The Solar System Distances Network Theory (Revised)

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
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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 Distances Network Theory (Revised)
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 –24th
August 2021
Abstract
Paper hypothesis.
- There's a light beam its velocity (1.16 mkm/) moves in parallel to Mercury Motion.
Paper Hypotheses Explanation
- Planet Motion be a cover for a light motion found behind. The light motion uses
Planet motion data for its own motion.
- The hypothesis creates one more universe in parallel to the universe of matter. And
by that, The Planets be similar to ships sail over a sea. But the sea is light motions
network.
- The hypothesis uses Mercury Motion as a examination point to prove that a motion
of a light beam its velocity (1.16 mkm/s) be behind Mercury Motion.
- The light motion causes to create the solar system distances in a network form and
depends on one geometrical design. Because of that, the solar system distances
analysis will prove the light supposed velocity (1.16 mkm/s) existence.
Paper Conclusion
- The Solar System Distances Be Created in a network form and depend on one
geometrical design which proves that these distances be created by a motion of a
light beam its velocity =1.16 mkm/sec.
Paper Objective
- The paper proves The Light Supposed Velocity (1.16 mkm/sec) Existence.

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- This proof disproves automatically Newton Theory Of The Sun Mass Gravity.
- Because
- The sun rays velocity = 300000 km/sec.
- Where
- The solar system distances prove that they are created by a light beam its velocity
=1.16 mkm/sec.
- means,
- The sun light velocity (300000 km/) is created as a side product from the light
supposed velocity (1.16 mkm/s),
- means
- The Sun Is Created After All Solar Planets Creation And Motion
- Which disproves Decisively Newton Theory Of The Sun Mass Gravity.
Please scan the figure (ORCID)

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Paper Contents
Subject Page No.
1- Introduction 3
2- Methodology 6
3-The Network Theory Details 8
3-1 Preface 8
3-2 The Motions Network Properties 10
3-3 Planet Motion Description 13
3-4 The Solar Planets Unified General Motion. 14
3-5 The Theory Point of Proof 16
3-6 The Paper Used Definitions 19
4- Light Beam (1.16 mkm/s) Existence Proves Discussion 20
4-1 Preface 20
4-2 The Light Beam (1.16 mkm/s) Proves Discussion 21
4-3 Jupiter Orbital Circumference Analysis 65
4-4 Mercury Jupiter Distance Analysis (720.7 mkm) 83
4-5 Saturn Motion data analysis. 102
4-6 The Solar System Distances As A Network (More Data) 111
5- The Sun Age Historical Proves 115
5-1 The Sun Circles The Earth 116
5-2 The Rate (1.0725) 117
5-3 The Sun Diameter Analysis 121
5-4 The Sun Rays Analysis 124
129
6- The Solar Planets Move One Unified General Motion 130
6-1 The Planets Unified General Motion Description 131
6-2 The Solar Planets Motions Are Complementary One Another 132
6-3 The Solar Planets Move An Unified Motion (A Team Motion) 137
7- The Solar Planets Motions Depend On Different Rates of Time 142
7-1 Preface 143
7-2 Earth And Pluto Rate Of Time 144
7-3 Mercury Motion Rate Of Time 153
7-4 Saturn Motion Rate Of Time 156
7-5 Uranus And Jupiter Rate Of Time 158
8- The Solar Planets Motions Analysis 160
8-1 Preface 161
8- 2 The Equal Distances Method 162
8-3 The Inner And Outer Planets Motions Be Distinguished 168
Appendix No.1 The Solar System Equal Distances List 177
Appendix No.2 Planet Orbital Distance Equation 179
Appendix No.3 Planet Gravity Equation 182
References and Biography 183

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1-Introdcution
- Paper hypothesis tells that,
- A light beam its velocity (1.16 mkm/s) moves in parallel to Mercury Motion.
- How to understand this hypothesis?
- Let's answer in following:
- The matter and space are created from energy provided by a light beam its velocity
(=1.16 mkm/s). The created matter doesn't separate from its light parent, On the
contrary, they move together one unified general motion. but this motion uses
different rates of time between light and matter motions.
- This is the original theory based on which the paper hypothesis be created.
- The difficulty is that,
- We need to prove (at first), a light beam its velocity (1.16 mkm/s) be found in the
solar system. where the hypothesis of light supposed velocity (1.16 mkm/sec) is a
very strange one to be suggested because of the Special theory of relativity.
- The original theory be a complex task if we try to prove it in the same paper with
the light supposed velocity (1.16 mkm/sec) existence.
- How To Manage This Situation?
- I picked up a piece of the theory which is the hypothesis, and hoped to prove the
light supposed velocity (1.16 mkm/sec) existence.
- The fact is that,
- A light beam its velocity (1.16 mkm/sec) be accompanying with Mercury motion.
This light beam uses Mercury motion data for its own motion and this using effect
on Mercury motion data.
- As a result for the light supposed velocity (1.16 mkm/s) motion in parallel with
Mercury motion, The Solar System Distances Be Created In A Network Form
By Energy Provided By A Light Beam Its Velocity =1.16 mkm/sec.
- Based on that,

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- If we analyze the solar system distances deeply, We Will Discover These
Distances Be Created In A Network Form By Energy Provided By A Light
Beam Its Velocity =1.16 mkm/sec
- Shortly
- This is a good approach because we try to catch directly the point of proof for the
light supposed velocity (1.16 mkm/sec) existence. In place to provide theoretical
explanations and answers for questions as (why light motion be accompanying
with planet motion?)… such questions have answers I can provide them… But
why I should do?
- We don't need to provide a complete theoretical description for light and planet
motions. Because I have a discovery need to prove it, which is the light supposed
velocity (1.16 mkm/sec) existence. If this light existence can be proved, Newton
Theory of the sun mass gravity will be just disproved decisively. And many other
effects can be produced as results.
- Briefly
- In place to provide a complete theoretical description I have provided one
hypothesis explains a case and through this case I hope to prove that a light
supposed velocity (1.16 mkm/sec) motion be behind Mercury motion which
proves its existence.
- By that I choose a direct approach for the proves discussion and the basic idea be
understandable where a light motion be accompanying planet motion because the
matter is created out of light and doesn't separate its parent light but move with it
in one unified general motion by using different rates of time.

