unified
general motion, or
- The solar planets be similar to knots on one robe or similar to points on the same
one trajectory of energy –
- The solar planets motions be similar to chess board pieces motions, each motion
depend on the other planets motions and be done by geometrical calculations.
- Clearly
- The Planets Move One Unified Motion Be Done In Equal Periods Of Time

I do this research
2nd
46
5- The Planets Orbital Distances Depend On Their Rates Of Time
5-1 Preface
5-2 The Planets Orbital Distances Test
5-3 Mars, Jupiter and Saturn Motions Analysis
5-4 One Equation controls The planets orbital periods and distances
5-5 Why Saturn And The Moon Use Equal Rates Of Time?
5-6 Why Mercury Use A Double Of Its Orbital Distance?
5-7 Pluto And Saturn Motions Rate of Time
5-8 The Rate (4.61) be used between Pluto and the moon motion
5-9 The Moon Orbital Motion Equation

I do this research
2nd
47
5-1 Preface
- In This Point We Have Defined Tasks Are:
- We prove, the solar planets distances be created based on their rates of time (Point
No. 5-2)
- We analyze Mars, Jupiter and Mercury Motions to explain how their motions data
be created (for example we answer, Why Saturn Orbital Distance = 2π x Mars
Orbital Distance) (Point No. 5-3)
- We prove, There's One Equation controls all planets orbital periods and distances
(Point No. 5-4)
- Then,
- We try to answer why Saturn rate of time equal the Earth moon rate of time and
what effect of this equality on both planets motions (Point No. 5-5).
- And
- We ask why Mercury use double values (Point No.5-6)
- Then
- We analyze Saturn and Pluto Motions Rate Of Time in (Point No.5-7)
- Then
- We analyze The basic rate between Pluto and the moon motion data (4.61), this
analysis is done (Point No. 5-8)
- The moon orbital motion equation we discuss (Point No. 5-9)

I do this research
2nd
48
5-2 The Planets Orbital Distances Test
- Let's refer to what we will do here
- We remember here Mercury rates of time with the other planets
- Then
- We suppose that, Mercury moves during one solar day
- And
- We will find the other planets periods of time in comparison with this 1 solar day
of Mercury motion based on these planets rates of time with Mercury
- Then
- We will discover the distances the planets pass during these periods
- Then
- We will compare these distances with Mercury motion distance for 1 solar day
=4.095 mkm
- Based on this comparison, we should test if the planets orbital distances be defined
based on these planets rates of time,
- Let's Do That In Following…

I do this research
2nd
49
I- Data
Mercury Rates Of Time
The Motions Distances Based On Mercury Rates Of Time
(1)
Mercury (4.095 mkm / solar day) moves during 1 day a distance =4.095 mkm
(2)
Venus (3.024 mkm / solar day) moves during 1.279 days a distance =3.867 mkm
(3)
Earth (2.574 mkm / solar day) moves during 2.078 days a distance = 5.35 mkm
(4)
Mars (2.082 mkm / solar day) moves during 3.91 days a distance = 8.14 mkm
(5)
Jupiter (1.13184 mkm / solar day) moves during 24.6 days a distance = 27.8 mkm
(6)
Saturn (0.838 mkm / solar day) moves during 61.1 days a distance = 51.2 mkm
(7)
Uranus (0.5875 mkm / solar day) moves during 174.8 days a distance = 102.7 mkm
(8)
Neptune (0.4665 mkm / solar day) moves during 344.8 days a distance = 160.8 mkm
(9)
Pluto (0.406 mkm / solar day) moves during 520.2 days a distance = 211.2 mkm

I do this research
2nd
50
II- Data Analysis
No. (2)
(Venus motion distance=3.867 mkm) /(Mercury Motion Distance 4.095 mkm) =0.944
Where
(Venus orbital circumference =680 mkm) /(Mercury Distance 720.7 mkm) =0.944
No. (3)
(Earth motion distance=5.35 mkm) /(Mercury Motion Distance 4.095 mkm) =1.3
Where
(Earth orbital circumference =940 mkm) /(Mercury Distance 720.7 mkm) = 1.3
No. (4)
(Mars motion distance=8.14 mkm) /(Mercury Motion Distance 4.095 mkm) = 1.98
Where
(Mars orbital circumference =1433 mkm) /(Mercury Distance 720.7 mkm) = 1.98
No. (5)
(Jupiter motion distance=27.8 mkm) /(Mercury Motion Distance 4.095 mkm) = 6.8
Where
(Jupiter orbital circumference =4900 mkm) /(Mercury Distance 720.7 mkm) = 6.8
No. (6)
(Saturn motion distance=51.2 mkm) /(Mercury Motion Distance 4.095 mkm) = 12.5
Where
(Saturn orbital circumference =9010 mkm) /(Mercury Distance 720.7 mkm) = 12.5
No. (7)
(Uranus motion distance =102.7 mkm) /(Mercury Motion Distance 4.095 mkm) =25.1
Where
(Uranus orbital circumference =18048 mkm) /(Mercury Distance 720.7 mkm) = 25.1

I do this research
2nd
51
No. (8)
(Neptune motion distance =161 mkm) /(Mercury Motion Distance 4.095 mkm) =39.3
Where
(Neptune orbital circumference =28255 mkm) /(Mercury Distance 720.7 mkm) = 39.3
No. (9)
(Pluto motion distance =211.2 mkm) /(Mercury Motion Distance 4.095 mkm) =51.6
Where
(Pluto orbital circumference =37100 mkm) /(Mercury Distance 720.7 mkm) = 51.6
- Let's discuss the previous data in following:

I do this research
2nd
52
III- Discussion
(A)
- The data proves that
- The solar planets orbital circumferences are defined depending on these planets
motions rates of time.
- Based on that,
- THE PLANETS MOTIONS RATES OF TIME DEFINE THESE PLANETS
ORBITAL PERIODS AND DISTANCES.
- i.e.
- A huge number of data be controlled by a small group of data.
- Means,
- The solar planets periods and distances are defined as functions in the planets
motions rates of time.
- But
- The planets motions rates of time be defined based on The Required Periods For
The Planets Orbital Distances Motions.
- Shortly
- It tells, if we know the required periods we can calculate all solar planets orbital
distances and periods.
- Now, the question is,
- Are these required periods be defined based on one another or independently from
one another?
- The question = All Planets Orbital Distances And Periods
- Because
- If the required periods be defined based on one another and we can use One
Equation to calculate these required periods – that will summarize all solar planets
periods and distances into One Equation.
- That shows the importance behind this discussion.

I do this research
2nd
53
- Because
- One Equation controls all planets orbital periods and distances
(B)
- Usually The Planets Motions Data shows different puzzles,
- Let's remember some of them, (Saturn Orbital Distance = 2π x Mars Orbital
Distance),and (Uranus Orbital Distance = 2 Saturn Orbital Distance) Also
(Neptune Orbital Distance = π x Saturn Orbital Distance).
- All these rates can be concluded from the previous distances, shows that, these
rates be results of the planets motions rates of time.
- I want to say, many planets motions data can be understood and explained by this
analysis of the passed distances.
- The planets motions rates of time give us a chance for better understanding for
how the planets motions data be created.
- By that we will have more clear vision about the planets creation process.
- In next point (5-3) we analyze mars, Jupiter and Saturn Motions data to explain
how their data be created by effect of their motions rates of time and by that, we
understand how these planets motions data be created, which solves many old
unanswered questions.
- Notice
- Mercury uses the distance 720.7 mkm in place of its orbital circumference 360
mkm, and uses its day period 175.95 solar days in place of its orbital period (88
days) and uses the required period 671 hours to move a distance = 2 x 57.2 mkm
(Mercury orbital distance) (error 1%)
- Shortly Mercury uses a double value of its data in comparison with the other
planets. Although if Mercury use the ordinary data as other planets it works by
Mercury uses double value because of a basic geometrical reason we should
discuss in point no. (5-6).

I do this research
2nd
54
5-3 Mars, Jupiter And Saturn Motions Analysis
- In this point we use the planets motions rates of time to explain the puzzled data of
Mars, Jupiter and Saturn Motions
- Let's take a look on this puzzled data in following:
- 4331 days (Jupiter Orbital Period) =2π x 687 days (Mars Orbital Period)
- 1433 mkm (Saturn Orbital Distance) =2π x227.9 mkm (Mars Orbital Distance)
- 10747 days (Jupiter Orbital Period) =2.5 x 4331 mkm (Jupiter Orbital Period)
- 24.1 km/s (Mars Velocity) =2.5 x 9.7 km/s (Saturn Velocity)
- 24.6 h (Mars rotation period) =2.5 x 9.9 h (Jupiter Day Period)
- ((10747 days x 24 h) /24.6 h) = ((4331 days x 24 h) /9.9h) = 10485
- 3.1 deg (Jupiter Axial Tilt) x 2 = (2.5 deg = Saturn orbital inclination)2
- Let's use the planets motions rates of time to help us to answer how the data be
created…

I do this research
2nd
55
I - Data
(5-3-a)
- 1 hour of Mars Motion = 6.3 hours of Jupiter Motion = 15.56 hours of Saturn
Motion = 44.6 hours of Uranus Motion = 88 hours of Neptune motion = 132.7
hours of Pluto motion
(5-3-b)
- 4331 days (Jupiter Orbital Period) =2π x 687 days (Mars Orbital Period)
- 1433 mkm (Saturn Orbital Distance) =2π x227.9 mkm (Mars Orbital Distance)
- 10747 days (Jupiter Orbital Period) =2.5 x 4331 mkm (Jupiter Orbital Period)
- 24.1 km/s (Mars Velocity) =2.5 x 9.7 km/s (Saturn Velocity)
- 24.6 h (Mars rotation period) =2.5 x 9.9 h (Jupiter Day Period)
- ((10747 days x 24 h) /24.6 h) = ((4331 days x 24 h) /9.9h) = 10485

I do this research
2nd
56
II –Discussion
Equation no. (5-3-a)
- 1 hour of Mars Motion = 6.3 hours of Jupiter Motion = 15.56 hours of Saturn
Motion = 44.6 hours of Uranus Motion = 88 hours of Neptune motion = 132.7
hours of Pluto motion
- These rates of time we have concluded in point no. (4) of this paper.
- i.e.
- 1 Solar Day of Mars Motion = 6.3 Solar Days of Jupiter Motion = 15.56 Solar
Days of Saturn Motion = 44.6 Solar Days of Uranus Motion = 88 Solar Days
of Neptune motion = 132.7 Solar Days of Pluto motion.
- Let's define the distances which these planets pass during these periods of time
- Mars (2.08 mkm per solar day) moves during 1 solar day a distance = 2.082 mkm
- Jupiter (1.1318 mkm per day) moves during 6.3 solar days a distance = 7.13 mkm
- Saturn (0.838 mkm per day) moves during 15.56 solar days a distance =13.1 mkm
- Uranus (0.5875 mkm per day) moves during 44.6 days a distance=26.22 mkm
- Neptune (0.466560 mkm per day) moves during 88 days a distance = 41.05 mkm
- Pluto (0.406 mkm per day) moves during 132.7 solar days a distance =53.88 mkm
- Based on that
- Mars during (1 solar day) moves 2.082 mkm
- Jupiter during (6.3 solar day) moves 7.13 mkm
- Saturn during (15.5 solar day) moves 13.1 mkm
- Uranus during (44.6 solar day) moves 26.22 mkm
- Neptune during (88 solar day) moves 41.05 mkm
- Pluto during (132.7 solar day) moves 53.88 mkm

I do this research
2nd
57
Data Analysis
(1)
- Saturn motion distance = 13.1 mkm = Mars Motion Distance 2.08 mkm x 2π
- This data explains why
- Saturn orbital circumference 9007 mkm = Mars orbital circumference 1433
mkm x 2π
(2)
- Uranus motion distance = 26.22 mkm = Saturn Motion Distance 13.1 mkm x 2
- Uranus orbital circumference 18048 mkm = Saturn orbital circumference
9007 mkm x 2
(3)
- Neptune motion distance =41.05 mkm =Saturn Motion Distance 13.1 mkm x π
- Neptune orbital circumference 28255 mkm = Saturn orbital circumference
9007 mkm x π
(4)
- Pluto motion distance =53.9 mkm =Saturn Motion Distance 13.1 mkm x 4.11
- Pluto orbital circumference 37100 mkm = Saturn orbital circumference 9007
mkm x 4.11
(5)
- Pluto motion distance =53.9 mkm =Uranus Motion Distance 26.22 mkm x 2.06
- Pluto orbital circumference 37100 mkm= Uranus orbital circumference 18048
mkm x 2.06

I do this research
2nd
58
Discussion (Continued)
- Now we know that,
- Saturn moves a distance =13.1 mkm = 2π x Mars Distance 2.082 mkm
- Simply that rate (2π) is created here because of the motions rates of time among
the planets.
- We have explained one data
- Based on that, the rate (2π) isn't be created by any pure coincidence of numbers
but because,
- 1 day of Mars motion = 6.3 days of Jupiter motion = 15.56 days of Saturn motion
- These rates of time effect on the planets motions distances
- The period 15.56 days enables Saturn to move 13.1 mkm = (2π) x Mars Motion
Distance per a solar day (2.082 mkm).
- By that we explain clearly how these planets data be created.
- The result is important because, it tells we can explain all planets motions data
origin.
- Now
- We explained why Saturn orbital distance = (2π) Mars orbital distance
- Let’s try with another data
- Why Mars Velocity =(2.5) Saturn Velocity?
- Let's analyze the data no (4 and 5) to explain this one
(5)
- Pluto orbital circumference 37100 mkm= Uranus orbital circumference 18048
mkm x 2.06 = Saturn orbital circumference 9007 mkm x 4.1
- Pluto is a common point between the 2 planets (Saturn and Uranus), and the rates
are (4.1 and 2.06), but we know that, Mercury motion distance per a solar day
=4.095 mkm and Mars motion distance per a solar day =2.08 mkm (error 1%)
- The rates can be used as these planets velocities…

I do this research
2nd
59
- That tells Mercury Velocity = 2 x Mars velocity (approximately), because Uranus
orbital distance = 2 x Saturn orbital distance – that connects the solar system and
makes it as one machine.
- But
- Even if we accept that, the velocities are Mars and Mercury, where we search for
Mars and Saturn velocities rate,
- BUT
- Mercury velocity 47.4 km/s x 2 = (Saturn velocity 9.7 km/s)2
- That makes Saturn velocity as a function in Mercury velocity, the point is we have
seen this type of equation before let's remember it
- I want to say,
- It's a geometrical method used by Saturn with Jupiter and with Mercury
- Please Notice
- By discovery of the planets motions rates of time we enable to explain the rate
(2π) between Saturn and Mars orbital distances. And now we have a trustee
method to explain this rate (2π).
- But, there are many other geometrical methods needed to be discovered to explain
the other data because of that the form (Data)2
is a form we need to discover its
geometrical mechanism. But the data isn't created by pure coincidence of numbers
on the contrary base on a geometrical mechanism need to be discovered.
- Notice
- The planet motion has its own effect on its data for example (26.7 =2.5 x10.7)
(where 10.7 h = Saturn day period, 2.5 deg = Saturn orbital inclination, 26.7 deg =
Saturn Axial Tilt) There's a geometrical machine behind this data but we still need
to analyze Saturn motion data in more deep analysis.
- Also 687 days (Mars Orbital Period) + 2 x 3.1 = 2 x 346.6 days (the nodal year)
(where 3.1 deg = Jupiter Axial Tilt)

I do this research
2nd
60
5-4 One Equation controls The planets orbital periods and distances
- Let's summarize how this one equation can be created in following:
(Point No. I)
- The Solar Planets move their orbital circumferences in equal periods (The Paper
hypothesis)
- As a result, the solar planets motions use different rates of time, as we proved in
the previous points discussions
- The geometrical result for the planets motion rates of time is the solar system
distances creation and distribution in A Network Form.
- Means, the solar system distances be in one piece or one geometrical design as one
network, they are created and distributed based on one geometrical design. The
solar system distances Network Theory be explained and proved in point no. (12)
of this paper.
- We need here just to know that, the distances are distributed relative to one another
based on a network form and by one geometrical design.
- This fact is our basic point on which we conclude the one equation which controls
the planets orbital periods and distances
(Point No. II)
- Mars was the next planet after Mercury with original orbital distance =84 mkm
and Pluto was the Mercury moon revolves around it
- Mars and Pluto had migrated from their original orbital distances to their current
positions.
- Now
- The solar system distances are created in a network form, and
- The planets migration caused changes in the whole network, effected negatively on
the solar system general design and put it in risk
- The solar system geometrical design had tried to repair the negative results which
be produced by the planets migration

I do this research
2nd
61
- To do this job,
- The solar system geometrical design wanted to reform the network based on Mars
and Pluto new positions…
- This Task necessitated the solar system geometrical design to depend on Mars and
Pluto motions rate of time in the one equation creation.
- Means
- The one equation which controls all planets orbital distances and periods be
defined as a function in Mars and Pluto motions rate of time (133)
- The solar system geometrical design had to do that to save the planets motions
general harmony and to repair the negative effects of the planets migration on the
whole solar system motion and design.
(Point No. III)
- Why the solar system geometrical design needs to depend on Mars and Pluto
motions rate of time? Let's try to answer in following
- The planets orbital distances and periods are defined based on these distances
motions rates of time, and
- The planets motions rates of time be defined based these planets required periods
to move their orbital distances.
- The required periods we have calculated clearly and used them to define the
planets motions rates of time, let's remember them in following
- The Required Periods List
- For Mercury (27.9 Solar Days) – for Venus ( 35.8 Solar Days) – for Earth (58.1
Solar Days) – for the moon (1700 Solar Days) – for Mars ( 109.4 Solar Days) for
Jupiter (687 Solar Days )- for Saturn (1710 Solar Days) - for Uranus (4890 Solar
Days) - for Neptune (9635 Solar Days) – for Pluto (14547 Solar Days)
- Now
- Based on these required periods the planets motions rates of time be created and
based on the rates of time the planets orbital periods and distances be created….

I do this research
2nd
62
- Means,
- If we have one equation expresses these periods based on one another, by that
these periods will be summarized into one equation and as a result we can define
all planets orbital periods and distances based on this one equation.
- Now this equation depend on Mars and Pluto Rate Of Time (133) and we have no
explanation why except the planets migration reason.
- The point is that,
- The outer planets periods total = the inner planets periods total x 133
- And by that the rate (133) be found in the basic definition of the planets required
periods to pass their orbital distances.
- The data analysis discussion can make this discussion more clear,
- Let's analyze the data in following directly
- Notice
- (Mars Migration Theory be discussed in the next Point no.6 of this paper)

I do this research
2nd
63
(Point No. IV) One Equation Controls The Data
- Let's Summarize The One Law Idea In Following:
o The Solar Planets Required Periods Total Be = 31705 Solar Days
o The Inner Planets Required Periods Total Be = 235.8 Solar Days
o The Outer Planets Required Periods Total Be = 31469 Solar Days
o Notice
o The moon used required period is 4.61 days and not 1700 days because we
account the moon orbital distance to earth and not to the sun
(A)
o 31469 days (the outer planets) = 235.8 days (the inner planets) x 133
o Where 1 Mars Solar Day = 133 Pluto Solar day
o And
(B)
o 31705 solar day = 109.4 solar days x 133 x 2.18
o Where
o 31705 days = All Planets Total
o 109.4 solar days = Mars required period to move a distance = Mars orbital
period.
o 133 is used for second time
o 2.18 mkm = Mars Motion for a period 25.1 hours
o The using of the rate (133) two times shows that, the rate using is mentioned
(C)
o Let's summarize the idea in following
o The planets required periods are controlled by the planets unified motion,
that means, the periods total (31705 solar days) controls all data
o The periods total (31705 solar days) be divided into 2 parts (the inner) and
(the outer) these 2 parts be defined based on the rate of time between Mars
and Pluto.

I do this research
2nd
64
o These 2 planets are the 2 basic points to which the (total periods) be divided.
o The total period division means it creates one Equation, because the periods
total be divided into (2 parts) and each planet transports its period to its
group planets. Means Mars transports the periods to the inner planets and
Pluto transports to the outer planets.
o Based on that, all periods be defined based on one another and creates one
equation.
(D)
o Mars transports the period of time to the Earth and earth transported to
Venus and Venus to Mercury. The inner planets transport the periods of
time from one another by using the planets orbital inclination (And Earth
uses its axial tilt)
o Pluto transports the period of time to the (4) outer planets, by Pluto effect on
these 4 planets days periods total (53.9 hours).
o Let's analyze the data in following for better understanding.
o NOTICE
o The Motion Be Transported Among The Solar Planets By Time Periods
Transportation.
o This meaning is clear concluded from the data, but no geometrical machine
is clear yet. The basic question is that, why the planets motions rates of time
effect on the planets motions features. This fact which we have seen in all
planets orbital periods and distances – the rates of time cause to control this
data – but by what force the rate of time does this effect? Here also we have
a similar point. The motion transportation be done by transportation of a
period of time, how a period of time can be transported from a planet to
another? the data can help this question answer.

I do this research
2nd
65
(Point No. V) (The Data Discussion)
I- Data
Group No. (A)
(A-I)
31469 days (the outer planets) = 235.8 days (the inner planets) x 133
(A-II)
31705 solar day = 109.4 solar days x 133 x 2.18
Group No. (B)
(5-4-a)
109.4 solar days =1.88 x 58.1 solar days
(5-4-b)
23.6 x 1.622 = 35.7 solar days x 1.0725
(5-4-c)
27.9 solar days = 7π x 1.279
(5-4-d)
133 solar days x 0.406 mkm =53.9 mkm

I do this research
2nd
66
II- Discussion
Group No. (A)
Equation no. (A- I)
31469 days (the outer planets) = 235.8 days (the inner planets) x 133
- This equation we know, it tells
- A rate 133 be found between the inner planets required periods total the outer
periods total – why? because 133 is the rate between Mars and Pluto motions, by
that the periods total be divided into the inner and the outer based on (133), that
means, all other planets will define their required periods based on Mars and Pluto
required periods.
- Shortly
- The periods total = (T) for example, this (T) is divided between Mars and Pluto at
first and the 2 planets transported the rest period to their groups planets!
Equation no. (A-II)
31705 solar day = 109.4 solar days x 133 x 2.18
o Where
o 31705 days = All Planets Total
o 109.4 solar days = Mars required period to move a distance = Mars orbital
period.
o 133 is used for second time
o 2.18 mkm = Mars Motion for a period 25.1 hours
- Let's try to understand this equation
- Mars needs a period 109.4 solar day to pass its orbital distance (227.9 mkm) (Mars
moves a distance 2.082 mkm per a solar day 24 hours)
- 109.4 days of Mars be equal 109.4 days x 133 at Pluto
- Means,

I do this research
2nd
67
- If the equation be
- 109.4 days x 2.082 mkm/s x 133, this equation will be understood, it tells
- The distance mars moves during 109.4 solar days be (109.4 x 2.082 =227.9 mkm
= Mars orbital distance), this distance be seen by Pluto as a period of time (1 mkm
= 1days) mans it equals (227.9 solar days) but (1soalr day of Mars Motion =133
days of Pluto motion) as a result the period will be (30589 days = Uranus orbital
period error 1%)
- This manning we can't use now
- Because the velocity isn't 2.082 mkm but 2.18 mkm and Mars moves 2.18 mkm in
25.1 hours
- The difference be = 4.5% this is very interesting difference because
- 90560 seconds /86400 seconds = 1.045
- Means
- 90560 is greater with 4.5% than the solar day 86400 seconds , But
- 90560 days = Pluto orbital period
- For a geometrical reason Mars uses 25.1 hours in place of 24 hours!
- This subject we should discuss with the next point (5-5) which answers why
Saturn motion rate of time = the moon motion rate of time.
- There we should discuss why Mars uses 25.1 hours in place of 24 hours in this
equation
- Notice
- 366565 km (the outer planets diameters total) = 2390 km (Pluto diameter) x 153.3
- 153.3 h = Pluto day period

I do this research
2nd
68
Group No. (B)
109.4 solar days =1.88 x 58.1 solar days
- Where
- 109.4 days = the period Mars needs to pass a distance 227.9 mkm = Mars orbital
distance, by using a velocity =2.082 mkm pr solar day
- 1 hour of Earth motion =1.88 hours of Mars Motion
- But
- 1.88 degrees = Mars orbital inclination (error 1%)
- I claim that, Mars transports the period (109.4 days) to Earth and be (58.1 days) by
using Mars orbital inclination – that may explain why Mars orbital inclination =
1.9 degrees.
- Shortly, Mars transports the period to Earth by using Mars orbital inclination.
- Please note
- The system controls all data by using the planets required periods, that means, the
control method is a period of time. That's what I need to explain clearly. The solar
group is a machine works as a great mechanical clock, the main force in this
machine is the period of time, it's the method of control for the machine work, So,
while Mars transports to Earth a period of time, means, Mars gives Earth a
treasure.

I do this research
2nd
69
Equation no. (5-4-b)
23.6 solar days x 1.622 = 38.27 days = 35.7 days x 1.0725, Where
- 1 hour of Venus = 1.622 hours of Earth
- 35.7 solar days = Venus Required Period To Pass Its Orbital Distance
- 23.6 days of Earth be = 38.27 days of Venus
- But Venus required period is 35.7 days = 38.27 days/ 1.0725
- Means, Venus required period be defined based on the rate 1.0725
- We know this rate controls 40% of all solar planets distances and by that its effect
is a fact we don't know why it effects on data, but it's a fact we have no option to
escape.
- Just,
- We know that Venus data uses this rate 1.0725 for example
- 243 days (Venus rotation period) = 224.7 days (Venus orbital period) x 1.0725
- This behavior is perfectly similar to the moon behavior where
- 29.53 days (The moon day period) = 27.3 days (the moon rotation period) x 1.0725
- And
- 2.4 mkm (the moon daily distance) =2.574 mkm (Earth daily distance) x1.0725
- So, if Venus use the transported period from Earth by effect of the rate (1.0725),
that can be considered as a usual using.
- Now
- What's 23.6 days? Because earth axial tilt =23.4 degrees and even if we consider
each 1 degree = 1 solar day, the number is 23.4 and not 23.6?!
- 17.4 degrees = the inner planets orbital inclination
- 17.2 degrees = Pluto orbital inclination (error 1%)
- 23.6 degrees = the outer planets orbital inclination
- 23.4 degrees = Earth Axial Tilt (error 1%)
- But
- 40.8 degrees = 23.6 degrees +17.2 degrees = 23.4 degrees +17.4 degrees

I do this research
2nd
70
Equation no. (5-4-c)
27.9 solar days = 7π x 1.279 (error 1%)
- Where
- 27.9 solar days = Mercury required periods to pass it’s a distance = 2 x 57.2 mkm
(Mercury orbital distance error 1%)
- 1 h of Mercury motion =1.279 h of Venus Motion
- 7 degrees = Mercury orbital inclination and 7π =22
- Venus transported the period through Mercury orbital inclination.
- We understand that, the periods transportation is done by using the planets orbital
inclination but, Earth uses its Axial Tilt.
Equation no. (5-4-d)
133 solar days x 0.406 mkm =53.9 mkm
- Where
- 1 solar day of Mars = 133 solar days of Pluto
- 0.406 mkm = Pluto motion distance per a solar day
- Equation no. (5-4-d) tells, Pluto moves 53.9 mkm during the period 133 solar days.
- We see that, this period (133) is the cornerstone of our discussion and the
cornerstone of the planets required periods total. we deal with this number along
the data discussion.
- I want to say, the distance (53.9 mkm) is mentioned distance because the period
133 solar days is mentioned period, as we have discussed clearly
- The 4 planets days periods total (Jupiter 9.9 +Saturn 10.7 + Uranus 17.2 + Neptune
16.1) = 53.9 hours
- We need to find the geometrical machine by which 1 mkm be used as 1 hour,
based on this machine, Pluto motion distance will cause the period 53.9 h to be
created. by that Pluto transports the period to them.