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2-Methodology
- I use the planets data analysis to discover the solar planets creation and motion.
- This method is so useful because it creates comparisons between planets motions
data and the theories which try to explain these planets motions data. Also the
method creates comparisons between planets motions data with each other.
- Let's use one example for explanation in following
- Example no. (1)
- Light supposed velocity (1.16 mkm/sec) moves during a solar day (86400 sec) a
distance = 100733 mkm
- Where
- 100733 mkm = The Solar Planets Orbital Circumference Total
- The data shows that, some puzzles be found, because the light supposed velocity
(1.16 mkm/sec) travels during a solar day which is a period defined by Earth
motion a distance = the planets orbital circumferences total
- We need to know if this data be created by chance or there's a geometrical reason
behind.
- Let's see one more example in following
- Example no. (2)
- 1.16 mkm/s x 86400 sec x 2= 100733 mkm x 2= (37100 mkm – 4900 mkm) x 2π
= 28255 mkm + 2 x 86400 mkm.
- This equation is the paper main equation and it creates many discussions…
- Here I try to show that, the distance analysis can lead to important conclusions
- We interest here in the part no. (3) of the equation only, and the rest will be
discussed in the paper discussion…
- (37100 mkm – 4900 mkm) x 2π
- Where
- 37100 mkm = Pluto Orbital Circumference
- 4900 mkm = Jupiter Orbital Circumference

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- The difference between Pluto and Jupiter orbital circumference be a radius in a
circle its circumference be = 2 x 100733 mkm
- Why?
- The equation tells (37100 mkm - 4900 mkm) = 32200 mkm (this is a distance be
produced by planets motions) and
- 2 x 100733 mkm be a distance produced by light motion (during 2 solar day)
- Between them the rate (2π)
- Can this data be produced by pure coincidence of numbers? Or it can be created
based on a geometrical reason…
- Means,
- Is there a geometrical reason to create the distance (32200 mkm) to be = 2 x
100733 mkm / (2π)?
- The 2 examples tried to explain the paper method concept. The paper analyzes
planets motions data as deep as possible and compare between different data to
know if this data be created in proportionality with one another, and if that's the
case why the data be created in proportionality.
- The previous discussion gave us no conclusion because the ideas aren't clear yet.
We need to complete this equation discussion in the paper to see the ideas behind
it.
- Notice
- Light known velocity (0.3 mkm/sec) travels during a solar day (86400 s) a distance
(=25920 mkm) and we see this value as a period of time which is (25920 years =
The Precession Cycle)
- The notice aims to shows that the light (generally) uses the solar day period for the
solar system basic data.

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3-The Network Theory Details
3-1 Preface
3-2 The Motions Network Properties
3-3 Planet Motion Description
3-4 The Solar Planets Unified General Motion.
3-5 The Theory Point of Proof
3-6 The Paper Used Definitions
3-1 Preface
- The paper provides 3 basic points for discussion which are
- (1st
Point) Mercury Motion Be Accompanying With A Light Beam Motion
- (2nd
Point) The Light Velocity Be 1.16 mkm/sec
- (3rd
Point) Mercury And Light Motions Accompaniment Caused To Create The
Solar System Distances In A Network Form.
- These points of discussion aim to prove, A Light Beam Its Velocity =1.16 million
km per second Be Found In The Solar System Motion.
- This proof depends on the distances creation and distribution. Where the distances
be created in a Network form and this network be created by an energy of a light
beam its velocity =1.16 mkm/sec.
- In more clear words
- The light motion be accompanying with the planet motion. As a result, the light
uses the planet motion data in different forms. By this vision, the universe be
doubled and another universe be found in parallel to the universe of matter because
the light uses planet motion data for its own motion.
- The vision causes difficulties for our simple description about the solar system.
where Newton considered the solar planet as a rock moves by the sun mass gravity
effect independently from all other planets motions, and each planet distance be
created independently from the other distances. the fact is that, the solar system is
a network of motions distances designed geometrically and the solar planet is a

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point on this network. Where this network be created by light motion, the planet
motion be done as a following motion for the light motion.
- The idea is so complex because it creates one more universe be found in parallel to
the universe of matter. Where this hidden universe leads the matter universe
motion.
- That means,
- the solar planets be similar to ships sail over a sea and this sea is the light motions
network.
- This vision gives us completely different vision from what Newton had told about
the planet motion.
- The theory basic point is, The Light Motion Uses Planet Motion Data For Light
Own Motion. That Creates 2 Different Readings For All Planets Data.
- To prove the paper hypothesis,
- The paper uses the distances between Mercury and Pluto to prove the network
theory because they are the most clear distances can prove the network theory.
- Again
- The basic point is that, the light uses planet motion data for different reasons, and
by that, planets data which we know perfectly shows new meanings because of the
light motion using of it.
- By that, we will see the light motion uses Mercury motion data for light own
motion and we will discover that Mercury own data be created by reasons out of
Mercury Motion Range.
- For that reason, the paper uses 2 different readings for the same data. and because
of that, the same data be discussed 2 times in the paper.

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3-2 The Motions Network Properties
- The Network Theory tells that, a light beam and a planet move side by side, how
can we describe this motion?
- Let's summarize this motion features in following
- (1st
Feature) The Motion Uses Different Rates Of Time.
- It's a simple idea. Because light moves in parallel to a planet motion, logically they
use different rates of time
- The basic rate of time for the light and matter motion is the following rate (One
Second Of Light Motion = One Solar Day Of A Planet Motion)
- Or in more clear words
- Light motion energy for (1 second) causes a planet to move for (1 solar day)
- This is the basic general rate between the light and matter motion.
- But
- The solar planets use this (general) rate in parts, and as a result they create small
rates of time among themselves where these small rates of time be unified in the
general one (1second of light motion = 1 solar day of planet motion)
- Because of that, different rates of time be found among the solar planets… for
example
- 1 second of light motion = 1 hour of Mercury motion
- 1 hour of Mercury motion = 9.18 hours of Pluto motion
- 1 hour of Mercury motion = 24 hours of Jupiter motion
- 1 hour of Earth motion = 6.37 hours of Pluto motion
- These small rates of time be unified together to create the general one (1 second of
light motion = 1 solar day of planet motion).

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- (2nd
Feature) The Distances be Created in a network form
- Light motion passes a distance. And then uses this distance as a period of time to
pass another distance and uses the second distance as a period of time to pass a
third distance …..etc
- By that the distances be created in a network form
- The distances creation almost costs energy provided by light motion. for that
reason, the light creates the distances by its motion. as a result planet motion
through these same distances costs no energy because light created them before.
That shows a distinguish between light and planet motions through the same
distances.
- The light using of a distance as a period of time is understandable behavior for our
physics science. But this same behavior be used also by a planet motion. and we
have no way to explain how a planet low velocity motion can use distances values
as periods of time.
- But this behavior be a fact we should prove through the paper discussion.
- Just we should notice the rates because
- Light motion uses (1 million km = 1 second), but
- Planet motion uses (1 million km = 1 solar day).
- These rates be defined based on the rule (Light motion energy for (1 second)
causes a planet to move for (1 solar day)
- These rates are important to analyze the motions data and discover the network
form for these motions.
- (3rd
Feature) The Distances Be As Railways
- The distances creation process caused the distances to be similar to railways but
the planets be as the trains.
- That makes the distances and trajectories of planets motions be defined based on
geometrical design and be created depending on one another geometrically. Where
the planets move as trains on these railways.

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Notice
- While we are so confused to explain the solar planets order. The fact is that, the
solar planets order is a typical to the interference frame be created by light
coherence in the double slits experiment (Young Experiment).
- the frame which produced bright fringes and dark fringes. If we accept that the
bright fringes be the planets and the dark fringes to be the distances (space) by that
the planets order be a typical to the interference frame.
- Where the greatest fringe be in the middle (Jupiter), and the fringes be decreased
on both sides regularly (Mars is exceptional because of its migration). But still
Uranus is exceptional because Neptune should be in its position.

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3-3 Planet Motion Description
- Because the motions distances and trajectories be as a network
- And
- Because the motions distances and trajectories be similar to railways and the
planets be similar to trains.
- As A Result
- The Planet Motion be defined as obligatory motion. where the planet has no option
of freedom to move through any trajectory but has to move through specific
trajectory be defined by the distance network created by light motion energy.
- Shortly
- That makes the planet motion be an obligatory motion. where the planet has an
obligation to do specific motion be defined by the network of motions.
- Now
- The obligation of motion be extending for the planets are created already and also
for the planets which will be created later. That because the solar system distances
are created based on one geometrical design and all motions are defined to be in
harmony geometrically with the distances network. As a result, the planet (which
will be created later) has a defined motion (now).
- Shortly,
- The planet is created for a defined task found before this planet creation. the
motion is defined because it's a part of the network which controls all planets
motions.
- That means,
- The solar planets move one unified general motion.
- Let's explain this motion in more details in the next point.

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3-4 The Solar Planets Unified General Motion.
- The theory leads us to the following conclusion: (The Planets Motions Create
One Unified General Motion)
- This conclusion tells the planets motions be similar to Train Carriages Motion
- Also It tells the planets motions be similar to Chess Pieces Motions
- The conclusion 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.
- Because the planet motion be complementary with other planets motions. to enable
to create the unified general motion, 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
- Shortly
- The network of distances create planets motion trajectories to be similar to the
railways. And planet move on their trajectories as trains move on railways.
- But
- The planets move one unified general motion, and by that, the solar planets move
as carriages of the same one train. They move together one unified general motion.

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- Also
- The solar planets motions be similar to a machine of gears motion. or similar to the
mechanical clock. Where the machine moves one unified motion and this one
motion be divided into tasks for each gear one task (Means, for each planet one
task be done by its motion).
- The Planets Tasks Together Unify To Create One Unified General Motion.

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3-5 The Theory Point of Proof
- What's The Theory Major Point?
- The theory tells something very interesting, that
- The moving planet which we see is a puppet on a puppet theater…
- In behind this planet motion,
- There's a light beam its velocity (1.16 mkm/s) moves in parallel
- This light beam uses (the planet motion data) for its own motion and creates the
distances in geometrical design and also creates the planet matter itself to be in
harmony with the motion defined by this light beam.
- The player is the light beam motion and the planet covers it.
- This is the idea shortly
- But
- How this theory effects on our vision toward the planet motion description?
- The point is that,
- Planet Motion data be defined by light motion and the planet motion covers it
- Means,
- If we search behind planet motion data we will find that it's found because of light
motion. the planet here works as a blackboard but the light is the teacher
- For example
- Mercury day period =4222.6 hours. Why?
- Why Mercury day period isn't = 4224 hours =176 solar days? Why should be less
than this value with 1.4 hours (= 5040 seconds)?
- The answer is that, Because of
- The Light Supposed Velocity (1.16 mkm/s) Effect On Mercury Motion.
- Why do we need to know that?

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- Because by this method we can discover and prove the light supposed velocity
(1.16 mkm/s). that shows from where the proves are originated… from the light
supposed velocity (1.16 mkm/s) motion effect on the planets motions data.
- Now,
- The paper importance be more clear
- Logically
- It's so hard to prove the light supposed velocity (1.16 mkm/s) existence basically
because of the special theory of relativity…
- But the proves I provide is the planets motions data.
- I have no method to observe the light supposed velocity (1.16 mkm/s) and no way
to test its existence empirically and know nothing about its motion nature
- But
- What I have is planets motions data be created by a direct effect of this light
supposed velocity (1.16 mkm/s) motion.
- And
- Because these effects on planets motions data be so wide and prevalent, I have to
suppose that a light beam with suppose velocity (1.16 mkm/s) must be found
behind this planets motions data.
Notice
- The paper approach provides an interesting point of analysis for the planets
distances and their motions data. that means
- While the theory ideas can be considered as imaginary ideas and be neglected as
ideas have no real roots. The fact is that,
- The paper analyzes planet motion data deeply and in different interesting methods,
by that, planet motion data be examined as deep as possible and basic questions be
produced as a result.

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- The respectful reader will find many interesting comparisons between planets
motions data and the physics book theories which can create more clear vision for
planets motions data understating…
- The light supposed velocity (1.16 mkm/s) existence is the answer I have found
after analysis for hundreds of planets motions data. if this answer is the wrong one
the respectful reader can suggest the correct answer.
- In all cases, Planets Motions Data be analyzed and examined as deep as possible
where many interesting facts and puzzles suggests the question (How This Data
Be Created?) and the suggested answer be (By An Effect Of A Light Beam
Supposed Velocity 1.16 mkm/s)

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3-6 The Paper Used Definitions
- The following words have the meanings mentioned for them. This page provides
the paper basic vocabulary
(1) A Defined Distance: It's a planet orbital distance or a distance between 2 planets.
(2) A Defined Period Of Time: it's a period of time be defined by a planet motion (as
a planet orbital period, rotation period or day period).
(3) (The Distance = Energy). This is a rule be used in the solar system geometry
explanation. means, the distance is a word refers to energy always.
(4) 50% Of The Distances Are Equal: 50% of all distances in the solar system be
equal one another. Appendix no. 1 provides a list for these equal distances. The
feature requires explanation where a geometrical reason must be found behind
these distances equality
(5) 40% Of The Distances Are Rated With (1.0725): 40% of all distances in the
solar system be rated to one another by the same one rate (1.0725). Appendix no. 1
provides one more list for these rated distances.
(6) Mars Migration Theory : It's a theory suggested that, Mars was the next planet
after Mercury with ancient orbital distance (=84 mkm) and had migrated to its
current orbital distance point (=227.9 mkm). Mars Migration Theory be discussed
in my previous paper (Please review The Paper References)
(7) Equal Distances: these are 2 equal distances be passed in defined periods of time
by 2 different planets. (for example Earth moves during its day period a distance =
2.5747 mkm but Pluto moves during its day period a distance = 2.5938 mkm. The
2 distances are different with 1% that make them almost equal). The equal
distances is a method be used in the solar system geometry widely and the paper
discusses this method in details in point no. (8-2).