I do this research
2nd
71
- Notice
- 2815 mkm = Mercury Uranus distance, if 1 mkm = 1 hour this value 2815 mkm =
2815 h = 2 x 1407.6 hours (Mercury rotation period)
- Notice
- (153.3 h /53.9 h) = 1.16 mkm/0.406 mkm= 4890 days /1700 days= (30589 days
/10747 days)
- This equation creates an importation meaning we should discuss later
- Notice
- We should remember the error (4.5%),
- This error is found between (2.082 mkm = Mars velocity per a solar day) and the
value 2.18 mkm (Mars motion distance in 25.1 hours we have found in equation
no. (II))
- This error is very important error because it controls basic data
- For example
- The moon required period to pass 149.6 mkm (Earth orbital distance) by its daily
displacement 88000 km be (1700 solar days)
- Pluto moves during (1775 solar days ) a distance =720.7 mkm (Mercury Jupiter
Distance)
- These 2 values difference = (4.5%),
- Where (5906 mkm/149.6 mkm) = 39.23 = 4.61 x 8.51
- We should discus that in point no. (5-8) which discuss the rate (4.61) between
Pluto and the moon motions.
- I wanted to explain that, this rate (4.5%) be defined by a geometrical necessity.
NOTICE
Venus (3.024 mkm per solar day) moves during 10485 days a distance =31705 mkm

I do this research
2nd
72
5-5 Why Saturn And The Moon Use Equal Rates Of Time?
I-Data
(5-5-a)
- The moon needs accurately 1700 solar days to pass a distance = 149.6 mkm =
(Earth orbital distance) by using its daily displacement 88000 km
And
- Saturn needs accurately 1710.457 solar days to pass a distance = 1433.5 mkm =
Saturn orbital distance
- The difference between the 2 periods (1700 days and 171.457 days = 251 hours)
(5-5-b)
90560 seconds = 25.1 hours
(5-5-c)
4.095 mkm (Mercury velocity per solar day) x 1433.5 days = 5848 mkm
0.838 mkm (Saturn velocity per solar day) x 6939.75 days = 5848 mkm
1.16 mkm/sec x 5040 seconds = 5848 mkm (Mercury Pluto Distance)
(5-5-d)
(4900 days /1710 days) = (153.3 hours / 53.9 hours) = (1.16 mkm /0.406 mkm)
(5-5-e)
317050 days = 37100 days x 8.5
But
(for revision) (A- II)
31705 solar days = 109.4 solar days x 133 x 2.18

I do this research
2nd
73
II-Discussion
Equation No. (5-5-a)
- The moon needs accurately 1700 solar days to pass a distance = 149.6 mkm =
Earth orbital distance
And
- Saturn needs accurately 1710.457 solar days to pass a distance = 1433.5 mkm =
Saturn orbital distance
- The difference between the 2 periods (1700 days and 171.457 days = 251 hours)
- The planets rates of time depends on the required periods to move their orbital
distances, and Saturn required period is 1710.457 solar days but the moon required
period is 1700 solar days – That causes their rates of time to be almost equal, for
example 1 hour of Earth motion be = 29.4 hours of the moon motion = 29.3 hours
of Saturn motion. almost they are equal
- We have to ask why?
- Let's summarize the idea in following:
(I)
- Uranus Orbital Distance =19.2 Earth Orbital Distance,
- This data means, if Uranus and Earth velocities are equal, During Uranus revolves
around the sun one revolution, Earth revolves around the sun 19 revolutions (19
years)
- And
- The Earth moon moves a cycle called Metonic Cycle continues for 19 years. we
know that Earth Cycle is 4 years (365 +365 +365 +366 days =1461 days) for that
we have asked frequently, why the moon moves Metonic Cycle?
- A suggested answer tell
- The Moon Moves Metonic Cycle By An Effect Of Uranus Motion
- And

I do this research
2nd
74
- We know that the moon and Saturn be created after Mars and Pluto migration,
(II)
- The idea is that, Uranus Motion effects on Saturn and the moon and by this effect,
Uranus caused the 2 planets periods to be (1700 days and 1710.5 days) (the
difference between them be 251 hours).
- Why? because
- Mars moves during 25.1 hours a desistance = 2.18 mkm
- This we have seen in the equation
- (for revision) (II-A) (point 5-4)
- 31705 solar days = 109.4 solar days x 133 x 2.18
- This equation needs the distance 2.18 mkm to define the periods 31705 days which
is the total required periods of the solar planets. Mars moves 2.18 mkm in a period
25.1 hours and the difference between the 2 planets (251 hours) be found to
support Mars motion.
- The number 251 hours is so effective because
- (1)
- If 1 hour = 1 month based on that 251 hours will be =251 months but Metonic
Cycle is (254 months) (error 1%)
- (2)
- 25.1 hours = 90560 seconds where Pluto orbital period =90560 days
- (3)
- 251 hours = 25.1 hours x 10 but
- 10747 days (Saturn Orbital Period) x 24 h x 10 = 10.7 h x 2 x 120536
- (where 10.7 h = Saturn day period and 120536 km = Saturn diameter)
- I try to show that, the data is integrated in harmony with one another, we don't
know the geometrical mechanism but the data shows interesting integration ability
- Shortly

I do this research
2nd
75
- Uranus caused the 2 planets periods to be rated with the difference 251 hours
(error 0.5%) and by that Uranus effect on the planets motions caused their required
periods to be equal approximately
Equation No. (5-5-c)
4.095 mkm (Mercury velocity per solar day) x 1433.5 days = 5848 mkm
0.838 mkm (Saturn velocity per solar day) x 6939.75 days = 5848 mkm
1.16 mkm/sec x 5040 seconds = 5848 mkm (Mercury Pluto Distance)
- The idea tells, Uranus effect on the 2 planets be done through Metonic Cycle,
Equation no. (5-5-c) gives one proof for this claim, because
- Saturn (0.838 mkm per a solar day) moves a distance =5848 mkm during (6939.75
solar days = Metonic Cycle)
- The distance 5848 mkm = (Mercury Pluto Distance) is the basic Distance in the
solar system design and based on it many important data be created.
- We will explain the importance of this distance (5848 mkm) in Part (II) of this
paper, in which we can discuss how this distance be created.
- The distance 5848 mkm is Mercury motion target, as we will discuss in that
discussion
- The point is that, Mercury moves the distance 5848 mkm by using a period
=1433.5 days where 1433.5 mkm = Saturn Orbital Distance
- And Saturn moves equal distance 95848 mkm) during Metonic Cycle, while
Saturn an the moon use equal rates of time geometrically (that tells this data b
created based on a geometrical mechanism which we have to discuss in Part (II) of
this paper.
- The data shows a deep connection between Mercury and Saturn motions.

I do this research
2nd
76
- Notice
- Let's refer to Mercury data here and discuss them in Part (II)
- Mercury moves during 84 days a distance =346.6 mkm (error 0.5%)
- Mercury moves during 346.6 days a distance =1433 mkm = Saturn orbital distance
- Mercury moves during 1433 days a distance =5848 mkm
- Notice
- (The velocity 1.16 mkm passes 5848 mkm in a period 5040 while Mercury day
period needs 5040 seconds to be 176 solar days) this data also be discussed in part
(II) of this paper.
Equation No. (5-5-e)
317050 days = 37100 days x 8.5 (where 3710 mkm =Pluto orbital circumference)
But
(for revision) (A-II) (point 5-4)
31705 solar days = 109.4 solar days x 133 x 2.18
- The total periods are 31705 days based on the period 25.1 hours but we use a
period = 251 hours, as a result the period will be 317050 days
- Pluto orbital circumference (37100 mkm) be used as a period of time (37100 days)
- 1 hour of Saturn motion = 8.5 hours of Pluto motion.
- The data tells
- Equation no. (5-5-e) proves the period (317050 days) is mentioned and designed.
- That means, our conclusion is a correct one and the difference between Saturn and
the moon periods (251 hours) be created by Uranus to support mars motion
through the equation

I do this research
2nd
77
Equation No. (5-5-d)
(4900 days /1710 days) = (153.3 hours / 53.9 hours) = (1.16 mkm /0.406 mkm) =
(30589 days/10747 days) - Where
- 4900 Days Uranus Needs To Move Its Orbital Distance
- 1710.5 Days Saturn Needs To Move Its Orbital Distance
- 153.3 hours = Pluto Day Period
- 53.9 hours = the outer planets days periods total
- 1.16 mkm = Jupiter motion distance during mars rotation period
- 0.406 mkm = Pluto velocity Per A Solar Day.
- 30589 days = Uranus Orbital Period
- 10747 days = Saturn Orbital Period
- This equation tries to prove that Pluto day period (153.3 h) effects on the 4 outer
planets days total (53.9 hours).
- But
- This equation also can be clear just after studying the paper part (II), many data
can be explained only after Part (II) discussion, because
- The matter is created out of light and moves with its parent light beam by using
different rates of time, by that, light motion be an effective continuous player in
matter motion and planets motions. For that, whatsoever I do to avoid light motion
effect on planets motions it can't do the job perfectly, light motion be discovered
through any planet motion data because it’s the planet motion reason and source of
its energy
- At least we can prove that, Mars ad Pluto were migrated from their original orbital
distances to the current and this even is done before the moon and Saturn creation.
- Let's prove that in the next point (Point No. 6)
- Notice
- The sun is the last created piece in the solar system and is created after all planets
- creations and motions we should prove this claim also. In point no. (14)

I do this research
2nd
78
5-6 Why Mercury Use A Double Of Its Orbital Distance?
- In our discussion we have seen that, Mercury used its double values usually
- Because of that,
- Mercury required period is 27.9 solar days and Mercury moves during this period a
distance = 2 x 57.2 mkm (Mercury Orbital Distance)
- Mercury uses its day period (175.94 solar days) and not its orbital period (88 solar
days)
- Also Mercury uses the distance 720.7 mkm in place of its orbital circumference
360 mkm
- Shortly
- Mercury uses double values
- Why?
- If we use Mercury values as all other planets, means it we use (88 days, 360 mkm
and 14.4 solar days as required period), no change will be done for the rates of
time system, the system will work normally
- But
- Mercury motion depends on the distance 720.7 mkm
- Mercury moves during its day period 720.7 mkm and
- 720.7 mkm =Mercury Jupiter Distance
- These facts make this distance 720.7 mkm is the basic distance in Mercury motion.
Mercury all other data be created based on thi0s distance 720.7 mkm
- Why?
- A deep interaction is found between Mercury and Jupiter based on this distance
720.7 mkm –
- In more clear words –
- the distances network (and system) between Jupiter and the inner planets depends
on this distance 720.7 mkm –

I do this research
2nd
79
- we have to know why Mercury motion distance during its day period =720.7 mkm
= Mercury Jupiter Distance – this question is a cornerstone question we have to
answer to create more clear understanding concerning this distance 720.7 mkm
- for that reason we analyze the distance 720.7 mkm = Mercury Jupiter distance in
point no. (8) of this paper.
- No more explanation for mercury data can be added unless we analyze the distance
720.7 mkm which can help our discussion –
- So Please review the point no. (8)

I do this research
2nd
80
5-7 Saturn And Pluto Rate Of Time
I-Data
(5-7-a)
1 day of Saturn = 8.51 days of Pluto
And
5848 days = 687 days x 8.51
(5-7-b)
Saturn moves during 687 days a distance = 575.7 mkm
Pluto moves during 5848 days a distance = 2374 mkm
And
2374 mkm = 4.12 x 576
(5-7-c)
1 day of Jupiter =2.48 days of Saturn = 7.13 days of Uranus = 14 days of Neptune =
21.1 days of Pluto
But
Pluto moves during 21.1 days a distance = 8.51 mkm
(5-7-d)
37100 = 8.51 x 4360 (Jupiter orbital period 4331 days error 0.6%)
90560 =8.51 x 10747 (Saturn orbital period error 1%)
(5-7-e)
(708.7 h /10.7 h) = (655.7 h /9.9h) = 66.23 = (132.47/2)
(5-7-f)
4 x 153.3 days = 133 x 4.61
(5-7-g)
250 x 365.25 days (Earth Orbital Period)= 90560 days (Pluto Orbital Period) (1%)
250 x 149.6 mkm (Earth Orbital Distance)= 5906 mkm (Pluto Orbital Distance) (1%)
(5906 mkm /149.6 mkm) = 39.4 =8.5 x 4.61

I do this research
2nd
81
II-Discussion
- About what this discussion argue?
- The discussion shows how Pluto motion data be created. that means, we use the
rate of time between Saturn and Pluto (8.51) as a method to discover how Pluto
motion data be created.
- Notice 1 solar day of the Earth moon motion = 1solar day of Saturn motion =
(8.51) solar days of Pluto motion (error 0.6%)
1 day of Saturn = 8.51 days of Pluto
And
5848 days = 687 days x 8.51
Equation No. (5-7-b)
Saturn moves during 687 days a distance = 575.7 mkm
Pluto moves during 5848 days a distance = 2374 mkm
And 2374 mkm = 4.12 x 576
1 day of Jupiter =2.48 days of Saturn = 7.13 days of Uranus = 14 days of Neptune =
21.1 days of Pluto
But
Pluto moves during 21.1 solar days a distance = 8.51 mkm
- The previous 3 equations aim to discuss one question, let's move with it step by
step in following:
- We know, 1 solar day of Saturn motion = 8.51 solar days of Pluto motion
- And
- 1 solar day of Jupiter motion be = 21.1 solar days of Pluto motion.

I do this research
2nd
82
- But
- During 21.1 solar days Pluto moves a distance = 8.51 mkm
- We know that 1 mkm can be = 1 solar day
- That means
- The value (8.51 solar days) must be = the value (8.51 mkm)
- Shortly,
- Both values be = 8.51 days (or 8.51 mkm)
- Now
- We need to understand what's happening…!
- Because
- 1 solar day of Saturn motion be =8.51 solar days of Pluto motion
- And,
- 1 solar day of Jupiter motion caused 8.51 solar days of Pluto motion
- But
- 1 solar day of Jupiter motion = 2.48 solar days of Saturn motion.
- I want to say
- The Previous Calculations Caused Jupiter And Saturn Solar Days To Be
Equivalent To 8.51 Solar Days Of Pluto Motion.
- As a result the rate (2.48) between Jupiter day and Saturn day be inherited in Pluto
data and because of that
- 90560 days = 2.48 x 37100 days
- Where
- 90560 solar day = Pluto orbital period
- 37100 mkm = Pluto orbital circumference
- I try to show that, the rate 2.48 between Pluto data be inherited from Saturn and
Jupiter motions effect on Pluto motion.
- This analysis is useful because it tell how Pluto data be created.

I do this research
2nd
83
- Now we know that (90560 = 2.48 x 37100) because the effect of Jupiter and Saturn
motions on Pluto motion
- But
- A Question
- Why Jupiter day causes a motion distance =8.51 mkm and Saturn causes a period
8.51 solar day? What's the geometrical necessity which needed that to b done?
Equation No. (5-7-d)
37100 = 8.51 x 4360 (Jupiter orbital period 4331 days error 0.6%)
90560 =8.51 x 10747 (Saturn orbital period error 1%)
- Where
- 37100 mkm = Pluto Orbital Circumference
- 90560 days = Pluto orbital Period
- The rate 8.51 tells that Pluto deal with Saturn orbital period (10747 days) as a
period of time but deal with Jupiter orbital period (4331 days) as a distance (4331
mkm) and in fact
- 4345.5 mkm =Earth Neptune Distance
- And if
- 1 mkm = 1 solar day these 2 values (4331 and 4345.5 be equal error 0.3%)
- That means, the rate (2.48) be found between (90560 and 37100) be created here
by effects of Jupiter and Saturn motions on Pluto motion. But how that's done?
And why?
- Let's try to answer in the next equation discussion

I do this research
2nd
84
Equation No. (5-7-e)
(708.7 h /10.7 h) = (665.7 h /9.9h) = 66.23 = (132.5/2), Where
- 708.7 hours = The Moon Day Period
- 655.7 hours = The Moon Rotation Period
- 10.7 hours = Saturn Day Period
- 9.9 hours = Jupiter Day Period
- 133 = The Rate Between Mars And Pluto Motions Time
- (means 1 solar day of Mars =133 solar days of Pluto)
Equation No. (5-7-f)
4 x 153.3 days = 133 x 4.61
- The rate (4.61) is the basic rate between Pluto and the moon motions, we should
remember that, 1 day of the moon motion = 1 day of Saturn motion = 8.51 days of
Pluto motion.
- That explains why the Earth moon motion data be entered in this seen, that
because Saturn and the moon rate of time are equal and by that, Saturn behavior
with Pluto be very similar to the moon behavior with Pluto or at least effect on it.
- But we should notice that, the value (133) is the rate of time between Pluto and
Mars motions.
- By that the 2 equations depend on Mars and Pluto rate of time (133)
- Please notice that, the outer planets required periods total = 133 the inner planets
required periods total… by that Mars Pluto rate is the solar system main rate, and
if this rate effect on the moon motion, that tells the moon must be in a central point
of the solar system geometry.
- The rate (4.61) is discussed in details in the nest point no. (5-8)
- Notice
- Equation no. (5-7-e) is a very accurate equation. The four values (708.7,655.7,
10.7 and 9.9) are created together clearly by this equation accuracy.

I do this research
2nd
85
Equation No. (5-7-g)
250 x 365.25 days (Earth Orbital Period)= 90560 days (Pluto Orbital Period) (1%)
250 x 149.6 mkm (Earth Orbital Distance)= 5906 mkm (Pluto Orbital Distance) (1%)
(5906 mkm /149.6 mkm) = 39.4 =8.5 x 4.61
- 1 solar day of Earth motion j=250 solar days of Pluto motion
- Earth orbital period an distance be in proportionality with Pluto orbital period and
distance based on this same rate.
- Earth and Pluto data be in deep consistency and the previous equation is an
example for many others.
- We need here to refer to one fact in following:
- Earth orbital period 365.25 days = Earth orbital distance 149.6 mkm x 2.48 (error
1.3%)
- This rate (2.48) be found in Earth data by effect of Pluto motion interaction with
earth motion.
- I try to show that, we have one rate (2.48) be transported from a point to another
and we have the data based on it this rate is transported. The rate is started between
Jupiter and sat run (1 day of Jupiter motion =2.48 days of Saturn motion) but the
rate be transported to Pluto motion based on the rate (8.51) where (1 day of Saturn
motion =8.51 days of Pluto motion), and because of that the rate (2.48) be found
between 90560 days (Pluto orbital period) and 37100 mkm (Pluto orbital
circumference) and then this same rate be transported to Earth by interaction with
Pluto motion and by that (365.25 days/149.6 mkm) =2.48
- I wish the analysis shows its useful results (1st
) the analysis explains how Pluto
data be created (2nd
) the data be transported among the planets.
- Notice
- (5906 mkm /149.6 mkm) = 39.4 =8.5 x 4.61
- The rate (4.61) is found between Pluto and the moon motions data –we should
discuss it in the next point (5-8)

I do this research
2nd
86
5-8 The Rate (4.61) be used between Pluto and the moon motion
- The rate (4.61) is the main rate between the moon and Pluto motion
- To show this fact as clear as possible 3 groups of Data we have to see
- Let's provide the first group in following
- I- Data
Group No (I)
- Earth (29.8 km/s) moves during a solar day (24 h) a distance = 2574720 km
- Pluto (4.7 km/s) moves during its day (153.3) a distance = 2593836 km
- The moon displacements total during (29.53 days) = 2598693 km
- Pluto and the moon motion distances be equal but they are different from Earth
motion distance with (1%).
Group No (II)
(5-8-a)
406000 km (Pluto motion distance during a solar day) = 4.61 x 88000 km
(88000 km = the moon daily displacement)
(5-8-b)
708.7 h (The moon day period) = 4.61 x 153.3 h (Pluto Day Period)
(5-8-c)
Tan (12.2) x 708.7 hours = 153.3 hours
(5-8-d)
Tan (13.17) x 655.7 hours = 153.3 hours
(5-8-e)
(10.96 deg /1.7 deg) = (153.3 h /24 h) (error 1%)
- Notice
- 13.177 degrees = the moon daily motion degrees
- 12.19 degrees = 13.177 degrees – 0.9856 degrees (Earth motion daily degrees)
For revision (5-7-e)
(708.7 h /10.7 h) = (655.7 h /9.9h) = 66.23 = (132.47/2)

I do this research
2nd
87
II- Discussion
(406000 km /88000 km) = (708.7 h/ 153.3 h) = 4.61
- Where
- 406000 km = Pluto motion distance per a solar day
- 88000 km = The moon displacement per a solar day
Tan (12.2) x 708.7 hours = 153.3 hours – where
- The angle 12.2 degrees = 13.177 deg – 0.98562 deg
- 13.177 degrees = The Moon Motion Degrees Per A Solar Day
- 0.98562 degrees = The Earth Motion Degrees Per A Solar Day
- The Equation shows that the moon and Pluto days periods are created based on
each other geometrically. The fact can't be denied. Pluto day period (153.3 h) is
created as a function in the moon day period (708.7 h).
Tan (13.17) x 655.7 hours = 153.3 hours where
- 655.7 hours = The Moon Rotation Period.
- 153.3 hours = Pluto Day Period and (153.3- ) = Pluto Rotation Period
- The Equation shows, Pluto rotation period is created depending on the moon
rotation period based on the angle (13.177 deg).
- The proportionality in data between Earth and its moon on one side and Pluto on
the other side shows that a geometrical mechanism must be found behind them.

I do this research
2nd
88
- More Data
Group No (III)
778.6 mkm (Jupiter Orbital Distance) = 4.61 x 170 mkm
550.7mkm (Jupiter Mars Distance) = 4.61 x 119.7 mkm
2094 mkm (Jupiter Uranus Distance) = 4.61 x 2 x 227.9 mkm
2872.5 mkm (Uranus Orbital Distance) = 4.61 x 629 mkm
3033.5 mkm (Uranus Pluto Distance) =4.61 x 654.9 mkm
5127 mkm (Jupiter Pluto Distance) = 4.61 x 2 x 550.7 mkm
5906 mkm (Pluto Orbital Distance) =4.61 x 1284 mkm
1375.6 mkm (Mercury Saturn Distance) =4.61 x 2 x 149.6 mkm
3062 mkm (Saturn Neptune Distance) =4.61 x 671 mkm
4345.5 mkm (Earth Neptune Distance) =4.61 x 940 mkm
4267.2 mkm (Mars Neptune Distance) =4.61 x 929 mkm
Where
170 mkm = Mercury Mars Distance
119.7 mkm = Venus Mars Distance
227.9 mkm = Mars Orbital Distance
629 mkm = Jupiter Earth Distance 654.9 mkm = Jupiter Saturn Distance
550.7 mkm = Jupiter Mars Distance 1284 mkm = Earth Saturn Distance
149.6 mkm = Earth Orbital Distance 671 mkm = Jupiter Venus Distance
940 mkm = Earth Orbital Circumference
929 mkm = Earth Jupiter Distance when the 2 planets be on the sun 2 sides
(Max error 1%)
Notice
The previous distances are (22 distances) where all distances in the solar system are
(45 distances) –means these distances be around 50% of all distances in the solar
system and be controlled by this rate (4.61)

I do this research
2nd
89
II- Discussion (Continued)
- Let's summarize this discussion
(I)
- How the rate 4.61be found?
- Pluto orbital distance 5906 mkm = Earth orbital distance 5906 mkm x 39.4
- But
- 1 solar day of the moon motion be =8.51 days of Pluto motion
- 39.4 = 8.51 x 4.61
- That means, the rate (4.61) is the complementary one with the rate of time
(II)
- Why this rate (4.61) controls more than 50% of the solar system distances?
- Because of the equation no. (5-7-e) we brought it for revision here
For revision (5-7-e)
(708.7 h /10.7 h) = (655.7 h /9.9h) = 66.23 = (132.47/2)
- Where
- 708.7h = the moon day period
- 655.7h = the moon rotation period
- 10.7 h = Saturn Day Period
- 9.9 h = Jupiter Day Period
- The equation is so accurate,
- The rate 66.2 = 132.4/2
- 1 solar day of Mars motion =133 solar days of Pluto motion
- But
- The outer planets required periods total =133 The inner planets required periods
total, as we have seen in point no. (5-4)
- That makes the rate (133) is a basic one and by that the rate (4.61) is a basic one
also.

I do this research
2nd
90
5-9 The Moon Orbital Motion Equation
- Let's explain why we discuss the moon orbital equation in this point…
- The moon needs 1700 solar days to pass a distance = 149.6 mkm=Earth orbital
distance by using its daily displacement 88000 km
- If 1000 km = 1 degree, so 1700 days will be 1.7 degrees
- In fact the moon orbital equation uses the angle 1.7 deg as the moon motion angle
per a solar day
- Because of that, the angle 1.7 degrees found its source 1700 solar days.
- But
- Can 1000 km= 1 day or 1 degree?
- If that's happened so, the moon daily displacement 88000km will be equal 88 days
= Mercury orbital period
- And
- In many other situations we found that it's useful to use the rate 1000 km = 1 day
=1 degree and because of that we study the moon orbital equation to show how the
angle 1.7 degrees be used by the moon motion.
- Let's summarize the moon orbital motion equation in following

I do this research
2nd
91
I- Why Does The Moon Use Pythagorean Triangle In Its Motion?
- Let's summarize this question answer in following:
o The moon uses Pythagorean triangle basically to decrease its displacement
daily through its orbit
o The moon daily displacement = 88000 km and the moon has to move this
distance every day without any decreasing (later we will know why!)
o But
o If the moon moves by this displacement as its orbital displacement the moon
would revolve around Earth through its apogee orbit only (r=0.406 mkm)
o For that reason
o The moon creates an angle between its motion direction and its orbit
horizontal level to create a displacement through its orbit less than (88000
km)
o As a result of this technique, the moon can revolve around Earth through
more near orbits than apogee orbit (r=0.406 mkm)
o Simply, because the moon uses this technique the moon can revolve around
Earth through perigee orbit (r=0.363 mkm)
o Let's explain this intelligent technique with some details to show the useful
result of using Pythagorean triangle by the moon orbital motion….

I do this research
2nd
92
II- How Does The Moon Use Pythagorean Triangle In Its Motion?
- The moon moves daily (88000 km) on the right triangle hypotenuse (AC), but the
moon creates an angle (θ) between its motion direction and its orbit horizontal
level, by that the real displacement through the moon orbit will be (L= 88000 km
cos (θ)), and by that, spite the moon moves 88000 km, but the real orbital
horizontal displacement be less than (88000 km) and this is the objective for which
the moon uses Pythagorean triangle –
As an example,
- If (θ) =28.63 degrees, the real displacement (L== 88000 km cos (θ)) = 77237 km,
So, if the moon real displacement daily be (77237 km), during 29.53 days the
moon will pass a distance = 2.28 million km and this will be the moon orbital
circumference, where 2.28 mkm = 2π x (0.363 mkm)
- The Moon Orbital Perigee Radius =0.363 mkm
- That means, the moon by a real displacement =77237 km can move around Earth
through the perigee orbit (radius =0.363 mkm), this is the useful result the moon
performs by using Pythagorean triangle,
- Now let's suppose the moon doesn't use Pythagorean triangle, what would happen?
- The moon daily displacement = 88000 km, during 29.53 days the moon moves a
distance = 2.598 mkm where 2.598 mkm = 2π x (0.413 mkm)
- The Moon Orbital Apogee Radius =0.406 mkm
- So the moon will move along month revolving around Earth through its apogee
orbit (or even far from apogee orbit) because the total distance can't be passed
through any more near orbit around Earth…
- The data shows how Pythagorean triangle is so useful for the moon orbital motion.

I do this research
2nd
93
The Angle θ
- The angle (θ) should get our attention for its specific effect…let's summarize the
idea in following
o The angle (θ) changes the real displacement (L = 88000 cos (θ)), through the
moon orbit..
o We know that, when the real displacement (L) be shorter the moon can
move through near orbits to Earth and by that the moon can be near or at
Perigee radius (0.363 mkm)
o When the real displacement (L) be greater the moon has to move through
orbits far from Earth and by that the moon can be near or at apogee orbit
(r=0.406 mkm)
o That means, the angle (θ) changes the real displacement (L) and also
changes the distance between the moon to perigee or to apogee, shortly, the
angle (θ) defines the moon position (as a ship) between 2 river banks….
- The angle (θ) defines the moon orbital motion basic features and we have to
discuss is deeply with the moon orbital motion equation (θ1= θ0 + 1.7 degrees),
but before we need to analyze the moon orbital motion
Notice
o We know that (363000)2
+ (86000)2
= (373000)2
o In Pythagoras triangle with dimensions (363000 km, 373000km, 86000 km),
what's the angle (θ)? The angle (θ) = 13.33 degrees
o Also (396800)2
+ (86000)2
= (406000)2
the angle (θ) = 12.229 degrees
o I have used (363000 km and 406000 km) because they are the perigee and
apogee radiuses between which the moon moves.
o The difference between angles = 1.1 degrees
i.e.,
The angle (1.1 deg.) controls the moon motion from perigee to apogee, we will need
this notice later in our discussion

I do this research
2nd
94
III- The Moon Orbital Motion
- The moon moves per a solar day a motion typical to the Earth motion to avoid the
separation from Earth through their motions, based on this rule, the moon moves
per a solar day 2.573 million km with an angle declines on the horizontal level
0.98562 degrees as typical to Earth motion
- If there's no Lorentz Length Contraction Phenomenon effect on the moon motion,
the moon motion trajectory would to be a parallel line to Earth Motion Trajectory,
But Lorentz Length Contraction effects on the moon motion daily distance (2.573
mkm) with a rate 1.0725 and causes this distance to be contracted (2.399 mkm)
- The moon difficulties are started here, because the difference between both
distances (0.17 mkm) will cause the moon to be separated from Earth motion
inevitably
- We should notice that, these motions are done far from our observation, means, we
see nothing of this motion distance, because the moon moves on the Earth orbital
circumference revolving around the sun, but, even if we can't observe this motion
distance the motion is still fact and proved by its power, because the Earth moves
per a solar day 2.573 mkm and if the moon doesn't move this same distance every
solar day that necessities the moon to be separated from the Earth through their
motions course – based on that- the facts prove this motion regardless our
observation ability for it.
- Now the moon has an additional distance to be passed (0.17 mkm) and the moon
has to pass this distance on the same solar day to avoid the separation from the
Earth during their motions.
- Because of that, the moon moves its daily displacement (88000 km) depends on
Earth gravity force (by which we see the moon in the Earth sky), but the different
distance (0.17 mkm) to be covered still needs the moon to move one more
displacement (= 88000 km)

I do this research
2nd
95
- The previous explanation tells that, the moon has to move 2 displacements each =
88000 km, while we see one displacement only because it's done through the
moon orbital motion around Earth but the other displacement should be done also
because this total distance (0.17 mkm) is required to cover the different distance
and create the total (2.573 mkm) which saves the moon and Earth motions
accompanying.
- Now we have 2 basic information about the moon orbital motion
o (1st
information) the moon uses Pythagorean triangle in its orbital motion
o (2nd
information) the moon has to move 2 displacements each =88000 km
and their total distance =0.17 mkm which is a required distance necessary to
cover the difference between the moon and Earth motions distances.
- This explanation helps us to understand why the moon uses Pythagorean triangle
in its motion, because the moon can't decrease its daily displacement (88000 km)
because the moon needs this distance to cover the different distance between its
contracted motion distance (2.399 mkm) and Earth motion distance (2.573 mkm),
So the moon needs to move this displacement perfectly, but if it's used as a
displacement through the moon orbit, the moon would be always a prisoner in the
apogee orbit (r=0.406 mkm) as we have discussed before, because of that, the
moon creates Pythagorean triangle technique by which the moon moves actually
88000 km daily but the real displacement through the moon orbit became less (L =
88000 Cos θ) and by that the moon can achieve 2 objectives, First to pass the
required distance (88000 km) and Second to move in near orbits to Earth, that
shows the intelligent moon motion technique…
- (Notice, Lorentz Length Contraction Effect Discussion is in Appendix No. 1)

I do this research
2nd
96
The Moon Orbital Motion Needs One More Orbit
- The previous explanation tells that, the moon moves 2 displacements each =88000
km, we see one of these 2 displacements but where's the other displacement?!
- We know that, the moon original motion (2.573 mkm) which is contracted to be
(2.399 mkm) isn't seen by us because the moon moves this distance revolving with
Earth around the sun along the Earth Orbital Circumference
- We may accept that, the 2nd
displacement the moon does on this same trajectory
and isn't seen by us.
- So,
- There must be one more orbit for the moon to move through this 2nd
displacement.
means,
- There's 2nd
Orbit For The Moon Motion
- But
- How can we discover this second orbit if we can't observe the 2nd
displacement
motion?
- We can discover this 2nd
orbit by the moon orbit data analysis. So we should
depend on the moon orbital triangle data analysis to define this 2nd
orbit position.
- For that we have to discuss the moon 2nd
orbit in our deep analysis of The Moon
Orbital Triangle Geometrical Structure.

I do this research
2nd
97
IV- The Moon Orbital Motion Equation
A- The Equation Concept
B- The Equation Test and Accuracy
A- The Equation Concept
The Moon Orbital Motion Equation
(θ1= θ0 + 1.7 degrees)
- The moon orbital motion equation is created depending on the concept we have
discussed, which is (the moon uses Pythagorean triangle in its orbital motion)
- The moon uses Pythagorean triangle and by this intelligent technique the moon be
under control of the angle (θ) change
- The angle (θ) defines almost all the moon motion features.…
- The moon uses this technique, aiming to create a real displacement shorter than its
actual displacement (88000 km) based on the equation (L =88000 cos (θ)) and by
that while the moon moves a displacement =88000 km but the real displacement
(L) through its orbit be shorter than 88000 km and by that the moon can revolve
around Earth through more near orbits than its apogee orbit (r=0.406 mkm).
- The moon orbital motion equation depends on this concept and, the equation
uses (the constant) 1.7 degrees as the moon daily motion degrees, and the equation
uses the previous day angle (θ0) to produce the today angle (θ1)
- We have 3 questions in this equation study which are:
o How does this equation work?
o Is this equation trustee and correct?
o Why does the equation use the angle 1.7 degrees for the moon daily motion?
Let's try to answer….

I do this research
2nd
98
How to use this equation?
- Perigee Radius =0.363 mkm, so Its Orbital Circumference =2.28 mkm
- Suppose the moon will revolve around Earth through perigee orbit only during
29.53 days, so
- (2.28 mkm /29.53 days) = 77237 km
- This is (the real displacement = L = 88000 km Cos θ = 77237 km),
- What's the angle θ value? the angle θ = 28.63 degrees
- Suppose the moon stand on this point yesterday with the angle (θ) =28.63 degrees,
where the moon will move today?
- From Perigee (the most near point to Earth) the moon will move in Ascending
motion because it moves from perigee (0.363 mkm) to apogee (0.406 mkm)
- In Ascending motion we use (-1.7 degrees) because the angle (θ) is decreased
where the real displacement (L) is increased, So let's do that in following
o (θ1= θ0 - 1.7 degrees)
o (θ1= 28.63 degrees - 1.7 degrees) = 26.93 degrees
o L = 88000 Cos (26.93 degrees) = 78454 km
o During 29.53 days so (78454 km x 29.53 days = 2.316 mkm)
o 2.316 mkm = 2π x 368722 km
That means
o The moon was (before motion) on Perigee radius (r=0.363 mkm) and starts
its motion displacement 88000 km. For day motion the equation uses 1.7
degrees, that means, the moon on perigee uses Pythagorean triangle with
angle (28.63 degrees) and during one solar day the moon uses - 1.7 degrees
and by that the angle will be (26.93 degrees)…... The angle 1.7 degrees
expresses The Moon Daily Motion
o By using Pythagorean triangle its angle (θ) = 26.93 deg, the displacement
(88000 km) will create a real displacement through the moon orbit = 78454
km and the moon will finish its motion today at a distance 368722 km

I do this research
2nd
99
means the moon is far from perigee radius with (368722 km-363000 km
=5722 km )
o So, the moon after 1 day motion will be at the point 368722 km and will
have the Pythagorean triangle its angle 26.93 degrees.
The Descending Motion
o When the moon moves from apogee (0.406 mkm) to perigee (0.363 mkm),
so the angle (1.7 degrees) will be positive (+1.7 degrees) because the angle
(θ) is increased and the real displacement (L = 88000 Cos (θ)) be shorter.
So
o If the moon in apogee radius (r=0.406 mkm), what's the angle (θ)?
o The apogee orbital circumference = 0.406 mkm x2π =2.55 mkm = 29.53
days x 86400 km, the angle (θ) = 10.96 degrees (=11 deg approx.)
o The moon moves from apogee to perigee (descending motion)
o (θ1= θ0 + 1.7 degrees) means (θ1= 11 degrees + 1.7 degrees) = 12.7 deg.
o L = 88000 Cos (12.7 degrees) = 85847 km
o During 29.53 days so (85847 km x 29.53 days = 2.535 mkm)
o 2.535 mkm = 2π x 403467 km
So
o After one day the moon will be on 403467 km far from apogee (406000 km)
with 2540 km
Now let's see this equation test and efficiency in following

I do this research
2nd
100
B- The Equation Test and Accuracy
- I have tested the Equation with real data for 2 months June 2020 and October 2020
- The results are very good and I provide the results here for better vision
concerning the equation efficiency
1st
Test June 2020
Day Registered Data The Results (1.7) Difference
6-6-2020 369418 km
7-6-2020 373729 km 374772.5 - 1044
8-6-2020 378917 km 378821.5 96
9-6-2020 384534 km 383667.7 867
10-6-2020 390096 km 388890 1206
11-6-2020 395156 km 394000 1156
12-6-2020 399345 km 398604.2 741
13-6-2020 402395 km 402361.3 34
14-6-2020 404153 km 405052.8 -900
15-6-2020 404574 km ---- ---
16-6-2020 403718 km 401848.5 1870
17-6-2020 401733 km 400876.1 857
18-6-2020 398840 km 398640.7 200
19-6-2020 395303 km 395417.4 115
20-6-2020 391409 km 391521.2 -113
21-6-2020 387432 km 387273.4 159
22-6-2020 383607 km 382968.4 639
23-6-2020 380110 km 378852 1258
24-6-2020 377044 km 375107 1937
25-6-2020 374451 km 371836.5 2615
26-6-2020 372338 km 369077 3262
27-6-2020 370703 km 366855.6 3847
[

I do this research
2nd
101
The 1st
Test Results Analysis:
- The Total Results Are 20 Values
(1st
Category)
o 15 values, defines the moon position in range 1300 km (Error 3%)
(2nd
Category)
o 2 values, defines the moon position in range 1300-2000 km (Error 4.6 %)
(3rd
Category)
o 3 values, defines the moon position in range 2000-3500 km (Error 8 %)
- The Results Explanation
- The distance from perigee to apogee =43000 km…
o 1st
Category of results defines the moon position in error range (1300 km) =
error (3%), that means, (15 values of 20) defines the moon position with
error (3%) only (Small Error Range)
o 2nd
Category of results defines the moon position in error range from (1300
km to 2000 km) = error (4.5%), that means (2 values of 20) defines the
moon position with error (4.5%) (Average Error Range)
o 3rd
km to 3500 km) = error (8%), that means (3 values of 20) defines the moon
position with error (8%) (Great Error Range)
- The Equation Accuracy
o The previous explanation shows that, the equation has a good range of
accuracy and its error is in the acceptable error range
The Conclusion
The Equation Is correct and trustee
And
It's a useful tool to define the moon position daily

I do this research
2nd
102
2nd
Test October 2020
Day Registered Data Results (1.7) Difference
5-10-2020 405,690 km --- ---
6-10-2020 404,171 km 403125.3 km 1046 km
7-10-2020 401,649 km 401390 km 259 km
8-10-2020 398,073 km 398545.6 Km - 473 km
9-10-2020 393,464 km 394568.8 km -1105 km
10-10-2020 387,944 km 389510 km -1567 km
11-10-2020 381,763 km 383520 km -1758 km
12-10-2020 375,302 km 376875.3km -1574 km
13-10-2020 369,063 km 369981km -919 km
14-10-2020 363,617 km 363363.4km 254 km
15-10-2020 359,530 km 357612 km 1918 km
16-10-2020 357,269 km 353307 km 3962 km
17-10-2020 357,105 km ---- --
18-10-2020 359,048 km --- --
19-10-2020 362,851 km 364979.7 km - 2129 km
20-10-2020 368,058 km 368579.3 km -522 km
21-10-2020 374,101 km 373492.4 km 609 km
22-10-2020 380,412 km 379168.3 Km 1244 Km
23-10-2020 386,497 km 385059.3Km 1438 km
24-10-2020 391,989 km 390694.3 km 1295 km
25-10-2020 396,659 km 395729.5 km 930 km
26-10-2020 400,395 km 399958.7 km 437 km
27-10-2020 403,181 km 403299 km 112 km
28-10-2020 405,059 km 405738.5 km -680 km
29-10-2020 406,104 km 407359.4 km -1256 km
[

I do this research
2nd
103
The Test Results Analysis:
- The Total Results Are 22 Values
(1st
Category)
o 15 values, defines the moon position in range 1300 km (Error 3%)
(2nd
Category)
o 5 values, defines the moon position in range 1300-2000 km (Error 4.6 %)
(3rd
Category)
o 2 values, defines the moon position in range 2000-3500 km (Error 8 %)
- The Results Explanation
- The distance from perigee to apogee =43000 km…
o 1st
Category of results defines the moon position in error range (1300 km) =
error (3%), that means, (15 values of 22) defines the moon position with
error (3%) only (Small Error Range)
o 2nd
km to 2000 km) = error (4.5%), that means (5 values of 22) defines the
moon position with error (4.5%) (Average Error Range)
o 3rd
km to 3500 km) = error (8%), that means (2 values of 22) defines the moon
position with error (8%) (Great Error Range)
- The Equation Accuracy
o The previous explanation shows that, the equation has a good range of
accuracy and its error is in the acceptable error range
The Conclusion
The Equation Is correct and trustee
And
It's a useful tool to define the moon position daily

I do this research
2nd
104
The Value 1.7 degrees
- The 3rd
question was, why the equation uses 1.7 degrees?
Because
1.7 degrees = 0.98562 degrees + 0.712 degrees
Where
- 0.98562 degrees = Earth motion daily degrees, and it equals the moon daily
motion degrees because the moon has to move an equal distance to Earth motion
daily distance to save their motions accompanying
- This question and the angle 0.712 degrees is discussed in my previous paper.

I do this research
2nd
105
The Moon Motion Difficulties
- There are 2 basic difficulties are observed in the moon orbital motions, let's refer
to them in following:
o (1st
Difficulty) The moon moves per day different distances from perigee to
apogee…..
o We know the moon moves from perigee to apogee (go and back) during
Anomalistic month (27.55 solar days)
o (43000 km x 2) / 27.55 days = 3122 km
o The moon doesn't use this rate (3122 km) in its motion, instead the moon
can move (6000 km) on one day only and on another day may move only
2500 km (or even less)!
o The moon orbital equation tries to solve this difficulty by using the rate 1.7
degrees in the equation (θ1 = θ0 + 1.7 degrees), the value 1.7 degrees is a
great number and enables the moon to move around (5000 km) per solar day
and by that if the moon moves per solar day 4000 km the different distance
will be 1000 km and if the moon moves 6000 km the different will be
– 1000 km, it’s the same difference, and by that, the error be minimized as
possible enabling the equation to be more efficient..
o (2nd
Difficulty) The moon stays in perigee and apogee points long time….
o That means, while the moon be on perigee or apogee, the moon doesn't use
the equation and doesn't change its distance to perigee or apogee for long
days…we may notice that in the equation tests, when the moon reach to
perigee or apogee the equation stops its work and stays 2 or 3 days to return
to its work… because the moon consumes long time to leave the points
(perigee and apogee)…

I do this research
2nd
106
6- Mars Migration Theory
6-1 Mars Migration Theory
6-2 Pluto Migration Theory
6-3 Planets Migration Theories Proves

I do this research
2nd
107
6-1 Mars Migration Theory
- Mars was the next planet after Mercury with original orbital distance =84 mkm
and Mars had migrated into Mars current orbital distance (227.9 mkm)
- Through Mars Displacement from (84 mkm) to (227.9 mkm), Mars had collided
with Venus (at first) and then with Earth (at second) and then reach to its current
orbital distance point (227.9 mkm).
- Mars had moved from (84 mkm) to (227.9 mkm) in a direct displacement – and
had pushed with it all collisions debris in its direction
- From these collisions debris the Earth could create its own moon and also Mars
moons are created from them and the rest debris be attracted by Jupiter Gravity
and Created The Asteroid Belt.
- Means,
- Mars Migration Theory is in full harmony with (Giant- Impact Hypothesis), but
- Mars itself had collided with Earth and caused the moon creation and not the
ancient planet (theia) as supposed by the Giant-impact hypothesis.
- The point is that, the planets order (Mercury – Venus –Earth) shows that Mars was
the second planet after Mercury – and Mars had migrated from (84 mkm) to (227.9
mkm) –through this displacement – Mars has collided with Venus and Earth–
- Mars collisions with Earth and Venus left many proves behind which prove that
Mars did these collisions and not the planet (theia).
- Mars Migration theory answers many of the Giant-impact hypothesis questions.
- Let's do that in following
(1st
Question)
- Why has Venus no a moon? spite it suffered from a similar collision as Earth?
- Mars had moved its displacement from (84 mkm) to (227.9 mkm), and through
this displacement Mars pushed all debris with it in its motion direction and
because of that, Venus had found no debris around, And couldn't create its own
moon because of that.

I do this research
2nd
108
- But the debris lost some of their motion momentum at Earth position and Earth
gravity is more strong than Venus, by that Earth could attract some debris by
which Earth created its moon.
(2nd
Question)
- What's The Lunar Magma Ocean (LMO) Origin?– It's Venus
- Because
- The Earth moon rocks are created from the 3 planets debris (Mars- Venus – Earth)
and for that the moon has a Magma Ocean (LMO).
(3rd
Question)
- Why The Iron Oxide (Feo) of the Moon= (13%)?
- Because The rate (13%) is a middle rate between Mars rate (18%) and the
terrestrial mantle (8%).
Notice no.(1)
- Mars Moons Creation is a good proof for Mars Migration Theory.
- The collisions debris lost the great part of their momentum when reach to Mars
orbital distance point (227.9 mkm), because of that, Mars with its small mass could
attracted 2 moons, because the debris is too much around it and can attract them
easily.
- But Venus great gravity couldn't create even one moon because no debris be found
around it
Notice no.(2)
- The asteroid belt is one more proof for Mars Migration
- The collisions debris which were pushed with Mars motion to its current orbital
distance point (227.9 mkm) couldn't all be attracted by Mars gravity because of its
small mass, Mars had attracted only its 2 small moons
- But
- Jupiter great gravity attracted the rest debris and they created the asteroid belt.
- The whole trajectory is filled with references of the event and shows that it's a fact.

I do this research
2nd
109
Notice no.(3)
- Mars diameter before Migration was 7070 km and be 6792 mkm after migration as
a result for the collisions Venus and Earth
- We know that because
- The equation (7070 km x 1092
= 84 mkm), where 84 mkm=Mars original orbital
distance.
- The equation (d = r x 1092
) is the equation controls planet orbital distance based on
its diameter. The equation be used by Mercury, Earth and Saturn but the planets
migration effects negatively on the other planets using of it.
- From this analysis we conclude, Mars Diameter be decreased with 4.1%
- Additional Data
- Mercury Orbital Distance 57.9 mkm = Mercury Diameter 4879 km x 1092
- Earth orbital distance 149.6 mkm = Earth diameter 12756 km x 1092
- Saturn orbital distance 1433.5 mkm =Saturn diameter 120536 km x 1092
Notice no.(4)
- If Mars Mass be decreased as a result for the collisions with 8%, in this case
- Mercury mass (0.33) and its orbital distance (57.9 mkm) be in proportionality with
Mars mass (0.688) and Mars original orbital distance (84 mkm) based on Newton
gravitation Equation – this same proportionality be working also with the current
data of Jupiter and Saturn.

I do this research
2nd
110
6-2 Pluto Migration Theory
- Pluto Was The Mercury Moon
- One great force stroke Mercury, Pluto and Mars, with order
- Means,
- Mercury had received the most strong stroke directly, Mercury couldn't escape
from its fate. But Mercury axial tilt was (One Degree) and after the stroke be (0.01
Degree or even Zero)!
- The third stroke be received by Mars by which Mars had Migrated
- The second stroke be received by Pluto, which caused Pluto to be flying along the
solar system and reach to the end point of the solar system.
- The solar system end point with distance (5906 mkm) is a defined geometrically.
- Before Pluto migration –
- Neptune orbital distance was 5906 mkm
- Means,
- Neptune was in the position in which Pluto now be found. The distance 5906 mkm
was Neptune original orbital distance
- Pluto had flying along the solar system distance and reach to Neptune position, and
had collided with Neptune, pushed it out of its orbit and then put it out of (Pluto
Influence area) – That's why Pluto Neptune Distance = Pluto eccentricity Distance
Result No. 1
Mercury Pluto Distance = 5848 mkm = 101 x Mercury orbital distance 57.9 mkm
- This data is so important because the solar system distances be designed in
proportionality with it. based on that the rate (101) is one geometrical pillar in the
solar system design.
Result No. 2
Neptune Be Out Of The Network
- We will study that, the solar system distances be created in a network form in point
no. (12), with many proves we discuss this point…
- But

I do this research
2nd
111
- How the distances be created in a network form? We should answer this question
in details in that point no.(12) – but let's summarize the idea in following:
- Matter is created out of light and the planets be created of one light beam- light
motion creates the distances through which the planets move – by that –
- The Solar System Distances Be Similar To Railways
- And, the planets be similar to trains
- That shows why the planets motions be obligatory motions that because light
provides the required energy for distances creation and forces the planets to move
through these distances by this energy.
- Now, we have very (unclear) event – because –
- Pluto had pushed Neptune and put it Out Of The Railways
- That means, Neptune has no orbit and no definition of motion – and because of
that Neptune needed energy to build its orbit!
- Frankly
- I don't understand correctly the meaning of the network or how Neptune be out of
the network- because no space (I know) out of the network and I can't define any
point out of the network
- But
- This idea I have built on 2 data which are
- (1) Neptune Pluto Distance = Pluto Eccentricity Distance =1411 mkm
- (2) Neptune has seized 14% of Jupiter energy to create its orbital circumference
28255 mkm and this lost energy is registered clearly in the solar planets distances
as we will discuss in Part no.(II) of this paper.
- These 2 data pushed me to conclude that, Neptune had no orbit and had to create
one for which Neptune seized the energy, Now this result must be done by the
collision between Neptune and Pluto.
Result No. 3
- Pluto and Neptune collisions caused to create Pluto moons and the Kuiper belt.

I do this research
2nd
112
6-3 Planets Migration Theories Proves
(Point No. 1) (Mars Migration Proves)
Mars Migration proves depends on the interaction of motion between Mercury and
Mars supposing that Mars was Mercury neighbor.
I- Data
(6-3-a)
(Mars Mass / Mercury Mass) = (6792 km/4879 km)2
(6-3-b)
(687 days /175.94 days) = (227.9 mkm /57.9 mkm)
(6-3-c)
7 degrees = 5.1 degrees + 1.9 degrees
II- Discussion
(Mars Mass / Mercury Mass) = (6792 km/4879 km)2
= 0.524
- Where
- 6792 km = Mars diameter
- 4879 km = Mercury diameter
- Equation no. (6-3-a) tells that, Mercury And Mars Masses And Diameters Are
Rated With Each Other.
Notice
- Mars motion daily =0.524 degrees
Equation No. (6-3-b)
(687 days /175.94 days) = (227.9 mkm /57.9 mkm)
- Where
- 687 days = Mars Orbital Period
- 175.94 days = Mercury Day Period
- 227.9 mkm = Mars Current orbital distance
- 57.9 mkm = Mercury Current orbital distance

I do this research
2nd
113
- More data is rated between Mars and Mercury
- We should notice that, the data proportionality is done mostly between the
neighbor planets… for example
- Mars orbital period (687 days) = the moon orbital period (27.3 days) x 25.2
- Where 25.2 deg = Mars Axial Tilt. Also
- 25.2 deg = 1.9 deg x 13.17 deg
- Where, 13.17 deg = The Moon Daily Degrees
- And 1.9 deg = Mars orbital inclination
7 degrees = 5.1 degrees + 1.9 degrees
- Where
- 7 deg = Mercury orbital inclination
- 5.1 deg = The moon orbital inclination
- 1.9 deg = Mars orbital inclination
- Mercury motion data shows a strong proportionality with Mars motion data which
gives reference that Mars once was a neighbor to Mercury.

I do this research
2nd
114
(Point No. 2) (Pluto Migration Proves)
Pluto Migration proves depends on the fact that Pluto was the Mercury moon and by
that Pluto be longed to the inner planets.
I- Data
(6-3-d)
Pluto is a planet belonged to the inner planet (small mass and long day period)
(6-3-e)
The rate (101) controls the solar system basic data
(Mercury Pluto Distance/ Mercury orbital distance) = 101
II- Discussion
- In Part no.(II) there are many other proves supports Pluto Migration Theory and
proves that Pluto be belonged to Mercury in its original case
- Notice
- 17.4 deg = 7 deg (Mercury orbital inclination) x 2.5 deg (Saturn orbital inclination)
- 17.4 deg = 5.1 deg (the moon orbital inclination) x 3.4 deg (Venus orbital
inclination)
- 17.2 deg = Pluto orbital inclination (error 1%)
- Notice
- Pluto Migration explains why bode law couldn’t predict Neptune Orbital Distance
– Because Neptune orbital distance be created as additional trajectory in the solar
system distances network – means- the equations be built on the original design
couldn't predict Neptune orbital distance because this distance be defined after The
Planets Migration.

I do this research
2nd
115
7- Planets Unified General Motion
7-1 The Planets Unified General Motion Description
7-2 The Solar Planets Motions Are Complementary One Another
7-3 The Solar Planets Move An Unified Motion (A Team Motion)

I do this research
2nd
116
7-1 The Planets Unified General Motion Description
- I claim that (The Planets Motions Create One Unified General Motion)
- This claim tells the planets motions be similar to Train Carriages Motion
- Also It tells the planets motions be similar to Chess Pieces Motions
- The idea discovers One Law controls the solar system motion and data.
- The planets unified general motion forces each planet motion to be complementary
with other planets motions to perform The Unified General Motion. as a result,
The Planet motion be An Obligatory Motion.
- The Solar Planet creates its data to be in consistency with its motion.
- The planet can't change its motion, but can change its data to be in consistency
with its motion. And the planet data be created in consistency with its motion.
- Because the planet motion be complementary with other planets motions.
- The planet data be created complementary with other planets data
- Based On This Vision
- The solar planets data be created complementary to other planets data and based
on that the planets data be created depends on One Geometrical Design.
- And
- The Planets Creation And Motions Data Be Controlled By One Equation Only

I do this research
2nd
117
- Please remember
- The Solar Group is similar to a machine of gears each planet is a gear in it, or
- The Solar Group is similar to one building and each planet is a part of this same
building, or
- The Solar Group is similar to puppets theater, all puppets are connected together
by one force and causes their motions, or
- The Solar Group is similar a canal water moves through the canal and causes the
rotation of 9 waterwheels. The Motion Direction of The waterwheels is different
from the water motion direction but no waterwheel can rotate without the water
motion, or
- The solar planets are similar to train carriages, they move together one unified
general motion, or
- The solar planets be similar to knots on one robe or similar to points on the same
one trajectory of energy –
- The solar planets motions be similar to chess board pieces motions, each motion
depend on the other planets motions and be done by geometrical calculations.
- Notice
- The planets motions data analysis in this point aims to prove this feature which is
(The Solar Planets Motions Be Integrated Into One Unified General Motion)
- Let's do that in the next points(5-2 and 5-3)

I do this research
2nd
118
7-2 The Solar Planets Motions Are Complementary One Another
I-Data
(The Interaction between Mercury, Mars and Jupiter)
(No. 1) (Mercury and Mars Motions)
- Mercury moves during its day period (4222.6 h.) a distance = 720.7 mkm
- Mars moves during (2802 hours) a distance = 243 mkm
- Mercury moves during (1407.6 hours) a distance = 243 mkm (error 1%)
- Mars moves during 346.6 d. a distance =720.7 mkm (Mercury Jupiter Dis.)
- Mercury moves during 346.6 d. a distance =1419 mkm (with 1433 error 1%)
- Mars moves during 687 d. a distance = 1433 mkm (Mars orbit. Circum)
- Mercury moves during 687 d. a distance =2815mkm (Mercury Uranus Dis.)
- Mars moves during 4331d. a distance = 9010mkm (Saturn orbit. Circum)
- Mercury moves during 4331d. a distance = 2815 mkm x 2π
- Mars moves during 224.7 d. a distance = π x 149.6 mkm (Earth orb. Dis)
- Mercury moves during 224.7 d. a distance = 920 mkm (with 928 error 1%)
- Mars moves during 365.25 d. a distance = π x 243 mkm (0.5%)
- Mercury moves during 365.25 d. a distance = 2 x 748 mkm
- Mars moves during 5040s. a distance= 121464 km= Saturn Diameter (+1%)
- Mercury moves during 5040s. a distance= 238896 km= 2 Saturn Diameters (-1%)
Where
1407.6 h = Mercury Rotation Period 2802 h = Venus Day Period
346.6 days = the nodal year 224.7 days = Venus orbital period
687 days = Mars Orbital Period
4331 days = Jupiter orbital period = 2π x 687 days

I do this research
2nd
119
(No. 2) (Mars And Jupiter Motions)
- Mars moves during 636.7 days a distance =1325 mkm= (Venus Saturn Dis.)
- Jupiter moves during 636.7 days a distance =720.7 mkm
- Mars moves during 687 d. a distance= 1433 mkm= Mars Orbital Circumference
- Jupiter moves during 687 d. a distance= 778.6 mkm= Jupiter Orbital Distance
- Mars moves during 4331d. a distance= 9010 mkm= Saturn orbital circumference
- Jupiter moves during 4331 d. a distance= 4900 mkm= Jupiter Orbital Circum.
- Mars moves during 778.6 d. a distance = 1622 mkm = Uranus Neptune Dis.
- Jupiter moves during 778.6 d a distance= 881 mkm =58 days x 15.19 mkm.
- ( 15.19 mkm = the 9 solar planets motions distances total per a solar day).
- Notice
- 2.082 mkm / day x 778.6 days = 1622 mkm = 1.1318 mkm /day x 1433 days