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4- Light Beam (1.16 mkm/s) Existence Proves Discussion
4-1 Preface
4-2 The Light Beam (1.16 mkm/s) Proves Discussion
4-3 Jupiter Orbital Circumference Analysis
4-4 Mercury Jupiter Distance Analysis (720.7 mkm)
4-5 Saturn Motion data analysis.
4-6 The Solar System Distances As A Network (More Data)
4-1 Preface
- What Do We Try To Do In This Point?
- We analyze the distances between Mercury and Pluto as deep as possible to prove
that, these distances are created by energy of a light beam its velocity =1.16 mkm/s
- Let's summarize the idea in following:
- The Light supposed velocity (1.16 mkm/sec) started its motion from Mercury, and
moves in 2 basic trajectories.
- The First trajectory be from Mercury to Jupiter where the light supposed velocity
(1.16 mkm/sec) created Jupiter orbital circumference.
- And then
- This light beam travels from Jupiter to Pluto
- The second trajectory be from Mercury to Pluto Directly
- There's also one more trajectory, which is Mercury motion trajectory.
- We should discuss each one in details..
- Our main task here is that, to analyze different distances to prove that these
distances can't be created except by a light beam its velocity = 1.16 mkm/sec.
- We should keep an eye on Jupiter motion data and distances because Jupiter data
gives a strong proof for the light supposed velocity (1.16 mkm/sec) existence

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4-2 The Light Beam (1.16 mkm/s) Proves Discussion
I - Data
(1)
1.16 mkm/s x 86400 s = 100733 mkm
And
0.3 mkm/s x 86400 s = 25920 mkm
(2)
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)
(3)
1.16 mkm/s x 4224 seconds = 4900 mkm
(4)
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
(5)
1.16 mkm/s x 5040 seconds = 5848 mkm (Mercury Pluto Distance)
And
5848 mkm (Mercury Pluto Distance) – 4900 mkm = 948 mkm
(6)
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

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(7)
(8)
1.16 mkm/sec x 542 seconds =629 mkm (Earth Jupiter Distance)
(9)
1.16 mkm/sec x 3600 seconds =2 x 2088 mkm (Jupiter Uranus Distance)
(10)
(1.16 mkm/sec)2
x 2x 243 seconds = 654.9 mkm (Jupiter Saturn Distance)

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II- Discussion
Equation no. (1)
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. (1) tries to prove that, 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.

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- 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).

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Equation no. (2)
1.16 mkm/s x 500 s = 580 mkm
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)
Where
- 629 mkm = Earth Jupiter Distance
- 5127 mkm = Pluto Jupiter Distance
- The data show specific features of motions which are:
- 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. That means, the distances be created
in a network form of distances and not as independent distances from one another.
- The data shows clearly that, a velocity (1.16 mkm/s) be used to create these
distances or a rate (1.16) be used in their creation, which shows that the distances
be created as a network of distances by using of energy provided by a light beam
its velocity 1.16 mkm/s (notice, the network meaning will be explained)
- More data can help our analysis
More Data
(A)
- 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).

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(B)
- 50% of all distances in the solar system are equal. and 40% of all distances are
rated with (1.0725). appendix no. 1 provide these distances lists.
Network of Distances
- The Planets Motions distances and trajectories are similar to railways but the
planets are similar to trains move on these railways.
- The previous data provides a very good example to explain how the distances be
created in a network form.
- The previous explanation shows clearly that, the distances can't be created
independent from one another. because the distances be created by using light
supposed velocity (1.16 mkm/sec) motion trough these distances
- Mercury uses the same strategy, in next point we analyze Mercury motion to
prove this point.
- But, let's see the next notice
Notice
1.16 mkm/s x 500 s = 580 mkm
- The previous data shows that, some relationship be created by these motion
between Earth and Venus. Because
- 500 seconds, the light known velocity (0.3 mkm/s) needs to pass Earth orbital
distance 149.6 mkm. And based on this period (500 s) the light supposed velocity
(1.16 mkm/s) travels (671 mkm =Venus Jupiter distance).
- That explains the meaning of the (2nd
reading of data)
- It's a light motion but we have a connection between Venus and Earth.
- For example (108.2 sec x 4 x 1.16 mkm/sec = 502 mkm)
- 500 seconds be produced as 502 mkm based on (108.2 mkm = Venus orbital
distance). The data connects only between Venus and Earth…

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Mercury Motion
- Mercury motion uses the light motion strategy of distances creation process. The
following data shows that.
o Mercury moves during 84 days a distance = 344 mkm
o Mercury moves during 346.6 days a distance = 1419 mkm
o Where
o 5848 mkm = (Mercury Pluto distance)
o 1433.5 mkm = Saturn orbital distance
o 346.6 days = the nodal year.
o Mercury uses the rate (1 mkm = 1 solar day). It's different from the light
motion rate (1 mkm = 1 second) .
o The values 344 mkm and 14919 mkm be different with 346.6 days and
1433.5 days with max error (1%)
- The data gives 2 information which are
- The distances be created in a network form, where Mercury moves a distance and
use it as a period of time to move another distance. By that the distances be created
in a network form.
- Mercury uses the rate (1 mkm = 1 solar day)
- that because (1 second of light motion = 1 solar day of a planet motion).
- this data will be discussed deeply later.

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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. (3) and (4) as basic
equations. And through discussion many other data be added
- Notice
- Equation no. (4) is the solar system geometry main equation and it's the paper
main one we discuss it along the paper,

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Equation no. (3)
- Where
- Equation no. (3) 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. 3 provides 3 braches of data! let's summarize them in following:
- (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

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light and planet motions but in our equation the rate (1.0725) causes some
difficulty for this comparison vision.
- Notice
- The distance 720.7 mkm (Mercury Jupiter Distance) be discussed in point no. (4-4)
of this paper.
- (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 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

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- 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
- 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.
- Notice
- The accompaniment of The Light supposed velocity and Mercury Motions caused
Mercury day Period to be (4222.6 hours) to create a connection between Mercury
and Pluto based on the distance (5848 mkm = Mercury Pluto Distance)
- Is There any Proof for this conclusion?
- Yes, There's a proof in Mercury Motion data,
- Let's analyze Mercury Motion Data in following:

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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
- But
- Mercury (4.095 mkm/ daily) moves during 1433.5 solar days a distance = 5870 mkm
Pluto Orbital Distance
- NASA Planetary Fact sheet had defined Pluto orbital distance as 5870 mkm
- And
- From 5 years NASA had modified this distance to be 5906 mkm
- In the theoretical calculations both distances are required!
- Let's explain that in following
o 5906 mkm – 57.9 mkm (Mercury orbital distance) = 5848 mkm
o This distance 5848 mkm accurately = 1.16 mkm/s x 5040 seconds
But
o Mercury motion during 1419.5 days passes a distance =5813 mkm = 5870
mkm – 57.9 mkm - And
o Mercury motion during 1433.5 days passes a distance =5870 mkm (Pluto
orbital distance)
Shortly
o Light motion depends on the distance (5906 mkm as Pluto orbital distance)
but Mercury motion depends on the distance (5870 mkm as Pluto orbital
distance).
o That tells both distances are found geometrically but also tells some
difference be found between light and planet motions.