I do this research
2nd
120
(No. 3) (Mercury And Jupiter Motions)
- Mercury moves during 687 d. a distance= 2815 mkm= Mercury Uranus Distance
- Mercury moves during 1433 d. a distance =5848 mkm= Mercury Pluto Dis.
- Jupiter moves during 1433 d a distance =1622 mkm=Uranus Neptune Dis.
- Mercury moves during 4331d. a distance= 2815 mkm x 2π
- Mercury moves during 4222.6 h a distance = 720.7 mkm
- Jupiter moves during 4222.6 h a distance = 200 mkm = 629 mkm/π
- Mercury moves during 550.7 d a distance = π x 720.7 mkm (0.4%)
- Jupiter moves during 550.7 d a distance = 629 mkm (error 1%)
Please Remember,
Data Group No. (1)
- Mars moves during 687 d. a distance= 1433 mkm= Mars Orbital Circumference
- Mars moves during 4331d. a distance= 9010 mkm= Saturn orbital circumference
- Mars moves during 4331 x π d a distance= 28255 mkm= Neptune orbital circumference
Data Group No. (2)
- Mercury moves during 687 d. a distance= 2815 mkm= Mercury Uranus Distance
- Mercury moves during 4331d. a distance= 2815 mkm x 2π
Data Group No. (3)

I do this research
2nd
121
II-Discussion
- The planets move defined distances in defined periods of time.
- The Defined Distance means a distance is known in the solar system, as to be any
planet orbital distance or a distance between any 2 planets.
- The Defined Periods Of Time means a period of any planet cycle, as 365.25 days
(Earth orbital period) or any planet orbital period. Or any planet rotation period or
any planet day period. All these periods are defined periods of time
- The argument tells that
- If the solar planets move their motions independently from one another, based on
that, the planets motions during defined periods of time should pass (random)
distances. Because these Planets are independent in their motion from one
another. (For example) Mercury motion depends on its orbital period (88 days) and
doesn't interest neither for Mars orbital period (687 days) nor for Jupiter orbital
period (4331 days) and by that, Mercury during these periods (687 days and 4331
days) should move some random distances (not defined in the solar system
distances).
- The data disproves Planet Motion Independency Concept. because the (3)
planets move defined distances in defined periods of time. These planets motions
aren't independent from one another. These motions are done based on One
Geometrical Design. And these motions be are similar to Chess Board Pieces
Motions. Each Motion is calculated geometrically and be obligatory.
- Notice
- Jupiter orbital distance analysis can support the discussion of this point greatly, we
discuss it in a separated point No. (9)

I do this research
2nd
122
7-3 The Solar Planets Move An Unified Motion (A Team Motion)
- Let's Summarize The Proof Idea In Following:
- This proof depends on 2 planets motion cycles are discovered later. Which are
(planet 8 days cycles) and (planet 6 days cycles). We study one Cycle only in this
paper because both cycles provide the same argument.
- The Cycle Shows That Different Planets Motions Be Used In The Same One
Cycle To Create One Final Result.
- Means,
- Many planets move relative one another to create one result.
- For Example
- Jupiter moves a distance (A) and Uranus moves a distance (B).
- The different distance between (A) and (B) = Jupiter Diameter
- Now
- The machine uses this behavior frequently, and (the different distance = Jupiter
Diameter) be created frequently while the cycle uses its different periods (8 days,
16 days, 24 days….etc).
- Based on that,
- We can't consider these planets motions are independent from one another because
the different distance be defined by the 2 planets motions. we have to consider
these planets motions as a team motion. They move relative to one another which
proves The Unified General Motion Concept.
- Based on that,
- (Planet 8 and 6 Days Cycle) disproves Planet Independent Motion Concept.
- In following we study Planet 8 Days Cycle

I do this research
2nd
123
(Planet 8 Days Cycle)
(I)
- Jupiter (13.1 km/s) moves during its day period (9.9 h) a distance = 466884 km
- But
- 466884 km = 449197 km (Jupiter Circumference) + 17687 km
- Where
- (8 x 17687 km = 141496 km (Jupiter Diameter) (error 1%)
- Based on that, we have concluded that, Jupiter has a cycle of 8 days
- Jupiter (13.1 km/s) moves during 8 Jupiter days (79.2 h) a distance = 3735072 km
- (3735072 km= 8 Jupiter circumferences + 141496 km (Jupiter diameter) (1%)
(II)
- The distance 3735072 km be passed also by Saturn and Neptune with a rate 80%
depends on one another as following:
- Saturn (9.7 km/s) moves during 10 Saturn days (107 h) a distance = 3736440 km
- (10 Saturn Circumferences = 3786750 km, the difference =50310 km = Uranus
Diameter error 1.5%)
- Neptune (5.4 km/s) moves during 12 Neptune days (193.2 h) a distance = 3755808
km
- (24 Neptune Circumferences = 3734323 km, the difference =21485 km = Mars
Circumference (error 0.6 %))
(III)
- Uranus (6.8 km/s) moves during Pluto day period (153.3 h) a distance = 3752784
km
- The distance 3752784 km = Jupiter motion distance during 8 days + 17687 km
- And because
- 17687 km x (8) = 141496 km (Jupiter Diameter) (error 1%)
- That tells another Cycle is found between Uranus and Jupiter based on 8 Pluto
days

I do this research
2nd
124
- That means, the distance be passed by Uranus during 8 Pluto days equal the
distance be passed by Jupiter during 64 Jupiter days and equal the distance be
passed by Saturn during 80 Saturn days and equal the distance be passed by
Neptune during 100 Neptune days
Let's see that in following
(1)
Jupiter (13.1 km/s) moves during (64 Jupiter days) a distance =29880756 km
(2)
Saturn (9.7 km/s) moves during (80 Saturn days) a distance =29891520 km
(3)
Neptune (5.4 km/s) moves during (100 Neptune days) a distance =31298400 km
(4)
Uranus (6.8 km/s) moves during (8 Pluto days) a distance =30022272 km
Comments
- Uranus motion distance (30022272 km) – Jupiter motion distance (29880756 km)
= 141496 km (Jupiter Diameter)
- The differences between these distances are less than 1 % (generally) and based on
that we can't consider they are different distances but we have to consider they are
equal distances.
- Although still there are small differences which are found for geometrical reasons
for example the difference between Jupiter and Saturn motions distances =
29880756 km – 29891520 km = 10921 = the moon circumference
- The data shows Planets Motions Dependency, because the different distances are
defined geometrically and that means these aren't 2 different distances of 2 plants
independent motions. On the contrary, the 2 distances are planned geometrically
and the 2 planets are 2 players to perform one different distance.

I do this research
2nd
125
The Discussion
- Let's discuss the previous data
(A)
- The outer planets are 5 planets, they consist 2 teams,
- The first team is consisted of Jupiter, Saturn and Neptune, these 3 planets move
based on a cycle (8 days cycle) depends on Jupiter motion with the rate 80%,
- That means
- The distance be passed by Jupiter in 8 Jupiter days be equal the distance be passed
by Saturn in 10 Saturn days and equal the distance be passed by Neptune during 12
Neptune Days
- The (small) difference between these 3 distances have geometrical necessities, as
we have seen in the difference between Jupiter and Saturn motions distances which
= 10921 km = The Earth Moon Circumference
- The moon circumference itself tells that it's a cycle because if it's not a cycle we
would find a part of the moon circumference
(B)
- The second team is Uranus and Pluto….
- Uranus uses Pluto day period (153.3 hours), and by that, Uranus (6.8 km/s) moves
during Pluto day period (153.3 hours) a distance = 3752784 km
- Because
- 3752784 km = Jupiter motion distance during 8 Jupiter days +17687 km
- Because of this data, we have concluded that, these motions depends on (8 days
Cycle), because
- Uranus needs to move during a period (= 8 Pluto days) to cause this value (17687
km) be = (141496 km (Jupiter Diameter) (1%)
- Because of Jupiter diameter we conclude that Uranus has a cycle of (8 Pluto days)
- Based on that

I do this research
2nd
126
- Uranus motion distance during 8 Pluto days = Jupiter motion distance during 64
Jupiter days = Saturn motion distance during 80 Saturn days = Neptune motion
distance during 100 Neptune days.
- How (Planet 8 Days Cycle) Can Prove The Unified Motion?
- Because
- Many planets motions be done to produce One Result
- This result be the different distance 141496 km (Jupiter Diameter) (1%)
- If we deal with planets independent motions this different distance can't be created
regularly and the Cycle can't be defined.
- Because (Planet 8 days Cycle) be defined, that means, the different distance
141496 km (Jupiter Diameter) be defined regularly which can be done only if
we deal with a team motion and NOT Planets independent Motions.
- Why does Uranus depend on Pluto Day Period?(additional question)
- Pluto day period is so long (153.3 h) in comparison with the outer planets days
periods. We suppose that, Uranus Motion effect on Pluto motion causes Pluto day
extension. We know Uranus did this effect because Pluto orbital inclination = 17.2
deg but Uranus day period =17.2 hours, even if we can't catch the mechanism of
this process yet, but the data shows that's Uranus effect on Pluto motion.
A Conclusion
Planet 8 Days Cycle disproves The Planet Independent Motion Concept,
On The Contrary, The Planets Move As A Team. (A Unified General
Motion)

I do this research
2nd
127
8- Mercury Motion Distance During Its Day Period
8-1 Preface
8-2 Why Mercury Jupiter Distance =720.7 mkm?
8-3 Jupiter Motion In A Comparison With Light Motion

I do this research
2nd
128
8-1 Preface
- As we have discussed in the planets motions rates of time
- Mercury uses its day period (175.94 days) in place of its orbital period (88 days)
and for that it consumes the period 671 h (as its required period) to pass a distance
= 2 x 57.2 mkm (Mercury orbital distance = 57.9 mkm error 1%)
- Also Mercury uses 720.7 mkm in place of its orbital circumference 360 mkm
- We have observed that in our discussion about the rates of time (points no. 3 and
4)
- The question was why?
- And the answer told that, because of the distance 720.7 mkm where Mercury
moves during its day period a distance =720.7 mkm = Mercury Jupiter distance
- The answer told that, Mercury Jupiter distance is the reason behind Mercury using
of its double values.
- In this point we analyze this distance to discover its effect on Mercury motion
- Please notice
- The distance 720.7 mkm has a wide discussion and be discussed in this point
No.(8), and also in points no. (9) and no. (11).

I do this research
2nd
129
8-2 Why Mercury Jupiter Distance =720.7 mkm?
I - Data
(8-2-a)
Mercury moves during its day period (4222.6 h) a distance =720.7 mkm
(8-2-b)
Mercury day period =2 Mercury orbital period = 3 Mercury rotation period
(8-2-c)
Jupiter moves during Mercury day period (4222.6 h) a distance =199.2 mkm
(8-2-d)
(240 mkm/158 mkm) =(88 days /58.65 days) (error 1.2%)
II – Discussion
(240 mkm/158 mkm) =(88 days /58.65 days) (error 1.2%)
- Where
- 240 mkm = Mercury Motion Distance during its rotation period (1407.6 hours)
- 158 mkm = π x 50.3 mkm (Mercury Venus Distance)
- 88 days = Mercury Orbital Period
- 58.65 days = Mercury Rotation Period
- Let's remember the question?
- Why Mercury Jupiter Distance =720.7 mkm? We have 2 reasons …. one reason is
related to Mercury Motion and the other is related to Jupiter motion.

I do this research
2nd
130
- (1st
Point Mercury Motion)
- Mercury Day period 4222.6 hours = 2 x Mercury orbital period 2112 hours, and
- Mercury Day period 4222.6 hours = 3 x Mercury rotation period 1407.6 hours…
- And
(I)
- Mercury moves during its rotation period a distance = 240 mkm (which is related
to Mercury Venus Distance) with the rate (1.5= Mercury orbital period / Mercury
rotation period).
- Means, Mercury Motion distance during its rotation period is related to Mercury
Venus Distance (it's related to Venus Basically because 243 days = Venus rotation
period)
(II)
- Mercury moves during its orbital period a distance = 360 mkm (Mercury orbital
circumference). (means this motion is related to Mercury Position)
(III)
- Mercury moves during its day period a distance = 720.7 mkm= Mercury Jupiter
distance
- By that
- Mercury has 3 cycles
- By its rotation period creates a connection with Venus (the most inner planet to it)
- By its orbital period does its revolution around the sun
- By its day period creates a connection with Jupiter (the nearest outer planet to it)
- There's some geometrical design on which Mercury moves
- And if this idea is a correct one, we can conclude simply that, Mercury has to
move a distance = 720.7 mkm= Mercury Jupiter distance by using on of its cycles
periods (and of course no better choice than Mercury Day Period)
- Geometrically, if Mercury connects with Venus by its motion during rotation
period, So mercury should connect with Jupiter with its day period.

I do this research
2nd
131
- If this description is a correct one, the question should be (why mercury day be
=4222.6 h) and this question we discuss in Jupiter orbital circumference analysis.
- Notice
- Mercury Venus distance is multiplied by (π) and then the rate (1.5)… while
Mercury Jupiter distance (720.7) be used directly… that because Mercury day
period is a period = (1 period) but mercury orbital period (0.5 period) and Mercury
rotation period be (1/3 Period). That means, the fractions be found with (orbital
and rotation periods) while day period uses the integer number… So the rates of
periods be seen in the distances.

I do this research
2nd
132
- (2nd
Point Jupiter Motion)
- Mercury Jupiter Distance =720.7 mkm = Mercury Motion distance during Mercury
day period (4222.6 h).
- But
- Jupiter moves during Mercury day period (4222.6 h) a distance = 199.2 mkm
- Because
- 720.7 mkm is Mercury Jupiter Distance, both planets may interest for this same
distance and we have seen that Mercury moves during its day period 720.7 mkm
and the question – Why Jupiter doesn't take the distance 720.7 mkm into account
in its motion?
- In fact Jupiter motion takes into consideration the distance 720.7 mkm,
- But the geometrical mechanism is hard for me to catch, let's refer to its data in
following:
- Jupiter moves during Mercury day period (4222.6 h) a distance = 199.2 mkm
- But
- 199.2 mkm x π = 629 mkm (Earth Jupiter Distance)
- And
- (629 mkm /929 mkm) = (637 mkm/ 940 mkm)
- Where
- 629 mkm = Earth Jupiter Distance
- 929 mkm = Earth Jupiter Distance when the 2 planets be on the sun 2 sides
- 940 mkm = Earth Orbital Circumference.
- 637 mkm = The Required Distance
- Jupiter moves during 637 days a distance = 720.7 mkm
- Based on that, Jupiter motion for 199.2 mkm causes its motion for 720.7 mkm,
means, some connection is found between the 2 motions
- Notice, this explanation can be more clear when we compare Jupiter motion with
light motion –we do that in point no. (8-3) for better explanation for this motion.

I do this research
2nd
133
8-3 Jupiter Motion In Comparison With Light Motion
I - DATA
(8-3-a)
4222.6 = 2π x 671
But
0.3 mkm /sec x 671 s = 201 mkm (light moves during 671s a distance 201 mkm)
13.1 km/s x 4222.6 h x 3600 s = 199.2 mkm (Jupiter moves 199.2 mkm)
(8-3-b)
4900 mkm = 0.3 mkm/sec x 16333 seconds
(8-3-c)
4900 mkm = 1.16 mkm/sec x 4224 seconds
(8-3-d)
4900 mkm = 199.2 mkm x 24.6
(8-3-e)
24.6 = (1.16/0.3) x 2π (error 1.2%)
(8-3-f)
(10747 x 24)/24.6 = 10485 = (4331 x 24)/ 9.9 h
(8-3-g)
(10747 /4331) =(24.6/9.9) = (24.1/9.7) = (17.4/7) =2.48

I do this research
2nd
134
II – Discussion
4222.6 h = 2π x 671 h
- Where
- 4222.6 h = Mercury Day Period
- 671 mkm =Venus Jupiter Distance
- But
- Light known velocity (0.3 mkm /s) during 671 s moves a distance (201 mkm)
- Jupiter (13.1 km /s) during 4222.6 h a distance (199.2 mkm) (error 1%)
- Where 4222.6 hours =Mercury Day Period
- The equation tells, Jupiter moves during Mercury day period (4222.6 h) a distance
= 199.2mkm = light motion distance during 671 seconds (671= 4222.6 /2π)
- We should notice that
- We deal with strict geometrical design, and no mistake is allowable
- Light known velocity (0.3mkm/sec) moves during 671 seconds a distance 201
mkm. We consider that light motion is the master one here
- Light motion effects on planet motion with the rate (2π) and both motions pass
equal distances based on this rate
- But
- 1 second of light motion = (2π) hours of Jupiter motion
- This is the rule we should keep in mind
- 1 second of light motion = 1 hour of planet motion but with the rate (2π) between
the passed distances. Means if the distances be equal so 1second be = (2π) hours
and if 1second be = 1hour the passed distance will not be equal but rated with this
rate (2π).
- Notice
- The rate (2π) rule we discuss in part (II) of this paper, but the rule 1 second = 1
hour isn't the general one, because the general one is 1 second = 1 day, but Jupiter

I do this research
2nd
135
here uses a part of this rate and we try to analyze its motion to see the depth behind
this rate.
- The conclusion is
- 1 second of light motion = (2π) hours of Jupiter motion
- We should see that as clear as possible
- Light does the motion at first, that means, we don't deal here with (720.7 mkm) but
we deal with light motion for a distance (201 mkm)
- This distance (201) will give a birth for planets motions distance (720.7 mkm) but
the distance (201 mkm) is the original one.
- Light And Jupiter Move Equal Distances
4900 mkm = 199.2 mkm x 24.6
- Where
- 199.2 mkm = Jupiter motion distance during Mercury day period (4222.6 h)
- 24.6 hours = Mars Rotation Period
- How to understand this hard equation (8-3-d)? we have to analyze more deep Mars
rotation period, let's try to do that in following
Equation no. (8-3-e)
- 24.6 = (1.16/0.3) x 2π (error 1.2%)
- Where
- 1.16 mkm/sec = light supposed velocity (will be discussed later)
- 0.3 mkm/sec = light known velocity
- Equation no. (8-3-e) tells a great information that, Mars rotation period is a point
inside which light supposed velocity (1.16 mkm/sec) interacted or cohered with
light known velocity (0.3mkm/ec)!
- We can't accept such great conclusion based on one simple equation, we have to
go to Mars and examine if its motion be related in this same conclusion

I do this research
2nd
136
- MORE DATA
(1)
- (300000 km /24.1 km) = (120536 km/9.7 km)
- Light known velocity (0.3 mkm/sec) / Mars velocity 24.1 km/sec = Saturn
diameter 120536 km / Saturn velocity for 1 second.
(2)
- 1160000 km = 24.1 km/sec x 48133 seconds
- Saturn (9.7km/s) moves during 48133 seconds a distance = 466556 km= Jupiter
motion distance during Jupiter day period = Neptune motion distance during a
solar day.
- But
- Saturn (9.7km/s) moves 300000 km during a period 30589 seconds (error 1%)
(30589 days = Uranus Orbital Period)
- Saturn (9.7km/s) moves 1160000 km during a period 120536 seconds (error 0.8%)
(120536 km = Saturn diameter)
- But
- (120536 /30589) = 3.94 but (1h of Mercury motion =3.91 h of Mars Motion)
- How to understand this data?
- The interaction between light known velocity (0.3 mkm/s) and light supposed
velocity (1.16 mkm/sec) be done in Saturn motion, and Saturn transports its
motion to Mars where the interaction between the 2 light beams velocities be seen
in Mars rotation period – that explains why Mars rotation period is so important as
such.
Equation no. (8-3-f)
(10747 x 24)/24.6 = 10485 = (4331 x 24)/ 9.9 h
Equation no. (8-3-g)
(10747 /4331) =(24.6/9.9) = (24.1/9.7) = (17.4/7) =2.48

I do this research
2nd
137
- 4331 days = Jupiter Orbital Period
- 9.9 hours = Jupiter Day Period
- 24.6 hours = Mars Rotation Period
- 24.1 km/s = Mars Velocity
- 9.7 km/s = Saturn Velocity
- 17.4 deg = the inner planets orbital inclinations total
- 7 deg = Mercury orbital inclination.
Additional Equation no. (1)
10 x 10747 days x 24 h = 10.7 h x 2 x 120536 = 2579280 hours - Let's remember
- Where
- 10.7 h = Saturn day period
- (1)
- Saturn (0.838 mkm /daily) moves during 120536 days a distance =100733mkm =
the distance be passed by light supposed velocity (1.16 mkm/s) during one solar
day (where 120536 km = Saturn Diameter)
- (2)
- The moon (2.4 mkm/ daily) moves during 10747 days a distance =25920mkm =
light known velocity (0.3 mkm/s) motion distance during a solar day. And
Additional Equation no. (2)
- 1433.5 mkm (Saturn Orbital Distance) = 17.75 mkm x (511.1 /2π)
- 17.75 mkm= the planets motions distances total per a solar day (this total contains
the moon velocity per a solar day 2.57 mkm = Earth velocity because they aren't
separated from one another)
- 511 deg = The Planets Axial Tilts Total
- Notice: these data will be more clear with the discussion of Part (II) point no (11)

I do this research
2nd
138
9- Jupiter orbital Circumference Analysis
9-1 Preface
9-2 Jupiter orbital Circumference Analysis

I do this research
2nd
139
9-1 Preface
- We analyze Jupiter orbital circumference (4900 mkm) here ,
- because
- it's very rich circumference in its discussion and we need to avoid the confusion
with the other parts of Jupiter motion data discussion
- for that reason we discuss this circumference here to provide the data analysis as
clear as possible
- we need Jupiter orbital circumference analysis in our discussion for Jupiter motion
data.

I do this research
2nd
140
9-2 Jupiter orbital Circumference Analysis
I –Data
(9-2-a)
4900 mkm = 1.16 mkm x 4224 = 1.1318 mkm x 4331 days
(9-2-b)
4900 mkm = 10921 km x 449197 km
And
9010 mkm = 10921 km x 2 x 413600 km
(9-2-c)
(708.7 h /10.7 h) = (655.7 h/ 9.9h) = 66.2
(9-2-d)
(9010 mkm /1433 mkm) = (4331 days /687 days) = 2π
(9-2-e)
(24.6 h/9.9h) = (24.1/9.7) =(10747 days /4331 days) = 2.48
But
(24.6/24.1) = (9.9/9.7) = (9.7/9.5)
(9-2-f)
0.3mkm/sec x 16333 sec =4900 mkm
But
27.78 km/sec x 16333 sec =449197 km (error 1%)
(9-2-g)
(37100 mkm - 4900 mkm) x 2π =28255 mkm +2 x86400 mkm
(9-2-h)
4900 mkm = 360 mkm +680 mkm +940 mkm +1433 mkm +1433 mkm
(9-2-i)
(90560 /88000) =(24.7/24) = (9.9/9.5) = ( 51118/49528) =(5040/4900)

I do this research
2nd
141
II-Discussion
- Where
- 4900 mkm =Jupiter Orbital Circumference
- 4222.6 hours = Mercury Day Period
- 1.16 mkm =Jupiter Motion Distance during 24.6h (Mars Rotation Period)
- 1.1318 mkm = Jupiter Motion Distance during 24 h ( a solar day)
- 4331 days = Jupiter orbital Period
- This equation cornerstone is the number (4224) because it's so near to (4222.6) =
Mercury day period.
- Mercury Day period = 4222.6 hours if (24.6 hours) of Jupiter Motion = 1 hour of
Mercury Motion- the periods (4331 days) and (4222.6 hours) will be equal
- Then
- The difference between (4224) and (4222.6) we can deal it later
- Now, the idea is that,
- Jupiter orbital circumference (4900 mkm) is designed based on the distance (1.16
mkm). So the distance (4900mkm) is designed to be 4224 distances each distance
=1.16 mkm.
- Jupiter (13.1 km/s) moves during 24.6 h (Mars rotation period) =1.16 mkm
- Means,
- The distance is designed based on 2 periods (4222.6 h and 24.6h) but the design
takes into consideration Jupiter velocity – means
- By using Jupiter motion and based on Jupiter velocity, the distance 1.16 mkm will
be passed on (24.6 h) and by that the circumference 4900 mkm will be = 4224parts
each part =1.16 mkm, means
- We have 3 players (Jupiter velocity, Mars rotation period, Mercury day period)
- Based on these 3 players Jupiter orbital circumference be design!

I do this research
2nd
142
Equation no. (9-2-a) (continued)
- The difference between 24 h and 24.6 h is (2.5%), and that causes the difference
between (1.16 mkm) and (1.1318 mkm) to be (2.5%) (= 28160 km)
- We realize now how this data be created, let's summarize it again in following:
- Jupiter velocity (13.1 km/s) be used by Mars rotation period (24.6 h) to pass a
distance = 1.16 mkm and Jupiter orbital circumference be design as a function in
the distance 1.16 mkm by the rate (4224) which is (Mercury Day Period 4222.6h)
- Means, Jupiter motion depends on 2 cycles (24.6 h and 4222.6 h)
- But
- Jupiter moves during 687 days a distance = 778.6 mkm = Jupiter orbital distance,
where 687 days = Mars Orbital Period
- That add one more player to the picture – because – We know– Jupiter motion
depends on 2 cycles (24.6h and 4222.6h) but now a third cycle be used also! Why?
- 687 days x 24 h = 24.6 h x 671 but (4222.6 h = 2π x 671)
- That means, some interaction is found inside Mars cycles periods found by some
effect of Mercury motion on Mars Cycles, this effect causes the period (687 days)
be dependent on the same 2 periods (24.6 h and 4222.6 h)
- Because of that Jupiter uses Mars Cycles Periods as its own periods!
- Please Note
- The data is created based on a geometrical system which we try to discover, But,
because we still don't know this geometrical system I try to arrange the data in best
from to discover it – for that reason I depend on the data arrangement – and
because of that – I add the data step be step to show clearly their effects – when we
discover the geometrical system behind this data – in this case- we will not be so
careful with data because we know the rule behind but now we don't this rule and
we should keep our eyes on the data arrangement because by the correct
arrangement the geometrical rule will be discovered.