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Mercury Motion Data Analysis
- The previous data shows that,
- Mercury uses light motion strategy. By that, Mercury moves a distance and then
uses this distance as a period of time to move another distance. By that the
distances be created in a network form.
- We don't know how a planet moves with low velocity can use distances values as
periods of time. The data needs more analysis to discover the rule on which this
using is done.
- Mercury motion aims to pass a distance = 5848 mkm= Mercury Pluto Distance,
this distance is the distinguish one in the 1st
trajectory of motion and also the 2nd
trajectory of motion. Many data be created in proportionality with this distance
which proves some geometrical machine be found behind these motions.
- Mercury moves the distance (5848 mkm) following light motion
- Mercury uses the rate (1 mkm= 1 solar day)
- Notice
- Jupiter Orbital Circumference (4900 mkm) is a very rich distance of discussions
and arguments. Because of that,
- Jupiter orbital circumference analysis be discussed in a separated point no. (4-3) .
- The idea is that, the light supposed velocity (1.16 mkm/s) creates a connection
between Mercury and Pluto by using Jupiter orbital circumference (4900). That
means, Jupiter orbital circumference be used as a central point in the light
trajectory of motion from Mercury to Pluto, because of that, the light supposed
velocity (1.16 mkm/s) created Jupiter orbital circumference (4900 mkm) by
different geometrical features which be required for this task. As a result we find
Jupiter orbital circumference is so complex distance and filled of questions.

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The Conclusions
- The data tells,
- Mercury is the origin point of the light supposed velocity (1.16mkm/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.
- We should discuss Saturn motion effect on Mercury motion. but that will be done
in another reading for the same data. means this data will be discussed again from
Saturn motion data point of view.

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Mercury Motion Data Analysis
- 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 belong 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)
- 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.

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(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, the 2 values be
equivalent)

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- (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 this data in separated point because Saturn creates another
reading for Mercury Motion Data.
- That means, the data be used by light motion for reasons but the same data be used
again by planet motion for different reasons.
- For that reason we discuss this data once again by using Saturn motion data as a
approach.
- This discussion be in point no (4-5) (Saturn Motion data analysis)

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Equation no. (4)
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. (4) is the paper basic equation. Because it explains The Basic Energy
Trajectory Through The Solar System…
- First, 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
- Light known velocity (0.3 mkm/s) travels during (86400 s) a distance = 25920
mkm

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- 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. (4) 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.

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(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

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- 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 the story which tells that Mercury and Pluto be 2 planets
created out of one light beam (as electron and position) and because of that the
distances between them be created based on a geometrical design.
Notice
- (2π) This rate distinguish between light motion and planet motion as we have
discussed in Equation no. (3)
- 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.

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(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
- 680 mkm = Venus 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. (3) (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

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- 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
- 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
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.
- Please review Mars Migration Theory (See The References).

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Notice
- 202318 mkm = (449197 km)2
- 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).

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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.

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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. (5), let's discuss it in
following…

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Equation no. (5)
5848 mkm (Mercury Pluto Distance) – 4900 mkm = 948 mkm
- Where
- 5848 mkm = Mercury Pluto Distance
- 948 mkm = Earth orbital circumference (940 mkm) error (1%)
- Equation no. (5) 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
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)

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- 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 (0.3m,km/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.
- We need to analyze this data as possible with Saturn motion data…
- Let's start the 3rd
motion trajectory discussion and keep an eye on Saturn motion
data.

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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 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.

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- But the point we need to consider is, (Saturn Motion Data)
- Because in this trajectory of motion, Saturn motion data be used frequently and we
need to know why…
- This trajectory of motion uses 5 equations no. (6,7,8, 9 and 10)
- Let's start the discussion immediately …
- Notice
- Equation no. (5) shows that Saturn data effects on Mercury motion data
- The fact is that, Saturn motion data be accompanying for Mercury motion data
along the way. we found Saturn motion data frequently in Mercury motion data
analysis.
- This information we should discuss in point no. (4-5) of this paper because it
creates another reading for the same data.
- Saturn motion data creates frequently new meaning for the motion data of Mercury
and of Pluto. by that Saturn be a player of shadow found continuously but we don't
know why…
- Because of that this equation with many other data will be discussed in the point
no. (4-5 Saturn motion data analysis).

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Equation no. (6)
1.16 mkm/s x 4900 seconds = 5678.1 mkm (Mars Pluto Distance)
And
- Where
- 4900 mkm =Jupiter Orbital Circumference
- 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. (6) 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 Mercuy and
Pluto through the period 5040 second and another connection between Pluto and
Mars through the distance (4900 mkm)

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- 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)

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Equation no. (7)
- Where
- 687 days = Mars orbital period (light use it as 687 seconds)
- The distance 594 mkm be = 0.3 mkm/sec x 1980 seconds
- And
- 1980 mkm = The Orbital Circumference Total Of (Mercury +Venus + Earth)
- 928 mkm = Earth Jupiter Distance when the 2 planets be on the sun 2 sides.
- Light supposed velocity (1.16 mkm/sec) uses its same strategy to pass a distance
and use the passed distance as a period of time to pass another distance.

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Equation no. (8)
1.16 mkm/sec x 542 seconds =629 mkm (Earth Jupiter Distance)
But
580 mkm = 1.0725 x 542 mkm
And
580 mkm = 1.16 mkm/sec x 500 sec.
- Let's move with this data step by step
o Light known velocity (0.3 mkm/s) needs 500 seconds to travel from the sun
to Earth (149.6 mkm)
o Light supposed velocity (1.16 mkm/sec) uses this same period 500 seconds,
and during this period passes a distance =580 mkm
o And
o 580 mkm = 1.0725 x 542 mkm
o Why the rate (1.0725) is used here? what is a geometrical effect of this rate
(1.0725) on planets motions? no available answer. Shortly 40% of all
distances in the solar system be rated one another with this rate (1.0725).
that tells, this rate using is a fact regardless any explanation.
o The distance 542 mkm be used as a period of time (542 s) (similar to all
distances between Jupiter and the inner planets in addition to Jupiter Pluto
distance)
o So
o 1.16 mkm/sec x 542 seconds =629 mkm (Earth Jupiter Distance)
o How can we understand this data?
o The period 500 s is used for light known velocity (0.3 mkm/s) motion from
the sun to Earth, this same period (500 s) be used by light supposed velocity
(1.16 mkm/s) to define the distance between Earth and Jupiter (629 mkm).