I do this research
2nd
143
4900 mkm = 10921 km x 449197 km
And
9010 mkm = 10921 km x 2 x 413600 km
- Where
- 4099 mkm = Jupiter Orbital Circumference
- 9010 mkm = Saturn Orbital Circumference
- 10921 km = The Moon Circumference
- 449197 km = Jupiter Circumference
- 413600km = the moon supposed apogee radius!!
- (Notice the moon daily displacement =88000 km and during 29.53 days the total
distance = 2.598693 mkm = 2π x 413600km, means if the total distance (2.598
mkm) be the moon orbital apogee circumference (as should be) the apogee radius
should be (413600 km) –but the apogee real radius is 406000 km – for that I call
the radius 413600km (the supposed apogee radius)
- Equation no. (9-2-b) tries to show that, Jupiter and Saturn orbital circumferences
be defined as functions in the moon circumference – that creates more complexity
for our discussion – because –Jupiter orbital circumference is already effected and
designed based on 2 cycles (24.6 h and 4222.6 h) and now the moon circumference
is effected also which creates more difficulties for out discussion
- Equation no. (9-2-b) tries to show that, Jupiter and Saturn orbital circumferences
are designed taken into consideration the moon circumference – that because – the
2 planets effect greatly on the moon motion -
- I want to say
- Jupiter orbital circumference be created effected by 2 forces (at least), the first
force is the 2 cycles periods effect (24.6 h and 4222.6 h) and the second force is
the moon circumference

I do this research
2nd
144
- I add the connection with the moon circumference here to show that Saturn motion
effects on Jupiter orbital circumference design and creation – after we discuss this
point we should return again to Jupiter orbital circumference analysis to see how
it's designed –
(708.7 h /10.7 h) = (655.7 h/ 9.9h) = 66.2
- Where
- 10.7 h = Saturn Day Period
- 708.7 h = The Moon Day Period
- 655.7 h = The Moon Rotation Period
-
- This equation is the cornerstone in the relationship between the moon on one side
and Jupiter and Saturn on the other side.
- This equation is a treasure – it has no error – means its error = (ZERO)
- Means,
- These 4 numbers are created together in one process!
- This equation we have discussed with the planets motions rates of time

I do this research
2nd
145
(9010 mkm /1433 mkm) = (4331 days /687 days) = 2π
- Where
- 9010 mkm = Saturn Orbital Circumference
- 1433mkm = Mars Orbital Circumference
- Equation (9-2-d) shows, Mars orbital period and circumference be created based
on Jupiter and Saturn orbital periods and circumferences by the same rate (2π)
- This equation is provided to show that, Jupiter motion uses mars rotation period
(not by chance) but based on a geometrical necessity
- I want to say
- The planets are similar to gears in one machine of gears, but, the planets in fact
aren't gears, and we don't know how to attribute (the motion transportation) for
planets motions. because we can't attribute any mechanical features for the Space
between the planets and no direct touch causes the motion transportation from a
planet to another – but the data shows that-any planet motion can't be considered
as independent or individual motion –for that reason we analyze the planets
motions data to discover how the motions data be transported or be taken into
consideration in comparison with other planets data
- Also this equation we have discussed with the planets motions rates of time

I do this research
2nd
146
(24.6 h/9.9h) = (24.1/9.7) =(10747 days /4331 days) = 2.48
But
(24.6/24.1) = (9.9/9.7) = (9.7/9.5)
- Where
- 24.6 h = Mars Rotation Period
- 24.1 km /s = Mars velocity
- 9.7 km /s = Saturn velocity
- 2.5 degrees = Saturn Orbital Inclination
- Equation no. (9-2-e) shows that many data of Jupiter and Saturn be created in
proportionality with one another –
- I use this equation to show my deep effect be occurred between planets motion..
- The equation itself is straight for example
- (24.6 h/9.9h) = (24.1/9.7), these 2 parts are straight because the equation divides
period on period (24.6 h/9.9) and velocity on velocity (24.1/9.7)
- But
- We can create some exchange by the geometrical rules, and based on that
- (24.6/24.1) = (9.9/9.7), this equality is correct geometrically, but it's not straight,
because, it divides (period) on (velocity) to be = (period) on (velocity)
- For that we will forget the units of numbers and use them arithmetically
- Based on that
- (24.6/24.1) = (9.9/9.7) = (9.7/9.5), And So,
- (9.7)2
= 9.9 x 9.5
- This is the equation I mean to use – this is a very important one –let's explain why

I do this research
2nd
147
- Saturn (9.7 km/s) moves during its day period (10.7 h) a distance = 373644 km
which differs from Saturn Circumference (378675km) with (1.3%)
- This data creates some idea tells, The Planet Moves During Its Day Period A
Distance = Its Circumference ( a concluded rule)
- I don't know what result can be produced if the planet does this motion – but
- Jupiter moves during its day period a distance = 104% of its circumference
- Means, there's a difference =4%
- That can kill the concluded rule simply
- Jupiter during 9.5 hours moves a distance = its circumference, the problem is that,
Jupiter day period =9.9 hours, this difference causes the error (4%) between Jupiter
motion distance during its day period and Jupiter circumference
- But we know that
- The difference 4% between Jupiter motions distance and its circumference causes
to create planet 8 days cycle.
- Shortly
- (9.7)2
= 9.9 x 9.5
- Where
- 9.7 km/sec = Saturn velocity
- 9.5 h = the period during Jupiter moves a distance = its Circumference
- I try to show that, Jupiter doesn't move during its day period a distance = its
circumference because of a geometrical necessity. Means
- An interaction of motions between Jupiter and Saturn caused Jupiter day period to
be longer with 4%.
- I hope the data does the task well
- We have many clear proves that, Jupiter motion through its orbital circumference
depends on other planets motions and cycles periods.

I do this research
2nd
148
0.3 mkm/sec x 16333 sec =4900 mkm
But
27.78 km/sec x 16333 sec =449197 km (Jupiter Circumference) (error 1%)
- Where
- 0.3 mkm/sec = Light Known Velocity
- 27.78 km /sec = The Moon Velocity
- Equation no. (9-2-f) tells that, the period the light uses to pass Jupiter orbital
circumference =the period the moon uses to pass a distance =Jupiter circumference
- This equation importance is found because the light uses the same period of time
- I try to show that,
- The Data Is Created Based On A Complex Geometrical Machine
- Please remember
- Jupiter (13.1 km/sec) moves during 10921 seconds a distance = 142984 km =
Jupiter diameter.
- And
- 4900 mkm (Jupiter orbital circumference) = 449197 km x 10921 km
- Where
- 449197 km = Jupiter Circumference
- 10921 km = the moon Circumference
- There's a very complex interaction is found between Jupiter, Mars, Saturn the
moon and Mercury… these planets motions data be used frequently with Jupiter
motion data -
- Shortly
- There's a complex machine behind Jupiter orbital circumference design

I do this research
2nd
149
(37100 mkm -4900 mkm) x 2π =28255 mkm +2 x 86400 mkm = 2 x 1.16 mkm/sec
x 86400 seconds
Where
- 37100 mkm =Pluto Orbital Circumference
- 28255 mkm =Neptune Orbital Circumference
- 680 mkm =Venus Orbital Circumference
- (28255 mkm – 680 mkm) x 2π = 86400 mkm
- Equation no. (9-2-g) is the solar system main equation because it shows the energy
distribution trough the solar system – we have to discuss this equation in details in
our discussion for light motion accompanying planet motion in (Part II) (Point
no.11)
- The point we need to discuss here is the distances distribution based on Jupiter
orbital circumference.
- The distance between Jupiter and Pluto by some geometrical method defines the
distance between Neptune and Venus –also we can use Earth orbital circumference
in place of Venus' because the error in both equations be only 1%
- We don't explain the equation third part (1.16 mkm/sec x 86400 sec) because it
depends on light motion hypothesis – we discuss it in light motion accompanying
planet motion (Point no.11)
- The difficulties be more clear now, let's try to explain as possible in following:
- The Planet Motion be done based on integration of other planets motions – means-
No Independent Motion be done in the solar system
- Because of that –
- Planets Motions data be transported among the planets and be used by each other –
for that reason we have seen that Jupiter uses Mars rotation period (24.6h) to move

I do this research
2nd
150
a distance =1.16mkm based on which Jupiter orbital circumference be design taken
into consideration Mercury day period (4222.6 h).
- Shortly
- Planets Motions Data be Transported from planet to another
- Newton theory has no a method to explain how the data be transported from one
planet to another – as a result – the explanation be usually (pure coincidence of
numbers) which kills our chance for understanding and remove the priceless data
- On the other side the planets aren't gears and the motion transportation, or energy
transportation which causes the data transportation is a process not defined yet and
we can't catch its features spite the deep analysis
- Based on this situation,
- I have tried to divide my work into 2 different tasks
- The first task to prove that, the planets motions can't be created independently –
Simply Planet Motion Independency Concept is Mistaken (Newton Theory is
wrong) – and to prove this idea – I have arranged the planets data in best form to
prove that- the motions can't be done independently
- The previous data is example….
- Although I have no explanation how this data be created but the data prove clearly
that no motion can be done independently
- The second task is to try to explain how the planets motions be done depending on
one another – and this point is discussed in the paper theory of (light motion
accompanying planet motion) (Point no.11)
- As a result we conclude, for now, that Jupiter orbital circumference is designed
based on a geometrical design takes into consideration the other planets motions
data.

I do this research
2nd
151
Equation no. (9-2-h)
4900 mkm = 360 mkm +680 mkm +940 mkm +1433 mkm +1433 mkm (error 1%)
- Where
- 360 mkm = Mercury Orbital Circumference
- 940 mkm = Earth Orbital Circumference
- 1433 mkm = Mars Orbital Circumference
- Why (1433 mkm) is used 2 times? (my answer is)
- The data considers the moon orbital circumference around the sun = 1433 mkm=
Mars orbital circumference –
- Why? I don't know
- Can this data be invented one? let's see the following data:
(1)
- Mercury moves during its day period a distance = Mercury Jupiter distance
(2)
- Venus moves during its orbital period a distance = Venus Jupiter distance (1.2%)
- (because Venus orbital circumference = Venus Jupiter distance wit error 1.2%)
(3)
- Earth moves during its orbital period a distance = Earth Jupiter distance (1.2%)
- (because Earth orbital circumference = Earth Jupiter distance wit error 1.2%)
- (notice in this case should Earth and Jupiter be on 2differnt sides from the sun)
- Means,
- The 3 planets define their orbital circumferences to be equal or rated to their
distances to Jupiter – this data should be added to the previous and think about
them together – some specific relationship be found between the 3 inner planets
and Jupiter motions data

I do this research
2nd
152
II-Discussion (continued)
- In following we need analyze Jupiter data in more deep analysis – the point is that,
we can't catch the idea behind – the data is so rich but the understanding is limited
- Let's try again in following
Jupiter motion in 24 h in 24.6 h A distance
778.6 mkm 687 671 671 mkm = Jupiter Venus distance
720.7 mkm 637 621 629 mkm == Jupiter Earth distance
671 mkm 593 580
629 mkm 555.7 542 550.7 mkm = Jupiter Mars Distance
4900 mkm 4331 4224 4224 mkm =2π x 671 mkm
2094 mkm 2094 mkm = (1.16)2
x 778.6 x 2
- Distances
- 778.6 mkm = Jupiter Orbital Distance
- 720.7 mkm = Jupiter Mercury Distance
- 671 mkm = Jupiter Venus Distance
- 629 mkm = Jupiter Earth Distance
- 550.7 mkm = Jupiter Mars Distance
- 2094 mkm = Jupiter Uranus Distance
- Periods
- 4331 days =Jupiter Orbital Period
- I try to show that, Jupiter motion isn't explained by out analysis of data – it still
have great secrets need to be discovered –
- Jupiter motion data shows that, planet motion depends on a very complex machine
which uses different cycles periods in one motion and can used distances values as
periods of time.

I do this research
2nd
153
- Notice
- 413600 km=9.7 km/sec x 42640 seconds
- But
- 30589 days x 24 h = 17.2 h (Uranus day period) x 42682
- This data tells
- 1 second of Saturn motion = 17.2 hours of Uranus motion!
Equation no. (9-2-i)
(90560 /88000) =(24.7/24) = (9.9/9.5) = ( 51118/49528) =(5040/4900)
- 90560 days = Pluto Orbital Period
- 88000 km = The moon daily displacement
- 24.7 hours = Mars Day Period
- 24 hours = Earth Day Period
- 9.9 hours =Jupiter Day Period
- 51118 km = Uranus Diameter
- 49528 km = Neptune Diameter

I do this research
2nd
154
10- The Sun Age Description
10-1 The Sun Circles The Earth
10-2 The Rate (1.0725)
10-3 The Sun Diameter Analysis
10-4 The Sun Rays Analysis

I do this research
2nd
155
10-1 The Sun Circles The Earth
- The sun gives one face always to the Earth during the motion. although Earth
revolves around the sun a complete revolution, we always see the same face of the
sun, that tells the sun rotates with us giving The Earth The Same face always
- Can this motion refer to the Sun dependency on the period (365.25 days = Earth
orbital period)?
- Can that lead us to conclude, the sun be created after the Earth and because of that,
The Sun motion takes into consideration the period 365.25 days?
- Also
- The Sun Behavior is similar perfectly to the moon behavior, both give the same
face to the Earth during the motion.
- Why we accept that the moon be created after The Earth Creation but The Sun be
Created Before The Earth Creation? if the 2 behaviors are similar it may prove
both (the sun and the moon) be created after The Earth Creation.

I do this research
2nd
156
10-2 The Rate (1.0725)
I- Data
(A)
(1)
= Mercury Jupiter Distance x 2
(2)
Earth Neptune Distance = Mercury Saturn Circumference (error 0.5%)
(3)
Jupiter Venus Distance = Venus Orbital Circumference (error 1.5%)
Jupiter Earth Distance = Earth Orbital Circumference (error 1.2%)
(4)
Jupiter Mercury Distance = Mars Orbital Distance x π (error 0.6%)
Jupiter Uranus Distance = Venus Jupiter Circumference (error 0.8%)

I do this research
2nd
157
(B)
More Data
Why These Distances Are NOT Equal?
1. 0725
.
1
mkm
2.41
nce
Circumfere
Orbital
Moon
mkm
2.58
Motion
Daily
Earth
=
2. 1.0725
km)
(378500
radius
Eclipse
Solar
Total
km)
(406000
radius
orbital
Apogee
=
3. 0725
.
1
distance
Mercury
Jupiter
mkm
720.3
Distance
Orbital
Juppiter
mkm
6
.
778
= (Error 0.7%)
4. 1.0725
Distance
Venus
Jupiter
mkm
670
distance
Mercury
Jupiter
mkm
720.3
=
5. 1.0725
Distance
Earth
Jupiter
mkm
629
Distance
Venus
Jupiter
mkm
670
= (Error 0.6%)
6. 1.0725
mkm)
(1325.3
Distance
Venus
Sarurn
mkm)
(1433.5
Distance
Orbital
Saturn
= (Error 0.8%)
7. 1.0725
mkm)
(1205.6
Distance
Mars
Sarurn
mkm)
(1284
Distance
Earth
Saturn
= (Error 0.7%)
8. 1.0725
mkm)
(2644
Distance
Mars
Uranus
mkm)
(2872.5
Distance
Orbital
Uranus
= (Error 0.7%)
9. 1.0725
mkm)
(4495.1
Distance
Orbital
Neptune
mkm)
(4846
= (Error 0.5 %)
(10)
Notice
4846 mkm =the inner planets orbital circumferences total. it will be discussed with Jupiter
orbital circumference analysis
0725
.
1
T.
Axail
Earth
23.4
T.
Axail
Mars
25.2
T.
Axail
Mars
25.2
T.
Axail
Satrun
26.7
Tilt
Axail
Satrun
26.7
Tilt
Axail
Neptune
28.3
=
=
=

I do this research
2nd
158
II – Discussion
- We have discussed The previous data before
- 50% of all distances in the solar system be equal one another, and
- 40% of all distances in the solar system be rated one another with the same rate
(1.0725)
- We answered why 50% of distances are equal and 40% are rated
- Let's ask
-
- How this data can prove that (the sun I created after all planets creation and
motion)? let's answer in following
- Let's consider this question in following:
o The usual case is the equal distances as the data in Point no. (A), That
means, these rated distances were equal distances one another before and all
of them be effected by the rate 1.0725,
o Means,
o We have some historical order here, which is
o The distances were equal, but
o Some change caused these equal distances to be rated with (1.0725)
o And, 1.0725 = (100/100) + (7.25/100)
o We can conclude that,
o The distances were equal till The Sun Creation, and after the sun creation,
the sun obliquity on ecliptic (7.25 deg) caused the equal distances to be
rated with one another by the same one rate (1.0725).
o Means, the rate 1.0725 proves that, The Sun Is Created After All Solar
Planets Creation.
o One more feature supports this powerful argument, that

I do this research
2nd
159
o The same rate (1.0725) is used for different distances. Because the sun
effects on all solar planets. That shows the rate (1.0725) must be related to
the sun because it's the same one rate. If this rate be belonged to any planet,
it will not effect on different distances equally because the sun only effects
on the planets equally. That shows the rate 1.0725 is belonged to the sun effect.

I do this research
2nd
160
10-3 The Sun Diameter Analysis
I-Data
1- Diameters
- (6792 km Mars Diameter) / (4879 km Mercury Diameter) =1.392
- (4879 km Mercury Diameter) / (3475 km the moon Diameter) =1.392 (1%)
- (71492 km Jupiter Radius) / (51118 km Uranus Diameter) =1.392
2- Distances
- (57.9 mkm Mercury orbital distance) /(41.4 mkm Venus Earth distance)=1.392
- (108.2 mkm Venus orbital distance) /(78.3 mkm Earth Mars distance)=1.392
- (149.6 mkm Earth orbital distance) /(108.2 mkm Venus orbital distance)=1.392
- (4267 mkm Mars Neptune distance) /(3061 mkm Saturn Neptune distance)=1.392
- (778.6 mkm Jupiter orbital distance) /(550.7 mkm Mars Jupiter distance)=1.392
- (550.7 mkm Mars Jupiter distance x 2) /(778.6 mkm Jupiter orbital distance) =1.392
- (5127 mkm Pluto Jupiter distance) /(3716 mkm Jupiter Neptune distance) =1.392
(error 1%)
- (5906 mkm Pluto orbital distance) /(4267 mkm Mars Neptune distance) =1.392
- (2872.5 mkm Uranus orbital distance) /(2094 mkm Jupiter Uranus distance) =1.392
3- Periods
- (243 days Venus Rotation Period) /(175.94 days Mercury day period) =1.392
- 2.5 deg (Saturn orbital inclination) /1.8 deg (Neptune orbital inclination) =1.392
- 1.8 deg (Neptune orbital inclination) / 1.3 deg (Jupiter orbital inclination) =1.392
- 5.1 deg (The moon orbital inclination) / 3.66 =1.392
4- The Sun Diameter
- 1.392 mkm (The Sun Diameter) = Venus Diameter 12104 km x 115.2
- 1.392 mkm (The Sun Diameter) = Mars Diameter 6792 km x 205.2
- 1.392 mkm (The Sun Diameter) = Neptune Diameter 49528 km x 28.3

I do this research
2nd
161
II- Discussion
- The data groups no. (1,2 and 3) show that,
- The rate 1.392 is used frequently among planets different data.
- Why? because
- The solar system is similar to one machine of gears and each planet is one gear in
it. because the machine moves one unified motion, the gears data should be in
proportionality with one another.
- Similar to that, the planets creation and motion data be created in proportionality
with one another. we will prove that in details in thus paper discussion.
- Based on this vision the solar planets data be created depends on one geometrical
design.
- The Sun Diameter Be defined geometrically before the sun creation.
- Because
- The machine depends on one geometrical design and each part should be created
in harmony with this geometrical design otherwise the machine can't work.
- We should notice that, the rate 1.392 is defined with the solar system creation
geometrically, Before the sun creation and also before the planets migration.
- This definition is done geometrically, so, the value (1.392) in the data is very
ancient. But the sun diameter (1.392 mkm) be created later.
- Let's discuss the data no. (4)

I do this research
2nd
162
Group Data No. (4)
- The sun diameter 1.392 mkm = Venus diameter 12104 km x 115.2 = Mars
diameter 6792 km x 205.2, But
- 25.2 + 90 = 115.2 and 115.2 +90 =205.2 where (25.2 degrees = Mars Axial Tilt)
And
- The sun diameter 1.392 mkm = Neptune diameter 49528 km x 28.3
- And 28.3 degrees = Neptune Axial Tilt
Discussion
- The data shows that, Venus, Mars and Neptune diameters are rated with the sun
diameter based on Mars axial tilt and Neptune axial tilt, Means,
- We have 3 planets diameters are rated with 2 planets axial tilts based on the sun
diameter.
- More logical that, the sun diameter is created after the 3 planets diameters,
depending on their values, causing planets axial tilts be created in proportionality.
- The opposite choice can't be working. Because if the sun be created at first and all
planets later, that creates a rule for the sun effect on the planets axial tilts, by that
we have to find (not) 3 planets only their axial tilts be rated based on the sun
diameter proportionality with their diameters, but found all planets behave by this
same rule because the sun has equal effect on all of them.
- The first option is more logical, that the sun is created after the planets creation and
the sun depended on 3 planets (specifically) for its diameter creation. by that the
data be more understandable
- Shortly
- The sun diameter analysis gives support for the claim (the sun is created after all
planets creation and motion).
Notice (1)
- In kepler third law in which (D3
= P2
x constant), the constant used for the solar
planets orbital distance and periods should be = 25

I do this research
2nd
163
10-4 The Sun Rays Analysis
- Let's Summarize The Sun Rays Creation Idea In Following:
(Point No. A)
o The Solar Planets Move One Unified General Motion. This Idea we discuss
and prove clearly in this paper.
o The planets unified general motion causes to unify the planets velocities in
one unifies velocity = the planets velocities total.
o The 9 planets velocities total =176 km/sec,
o But
o The Earth moon is a player and its velocity should be added, where the
moon velocity be considered = Earth velocity because they aren't separated
from one another through their motions.
o The 10 planets velocities total be 176 +29.8 =205.8 km/s
o Based on this velocity (205.8 km/s) the sun rays be created
(Point No. B)
o 10921 km (The Moon Circumference) x 86400 Seconds = 940 mkm
o This Equation tells. If Earth revolves around the sun a complete revolution
(=940 mkm = Earth Orbital Circumference) in one solar day only (86400
sec), in this case, the moon circumference (10921 km) will be equal the
Earth motion distance during 1 second.
o But, because
o Earth revolves around the sun in one year (=365.25 days), so this equation
may refer to a rate of time is used by The Sun herself.
o Let's use this idea as a hypothesis. where
o 1 Day Of The Sun Motion = 365.25 Days Of Earth Motion
o We suppose that's The Sun Rate Of Time

I do this research
2nd
164
(Point No. C)
o The velocity (205.8 km/s) passes during a solar day (86400 s) a distance
=17.75 mkm
o 1 day of Earth motion = 237 seconds of the sun motion.
(This value depends on our hypothesis)
o Now
o The distance =17.75 mkm and the time =237 seconds what's the velocity?
o 75000 km/s = (a quarter Light Velocity) = 0.25 C
o But
o Earth has a cycle of 4 years (1461 days = 365 +365 +365 +366)
o Based on one year, the velocity be (0.25 C)
o And
o Based on the cycle (4 years), the velocity (0.25C) became (C light velocity)
o Based on this explanation the sun rays be created.
(Point No. D)
o There's one more method to create the sun rays
o 0.25 C x 4 C = C2
o Where
o The velocity 0.25 C is produced already in the previous discussion
o We need only the value (4C) to perform this equation
o Jupiter motion enjoys by light motion supposed velocity (1.16 mkm/s) this
value is very near to our required value (4C)
o Light supposed velocity travels for 1 second and passed 1.16 mkm
o And
o Earth rotates around its axis one per solar day and move a distance = its
circumference =40080 km in a solar day
o If 1 solar day of Earth motion be = 1 second of light supposed velocity
motion.

I do this research
2nd
165
o Based on that, the distances can be added
o 1.16 mkm + 40080 km = 1.2 million km = 4C
o By that the equation works
o 0.25 C x 4 C = C2
o That explains why, Jupiter and The Moon data are the 2 basic players in the
sun data definition. Because the sun creation depends on Earth and Jupiter
motion. in addition to Uranus motion which is seen in the Earth moon
motion.
o The data explains how the sun rays be created
(Point No. E)
o The rate of time (1 solar day of planet motion = 1 second of light motion) is
the original rate of time in the solar system. We should refer to that in the
planets unified general motion discussion.
o The idea is that, the light motion for 1 second provides the required energy
for a planet to move one solar day, because the light motion depends on the
solar day as we have discussed. By that the planets divides this great rate of
time (1 sec = 86400 sec) on their motions to work as a great clock and
produce this rate by their Unified General Motion.
o The basic part (1 sec =365.25 sec) be between the Earth and the Sun motion.
(Point No. F)
o Because the solar system be created out of a light beam its velocity 1.16
mkm/sec
o And the sun light velocity 0.3 mkm/sec
o That proves the light (0.3 mkm/sec)be created as a product from the original
one 1.16 mkm/sec
o And that proves
o The sun is created after all solar planets creation and motion.

I do this research
2nd
166
Extensive Discussion
- Why Does The Previous Investigation Prove The Claim?
- The claim tells that,
- The Sun Is The Solar System Last Created Piece. And that means, the sun is
created after All Planets Creation And Motion.
- If this fact can be proved, it will disprove Newton Theory of The Sun Mass
Gravity decisively.
- How to prove that the sun is created after all planets creation and motion?
- In addition to the historical proves there's one basic proof which is
- The sun rays velocity be = 300000 km
- While
- The solar planets matters and distances be created out of light beam its velocity
=1.16 mkm/sec.
- As a result, I have analyzed the planets data to prove that, this data be created
based on a light beam its velocity =1.16 mkm/sec
- After this analysis, we have to describe how the sun rays be created and the
previous data provides this description.
- Now
- Why Does This Analysis Provide A Sufficient Proof?
- The proof sufficiency depends on the arguments and discussions logic
- The discussions used hundreds of the planets data and put them in some form to
prove some idea. I can create the idea but can't create the data. that means if the
data be in harmony with the idea that supports the proof sufficiency.
- We have used hundreds of data in different discussion, if the discussion logic be
straight and the data be in harmony with it that proves the claim clearly
- The point is that, we searched for the rules based on which this data be created. so
as much as the used data be in harmony with these rules that prove these rules
sufficiency.

I do this research
2nd
167
- For example
- I claim (1 day of the sun motion = 1 year of Earth motion). this idea we have
accepted as a hypothesis and used it based on that
- But
- This idea is created because of the moon circumference equation (10921 km x
86400 seconds =940 mkm). Means, we have some data refer to this hypothesis.
- So we have used this idea as a hypothesis.
- We have found that, the data be in harmony and produce a velocity = (0.25 C)
- This velocity be created as a result for the hypothesis we suggested because of the
moon circumference
- But
- Earth Cycle (4 years) can change this velocity (0.25C) into ( 300000 km/s)
- I try to show that, w don't create the data
- We Rearrange The Data In New Positions Relative To One Another
- This is the basic method of the proof.
- Then we find that
- Jupiter (again) provides us a chance to create the value (C2
) (which can be a
source of the sun rays energy)
- So we used the velocity 1.16 mkm/s which we have learnt from Jupiter motion
data and added it to earth circumference to produce the required value (4C)

I do this research
2nd
168
Paper Part No. (II)
The Solar System Creation Theory

I do this research
2nd
169
11- The Solar System Creation Theory
11-1 Preface
11-2 The Solar System Creation Theory
11-3 The Inner Planets Motions Reason
11-4 Light Beam (1.16 mkm/sec) Proves Discussion
11-5 Saturn Motion More Analysis
11-6 Equal Distances Method

I do this research
2nd
170
11-1 Preface
- There's a light beam its velocity =1.16 million km per second
- The solar planets matters and distances are created of Energy provided by this light
beam whose velocity (1.16 mkm/sec) (The Theory hypothesis)
- Shortly
- The solar planets are created and moving based on one light beam its velocity is
1.16 million km per second
- Means,
- The light beam whose velocity 1.16 mkm/sec is The Real Motor found behind the
solar planets motions.
- In following I provide my theory for the solar system creation and motion. and I
provide many explanations for this theory
- But
- The theory is NOT the objective of this point discussion –
- We discuss this point to prove that (a light beam its velvety 1.16 mkm/sec be
found in the solar system motion and caused its creation)
- This point discussion interests basically for the light beam its velocity =1.16
mkm/sec
- I want to say
- Through the paper discussions I have provided data disproves Newton theory and
created astonishments and surprises for the respectful readers because they show
un-expectable using of data –
- Through the whole paper discussions, we have moved from planet data to another
trying to understand how this data be created – and left hundreds of questions
behind because no answers we could reach for our questions…
- Then

I do this research
2nd
171
- The light beam velocity 1.16 mkm/sec be supposed and then hundreds of data be
understandable –
- A great light came through our cave darkness and then we have one chance to
leave it ….
- A light beam its velocity 1.16 mkm/sec is found under the solar planets matters
and distances creation – this light beam is the responsible for the planets motions
and the distances creation and distribution – it's the main force behind and controls
every thing and saves the solar group
- I want to say
- I provide a good theory about the solar system creation and good explanations for
different features of planets motions- but –this is a cover –
- I make a cover for the existence proves discussion of the light beam velocity (1.16
mkm/sec).
- So,
- We keep our eyes on the data discussions because we –here – reread the same data
once again and will discover that the planets motions data be created as a result of
the light supposed velocity (1.16 mkm/sec) effect on these planets motions.

I do this research
2nd
172
11-2 The Solar System Creation Theory
(I)
- Matter is Created out of light beam, and the born matter isn't separated from its
light parent but moves with it one unified motion by using different rates of time
between the matter and light.
- The matter is created of energy and the space is created of energy and by that, the
matter is created in motion.
(II)
- The solar planets and their distances are created from one light beam its velocity
=1.16 mkm/sec (The Theory Hypothesis)
- The energy of the light supposed velocity (1.16 mkm/sec) be used for the solar
system creation and by that this light beam motion features be registered in the
planets matters and their distances.
- The planets move in comparison with their parent light beam in one unified
general motion. by using different rates of time
- The basic rate of time is (1 Second Of Light Motion = 1 Day Of Planet Motion)
- Or
- (Energy of light motion for 1 second be used to move a planet for 1 solar day)
- This great rate be divided among the planets motions (as we have discussed in
planets motions use different rates of time – Points no.3, 4 and 5 )
(III)
- Based on this vision
- The Solar System Be Similar To A Great Clock.
- The Clock input is an energy of light motion for 1 second and the clock output is
an energy enough for a planet motion for 1 solar day –
- The rates of time be used by the planets to create One Unified General Rate Of
Time.

I do this research
2nd
173
(IV)
- The solar planets be similar to points on the same one trajectory of energy – or
they are knots on the same rope –
- Also
- The solar planets be similar to gears in one machine of gears – or the planets be
similar to puppets on a puppets theater –
- The rope can show the best similarity for the solar planets – the matter and apace
be created of light beam energy – and by that – the light creates planet matter and
its distance by light energy – by that – light created planet matter and its distance
to be in proportionality with other planets data because all of them be created out
of the same one energy.
- The creation process of planet matter and distance isn't discovered yet but the light
motion supposed velocity (1.16 mkm/sec) be registered in the planets motions data
and by that we discovered that these planets be created by this light beam energy.
(V)
- Because all planets be created of one light beam, the planets move with this light
beam and by that we conclude simply a light beam motion be in accompanying
with planets motions –
- Light motion be accompanying with one motion– because of that
- The solar planets motions be integrated into One Unified General Motion.
- The solar planets move one unified general motion – This motion be similar to
Train Carriages Motion – they all move together one unified general motion– spite
the planets velocities are different from one another – but as a machine of Gears –
the gears move by different velocities but create one unified general motion.
(VI)
- A river water motion can be very good example for our explanation ….

I do this research
2nd
174
- Imagine 9 waterwheels be built on one canal, so the 9 waterwheels rotate because
of the water motion (the water motion here is similar to the light motion and the
waterwheels are similar to the planets)
- This example is a great help for our explanation – the water motion direction is
different from the waterwheels rotation direction but no waterwheel can rotate
without water motion – also – the waterwheel motions effect one another because
they effect on the water amount for one another – that perfectly shows how the
solar system works
(VII)
- Now we have one more great help,
- The solar planets move One Unified General Motion because the light beam deal
with the unified motion of the planets
- The planets unified motion forces each planet to move a complementary motion
which can be integrated with one another to create the planets unified general
motion.
- Now each planet motion became (some how) obligatory motion because it should
be integrated with other planets motions to create the unified general motion.
- By that the planets motions be obligatory motions
- Planet data should be in harmony with this planet motion –the motion is obligatory
to be complementary with other planets motions – based on that – the planets
motions data should be complementary data with one another.
- Based on this vision
- There's one law controls all planets motions data –that because – the data should
be created in harmony with their motions and the motions must be complementary
to create one unified general motion.
- By that, the light beam motion controls the solar planets motions and their data –
shortly we have One Law controls all data.

I do this research
2nd
175
(VIII)
- Light supposed velocity (1.16mkm/sec) starts its journey from Mercury to Jupiter
directly and then Jupiter Energy is sent to Pluto directly (1st
trajectory)
- There's one more trajectory of light supposed velocity (1.16 mkm/s) motion starts
from Mercury also to Pluto directly – light supposed velocity by this motion
caused to create Mercury Day period (4222.6 h) – as we will discus that deeply
later -
- These 2 trajectories are light motions, but
- We analyze in comparison Mercury Motion following the light motion, and also
we analyze Saturn motion in parallel.
(IX)
- Jupiter motion distance during Mars rotation period (24.6 hours) be = 1.16 mkm
and this is the reason of the important of Jupiter motion and Mars rotation period –
light supposed velocity moves for 1 second a distance = 1.16 mkm and that creates
the comparison of data with Jupiter motion.
(X)
- Light known velocity (0.3 mkm/sec) is found in the solar system and behaves as
the light supposed velocity effects on the planets motions data.
- But
- Light known velocity (0.3mkm/sec) is almost created later – because the sun is
created after all planets motions –and by that – the light known velocity (0.3
mkm/sec) be created almost as a side product from the original one the light
supposed velocity (1.16mkm/sec)
(XI)
- Spite the previous ideas – Jupiter motion data shows light supposed velocity (1.16
mkm/sec) effects more greatly than any other planet. This fact isn't solved yet –
Jupiter (all) distances be created based on the light supposed velocity (1.16

I do this research
2nd
176
mkm/sec) – and this feature isn't seen in any other planet – nor even Mercury –
let's provide Jupiter list here before out discussion to be a reference
- (1.16 mkm/sec) x 671 seconds = 778.6 mkm (Jupiter Orbital Distance)
- (1.16 mkm/sec) x 629 seconds = 720.7 mkm (Jupiter Mercury Distance)
- (1.16 mkm/sec) x 580 seconds = 671 mkm (Jupiter Venus Distance)
- (1.16 mkm/sec) x 500 seconds = 580 mkm
- (1.16 mkm/sec) x 542 seconds = 629 mkm (Jupiter Earth Distance)
- (1.16 mkm/sec) x 940 seconds = 2 x 550.7 mkm (Jupiter Mars Distance) (1%)
- (1.16 mkm/sec)2
x 2 x 778.6 seconds = 2094 mkm (Jupiter Uranus Distance)
- (1.16 mkm/sec) x 3717 seconds = 4331 mkm
- (1.16 mkm/sec) x 5127 seconds = 5906 mkm (error 0.7%)
- (1.16 mkm/sec) x 2644.5 mkm = 3033.5 mkm (Uranus Pluto Distance)
(error 1%)
- Where
- 940 mkm = Earth orbital circumference
- 3717 mkm = Jupiter Neptune Distance
- 5906 mkm = Pluto Orbital Distance
- 2644 mkm = Mars Uranus Distance
- The distances be used as periods of time to pass other distances – by this system
the distances be created in a network form -
- The distances rate is (1 mkm = 1 second) (this rate is used for light motion)
- And
- (1 mkm =1 solar day) this rate is the rate used by planets motions.
- Some very strange notice that Jupiter all distanced depend on (1.16 mkm/s)we
should analyze that as possible in this point discussion

I do this research
2nd
177
11-3 The Inner Planets Motions Reason
I- Data
(11-3-a)
Earth Mass = The Masses total of (Mercury + Venus + Mars)
(11-3-b)
Earth Circumference = The 5 inner planets diameters total
(Means the diameters total of (Mercury + Venus + The Moon + Earth + Mars))
(11-3-c)
2 x Earth Circumferences = 2 Mars Circumferences + Venus Circumference
(11-3-d)
Earth Circumference = The moon freedom space of motion
II- Discussion
- The double production experiment tells us
- From Gamma ray one electron and one positron can be created, 2 particles equal in
mass and opposite in charge.
- Equation no. (11-3-a) tells, Earth Mass = The 3 Inner Planets Masses Total (error
1%)
- Can we imagine that, a great gamma ray be used to create the inner planets.
- Earth is (The Great Electron) (means, Earth atom has a positive nucleus and
negative charges revolve around)
- The 3 Inner Planets Consist Together (The Great Positron) (means, the inner
planets Mercury – Venus – Mars, Their atoms have negative nucleus and
positive charges move around)
- If these hypotheses be working, the inner planets can create One Electric Field to
cause their motions around the sun.
- The error (1%) refers to a part of mass be consumed to create the required energy
for these planets motions through their orbits.

I do this research
2nd
178
- By that the error (1%) is found for a physical necessity and not be found as an
error in measurements.
- The previous description provides a suitable base to create an eclectic field
- The electric field is better for the planets motions than the sun gravity force.
- Newton has no one proof for the sun mass gravity force but the electric field
theory depends on the masses equality fact.
- The next discussion can support the electric field theory.
Earth Circumference = The 5 inner planets diameters total
(Means the diameters total of (Mercury + Venus + The Moon + Earth + Mars))
2 x Earth Circumferences = 2 Mars Circumferences + Venus Circumference
- For what we search? For Electric Charges
- If we can prove that, some electric charge be found in the inner planets. That can
support this idea greatly
- Why these 2 equations refer to charges in the inner planets data?
- Because of Gauss Law!
- Gauss law tells, The Charges Be On The Outer Surface, that means, the outer
surface can be effected Geometrically by electric charges. Because more wide
outer surface receives more charges. That tells The Charge Electric Field Effect on
the matter dimensions must be A Geometrical Effect
- I want to say
- If some planet be charged its outer surface formation and dimensions should be
effected Geometrically because of the electric charge.
- Means, the geometrical effects should be used as a proof for charges existence.

I do this research
2nd
179
- The previous 2 equations no. (b and c) can support this idea because the inner
planets circumferences and diameters be created in proportionality which can refer
to geometrical effects be produced by electric charges on their creation.
Earth Circumference = The moon freedom space of motion
- The moon orbital Perigee radius =363000 km
- The moon orbital apogee radius =406000 km
- The difference 43000 km
- But the space without the moon diameter =43000 km- 3475 km = 39500km= Earth
Circumference (error 1.3%)
- This data is used for the same argument, A geometrical effect be found on the
surface distances which can be used as a reference for an electric field effects on
the matter dimensions
- Notice
- 43000 km (the distance from perigee to apogee) x 3475 km (the moon diameter)
=149.6 mkm (Earth Orbital Distance). I try to show that, geometrical features are
found behind the dimensions and distances definition in the inner planets. And
these geometrical features supports the electric field effect claim.
- Notice
- If the planets move as chess board pieces - The electric field is the best method
can provide this type of motion (Chess Board Motion)
- That supports the claim, (The Inner Planets Motions Depend On Electric Field)

I do this research
2nd
180
A Comment
- The inner planets electric field idea tries to create a physical machine causes the
planets motions.
- I want to say,
- The solar system creation theory supposes that the light motions causes the solar
planets motions- and the data supports this description but
- We have no a real machine causes planets motions depend on light motion.
- For that
- I provide this idea as an example of such machine
- I don't confirm that, the inner planets motions depend on electric field – but I prove
an example of planets motions be caused by light motion- if this is the method
used by the inner planets so the theory is correct otherwise this example works as
an example for similar machines may work by more correct ideas than this –
- Shortly
- The machine aims to use light motion to cause planet motion.

I do this research
2nd
181
11-4 Light Beam (1.16 mkm/sec) Proves Discussion
I - Data
(11-4-a)
1.16 mkm/s x 86400 s = 100733 mkm
And
0.3 mkm/s x 86400 s = 25920 mkm
(11-4-b)
1.16 mkm/s x 500 s = 580 mkm
1.16 mkm/s x 580 s = 671 mkm (Venus Jupiter Distance)
1.16 mkm/s x 671 s = 778.6 mkm (Jupiter Orbital Distance)
1.16 mkm/s x 629 s = 720.7 mkm (error 1%) (Mercury Jupiter Distance)
1.16 mkm/s x 5127 s = 5906 mkm (error 0.7%) (Pluto Orbital Distance)
(11-4-c)
1.16 mkm/s x 4224 seconds = 4900 mkm
(11-4-d)
1.16 mkm per second x 2 x 86400 seconds =2 x 100733 mkm = (37100 mkm – 4900
mkm) x 2π = 28255 mkm + 2 x 86400 mkm
(11-4-e)
1.16 mkm/s x 5040 seconds = 5848 mkm (Mercury Pluto Distance)
And
5848 mkm (Mercury Pluto Distance) – 4900 mkm = 948 mkm
(11-4-f)
1.16 mkm/s x 4900 seconds = 5678.1 mkm (Mars Pluto Distance)
And
1.16 mkm/s x 5687.1 seconds = 6585.4 mkm
(11-4-g)
193 seconds x 0.3mkm/sec = 57.9 mkm (Mercury Orbital Distance)
196.5 seconds x 1.16 mkm/sec = 227.9 mkm (Mars Orbital Distance)
(11-4-h)
5040 sec = 88 sec x 57.27 (this data answers why Mercury day = double its period)

I do this research
2nd
182
II- Discussion
1.16 mkm/s x 86400 s = 100733 mkm
And
0.3 mkm/s x 86400 s = 25920 mkm
- Light known velocity (0.3 mkm/s) travels during a solar day (86400 s) a distance =
25920 mkm.
- Because it's a light motion we see this value (25920 mkm) as a period of time
(25920 years = the Precession Cycle)
- Also
- The distance 25920 mkm be = The Planets Motions Distances Total During A
Period =1461 days (where 1461 days =365 +365 +365 +366)
- Because the period 1461 days is (Earth Cycle) we consider that the distance 25920
mkm be mentioned by planets motions total
- Notice, the distances total added the Earth moon motion distance to the 9 planets
motions distance to create the total per a solar day (17.75 mkm) where the moon
motion daily distance be considered = Earth motion daily distance because they
aren't separated from one another through their motions.
- The light supposed velocity (1.16 mkm/s) travels during a solar day (86400 s) a
distance = 100733 mkm = The 9 Planets Orbital Circumferences Total
- Equation no. (11-4-a) tries to prove, Light Motion Uses The Solar Day Period
As Its Basic Period Of Motion.
- Light known velocity and light supposed velocity both use the solar day as the
basic period of time. Because of that, both light beams passes essential distances in
the solar system geometry.
- The next discussions will prove that these distances (25920 mkm and 100733
mkm) are essential distances in the solar system geometry.

I do this research
2nd
183
- Notice
- The rule tells
- (Light Motion For 1 Second = Planet Motion For 1 Solar Day)
- This rule which is a supposed one and we should prove it through the paper
discussions…
- This rule tells that,
- The light motion depends on the solar day as a period of time as similar to its
dependency on the second as a period of time.
- That means,
- These 2 periods of time (the second and the solar day) are the 2 basic columns for
light and planet motions accompaniment.
- And by that, light motion doesn't use the solar day as similar to any other planet
day period. On the contrary, the solar day is the basic period of time for light
motion (For both velocities, the known and supposed).

I do this research
2nd
184
1.16 mkm/s x 500 s = 580 mkm
1.16 mkm/s x 580 s = 671 mkm (Venus Jupiter Distance)
1.16 mkm/s x 671 s = 778.6 mkm (Jupiter Orbital Distance)
1.16 mkm/s x 629 s = 720.7 mkm (error 1%) (Mercury Jupiter Distance)
1.16 mkm/s x 5127 s = 5906 mkm (error 0.7%) (Pluto Orbital Distance)
- The data is apart of Jupiter series we have seen, and the data shows specific feature
which is:
- The light supposed velocity (1.16 mkm/s) uses the distances values as a periods of
time by the rate (1 mkm = 1 second).
- Based on that, the distances be created depends on one another and no distance be
created independent from the other distances. means, the distances be created in a
network form of distances and not as independent distances from one another.
- The data shows how the solar system distances be created in a network form
- In details
- Light supposed velocity (1.16 mkm/s) uses the period (500 sec) to pass a distance
580 mkm, and
- Light supposed velocity (1.16 mkm/s) uses the distance (580 mkm) as a period of
time (580 sec) to pass a distance 671 mkm (= Venus Jupiter Distance), and
time (671 sec) to pass a distance 778.6 mkm (=Jupiter Orbital Distance), and
time (629 sec) to pass a distance 720.7 mkm (=Jupiter Mercury Distance), …etc
- We have discussed this data before but now we have another reading for this same
data
- Notice (The Distances Network Theory Be Discussed In Point No.12)

I do this research
2nd
185
The Motion 1st
Trajectory
- The Trajectory Summary
- Light supposed velocity (1.16 mkm/s) travels starting from Mercury to Jupiter,
creating Jupiter orbital circumference
- And, then
- Light supposed velocity (1.16 mkm/s) travels from Jupiter to Pluto (Directly) and
doesn't touch any other planet in the way.
- The motion trajectory is built on 2 basic equations and many other additional data.
because
- Mercury Motion uses light supposed velocity (1.16 mkm/s) because of that,
different data of Mercury motion be created based on this light supposed velocity
(1.16 mkm/s) energy
- Shortly
The motion 1st
trajectory is defined based on Equation no. (11-4-c) and (11-4-d) as
basic equations. And through discussion many other data be added
Notice
Equation no. (11-4-d) is the solar system geometry main equation and it's the paper
main one we discuss it along the paper,

I do this research
2nd
186
1.16 mkm/s x 4224 seconds = 4900 mkm
- Where
- Equation no. (11-4-c) tells, the light supposed velocity (1.16 mkm/s) travels during
4224 seconds a distance = 4900 mkm = Jupiter Orbital Circumference.
- Now, let's suppose 1 second of light motion = 1 hour of Mercury Motion
- based on that, 4224 seconds be equivalent to 4224 hours (but 4222.6 hours =
Mercury Day Period).
- Equation no. (11-4-c) provides 3 branches of data! let's summarize them:
- (1st
Branch) Light supposed velocity (1.16 mkm/s) moves in (4224 s) a distance
=4900 mkm but Mercury moves during (4222.6 h) a distance = 720.7 mkm
(=Mercury Jupiter Distance) and we suppose that 1 second of light motion = 1
hour of Mercury Motion. Because of that we need to find some connection
between 4900 mkm and 720.7 mkm to create a comparison between light motion
and planet motion.
- The comparison is complex because, I suppose the rate (2π) distinguish between
light and planet motions. By that, if light moves 4900 mkm, a planet motion
should be 778.6 mkm but Mercury motion distance only be 720.7 mkm, where,
- 778.6 mkm (Jupiter orbital distance) =1.0725 x 720.7 mkm (Mercury Jupiter
Distance)
- The rate (1.0725) creates a negative effect on the rule of the value (2π). although
the rate (1.0725) be used between 40% of all distances in the solar system. But
because it be found here creates more complexity for the discussion and causes
some negative effect for the comparison between light and planet motions. because
always in other data we have found that the value (2π) distinguish clearly between
light and planet motions but in our equation the rate (1.0725) causes some
difficulty for this comparison vision.

I do this research
2nd
187
- (2nd
Branch) Mercury Day Period Creation.
- We have supposed that, 1 second of light motion be equivalent to 1 hour of
Mercury Motion. that means the period (4224 s ) be equivalent to (4224 h) and
Mercury Day Period = (4222.6 hours).
- This supposition suggests that, Light supposed velocity (1.16 mkm/) motion causes
to create Mercury Day Period be (= 4222.6 hours).
- The data is accurate and clear
- NOW
- The data tells that…
- Mercury is the source of the light supposed velocity (1.16 mkm/s)
- The light supposed velocity (1.16 mkm/s) started from Mercury and sent to Jupiter,
creating the distance (4900 mkm = Jupiter Orbital Circumference).
- This motion depends on Mercury Day Period.
- As a result, Mercury Day Period (422.6h) be defined by light supposed velocity
(1.16 mkm/s) motion effect on Mercury Motion.
- i.e.
- Mercury Day Period = 4222.6 hours And
- 4900 mkm= 1.16 mkm/s x 4224 seconds
- We suppose that, 1 hour of Mercury Motion = 1 second of light motion… means
- The difference (4224 -4222.6 = 1.4) is created by the light supposed velocity (1.16
mkm/s) motion effect. that means,
- The light motion effected on Mercury Motion and caused Mercury day period to
be 4222.6 hours and less than 4224 hours with 1.4 hour = 5040 seconds
- Why??
- Because
- Light supposed velocity (1.16 mkm/s) travels during (5040 s) a distance =5848
mkm = Mercury Pluto Distance

I do this research
2nd
188
- The light supposed velocity (1.16 mkm/s) created a connection between Mercury
and Pluto motions by creating the period 5040 seconds as a difference between
Mercury Day period (4222.6 h) and 176 solar days (=4224 hours).
- We can't discover the geometrical machine behind this data. But the data clearly
shows that, the light supposed velocity (1.16 mkm/s) travels during 5040 seconds a
distance = 5848 mkm.
- We have 3 information we got based on this analysis which are:
- (1st
) Mercury Is The Origin Point Of The Light Supposed Velocity (1.16 mkm/s)
- (2nd
) light supposed velocity (1.16 mkm/s) motion created Jupiter orbital
circumference 4900 mkm.
- (3rd
) Mercury Day Period be (4222.6 h) and less than (4224 h) with (5040 s)
because of light supposed velocity (1.16 mkm/s) effect on Mercury motion, to
create a connection between Mercury and Pluto motions.
- A Question
- The accompaniment of The Light supposed velocity and Mercury Motions caused
Mercury day Period to be (4222.6 hours)
- Is There any Proof for this conclusion? Let's see it in following …

I do this research
2nd
189
Mercury Motion Data
- 5040 seconds = 84 minutes
- Mercury (4.095 mkm/ daily) moves during 84 solar days a distance = 344 mkm
- And
- Mercury (4.095 mkm/ daily) moves during 346.6 solar days a distance = 1419.5 mkm
- And
- Mercury (4.095 mkm/ daily) moves during 1419.5 solar days a distance = 5813 mkm
- (the difference between 5813 mkm and 5848 mkm is 0.6%)
- The mechanisms is complex, but the data shows Mercury motion follows light
motion. Also Mercury uses light motion strategy. By that it moves a distance and
use it as a period of time to pass another distance. Nothing is clear in this process –
but we have 2 things (the period 5040 seconds which is related to Mercury day
period and the distance 5848 mkm = Mercury Pluto Distance) these 2 values are
used by light supposed velocity motion now be used by Mercury Motion.
- For that reason I suppose Mercury follows light motion
- PLEASE NOTE
- Based on what this paper discussions be created?
- I Depend On The Planets Data Arrangement
- This is the main force behind the paper arguments. For example, I claim there's a
light beam its velocity =1.16 mkm/sec, I have no empirical proves for this light
beam existence –so – based on what – such heavy claim can be raised? Based on
the planets motions data arrangement and order -
- We have seen that all Jupiter distances be rated with (1.16 mkm/sec), So the
hypothesis tells (A light beam its velocity 1.16 mkm/sec be found in the solar
system) is an intelligent answer for all Jupiter distances –
- And here, we have many data of Mercury Motions be arranged logically based on
the velocity (1.16 mkm/sec). By arrangement all this data beside each other we
catch one clear idea about how the solar system works

I do this research
2nd
190
The Conclusions
- The data tells,
- Mercury is the origin point of the light supposed velocity (1.16 mkm/s) motion.
that means, the light starts its motion from Mercury point. (A Conclusion No. 1)
- The light supposed velocity (1.16 mkm/s) passes the distance (4900 mkm = Jupiter
orbital circumference) in a period (4224 seconds). By that the light connects
between Mercury motion and Jupiter motion. (A Conclusion No. 2)
- Mercury Day Period be less than (4224 hours) by 5040 seconds because light
supposed velocity (1.16 mkm/s) travels during 5040s a distance = (5848 mkm) =
(Mercury Pluto Distance)
- Light supposed velocity (1.16 mkm/s) uses (5040s) to pass the distance (5848
mkm) and create a connection between Mercury motion and Pluto motion. (A
Conclusion No. 3)
- Notice
- 5848 mkm (Mercury Pluto Distance) –4900 mkm (Jupiter orbital circumference) =
946.5 mkm
- Where
- 940 mkm = Earth Orbital Circumference (error 0.7%)
- The value (946.5 mkm) is very important we will discuss in 2nd
Trajectory of
motion because it's related to Saturn motion data.
- Jupiter orbital circumference be analyzed in point no. (9) be seen a very rich
distance, and here we try to see the reason behind this distance richness.

I do this research
2nd
191
Mercury Motion Data Analysis (Continued)
- The previous data analysis concluded that, Mercury is the source of light supposed
velocity (1.16 mkm/s).
- So, let's ask,
- Is Mercury Motion data in harmony with light supposed velocity (1.16 mkm/s)?
(I)
- Light suppose velocity (1.16 mkm/s) needs only 50 seconds to pass Mercury
orbital distance (57.9 mkm).
- Can this (50 s) be a data belonged to Mercury Motion?
- Mercury (47.4 km/s) moves during 50.3 s a distance = 2390 km (Pluto diameter)
- Mercury (47.4 km/s) moves during 50.3 m x 60 s a distance = 142984 km (Jupiter
diameter)
- Mercury Venus Distance = 50.3 mkm ( 1mkm = 1 second)
- Mercury Neptune Distance (4437 mkm) = 50.3 mkm x 88
- (Mercury Orbital Period =88 days)
- Mercury Pluto Distance (5848 mkm) = 50.3 mkm x 116.2
- (57.9 mkm Mercury orbital distance x2 = 115.8 mkm)
- The data shows that, the period 50.3 seconds is a value be used frequently by
Mercury motion data. and mercury moves during (50.3 s) a distance = Pluto
diameter and moves during (50.3 m) a distance = Jupiter diameter, showing that
the 2 planets are the same planets mentioned by the 1st
trajectory motion.
- The basic difficulty is that, the light motion creates geometrical effects which we
can't discover because we don’t know neither their rule nor their effects. We have
only some data be transported from a point to another and that creates difficulties
for the analysis process.

I do this research
2nd
192
(II)
- Mercury Orbital Distance =57.9 mkm
- Mercury rotation period =58.65 solar days.
- The rate (1 mkm = 1 day) be used frequently in different motions data.
- The 2 values (57.9 mkm and 58.65 mkm) have an error (1%)
- Means, the 2 values (57.9 mkm and 58.66 days) can be created by the light
supposed velocity (1.16 mkm/s) motion during 50 seconds
- I try to show that, Mercury motion different data be in proportionality with light
supposed velocity (1.16 mkm/s) motion. which proves that Mercury is the origin
point of the light supposed velocity (1.16 mkm/s) motion.
- Because not only the period (5040 mkm) can be used or the period 4222.6 h
(which us equivalent to 4224 s) but also the period (50s)
- Notice
- 153.3 hours = Pluto day period
- 151.7 hours = 0.99 x 153.3 h (Pluto day period)
- But
- 176 mkm = 1.16 mkm/s x 151.7 seconds
- Where
- Mercury Day period =175.94 solar days
(If 1 day = 1 mkm, So 175.94 days be = 175.94 mkm)

I do this research
2nd
193
- (3rd
Branch) Saturn Motion Effect.
- Saturn (0.838 mkm / daily) moves during = 5848 days, a distance = 4900 mkm
- This data shows Saturn motion effect on Mercury motion data
- And
- We have to discuss Saturn data in a separated point because Saturn creates one
more reading for Mercury Motion Data
- We discuss Saturn motion data in point no. (11-5)

I do this research
2nd
194
1.16 mkm per second x 2 x 86400 seconds =2 x 100733 mkm = (37100 mkm –
4900 mkm) x 2π = 28255 mkm + 2 x 86400 mkm = 202318 mkm (error 1%)
- Where
- 100733 mkm = The 9 Solar Planets Orbital Circumferences Total
- 28255 mkm = Neptune orbital Circumference
- Equation no. (11-4-d) is the paper basic equation. Because it explains The Basic
Energy Trajectory Through The Solar System…
- First, We remember that we have discussed this equation before but here we
discuss it through another approach which shows a different reading for the data
- Also we notice that,
- The equation uses the rate (2) because of Neptune orbital circumference. Where
the equation 3 parts can be used without the rate (2), But the last part (28255 mkm
+ 2 x 86400 mkm) needs the rate (2) because of Neptune orbital circumference
(28255 mkm). That explains why the equation uses the rate (2).
- Now let's divide the equation into 4 Parts.
(Part No. 1)
- 1.16 mkm per second x 2 x 86400 seconds = 2 x 100733 mkm
- Part no. (1) tells us a known information which is
- Light supposed velocity (1.16 mkm/s) travels during one solar day (86400s) a
distance = 100733 mkm = The 9 Solar Planets Orbital Circumferences Total
- The rate (2) is used for Neptune Orbital Circumference.
- This part of equation tells that, light motion depends on the solar day period. This
idea we have discussed in equation no. (1) of this discussion.
- Also we know that

I do this research
2nd
195
- Light known velocity (0.3 mkm/s) travels during (86400 s) a distance = 25920
mkm
- Where
- 25920 mkm = 17.75 mkm x 1461 days
- 17.75 mkm = the planets distances total during a solar day in addition to the Earth
moon motion distance during a solar day, and
- 1461 days = Earth Cycle (365 +365 +365 +366)
- The data tries to prove that, the light motion depends on the solar day period.
- Notice
- (25920 mkm x 2) =37100 mkm x 1.392
- Where
- The sun diameter =1.392 mkm
- The data uses the rate (2) because this rate be used in equation no. (11-4-d) which
expresses the solar system basic energy trajectory.
- That means the value (25920 mkm x 2) refers to the total energy according to light
known velocity (0.3 mkm/s) range of motion.
- The data tells the value (25920 mkm x 2) expresses really the total energy because
37100 km (Pluto orbital circumference) is the greatest orbit
- That means, when the total energy be expresses by the value (2 x 100733 mkm)
this energy be transported to the light known velocity based on the same rate
because it's one machine.
- The data proves the energy unification through the solar system spite of using 2
different velocities of light.