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o That may explain why Earth orbital circumference 940 mkm = Earth Jupiter
distance 928 mkm (error 1.3 %)
o (notice 928 mkm =778.6 mkm +149.6 mkm, means, Earth and Jupiter
should be on 2 different sides from the sun to create the distance 928 mkm)
o This equality of distances is found because of using the same period of time
(500 s) in both directions (the sun and Jupiter)
Notice
o 928 mkm /542 mkm = 1.7
o The angle (1.7 deg) we have discussed in the moon orbital motion equation.
o This data shows that, there's a deep interaction behind this rate (928 mkm/
542 mkm). Because it effects on the earth moon orbital motion and causes
the moon to use the angle 1.7 deg in the equal (θ1 = θ0 +1.7 deg)
o (The Moon Orbital Motion Equation Is Discussed In My Previous Paper-
Please Review The References)

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Equation no. (9)
1.16 mkm/sec x 3600 seconds =2 x 2088 mkm (Jupiter Uranus Distance)
- Equation no. (12) tells that during 1 hour (3600 sec) light supposed velocity travels
a distance =2 x 2088 mkm
- Where
- 2094 mkm = Jupiter Uranus Distance (the difference with 2088 mkm is 0.3%)
- We notice clearly that, the light supposed velocity uses the distances between
Jupiter and the other planets which is a noticeable behavior, Why?
- But
- What's the value 3600s (=1 hour)? Why this value belongs to Jupiter motion?!
- in our discussion for the planets motion using for different rates of time, we will
discover that the rate of time between Mercury and Jupiter is (1 h = 24 h) means
- (1 hour of Mercury motion) = (24 hours of Jupiter motion)
- And
- This same rate be used with Uranus motion, means
- (1 hour of Mercury motion) = (24 hours of Uranus motion)
- For that reason, the period (3600s) which is used in the Equation no. (9) can refer
for a motion during 1 solar day (=24 hours). That makes the motion depends on a
defined period of time.
- That shows the distance (2094 mkm = Jupiter Uranus Distance) has a deep effect
on the solar system geometry.

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- Notice
- Light known velocity (0.3 mkm/s) travels during a period (2094 sec) a distance
=629 mkm (Earth Jupiter Distance).
- That means, the light known velocity (0.3 mkm/s) uses Jupiter Uranus Distance
(2094 mkm) as a period of time (2094 sec). And during this period the light travels
a distance = 629 mkm (Earth Jupiter Distance)
- The data shows that, Jupiter is found in light motion trajectory from Earth to
Uranus.
- That makes the 2 distances (Earth Jupiter distance and Jupiter Uranus distance)
consist together one trajectory of light motion. means geometrically they be found
together.
- Notice
- Light known velocity (0.3 mkm/s) travels during a period (720 sec) a distance
=216 mkm (Venus Orbital Diameter).
- Where 720.7 mkm= the distance from Mercury to Jupiter
- I don't know how this data be created. but I try to prove the theory of the railway.
The idea tells that, the light motions (by both velocities 0.3 mkm/s and 1.16
mkm/s) use the planets motions data in their creation of the motions network
which be found under the planets motion.

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Equation no. (10)
(1.16 mkm/sec)2
x 2x 243 seconds = 654.9 mkm (Jupiter Saturn Distance)
- Where
- 243 solar days =Venus rotation period.
- During (2 x 243 sec), the light supposed velocity (1.16 mkm /sec) travel a distance
= 563.8 mkm
- Again the distance 563.8 mkm be used as a period of time (563.8 sec),
- During (563.8 seconds) light supposed velocity (1.16 mkm/sec) travels a distance
=654.9 mkm = Jupiter Saturn Distance
- Notice
- During 654.9 seconds, light supposed velocity (1.16 mkm/s) travels a distance
=760 mkm
- But
- light known velocity (0.3 mkm/s). During 760 seconds travels a distance =227.9
mkm (= Mars orbital distance)
- I try to prove that, the planets motions depend on a network of light motions. by
that, the classical description of planet independent motion be refuted. Because
planet motion be found depends on light motions and accompanying with it.
- We note, the period (760 seconds) is used by light known velocity (0.3 mkm/s) to
define Mars New Orbital Distance (227.9 mkm) after its Migration.
- But
- We see clearly, the light supposed velocity (1.16 mkm/s) is the original light beam
by which all planets data be defined.
- The planets motion analysis will be discussed in this paper to prove that the planets
motions be integrated into One Unified General Motion, which disproves Planet
motion independence concept.

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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) 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)
- We have to ask why Saturn is the basic player in Mercury and Pluto motions data?
let's try to find an answer in Saturn Motion Data analysis.

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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 paper hypothesis gives a chance for better explanation….
- The hypothesis supposes that Mercury and Pluto be 2 particles similar to an
electron and a positron be created from one gamma ray, but the 2 planets be
created out of one light beam its velocity =1.16 mkm/s.
- 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 the inner planets and with Pluto, 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)?

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- 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

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- Where
- 406000 km = Pluto Motion Distance during a solar day
- 88000 km = The moon displacement daily
- 708.7 hours = The Moon Day Period
- 153.3 hours = Pluto Day Period
- And
- 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
- 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 (58.66days) a distance = 243 mkm(1%)
- Mars moves during Venus day period ( 116.75 days) a distance = 243 mkm

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- 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 completely
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.

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o That means, this method works as a waterwheel or as a lever. A tool works
by a geometrical 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 known velocity (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 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.

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4-3 Jupiter Orbital Circumference Analysis
I-Data
Group No. (1)
(A)
1.16 mkm/s x 4224 seconds = 4900 mkm (Light Motion Reading)
(B)
(449197 km / 27.78) = (4900 mkm /0.3 mkm)
(C)
1.16 mkm/s x 4224 seconds = 4900 mkm (Planet Motion Reading)
(D)
4900 mkm = 360 mkm +680 mkm +940 mkm +1433 mkm +1433 mkm
(E)
1.1318 mkm/s x 4222.6 days = 4780 mkm
(F)
4900 mkm = 449197 km x 10921 km
But
4846 mkm =1.392 mkm x 3475 km
(G)
9.7 km/s x 142984 seconds = 1386945 km (notice 1.392 mkm = the sun diameter)
(H)
4900 mkm = 0.384 mkm x 12756 km =0.406 mkm x 12104 km
(I) (please remember)
(37100 mkm – 4900 mkm) x 2π =28255mkm + 2 x 86400 mkm =2 x 1.16 mkm/s x 86400 s
Group No. (2)
(J)
(2 x The Sun Mass) = Jupiter Mass x 2094

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II-Discussion
- Jupiter orbital circumference is a very rich distance for discussion and analysis,
also it provides surprised using of data.
- There are 3 basic features of this distance makes it very specific distance, let's refer
to them in following:
- (1) The Distance 4900 mkm Is Defined By Many Planets Motions
- (2) The Distance 4900 mkm be in proportionality with The Sun Diameter.
- (3) The Distance 4900 Is Defined also by Light Motion Also.
- Let's analyze the distance 4900 mkm in following…
Equation No. (A)
1.16 mkm/s x 4224 seconds = 4900 mkm (Light Motion Reading)
- We have studied this equation deeply in the previous discussion.
- This equation is provided here to remember its meaning only because we need it in
the following discussion
- Equation no. (A) tells,
- Light supposed velocity (1.16 mkm/s) travels during the period 4224 seconds a
distance = 4900 mkm = Jupiter orbital circumference
- We consider that, 1 second of light motion be = 1 hour of Mercury motion and by
that the period (4224 s) be = (4224 hours)
- This analysis lead us to conclude that light supposed velocity (1.16 mkm/s) motion
effects on Mercury motion and causes Mercury day period to be 4222.6 hours and
decreased it from (4224 h) by (5040 seconds) because the light supposed velocity
(1.16 mkm/s) travels during 5040 seconds a distance = 5848 mkm (= Mercury
Pluto Distance). And by that the light supposed velocity creates a connection
between Mercury and Pluto motions.
- This is a summary for this equation which we have discussed deeply before
(Equation No. 3) of the light supposed velocity proves discussion.