I do this research
2nd
196
(Part No. 3)
- 2 x 100733 mkm = (37100 mkm – 4900 mkm) x 2π =
- This part is very important because
- The data shows the energy motion (or light motion) started from 4900 mkm
- That means,
- To create Pluto orbital circumference 37100 mkm, the light starts from the point
4900 mkm and measures a distance = 100733 mkm by geometrical calculations.
- We have 3 points here
- The start point which is (4900 mkm =Jupiter orbital circumference)
- The direction of motion (toward Pluto)
- The measurement unit is 100733 mkm (defined by light supposed velocity motion
during a solar day)
- The data tells, Pluto orbital circumference is created depending on (4900 mkm )
Jupiter orbital circumference by using the light motion.
- We remember
- The distance 4900 mkm be passed by light supposed velocity (1.16 mkm/s)
depends on the period (4224 s) which is defined in comparison with Mercury day
Period (4222.6 hours)
- As A Result, we conclude that,
- Light supposed velocity (1.16 mkm/s) starts from Mercury to Jupiter and continues
from Jupiter to Pluto. that shows one clear trajectory of motion from Mercury to
Jupiter to Pluto.
- That means, the motion from Mercury to Jupiter considers Mercury is the head of
the inner planets and by that the connection between Mercury and Jupiter express
the connection between Jupiter and all inner planets.
- And
- The connection between Jupiter and Pluto doesn't contain any other planet. It’s one
directed motion from Jupiter to Pluto immediately

I do this research
2nd
197
- Shortly
- Jupiter and Pluto orbital circumferences be created by light supposed velocity
(1.16 mkm/s) motion depends on Mercury motion data by creating different steps
of motions to connect between Mercury and Pluto
- This data confirms that the distances be created based on network form and
according to one geometrical design.
Notice
- (2π) This rate distinguish between light motion and planet motion as we have
discussed before.
- The rate (2π) distinguishes here clearly between the planet and light motions.
where the difference (37100 mkm – 4900 mkm) refer to the planets motions where
the value 2 x 100733 mkm refers to the light motion and the rate (2π) distinguish
between the 2 motions.
- Venus equation which we will study with the part no. (4) of this equation uses also
the rate (2π) between the energy source and the planet received energy. we should
refer to this meaning in its discussion.
- And
- The data gives a new meaning for the rate (2π).
- It's the rate between light motion and planet motion or
- It's the rate between the giver energy motion and the receiver energy motion
- That tells Mars energy be provided by Jupiter motion and Pluto energy be
provided by Earth motion.

I do this research
2nd
198
(Part No. 4)
- 2 x 100733 mkm = 28255 mkm + 2 x 86400 mkm
- The total energy be reflected from Pluto to Neptune. The distances behave
perfectly as light beams.
- Neptune had seized 14% of the total energy and the rest of energy be reflected
toward the inner planets into 2 trajectories of energy each has (86400 mkm)
- We suppose that (Distance = Energy)
- The energy be sent from Neptune to Venus and Earth, let's see their equations in
following
- 2 x 86400 mkm = (28255 mkm – 680 mkm) x 2π (Venus Equation)
- Where
- 28255 mkm = Neptune Orbital Circumference
- Earth equation is perfectly similar to Venus equation where between both the error
is less than 1%
- Also we see the rate (2π) which distinguish between light and planet motion as we
have seen in our equation no. (11-4-c)
- (or it distinguish between the source of energy and the planet received energy).
- Part (4) of the equation tells that, Neptune had seized 14% of the total energy and
the rest of energy is sent to Venus and Earth into equal trajectories of energy each
has (86400 mkm).
- That means, the inner planets energy be received from Neptune. And because of
that, the inner planets distance be controlled by Neptune orbital distance (4495.1
mkm) as the data shows that clearly.
- 4495.1 mkm (Neptune orbital distance) = 78.3 mkm x 57.9 mkm
- = 108.2 mkm x 41.4 mkm
- = (149.6 mkm x 120 mkm)/4

I do this research
2nd
199
- Where
- 57.9 mkm = Mercury Orbital Distance
- 108.2 mkm = Venus Orbital Distance
- 149.6 mkm = Earth Orbital Distance 41.4 mkm = Earth Venus Distance
- 78.3 mkm = Earth Mars Distance 120 mkm = Venus Mars Distance
- Notice
- Because the distance=energy and the inner planets receive their energy by Neptune
for that reason, Neptune orbital distance control all most distances among the inner
planets.
- Now let's ask
- Why Neptune had seized 14% of the total energy but Pluto seized nothing?
- Because
- Mars Migration Theory tells us…
- Pluto was The Mercury Moon and had migrated with Mars Migration.
- Neptune, in that time, was occupied the point (5906 mkm = Pluto current orbital
distance). And
- Pluto be migrated and thrown to the end of the solar group, flying along way and
had collided with Neptune at the point (5906 mkm).
- Pluto had pushed Neptune and put it out of Pluto (influence area!), for that reason
Pluto eccentricity distance = 1410 mkm= Pluto Neptune Distance
- The collision between Pluto and Neptune produced Pluto moons and the Kuiper
belt.
- Neptune be forced to be more near to the sun with orbital distance =4495.1 mkm
and without orbit.
- Neptune be out of the distances network (the railways) generally and needed
energy to built its orbit
- That explains why Bode Law Couldn't predict Neptune orbital distance, because
it's created (additionally) on the original geometrical design to repair the planets

I do this research
2nd
200
migration negative effects on the solar system motion. means, it's a distance not
designed geometrically with the original design. So the equations depend on the
original design couldn't predict Neptune orbital distance.
- Mars Migration Theory Is Discussed in Point No.(6)
Notice
- 202318 mkm = (449197 km)2
= 1.16 km/s x 2 x 86400 sec
- Where
- 449197 km =Jupiter Circumference
- Accurately
- (202318 mkm) / (449197 km)2
= (361/360)
- What's (361 degrees)?
- The cycle is 360 degrees but the Earth moon orbital regresses 19 degrees per year
and during 19 years moves Metonic Cycle (19 years) (=361 degrees).
- The moon orbit regression effects on many of the planets motions data and
because of that the usual rate (360) be changed into (361).

I do this research
2nd
201
The Motion 1st
Trajectory (A Summary)
- Mercury is the origin point of this motion trajectory
- The energy be transported from Mercury to Jupiter by the light supposed velocity
(1.16 mkm/s) motion and by this energy Jupiter orbital circumference be created
- The energy be transported from Jupiter to Pluto by the light supposed velocity
(1.16 mkm/s)
- The energy be reflected from Pluto to Neptune and
- The energy be reflected from Neptune to the inner planets
- The Motion 1st
Trajectory Shows the energy trajectory from Mercury to
Jupiter to Pluto, which creates the solar system basic energy trajectory.

I do this research
2nd
202
The Motion 2nd
Trajectory
- Light supposed velocity (1.16 mkm/s) travels starting from Mercury to Pluto
directly
- In this trajectory of motion the light doesn’t use a central point as did with Jupiter
orbital circumference. Instead the light travels from Mercury to Pluto directly.
- As a result, this Trajectory of motion be formed based on the following data,
- Light supposed velocity (1.16 mkm/s) travels during (5040 seconds) a distance =
(5848 mkm= Mercury Pluto Distance)
-
- This trajectory of motion we have studied deeply. Because the period (5040 s) is
distinguish Mercury day period as we have discussed.
- But this trajectory of motion still have more important data
- This trajectory of motion uses one equation only which is no. (11-4-e), let's discuss
it in following…

I do this research
2nd
203
5848 mkm (Mercury Pluto Distance) – 4900 mkm = 948 mkm
And
5848 mkm (Mercury Pluto Distance) =5040 seconds x 1.16mkm/sec
- Where
- 5848 mkm = Mercury Pluto Distance
- 948 mkm = Earth orbital circumference (940 mkm) error (1%)
- Equation no. (11-4-e) shows Earth orbital circumference be taken into account
with the light supposed velocity motion for the 2 distances (5848 mkm and 4900
mkm)
- But
- Why the distance (948 mkm) be important here?
- The Earth moon daily displacement (=88000 km), and if the moon passes Earth
orbital circumference by using its displacement (88000 km), the moon needs a
period =10747 days to pass a distance = (945.5 mkm)
- Where 10747 days = Saturn Orbital Period
- The important information isn't (Saturn orbital period) but its (Saturn data),
because Saturn data be used frequently in the motions trajectories data between
Mercury and Pluto, for light and planet trajectories of motions.
- i.e.
- Through different data analysis we reach to Saturn data. this time we found (10747
days = Saturn orbital period), and in previous discussion we have found 3 data be
related Saturn motion defined by the light and mercury motions by using the
period (5040 seconds) which are:
o Saturn moves during (5848 days) a distance = 4900 mkm
o Mercury moves during (5040 seconds) a distance = 2 x Saturn diameters

I do this research
2nd
204
o Mercury moves during 1433.5 days a distance = 5870 mkm = Pluto orbital
distance (old definition) or near to (5848 mkm = Mercury Pluto Distance)
Where (1433.5mkm = Saturn Orbital Distance)
- Why (5848 mkm) – (4900 mkm) = 948 mkm (Earth Orbital Circumference)?
(error 1%)
- It's A Geometrical Distribution For Distances,
- But not only,
- Because this distribution contains data (10747 days) be used by Saturn and the
Earth moon motions…
- This data we should discuss with Saturn motion data because Saturn motion is
related to the light known velocity (0.3mkm/s) motion.
- The period 10747 days also be used by the moon to pass a distance = the distance
be passed by light known velocity motion during a solar day.

I do this research
2nd
205
The Motion 3rd
Trajectory
- Light supposed velocity (1.16 mkm/s) uses 2 different trajectories of motions (no.
1 and no. 2) to pass a distance =5848 mkm (Mercury Pluto Distance)
- And these motions be done by light velocity
- The first trajectory, the light travels from Mercury to Jupiter and then to Pluto
- The second trajectory, the light travels from Mercury to Pluto Directly
- The third trajectory expresses a motion from Mercury to Pluto (5848 mkm) but by
using mercury motion (Mercury velocity per a solar day =4.095 mkm)
- We have seen this data before, let's remember them:
o Mercury moves during 84 days a distance = 344 mkm
o Mercury moves during 346.6 days a distance = 1419 mkm
o Mercury moves during 1433.5 days a distance = 5848 mkm
o Where
o 5848 mkm = (Mercury Pluto distance)
o 1433.5 mkm = Saturn orbital distance
o 346.6 days = the nodal year.
- This trajectory of motion refers (frequently) to Saturn motion data among mercury
motion data. we don't know why that's happening…
- We can conclude that, the planet (Mercury) follows the light motion but with
different rate of time.
- Because this behavior is described by the matter creation theory. Where the theory
tells, the matter is born out of light but the born matter doesn't separate from its
light parent. On the contrary, they move together in one unified motion but by
using different rates of time.
- That's happening here. Mercury moves with light suppose velocity (1.16 mkm/s)
by using a different rate of time. (Saturn motion is discussed in 11-5)

I do this research
2nd
206
1.16 mkm/s x 4900 seconds = 5678.1 mkm (Mars Pluto Distance)
And
- Where
- Light supposed velocity (1.16 mkm/sec) travels during 4900 seconds a distance =
5678.1 mkm (Mars Pluto Distance).
- We remember that,
- The light supposed velocity (1.16 mkm/sec) depends on the period (4224 sec) to
pass the distance 4900 mkm.
- And depends on the period (5040 second to pass) the distance 5848 mkm (Mercury
Pluto distance)
- And we have asked why light supposed velocity (1.16 mkm/sec) created 2
trajectories between Mercury and Pluto positions? Or in more clear words why
light supposed velocity creates Jupiter orbital circumference (4900 mkm)by using
(4224 seconds)?
- Equation no. (11-4-f) gives an answer.
- Because light supposed velocity creates a connection between Mars and Pluto
positions with Mercury.
- Mars migration theory tells, Pluto was the Mercury moon and Mars was the next
planet to Mercury. And the 2 planets were migrated from their original positions.
- Light supposed velocity (1.16 mkm/sec) creates a connection between Mercury
and Pluto through the period 5040 second and another connection between Pluto
and Mars through the distance (4900 mkm)

I do this research
2nd
207
- Notice
- Light supposed velocity (1.16 mkm/sec) travels during 5687.1 seconds a distance
= 6585.4 mkm
- By the rate (1 mkm = 1 day) this distance 6858.4 mkm be = 6858.4 solar days
- Where (Saros Cycle =6858.4 solar days)
- Please remember
- Light known velocity (0.3 mkm/sec) moves during a solar day (86400 s) a distance
=25920 mkm,
- Where we see this distance (25920 mkm) as a period of time (25920 years = The
Procession Cycle)
- I try to show that, the period of Saros Cycle (6585.39 days) is similar to the period
(25920 years), both be created by light motion behavior, which proves it's usual a
behavior in the solar system
- But that creates one more connection between Mars motion and the Earth moon
motion because Saros is a cycle produced by the moon motion effect.

I do this research
2nd
208
193 seconds x 0.3 mkm/sec = 57.9 mkm (Mercury Orbital Distance)
196.5 seconds x 1.16 mkm/sec = 227.9 mkm (Mars Orbital Distance)
- Equation no.(11-4-g) tries to explain why Mercury and Mars Motions have
interaction as we have proved in Jupiter Motion Analysis
- The reason is that, the 2 planets orbital periods are defined based on equal periods
approximately (193 seconds and 196.5 seconds the difference 2%), by using the 2
light velocities
- That shows the period (193 sec or 196.5 sec) are the reason of the 2 planets
motions integration or interaction
- But
- As we have seen Jupiter motion also be integrated with Mercury and Mars motions
- Can this period (196.5 sec) effect on Jupiter motion also?
- Yes
- 8 Jupiter days = 8 x 9.9 x 3600 = 285120 seconds = π x 90756.5 seconds
- Pluto orbital period =90560 days
- 90756.5 seconds - 90560 seconds = 196.5 seconds
- Where
- Light motion uses any period (of 1 solar day) as (1 second)

I do this research
2nd
209
The Motion Trajectories Summary
- As we have discussed in the previous analysis,
- There are 3 motions started from Mercury moving toward Pluto, and these motions
are:
o Light supposed velocity (1.16 mkm/) motion during (5040 s) to pass (5848
mkm = Mercury Pluto Distance)
o Light motion from Mercury to Jupiter by using (4224s)
o 1.16 mkm/s x4224 seconds =4900 mkm (= Jupiter orbital circumference),
where Mercury uses Jupiter orbital circumference as its central point to
connect with Pluto.
o Mercury Motion depends on (5040) to pass in three steps toward Pluto by
the data
o 4.095 mkm x 84 days =344 mkm
o 4.095 mkm x 346.6 days =1419 mkm
o 4.095 mkm x 1433.5 days =5870 mkm (very near to 5848 mkm)
o Notice
o Light supposed velocity (1.16 mkm/s) motion depends on the rate (2π R) but
Planet motion depends on (R). So if Mercury want to connects with Jupiter,
light motion doesn't use the distance (778.6 mkm Jupiter orbital distance)
but use (4900 mkm =Jupiter orbital circumference)
o 778.6 mkm =1.0725 x 720.7 mkm (Mercury Jupiter Distance)

I do this research
2nd
210
Equation no. (11-4-h)
5040 sec = 88 sec x 57.27
5040 seconds are required for Mercury Day Period to be 176 solar days
88 days = Mercury Orbital Period
57.27 =Mercury Rotation Period / Mars Rotation Period
Let's remember our question
Why Mercury Day Period = 2 Mercury Orbital Periods?
4331 solar days = 24.6 hours x 175.94 = 687 solar days x 2π
Where
- 4331 days = Jupiter Orbital Period, and 175.94 days = Mercury Day Period, and
- 687 days = Mars Orbital Period , and 24.6 hours = Mars Rotation Period
- Based on this data we have concluded the following rates of time
- And
- 671 hours of Mercury Motion = 687 days of Jupiter motion = 109 days of
Mars motion
- The we have discovered that,
- During 671 hours Mercury moves a distance 57.2 mkm= 2 x Mercury Orbital
Distance (error 1%)
- And
- During 687 days Jupiter moves a distance 778.6 mkm = Jupiter Orbital
Distance
1 hour of Mercury Motion = 24.6 hours of Jupiter Motion
And
1 hour of Mercury Motion = 3.9 hours of Mars Motion
And
1 hour of Mars Motion = 2π= 6.3 hours of Jupiter Motion

I do this research
2nd
211
- And
- During 109 days Mars moves a distance 227.9 mkm = Mars Orbital Distance
- The question was
- Why Mercury moves distance = (2 x Mercury Orbital Distance), But Mars And
Jupiter Move Distances Only Equal Each Planet Orbital Distance?
- Then
- We have found, 5040 s are required for Mercury day to be =176 solar days
- And
- During 5040 seconds Mercury (47.4 km/s) moves a distance = 2 x 119448 km =
Saturn diameter (= 120536 km) (error -1%)
- But
- During 5040 seconds Mars (24.1 km/s) moves a distance =121464 km = Saturn
diameter (= 120536 km) (error +1%)
- This data gives one more reference that, the (one value for Mars) produces (2
values for Mercury)!
- We have a reason to find the rate (2) in data and because Mercury day = (2)
Mercury orbital period –some interesting feeling pushed to dig under this data may
we find answer why Mercury Day Period = 2 Mercury Orbital Period
- Let's return to the equation

I do this research
2nd
212
(Equation no. (11-4-h) (continued)
5040 sec = 88 sec x 57.27
- Light motion for 1 second causes a planet motion for 1 solar day,
- Because of this rule we couldn't discuss this equation in Jupiter Motion analysis
- Light moves 88 seconds and caused Mercury to move 88 solar day
- That means, this equation is Mercury orbital revolution equation… because of that
the rate (57.2) be produced, where 57.9 mkm =Mercury orbital distance (error 1%)
- Let's remember this data
- Mercury rotation period 1407.6 h / Mars rotation period 24.6 h = 57.2
- And
- Mercury day period 4222.6 h / Mars rotation period 24.6 h = 171.7
- Both values (57.2 and 171.7) are Mercury distances with the same error (1%)
- (57.9 mkm= Mercury orbital distance, and 170 mkm = Mercury Mars Distance)
- How Mercury orbital period 88 days can produce Mercury day period?
- 57.2 =2 x 28.6 (Neptune axial tilt =28.3 degrees error 1%)
- It's Neptune effect on Mercury Motion
- Why Neptune?? Because
- Neptune Orbital Distance 4495.1 mkm = π x Saturn Orbital Distance 1433.5 mkm
- Notice
- Neptune orbital period 59800 days = Mercury day period 175.94 days x 339
- But, Venus rotation period 5832.5 hours = Uranus rotation period x 339
- It's the machine by which the rate (5.1) be transported, let's remember it here
o Venus Velocity = Uranus Velocity x 5.1
o Mars Velocity = Pluto Velocity x 5.1
o The Moon Velocity = Neptune Velocity x 5.1
- That tells why
- Mercury moves during its rotation period a distance 243mkm (error 1%) and
- Mars moves during Venus day period a distance 243mkm (error 1%)
-

I do this research
2nd
213
The General Discussion
- I have tried to arrange the data in best form to show that, the data be created based
on a geometrical design for a light beam motion by which the solar planets and
their distances be created.
- The Theory hypothesis gives a chance for better explanation….
- The hypothesis tells a light beam its velocity (1.16 mkm/sec) caused the solar
system creation and motion, and by effect of the light motion the distances be
created in a network form.
- The hypothesis shows itself so complex because we need to prove (at first) that
there's a light beam its velocity =1.16 mkm/s
- By the planets motions data arrangement I have tried to show that the planets
motions data be created depends on a light beam its velocity =1.16 mkm/s and if
this light beam isn't found there's no chance to create the solar planets motions data
and their distances as their values. This discussion is done in so deep details where
Jupiter motion data and its distances with other planets, this all data will be simply
removed if there's no light beam its velocity =1.16 mkm/s.
- The data still have more support for the hypothesis but needs at first to discover
the geometrical design based on which the planets motions data be created.
- Although, this try is a strong one, but many basic questions are still left behind, we
have to face them as a part of our effort searching for the truth
- (A Question) what's the relationship between Planet motion distance and a
distance be found between 2 planets? If these 2 distances be equal why does this
give any geometrical meaning?
- For example
- Light supposed velocity (1.16 mkm/s) travels during 5040 seconds a distance =
5848 mkm = Mercury Pluto Distance
- What's the relationship between the distance be passed by light motion (5848
mkm) and Mercury Pluto Distance (5848 mkm)?

I do this research
2nd
214
- If these 2 distances are equal why this is important?
- My answer for this question is a simple one ( I Don't Know How The Machine
Works).
- For example
- Earth moves during a solar day (24 h) a distance = 2574720 km
- Pluto moves during its day period (153.3 h) a distance = 2593836 km
- The moon displacements total during (29.53 days) = 2598693 km
- These 3 distances are equal with max error only (1%)
- Can these equal distances effect on any planet motion data? yes
- As we discuss in (The Planets Motions Depend On Different Rates Of Time)
the three planets motion data be created in full proportionality with one another.
can this data be in proportionality because of the distances equality or the distances
equality be a result of the data proportionality, I can't give a decision.
- The fact is that, the planets data be in proportionality and the distances are equal.
- To prove this point of view let's bring the data here for better understanding…
- More Data
Group no. (1) (Pluto And Earth Rate Of Time)
- (Earth velocity / Pluto velocity) = (Pluto Day period/ earth day period) = 6.37 (0.8%)
- Pluto orbital distance 5906 mkm = Earth orbital circumference 940 mkm x 2π
- Pluto orbital period 90560 days = Earth Cycle (1461 days) x 2π3
Group no. (2) (Pluto And The moon Rate Of Time)
(1)
(406000 km /88000 km) = (708.7 h/ 153.3 h) = 4.61
(2)
Tan (12.2) x 708.7 hours = 153.3 hours
(3)
Tan (13.17) x 655.7 hours = 153.3 hours

I do this research
2nd
215
- Where
- 406000 km = Pluto Motion Distance during a solar day
- 88000 km = The moon displacement daily
- And
- The angle 12.2 degrees = 13.177 deg – 0.98562 deg
- 0.98562 degrees = The Earth Motion Degrees Per A Solar Day
- And
- 655.7 hours = The Moon Rotation Period.
- 153.3 hours = Also (153.3-) Pluto Rotation Period
- The data confirms the point of view, the equal distances effects on the planets
motion data be caused it to be in proportionality with one another. why or how?
We still search for the machine geometrical details.
- The discussion is limited and can't receives its power, as a creature crippled and
can't walk! Why? because we don't know the used geometrical rules based on
which the solar system be created and moving.
- For example
- The equal distances (as we have seen in Earth, its moon and Pluto data) is a
method used widely in the solar system. That means, all planets use this method.
All planets create equal distances with one another and based on these equal
distances the planets motions data be in proportionality with one another.
- Let's use one more example:
- Mercury moves during its rotation period (58.65 days) a distance = 243 mkm(1%)
- Mars moves during Venus day period ( 116.75 days) a distance = 243 mkm

I do this research
2nd
216
- These are the equal distances, where is the proportionality of data?
- (Mercury Day Period/ Venus Day Period) =(Mercury orbital period/ Mercury
rotation period)
- The data shows that, Venus Day period be created in proportionality with Mercury
cycles periods. Can that be a result for the equal distances?
- The equal distances is a method used thousands of times in the solar system, where
many planets motions distances be passed in defined periods of time be equal one
another. we don't know what effect be done from this distances equality and by
that a great part of the machine be out of our sight…
- Shortly,
- We have to depend on some conclusions (and even guess) to conclude how the
solar group be created and moving
- The other option is the imaginary ideas. As Newton idea of the sun mass gravity,
where he imagined that the sun is the planets motions reason while the fact is that
the sun is created after all planets creation and motion. and in fact the sun rays be
created by using the planets motions energies total. means, based on the planets
motions the sun be created, and that tells the planets can be found without the sun
but the sun can't be found without the planets. Newton idea is mistaken.
- Our situation is not so better. The point is that, the geometrical rules book is inside
the light nature, and we can't catch how the geometrical effects be done but we can
discover them in every where.
- Let's summarize the machine basic difficulties
o Planet moves side by side with a light motion by that the planet motion
refers to light motion features but we can't catch the velocity which creates
these features
o The equal distances method is the method caused the planets motions. this
method works geometrically.

I do this research
2nd
217
o That means, this method works as a waterwheel or as a lever. A tool works
by a geometrical rule and effect. But we don't understand what benefit be
produced by the equality of distances. This equality causes the planets
motions but how?
o We notice also that,
Jupiter motion depends on light supposed velocity (1.16 mkm/s) and
all distances around Jupiter be defined based on this velocity (1.16
mkm/) but
Saturn motion depends on the light (0.3 mkm/s) where different data
belong to Saturn depends on light known velocity.
The Earth moon motion be in proportionality with both light beam
velocities! The data confirms this meaning
10921 km x 27.3 days =300000 km (error 0.5%) the equation tells if
the moon rotates around its axis one time per solar day as Earth does,
the moon would pass by its circumference during 27.32 days (the
moon orbital period) a distance =300000 km = light known velocity
(0.3 mkm/s) motion distance during 1 second.
43000 km x 27.3 days =1160000 km (error 1.3%) the equation tells if
the moon is a point without circumference and moves from perigee
point to apogee point in a solar day and in the next day moves from
apogee to perigee … during the moon orbital period (27.3 days)the
moon would move a distance =1.16 mkm
Notice (43000 km = the distance between perigee and apogee points)
10921 km x 86400s =940 mkm, the equation tells if Earth revolves
around the sun a complete revolution (940 mkm) in one solar day
only (86400 s), the moon circumference (10921 km) will equal a
distance of Earth motion during 1 second period of time. (notice, the
geometrical design can create the effect without planet real motion)

I do this research
2nd
218
11-5 Saturn Motion More Analysis
I- Data
(11-5- a)
10747 days x 10x 24 h = 10.7 h x 2 x 120536
(11-5- b)
120536 days x 0.838 mkm /day = 100733 mkm = 1.16 mkm/sec x 86400 sec.
(11-5- c)
5848 days x 0.838 mkm /day = 4900 mkm (Jupiter orbital circumference)
(11-5- d)
929 days x 0.838 mkm /day = 778.6 mkm (Jupiter orbital distance)
(11-5- e)
1106 days x 0.838 mkm /day = 929 mkm
(11-5- f)
(2 x 120536 km)/ (2390 km) = 101 = (5848 mkm /57.9 mkm)
(11-5- g)
5848 mkm- 4900 mkm = 946.6 mkm
(11-5- i)
(300000 km/120536 km) = (24.1/9.7) = (10747/4331)
(11-5- j)
25920 mkm= 0.838 mkm/day x 30589 days = 2.4 mkm x 10747 days
(11-5- k)
(2 x 120536/24.1) = (300000 /29.8)
(11-5- l)
9.7 km/s x 30589 sec =300000 km
And
9.7 km/s x 120536 sec =1160000 km (error0.8%)
(11-5- m)
2.082 mkm x 778.6 days = 1622 mkm = 1.1318 mkm x 1433 days

I do this research
2nd
219
II- Discussion
Equation no. (11-5- a)
10747 days x 10x 24 h = 10.7 h x 2 x 120536
- Where
- 120536 km = Saturn Diameter
- 10.7 hours = Saturn Day period
Equation no. (11-5- b)
- Where
- 0.838 mkm/day = Saturn Velocity per a solar day
- 100733 mkm = The Solar Planets Orbital Circumferences Total
Equation no. (11-5- c)
- Where
- 5848 mkm =Mercury Pluto distance (be used as period 5848 days)
Equation no. (11-5- d)
- Where
- 929 mkm = Earth Jupiter distance when the 2 Planets be in the sun 2 sides

I do this research
2nd
220
Equation no. (11-5- e)
1106 days x 0.838 mkm /day = 929 mkm
- Where
- 1106 mkm = the distance Venus passes in 365.25 days
Equation no. (11-5- f)
(2 x 120536 km)/ (2390 km) = 101 = (5848 mkm /57.9 mkm)
- Where
- 2390 km = Pluto Diameter
Equation no. (11-5- g)
5848 mkm- 4900 mkm = 946.6 mkm
- Where
- 4900 mkm = Jupiter orbital circumference
Equation no. (11-5- i)
(300000 km/120536 km) = (24.1/9.7) = (10747/4331)
- Where
- 300000 km = light known velocity motion for 1 second
- 24.1 km/sec = Mars velocity
- 10747 days = Saturn orbital period
- 4331 days = Jupiter orbital period

I do this research
2nd
221
Equation no. (11-5- j)
25920 mkm= 0.838 mkm/day x 30589 days = 2.4 mkm x 10747 days
- Where
- 30589 days =Uranus Orbital Period
- 10747 days = Saturn orbital period
- 2.4 mkm/day = The moon Velocity per a solar day
Equation no. (11-5- k)
(2 x 120536/24.1) = (300000 /29.8)
- Where
- 24.1 km/sec = Mars velocity
- 29.8 km/sec = Earth velocity
Equation no. (11-5- l)
9.7 km/s x 30589 sec =300000 km And
9.7 km/s x 120536 sec =1160000 km (error0.8%)
- Where
- 30589 days =Uranus Orbital Period
Equation no. (11-5- m)
2.082 mkm x 778.6 days = 1622 mkm = 1.1318 mkm x 1433 days
- Where
- 2.082 mkm/day = Mars Velocity per a solar day
- 778.6 mkm = Jupiter orbital distance (be used as a period)
- 1.1318 mkm/day = Jupiter Velocity per a solar day
- 1433 mkm = Saturn orbital distance (be used as a period)

I do this research
2nd
222
Equation no. (11-5- n)
Mercury moves during (5040 sec) a distance = 2 Saturn diameter (error -1%)
Mars moves during (5040 sec) a distance = Saturn diameter (error +1%)
II- Discussion (continued)
- There are 3 reasons for Saturn motion data importance which are
(I)
- Saturn diameter be used in double value, let's see that in following:
(a)
Mercury moves during (5040 sec) a distance = 2 Saturn diameter (error -1%)
Mars moves during (5040 sec) a distance = Saturn diameter (error +1%)
(b)
Pluto (4.7 km/sec) move during (51118 sec) a distance = 120536 km x2
(c)
Pluto (4.7 km/sec) move during (2 x 120536sec) a distance = 1.1318 mkm = Jupiter
motion distance per a solar day
That tells a geometrical mechanism controls Saturn motion and data
And
(2 x 120536 km)/ (2390 km) = 101 = (5848 mkm /57.9 mkm)
- Where
- 2390 km = Pluto Diameter

I do this research
2nd
223
(II)
Saturn motion be effective as light motion. fro example
Equation no. (11-5- b)
- Where
- 100733 mkm = The Solar Planets Orbital Circumferences Total
Equation no. (11-5- c)
- Where
- 5848 mkm =Mercury Pluto distance (be used as period 5848 days)
Equation no. (11-5- d)
- Where
Equation no. (11-5- e)
1106 days x 0.838 mkm /day = 929 mkm
- Where
- 1106 mkm = the distance Venus passes in 365.25 days
-

I do this research
2nd
224
- All these values are light motion values, for example
- Saturn moves during 120536 days a distance = light supposed velocity motion
distance during a solar day =100733 mkm= the 9 planets orbital circumference
total.
- That tells Saturn motion effects greatly on the solar system because the interaction
with light motion.
(III)
- Saturn data has a relationship with light motion wit both velocities (1.16 mkm/sec
and 0.3 mkm/sec). for example
- Saturn (9.7 km/sec) moves during 30589 seconds a distance 300000 km
- Also
- The moon (27.78 km/sec) moves during 10747 seconds a distance 300000 km
- And
- (300000 km / 24.1) = (120536 km/9.7)
- The relationship with light motion causes great effect of Saturn motion on the solar
system geometry –

I do this research
2nd
225
11-6 Equal Distances Method
- The Equal distances Method is A Feature be discovered in the solar system motion
- 2 Planets move equal distances in defined periods of time. And as a result this 2
planets data be in proportionality with one another.
- This feature be analyzed as deep as possible to know why it's used and what
geometrical effect it creates. Let's explain that by using examples in following:
Example No. (1) (Earth and Pluto Data)
(A)
- Earth moves during its day period a distance = Pluto motion distance during its day
period = the Earth moon displacements total during its day period (error 1%).
(B)
- (Earth velocity / Pluto velocity) = (Pluto Day period/ Earth day period) = 6.37 (0.8%)
- Pluto orbital distance 5906 mkm = Earth orbital circumference 940 mkm x 2π
- Pluto orbital period 90560 days = Earth Cycle (1461 days) x 2π3
(C)
- (Pluto motion distance daily406000 km/ the moon displacement daily88000 km)=4.61
- (The Moon Day Period 708.7 h / Pluto Day Period 153.3 h) = 4.61
- Tan (12.2) x 708.7 h (The Moon Day Period) = 153.3 hours (Pluto Day Period)
- Tan (13.17) x 655.7 h (The Moon rotation Period) = 153.3 hours (Pluto Day Period)
- (Where 13.17 deg = the moon daily motion degrees, and 12.2 deg =13.17 deg – 0.9856
deg Earth motion daily degrees)
- We have discussed this data before.
- The proportionality of data may be created by the equal distances of motions or by
the rate of time. Or may both be created based on each other.
- The fact is that,
- The 2 planets move equal distances in defined periods of time and their data be
created in proportionality with one another.