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Equation No. (B)
(449197 km / 27.78) = (4900 mkm /0.3 mkm) (error 1%)
- Where
- 449197 km = Jupiter Circumference
- 0.3 mkm/sec = light known velocity (300000 km /s)
- 27.78 km/s = The Moon Velocity.
- Equation no. (B) tells, the period the Earth moon needs to move a distance =
449197 km = Jupiter circumference = 16330 sec
- Light known velocity (0.3 mkm/) needs this same period of time (16330 sec) to
pass a distance =4900 mkm (= Jupiter orbital circumference)
- i.e.
- during the moon passes Jupiter circumference, the light known velocity passes
Jupiter orbital circumference! Why?
- Equation no. (f) causes more puzzles for this discussion, where (4900 mkm =
449197 km x 10921 km), it tells Jupiter orbital circumference be defined as a
multiplication between the moon circumference and Jupiter circumference!
- Geometrically we considered Jupiter orbital circumference (4900 mkm) be built
based on the 2 measurements (Jupiter circumference and the moon circumference!)
- This idea be supported by the fact
- Jupiter (13.1 km/s) moves during 10921 sec a distance = 142984 km =Jupiter
diameter.
- This data gives a feeling, the geometrical structure of Jupiter orbital circumference
depended on Jupiter circumference and the moon circumference…. But
- What's the relation of this geometrical structure with light known velocity? How
light known velocity (0.3 mkm/s) be in proportionality with this same data?!
because light known velocity (0.3 mkm/s) should be an independent velocity and
has no proportionality with the planets motions velocities!

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- Can we understand this data?!
- Jupiter orbital circumference be created by a motion of light supposed velocity
(1.16 mkm/s) during the period (4224 seconds). But the distance (4900 mkm) be in
proportionality with light known velocity (0.3 mkm/s) in comparison with planets
motions velocities…
- Shortly why (449197 km / 27.78) = (4900 mkm /0.3 mkm)?
- Or
- (light velocity 0.3 mkm/ the moon velocity 27.78 km) = (4900 mkm /449197 km)?
- The rate (0.3 mkm /27.78) = 10800
- And
- The rate (4900 mkm/449197 km) =10921 (the moon circumference =10921 km)
- The difference between 10800 and 10921 is 1%
Notice
- Why the moon velocity be =27.78 km/s? because the moon moves daily a
distance = Earth motion distance daily because they aren't separated from one
another. means the moon daily motion distance =2.574 mkm but this distance be
effected by the rate 1.0725. (we have referred to that frequently before). We don't
know why the distance 2.574 mkm be effected by the rate (1.0725) but 40% of all
distances in the solar system be effected by this same rate 1.0725 and appendix
no.1provides a list for these distances. By that the moon distance 2.574 mkm be
=2.4 mkm daily which =27.78 km/sec.
Notice
- Earth (29.8 km/s) moves during (16330 sec) a distance = 4 x Saturn diameter (1%)
- Saturn (9.7 km/s) moves during (16330 sec) a distance = Uranus Circumference (1.3%)
- Neptune (5.4 km/s) moves during (16330 sec) a distance = 88000 km (the moon
daily displacement).

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Equation No. (C)
1.16 mkm/s x 4224 seconds = 4900 mkm (Planet Motion Reading)
- This equation we know perfectly
- Light supposed velocity (1.16 mkm/s) travels during 4224 seconds a distance =
4900 mkm = Jupiter Orbital Circumference
- But
- This equation has another reading, let's try with it in following
(1)
o Jupiter (13.1 km/s) moves during 24.6 h a distance = 1.16 mkm.
o Where (24.6 hours = Mars rotation period)
(2)
o Jupiter orbital circumference (4900 mkm) be divided on the distance 1.16
mkm and the rate 4224 (or 4222.6).
o Means, if each 1 hour of Mercury Day period be = 24.6 hours, based on that,
Jupiter during the period (4222.6) x (24.6 h) moves a distance =4900 mkm =
Jupiter orbital circumference.
o Can that refer to some interaction of motions among the 3 planets (Mercury,
Mars and Jupiter)? Of course yes, this fact is proved in the planets unified
motion discussion in this paper (point no.6 of this paper) where huge
number of data shows clearly the motions interactions among the 3 planets.
(3)
o Why Jupiter moves 1.16 mkm in a period 24.6 hours? The planets machine
is so complex. We understand that, light supposed velocity (1.16 mkm/s)
motion caused the motions interactions among the 3 planets by its energy.
because the planets are created and moving by this energy.
o But geometrically the distance 4900 mkm depend on (1.16 mkm and 4224)
where the distance 1.16 mkm depends on (24.6 hours). That causes the
interactions between the 3 planets motions.

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(4)
o I want to say that,
o We have 2 pictures for the same scene. As if we use 2 covers for the same
place. One cover is put on another. the 2 covers are equal in length. But they
are different in nature. Similar to that, we have equal distances (4900 mkm)
be passed by (light motion) and by (planet motion). the motions create
interactions in data. while we don't see the light motion but we see only the
planet motion. but we discover the interactions which are done as a result
for light motion effect on planet motion.
(5)
o The real data causes even more confusion, because light moves 1.16 mkm in
one second. but Jupiter moves 1.16 mkm in 24.6 hours, by that the simple
interactions which depend on (1 second) should be explained in some
complex form based on the period (24.6 h) and then we have to ask (why
Jupiter uses Mars rotation period 24.6 h?). Because we know that some
relationship is found between Mercury Day Period and Jupiter orbital
circumference. And now we have a relationship between Jupiter orbital
circumference (4900 mkm) and Mars rotation period (24.6 h) through the
distance 1.16 mkm. That simply creates a relationship (or a proportionality)
between Mercury day period and Mars rotation period! (why?)
o Notice/
o (Mercury Day period 4222.6h) / (Mars rotation period 24.6 h) =170 (error 1%)
o Mercury Mars Distance = 170 mkm
o But
o (1407.6 h/ 24.6 h) = 2 x ( (4900 mkm) /(170 mkm)), where
o 1407.6 h = Mercury Rotation Period
o 24.6 h = Mars Rotation Period
o 4900 mkm = Jupiter Orbital Circumference

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o I wish the argument logic is clear.
o Jupiter moves a distance 1.16 mkm during a period (24.6 h =Mars Rotation
Period) because (1st
, Jupiter motion and Jupiter orbital circumference 4900
mkm have a relationship with Mercury Day Period). And (2nd
, Mercury
motion has a relationship with Mars Motion as the previous data refers).
We deal with a machine of gears, the main motor is the light motion and the planets
are gears, because of that, their data be in proportionality with one another according
to the motions interactions.
Notice
Mars moves during 13.37 hours a distance = 1.16 mkm