I do this research
2nd
226
- Let's analyze the previous example as possible:
o Equal Distances Be Passed By Earth And Pluto Motions,
o We should pay attention for the equal distances nature, because the distances
are equal. But these distances are not random distances. On the contrary,
these are mentioned distances because the planets pass these distances
during their days periods.
o It's necessary to put this data in front of our eyes (always). Because the
equal distances aren't any random equal distances but distances be passed
during defined periods of time and by that these distances be mentioned to
be equal.
o Defined period of time means a period be defined by a planet motion. as a
planet orbital period or day period or rotation period …etc
o Shortly a period depends on a planet motion
Example No. (2) (Mercury and Mars Data)
- Mercury moves during its rotation period a distance =243 mkm (error 1%)
- Mars moves during Venus day period a distance =243 mkm
- We have 2 equal distances! Where's the data proportionality?
- (Mercury Day Period)/ (Venus Day Period) = (Mercury Orbital Period / Mercury
Rotation Period)
- That tells, Venus day period is created as a function in Mercury cycles periods.
Can Mars motion for the distance 243 mkm be the reason of this proportionality?
Notice
243 solar days = Venus Rotation Period (if 1 mkm = 1 solar day), so this distance
(243 mkm) may refer to Venus Rotation Period
On the other side
Venus moves during (365.25 days) a distance = 1106 mkm =2π x 175.94 mkm
175.94 solar days = Mercury Day period (means, 1 mkm = 1 solar day is a rate
known by Venus and Mercury).

I do this research
2nd
227
The Equal Distances Using Range
- Why Do We Study The Equal Distances Method?
- Because
- This Method is almost the planets motions method. That means the solar planets
don't move by the sun mass gravity but move by the equal distances method (or at
least the outer planets move by the equal distances method).
- That means,
- The solar system motion depends on the equal distances method greatly.
- Why Do We Need to Know That? Even If Be A Fact
- Because
- This explains the heavy using of the equal distances methods in the solar system
motion. and by that we explain the following facts:
o 50% Of All Distances In The Solar System Be Equal One Another
o And
o 40% of all distances in the solar system be rated one another with the same
one rate (1.0725)
o This data be in appendix no. 1 of this paper
- Notice
- We have discussed this data before but the data of (50% of all distances are equal)
and (40% of all distances are rated) be put in the appendix no. 1 to be used as
reference

I do this research
2nd
228
Let's summarize the equal distances using range in following:
(1)
50% of all distances in the solar system are equal one another
and
40% all distances in the solar system are rated with the same rate (1.0725)
(2)
Planet 8 days Cycle depends on the equal distances using
We have discussed Planet 8 days Cycle in The planets Unified General Motion
(3)
Different examples we have discussed as Earth and Pluto equal distances
And also Mercury and Mars equal distances
(4)
Planet diameter or circumference using as a period of time depends on equal distances
let's provide some examples in following:
o Jupiter (13.1 km/s) moves during 10921 second a distance =142984 km =
Jupiter diameter.
o Uranus (6.8 km/s) moves during 7510 second a distance =51118 km =
Uranus diameter.
o Pluto (4.7 km/s) moves during 10921 second a distance =51118 km =
Uranus diameter.
o Pluto (4.7 km/s) moves during 51118 seconds a distance =2 x 120536 km =
Saturn Diameter.
o Pluto (4.7 km/s) moves during 2 x 120536 seconds a distance =1.1318 mkm
= Jupiter motion distance during a soar day.
o Where
o 10921 km = The Earth Moon Circumference
o 7510 km = Pluto Circumference

I do this research
2nd
229
The Equal Distances Using Reason
- I provide here an explanation why the equal distances method be used
- I want to say,
- This explanation (whether be correct or incorrect) can't effect on the fact that (the
Equal distances is the reason of the solar planets motions) because this fact
depends on The Equal Distances Method Wide Range Of Using.
- So, let's ask
- Why Be The Equal Distances Used As The Method Of The Planets Motions?
- Because
- This is the method be used between light motion and planet motion. and between 2
planets motions.
- That means,
- If a light beam energy be used to cause a planet motion, this energy be used
through the equal distances method
- And
- If a planet motion energy be used to cause another planet motion, this energy be
used through the equal distances method
- It's the same one method by which the motion be transported from a light beam to
a planet or from a planet to a planet. One method be used for 2 jobs
- As a result, we find distances be passed by light motion and be passed (again) by
planets motions.
- That means, because The Matter Is Created Out Of Light
- The Solar Planets motions depend on the equal distances method

I do this research
2nd
230
Example
- Light known velocity (0.3 mkm/s) travels during a solar day (86400 seconds) a
distance= 25920 mkm
- But
- All planets motions distances per a solar day (=17.75 mkm) x 1461 days = 25920
mkm
- If the period of time is not the cycle (1461 days), this equation would have no any
value. But because the period of time is a cycle (1461 days =365 +365 +365 +366)
defined by Earth motion. that tells the distance 25920 mkm be passed by 2 players.
(A light beam and all planets motions total).
- To prove this is a light motion. we see the value 25920 mkm as a period of time
which is (the precession cycle 25920 years)
- Also
- The moon (2.4 mkm/ daily) moves during 10747 days a distance =25920 mkm
- The data tries to show that, the equal distances be used by light and planet motions
or by planet and planet motions.

I do this research
2nd
231
12-The Solar System Distances Be Created In A Network Form
12-1 Preface
12-2 The Continuum effect Through the Solar System Distances
12-3 The Solar System Distances Distribution
12-4 The Solar System Distances Dependency On One Another

I do this research
2nd
232
12-1 Preface
- The Solar Planets data analysis shows that, The Planets Distances Be Created In A
Network Form.
- Means,
- The solar system distances are considered as one geometrical design or one
geometrical system. There's no one distance defined as independent distance or
created out of this one geometrical design.
- Let's suppose, this geometrical design be a square. So all dimensions in this
square be defined geometrically based on this square angles and data. means no
one dimension can be created independently from this square data because all of
them are parts in the same one geometrical design.
- Suppose we have a triangle its 2 angles are (60 degrees and 80 degrees) what's the
third one? (40 degrees). Can this angle be any thing else than (40 degrees)? NO it's
obligatory value defined by the geometrical basic rules
- Similar to that,
- The Solar System Distances Be Created Based On One Geometrical Design.
- But
- By using this vision we will have surprises. Because this vision tells (for example)
Venus Jupiter Distance be= 671 mkm Because Earth Jupiter distance = 629 mkm
- That creates another vision
- While we consider Venus is a planet and Earth is another planet and they move
independently from one another. We are astonished now with this fact that, their
distances to Jupiter be defined based on one another! How? And Why?
- How (Because The Solar System Distances Be Created In A Network Form)
- Why (this answer was discussed in Mercury Motion Analysis)
- The fact is proved strongly that
- The Solar System Distances Be Created in A Network

I do this research
2nd
233
- There Are Many Basic Proves For This Fact which are:
- (1) The Continuum effect Through the Solar System Distances
- (2) The Solar System Distances Distribution
- (3) The Solar System Distances Dependency On One Another
- Let's discuss these proves in their details in following
- Notice
- The idea tells that,
- By energy the distances be created in a network form
- means
- The Distances Be Similar To The Railways And The Planets Be Similar To The
Trains Move On The Railways.
- The important point is that, the distance be created by energy and because of that
it's a defined trajectory of motion. that makes the planet moves in some obligatory
motion because the energy created the distances.
- Means, the distances aren’t common space but defined distances be built for
specific reason. Based on that (Distance = Energy)
- And
- The created distance defined a trajectory of motion
- As a result
- The solar system distances be as a map created by energy

I do this research
2nd
234
12-2 The Continuum effect Through the Solar System Distances
Point No. (I)
The Continuum Effect Description
I- Data
(1)
50% of all distances in the solar system be equal one another
(2)
40% of all distances in the solar system be rated one another
Let's provide this data in details in following:
Data Group No. (A)
(1)
(2)
(3)
(4)
(These Equal Distances Be Around 50% Of All Distances In The Solar System)

I do this research
2nd
235
Data Group No. (B)
10. 0725
.
1
mkm
2.41
nce
Circumfere
Orbital
Moon
mkm
2.58
Motion
Daily
Earth
=
11. 1.0725
km)
(378500
radius
Eclipse
Solar
Total
km)
(406000
radius
orbital
Apogee
=
12. 0725
.
1
distance
Mercury
Jupiter
mkm
720.3
Distance
Orbital
Juppiter
mkm
6
.
778
= (Error 0.7%)
13. 1.0725
Distance
Venus
Jupiter
mkm
670
distance
Mercury
Jupiter
mkm
720.3
=
14. 1.0725
Distance
Earth
Jupiter
mkm
629
Distance
Venus
Jupiter
mkm
670
= (0.6%)
15. 1.0725
mkm)
(1325.3
Distance
Venus
Sarurn
mkm)
(1433.5
Distance
Orbital
Saturn
= (0.8%)
16. 1.0725
mkm)
(1205.6
Distance
Mars
Sarurn
mkm)
(1284
Distance
Earth
Saturn
= (0.7%)
17. 1.0725
mkm)
(2644
Distance
Mars
Uranus
mkm)
(2872.5
Distance
Orbital
Uranus
= (0.7%)
18. 1.0725
mkm)
(4495.1
Distance
Orbital
Neptune
mkm)
(4864
= (0.5 %)
(These Rated Distances Be Around 40% Of All Distances In The Solar System)
Additional Data
(10)
0725
.
1
T.
Axail
Earth
23.4
T.
Axail
Mars
25.2
T.
Axail
Mars
25.2
T.
Axail
Satrun
26.7
Tilt
Axail
Satrun
26.7
Tilt
Axail
Neptune
28.3
=
=
=

I do this research
2nd
236
II- Discussion
- Let's suppose that,
- The solar planet moves independently from all other planets motions depending on
the sun mass gravity, in this case how these equal distances (50%) and the rated
distances (40%) be created?
- No logic can imagine a different reason for each 2 equal distances? We conclude
logically that (One Reason Caused These Distances To Be Equal)
- Based on that, we don't deal with independent distances instead we deal with one
machine be created by all distances integration. and As A Result, ONE REASON
effects on (the machine) caused (all distances to be effected by this one reason)
- I want to say,
- These 2 groups of data disproves Planet Motion Independency Concept.
- There's no way to create the equal and rated distances in huge percentage if the
planets motions be independent from one another.
- But
- Why These Distances Are Equal?
- The answer is somehow complex….. let's provide a short answer in following:
- The solar planets don't move by the sun mass gravity. But they move by equal
distances method (or at least the outer planets move by this method)
- The equal distances method is a method be used by many planets. Let's explain
how this method be used…. 2 Planets move equal distances in defined periods of
time (as example, Earth moves during its day period a distance =2.574mkm and
Pluto moves during its day period a distance = 2.598 mkm, these 2 distances are
different with 1% only and by that they can be considered equal distances. As a
result for these 2 motions equal distances Earth and Pluto motions data be in
proportionality with one another)

I do this research
2nd
237
- This feature be used hundreds of time by the solar planets, I guessed that, because
this method is used hundreds or even thousands of times in the solar system, this
method causes the planets motions (but I don't know how yet). The idea is that,
this method (the equal distances) works by a geometrical rule, as a waterwheel or a
lever, these methods work by geometrical rules. Similar to that the equal distances
method caused the planets motions
- As a result for using the equal distances method, 50% of all distances in the solar
system be created equal one another and also 40% of all distances be rated with
one another by the same one rate (1.0725)
- But , Please note
- The 50% equal distances be created because the solar system uses the equal
distances method as I have explained it before.
- But
- Why Does The Rate (1.0725) effect On 40% Of All Distances?
- The rate (1.0725) effect on (40% distances) is a proof for a continuous effect
through the distances. There's Some Effect Like A Continuum effects on the
solar system distances!
- How that can be done if the planets move independently from one another? and
- If the distances be created independently from one another how one effect can be
continuous through (40%) of all distances?
- I try to prove, the distances are created in a network or by one geometrical design.
- means,
- the distances be created together in one geometrical system as one machine
because of that one effect can continue through (40%) of all distances
- the explanation of using this rate (1.0725) isn't our question…. Our basic notice is
that this effect has a continuous task through the solar system distances which
can't be done except if these distances be created based on one geometrical system
or in one Network

I do this research
2nd
238
- I wish the argument is powerful and proved strongly
- The data gives us a chance to show that, the planet motion can't be independent
from other planets motions. we here don't deal with independent data but with one
system of data.
- Notice
- The Continuum Effect Through The Solar System Is A Powerful Proof For The
Planets Creation And Motion Depend On A Light Beam
- Because
- The Continuous Motion And Effect Is A Feature Of Light Motion – and by
that – the solar planets can't be described as separated rigid bodies.
- Based on that,
- The solar system distances as a network should be used as a proof for the solar
planets creation out of light.

I do this research
2nd
239
12-3 The Solar System Distances Distribution
(1st
Point) The Idea Explanation
- There are 2 basic features we should discuss about the solar planet motion, thee 2
features are:
- (1) The Planets Motions Use Different Rates Of Time
- (2) The Planets Motions Use The Distances As Periods Of Time
- These 2 features we have discussed before in this paper discussion
- I need here just to explain how these features be used to test their effects on the
distances distribution
- There are different rates of time be used by the solar planets example:
- 1 hour of Mercury motion = 24.6 hours of Jupiter Motion
- And
- We have seen the rate (4.61) be the working rate between Pluto and the Earth
moon motions.
- Because the distances be used as periods of time and vice versa, the causes the rate
of time to be used as rates among the distances – as we have seen that the rate
(4.61) controls 50% of all distances in the solar system.
- Let's remember this rate in following
- 1 hour of Pluto motion =8.5 hours of the moon motion
- 8.5 =2 x 4.25 but (4.61 = 1.0725 x 4.25) (error 1%)
- By that the rate (4.61) be produced based on the rate of time (8.5) between Pluto
and the moon motion. and for geometrical reason this rate is the working one – for
that,
- Pluto move during a solar day 406000 km = 4.61 x (88000 km)
- (The moon displacement per a solar day =88000 km),
- and
- 708.7 hours (The Moon Day Period) = 4.61 x 153.3 hours (Pluto Day Period)

I do this research
2nd
240
- Now this rate (4.61) we see in Pluto and the moon motion data, but because it
works as a rate of time, it effects great on the distances distribution, let's remember
its data in following…
(2nd
Point) The Rate (4.61) Effect On The Distances Distribution
I- Data
778.6 mkm (Jupiter Orbital Distance) = 4.61 x 170 mkm
550.7mkm (Jupiter Mars Distance) = 4.61 x 119.7 mkm
2094 mkm (Jupiter Uranus Distance) = 4.61 x 2 x 227.9 mkm
2872.5 mkm (Uranus Orbital Distance) = 4.61 x 629 mkm
3033.5 mkm (Uranus Pluto Distance) =4.61 x 654.9 mkm
5127 mkm (Jupiter Pluto Distance) = 4.61 x 2 x 550.7 mkm
5906 mkm (Pluto Orbital Distance) =4.61 x 1284 mkm
1375.6 mkm (Mercury Saturn Distance) =4.61 x 2 x 149.6 mkm
3062 mkm (Saturn Neptune Distance) =4.61 x 671 mkm
4345.5 mkm (Earth Neptune Distance) =4.61 x 940 mkm
4267.2 mkm (Mars Neptune Distance) =4.61 x 929 mkm
Where
170 mkm = Mercury Mars Distance
119.7 mkm = Venus Mars Distance
227.9 mkm = Mars Orbital Distance
629 mkm = Jupiter Earth Distance
654.9 mkm = Jupiter Saturn Distance
550.7 mkm = Jupiter Mars Distance
1284 mkm = Earth Saturn Distance
149.6 mkm = Earth Orbital Distance
671 mkm = Jupiter Venus Distance
929 mkm = Earth Jupiter Distance when the 2 planets be on the sun 2 sides
940 mkm = Earth Orbital Circumference (Max error 1%)

I do this research
2nd
241
II- Discussion
- The solar system orbital and internal distances total be (45 distances)
- The previous data shows that (22 distances) are rated with the rate (4.61)
- Means
- 50% of all distances in the solar system be rated with one another based on this
same rate (4.61).
- The data proves the distances distribution depends on the planets motions rates
time.
- The data proves, The Solar System Distances Be Created As A Network.

I do this research
2nd
242
(3rd
Point) The Distances Distribution Proof.
I - Data
Equation no. (A)
1.16 mkm per second x (2) x 86400 seconds =(2) x 100733 mkm = (37100 mkm –
4900 mkm) x 2π = 28255 mkm + (2) x 86400 mkm
- Where
- 100733 mkm = The 9 Planets Orbital Circumferences Total
- This equation is the paper main equation and we have discussed it deeply before
(Equation no. 11-4-d)
- The equation shows that, the solar system distances be distributed based on
geometrical design.

I do this research
2nd
243
12-4 The Solar System Distances Dependency On One Another
I-Data
(1)
1.16 x 580 mkm = 671 mkm (Venus Jupiter Distance)
1.16 x 671 mkm = 778.6 mkm (Jupiter Orbital Distance)
1.16 x 629 mkm = 720.7 mkm (error 1%) (Mercury Jupiter Distance)
1.16 x 542 mkm = 629 mkm (Earth Jupiter Distance)
1.16 x 5127 mkm = 5906 mkm (error 0.7%) (Pluto Orbital Distance)
(2)
1.16 x 778.6 mkm x 2 = 1806 mkm
1.16 x 1806 mkm = 2094 mkm (Jupiter Uranus Distance)
1.16 x 2094 mkm = 2 x 1205 mkm (error 0.8%) (Mars Saturn Distance)
(3)
(1.16)2
x 170 mkm = 227.9 mkm (error 0.4%) (Mars Orbital Distance)
Where 170 mkm = Mercury mars Distance
(5)
1.16 x 1980 mkm = 2296.8 mkm
1.16 x 2296.8 mkm = 2644.6 mkm (error 0.7%) (Mars Uranus Distance)
1.16 x 2644.5 mkm = 3033.5 mkm (error 1%) (Uranus Pluto Distance)
We have discussed this data before, let's summarize its idea

I do this research
2nd
244
II-Discussion
We have discussed this data before, let's summarize its idea
- The previous distances are example for similar huge number of distances which
depend on the rate (1.16)
- We should analyze this rate (1.16) later
- But
- We need here to see how the distances be distributed base don this rate (1.16)
- The next data can help our analysis greatly
More Data
- 778.6 mkm (Jupiter Orbital Distance) = 1.0725 x 720.7 mkm (error 1%)
- 720.7 mkm (Jupiter Mercury Distance) = 1.0725 x 671 mkm
- 671 mkm (Jupiter Venus Distance) = 1.0725 x 629 mkm
- 629 mkm (Jupiter Earth Distance) = 1.0725 x 580 mkm (error 1%)
- This data also proves, these distances be created based on a geometrical design.
Because the same distances be rated on (1.0725 and 1.16).
- Please remember 1.16 =(1.077)2
The analysis shows that the solar system distances can't be created interpedently from
one another but created in a network form. And because of that the distances be
distributed based on a geometrical design and as a result the distances be rated with
the rates (1.16) and (1.0725)

I do this research
2nd
245
Appendix no. (1)
The Solar System Equal Distances List
(A)
(1)
(2)
(3)
(4)
II- Discussion (A)
The previous distances form around 50% of all distances found in the solar system
(All orbital and internal distances)… Why These Distances Are Equal One Other?
We may notice that – the distances equality can be produced more easily by light
motion than the rigid body motion - for example – when we push a ball toward a wall
the ball after collision with the wall will return a distance (NOT) equal the original
one - because the collision causes to decrease the ball motion momentum – but the
light can be reflected at equal distances easily – means – equal distances can be
produced by light motion more easy than the Rigid Body Motion.

I do this research
2nd
246
(B)
19. 0725
.
1
mkm
2.41
nce
Circumfere
Orbital
Moon
mkm
2.58
Motion
Daily
Earth
=
20. 1.0725
km)
(378500
radius
Eclipse
Solar
Total
km)
(406000
radius
orbital
Apogee
=
21. 0725
.
1
distance
Mercury
Jupiter
mkm
720.3
Distance
Orbital
Juppiter
mkm
6
.
778
= (Error 0.7%)
22. 1.0725
Distance
Venus
Jupiter
mkm
670
distance
Mercury
Jupiter
mkm
720.3
=
23. 1.0725
Distance
Earth
Jupiter
mkm
629
Distance
Venus
Jupiter
mkm
670
= (0.6%)
24. 1.0725
mkm)
(1325.3
Distance
Venus
Sarurn
mkm)
(1433.5
Distance
Orbital
Saturn
= (0.8%)
25. 1.0725
mkm)
(1205.6
Distance
Mars
Sarurn
mkm)
(1284
Distance
Earth
Saturn
= (0.7%)
26. 1.0725
mkm)
(2644
Distance
Mars
Uranus
mkm)
(2872.5
Distance
Orbital
Uranus
= (0.7%)
27. 1.0725
mkm)
(4495.1
Distance
Orbital
Neptune
mkm)
(4864
= (0.5 %)
(10)
Discussion (B)
The same rate (1.0725) is used for all equations (around 18 distances = 40% of all
solar system distances) – why?
Suppose the equal distances are produced by light reflection and that cause these
distances to be equal – as I have supposed in the previous point (A).
Now suppose– part of these equal distances – is passed through another frame relative
to us – so this part of distances will suffer from Lorentz Length Contraction Effect
which is seen in the rate 1.0725
(Another frame can be found in the solar system because we deal with light motion) –
This explanation can answer why some distances are equal and others are rated with
the same rate (1.0725) – it's simply a feature of light motion.
0725
.
1
T.
Axail
Earth
23.4
T.
Axail
Mars
25.2
T.
Axail
Mars
25.2
T.
Axail
Satrun
26.7
Tilt
Axail
Satrun
26.7
Tilt
Axail
Neptune
28.3
=
=
=

I do this research
2nd
247
References and Biography
Planet Motion Features Disproves Newton Theory (Revised)
https://www.academia.edu/52675549/Planet_Motion_Features_Disproves_Newton_Theory_Revised_
(Preprint No. 1) Mars Migration Theory
https://www.academia.edu/49051037/_Preprint_No_1_Mars_Migration_Theory
(Preprint No. 2) The Moon Orbital Motion Equation
https://www.academia.edu/49051029/_Preprint_No_2_The_Moon_Orbital_Motion_Equation
The Solar Planets Motions Description (Revised)
https://www.academia.edu/49464125/The_Solar_Planets_Motions_Description_Revised_
The Moon Orbital Motion Geometry (II)
https://www.academia.edu/45181646/The_Moon_Orbital_Motion_Geometry_II_
Dr. Budochkina, Svetlana Aleksandrovna
Associate professor - Candidate of physico-mathematical sciences (2005)
http://www.mathnet.ru/eng/person22119
List of publications on Google Scholar
List of publications on ZentralBlatt
https://mathscinet.ams.org/mathscinet/MRAuthorID/757317
http://elibrary.ru/author_items.asp?spin=6087-3245
http://orcid.org/0000-0003-3447-0425
http://www.researcherid.com/rid/G-7453-2014
http://www.scopus.com/authid/detail.url?authorId=6507007003
https://www.researchgate.net/profile/Svetlana_Budochkina
Full list of
publications:
http://web-local.rudn.ru/web-
local/prep/rj/index.php?id=2944p=15209
Mr. Gerges Francis Tawdrous +201022532292
Physics Department- Physics Mathematics Faculty
Curriculum Vitae http://vixra.org/abs/1902.0044
E-mail mrwaheid@gmail.com
Twitter https://twitter.com/GergesTawdrous1
Facebook https://www.facebook.com/gergis.tawadrous/
Linkedln https://eg.linkedin.com/in/gerges-francis-86a351a1
Quora https://www.quora.com/profile/Gerges-F-Tawdrous
Researchgate https://www.researchgate.net/profile/Gerges-Tawdrous
Researcherid https://publons.com/researcher/3510834/gerges-tawadrous/
ORCID https://orcid.org/0000-0002-1041-7147
Google https://scholar.google.com/citations?user=2Y4ZdTUAAAAJhl=en
box https://app.box.com/s/47fwd0gshir636xt0i3wpso8lvvl8vnv
Academia https://rudn.academia.edu/GergesTawadrous
List of publications http://vixra.org/author/gerges_francis_tawdrous
Also https://www.slideshare.net/Gergesfrancis?utm_campaign=profiletrackingutm_medium=sssiteutm_source=ssslideview
Please scan the figure (ORCID)

The Solar System Design Disproves Newton Theory (Revised No.2)

More Related Content

What's hot (20)

Similar to The Solar System Design Disproves Newton Theory (Revised No.2) (20)

More from Gerges francis (20)

Recently uploaded (20)

The Solar System Design Disproves Newton Theory (Revised No.2)