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Equation No. (D)
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
- The data considers that, the Earth Moon Orbital Circumference around the sun
=1433 mkm = Mars Orbital Circumference.
- If we accept this idea, the data will give a clear meaning. Which tells that (Jupiter
orbital circumference = the inner planets orbital circumferences total)
- Then we have to ask.
Why does Jupiter Orbital Circumference=
= The Inner Planets Orbital Circumferences Total?
- More data can help our analysis,
(I)
- We accept that, Jupiter orbital circumference 4900 mkm be passed by a light
supposed velocity (1.16 mkm/s) in a period = 4224 seconds. And by this motion
the light caused to create Mercury day period (4222.6 h) where we accept that 1
second of light motion be = 1 hour of Mercury motion.
- Venus and Earth Data also be effected by (4224), Let's remember their data
o 4224 mkm = 2π x 671 mkm (Venus Jupiter Distance)
o And
o (5040 sec/4224 sec) = (928 mkm/778.6 mkm)
o Where

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o 778.6 mkm = Jupiter orbital distance
o 928 mkm = Earth Jupiter distance (notice 928 mkm= 778.6 mkm +149.6 mkm)
o 5040 sec = Mercury Day Period be less by 5040 seconds than 176 solar days
(II)
o 778.6 mkm = 1.0725 x 720.7 mkm (Mercury Jupiter Distance)
o 720.7 mkm = 1.0725 x 671 mkm (Venus Jupiter Distance)
o 671 mkm = 1.0725 x 629 mkm (Earth Jupiter Distance)
o 778.6 mkm = Jupiter Orbital Distance
(III)
o 1.16 mkm /sec x 500 sec = 580 mkm
o 1.16 mkm /sec x 580 sec = 671 mkm (Venus Jupiter Distance)
o 1.16 mkm /sec x 671 sec = 778.6 mkm (Jupiter Orbital Distance)
o 1.16 mkm /sec x 629 sec = 720.7 mkm (Mercury Jupiter Distance) (1%)
- The data shows, the distances between Jupiter and the inner planets are defined
based on Geometrical Calculations, because of that all distances are rated with
(1.0725) and (1.16).
- Means, these planets motions are not independent motions, on the contrary they
are (complementary motions) and be created based on One Geometrical Design.
- The previous data shows a complex geometrical machine behind. More analysis
provides even more complex interactions for example
- (778.6 mkm/ 629 mkm) = (720.7 mkm/580 mkm)
- These distances also we have used in the previous data. that means, the distances
between Jupiter and the inner planets be created by some geometrical system
makes these distances be in proportionality to one another mutually.
- Means,
- Not just, Jupiter orbital circumference be = the inner planets orbital circumferences
total, but also all used distances be created in proportionality with one another.

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Equation No. (E)
1.1318 mkm/s x 4222.6 days = 4780 mkm
- Where
- 1.1318 mkm /day = Jupiter motion distance during a solar day
- 4222.6 hours = Mercury Day Period
- Equation (E) supposes that 1 hour of Mercury Motion = 1 day of Jupiter motion.
by that, Jupiter uses Mercury Day Period as 4222.6 Solar Days
- During 4222.6 days Jupiter moves 4780 mkm. But during 4331 days Jupiter moves
4900 mkm.
- That means, the period 4222.6 days is incorrect period, because if we want to
measure the distance 4900 mkm we should use the period 4331 days
- But
- (4222.6 days /4331 days) = (654.9 mkm/ 671 mkm)
- Where
- 654.9 mkm = Jupiter Saturn Distance
- 671 mkm = Jupiter Venus Distance
- That tells the rate (4222.6 /4331) is defined geometrically and by that, the distance
4780 mkm which is less than 4900 mkm by 2.5% may refer to the distance 4846
mkm (with error 1.3%)
- Based on that
- Equation no. (E) tells, Jupiter moves during 4222.6 days a distance = (4780 mkm)
= 4846 mkm (the inner planets orbital circumferences total) (error 1.38%)
- Equation no. (E) can refer to that, 1 hour of Mercury Motion be = 24 hours of
Jupiter motion.
- This is the same rate of Uranus where 1 hour of Mercury Motion be =24 hours
of Uranus motion.
- Be we know that, 1 h of Mercury motion = 9.18 h of Pluto motion

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- Based on that
- 1 h of Pluto motion = 2.62 h of Uranus Motion = 2.62 h of Jupiter Motion
- We should discuss that in details in the planets unified motion discussion, we need
this data here because
- The rate 2.62 is so specific for both planets
- Uranus Circumference x 2.62 = Uranus motion distance during its day period
(421056 km)
- And
- Jupiter Circumference x 2.62 = 1.16 mkm (error 1.4%)
- I try to show that, the distance (1.16 mkm) is found again under some new data. it's
not some random number found by chance in Jupiter data. it's a basic distance
based on which Jupiter motion and orbital circumference be designed, because of
that we face this distance (1.16 mkm) in any place we move.
- The negative point is that, we can't discover the geometrical design based on which
this distance 1.16 mkm be defined. For that reason, we meet this distance (1.16
mkm) frequently but couldn't catch the machine which produces it.
- Notice
- Light known velocity (0.3 mkm/s) x 4780s =1433.5mkm (Saturn Orbital Distance)
- Light known velocity (0.3 mkm/s) x 4879s =1461 mkm (Saturn Orbital Distance)
- Where
- 4879 km = Mercury Diameter
- 1461 days = Earth Cycle (365 +365 +365 +366 days =1461 days)
- The distance 4780 mkm be used as (4780 seconds) where usually light uses the
rate (1 mkm = 1 second) but the planet uses the rate (1 mkm = 1 day).

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Equation No. (F)
4900 mkm = 449197 km x 10921 km
But
4846 mkm =1.392 mkm x 3475 km
- Where
- 449197 km =Jupiter Circumference
- 10921 km = The Moon Circumference
- And
- 1.392 mkm = The Sun Diameter
- 3475 km = The Moon Diameter
- 4846 mkm = The Inner Planets Orbital Circumferences Total
(4846 mkm = 360+680+940+1433+1433)
- Equation (F) tells, the multiplication of the sun diameter and the moon diameter =
accurately the value (4846 mkm)
- But
- The multiplication of the Jupiter Circumference and the moon Circumference =
accurately the value (4900 mkm)
- Can that refer to some relationship between the distance 4900 mkm and the sun
creation?
- Let's move step by step in following:
o Jupiter diameter x π2
= 1411196 km which is different from the sun diameter
1392000 km with the rate (1.38%)
o But
o Saturn (9.7 km/s) moves during 142984 seconds a distance = 1386945 km
(notice 1.392 mkm = the sun diameter and the difference 0.4%)
o Can this data tell (the sun is created in proportionality with Jupiter diameter)

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o We have 2 reasons for this conclusion. Because (1st
) the data refers to a
proportionality between the sun and Jupiter diameter and (2nd
) the data
refers to proportionality between the sun and Jupiter orbital circumference.
That shows 2 basic data be in proportionality with the sun diameter.
o The second player is the Earth moon
o The moon circumference (10921 km) be in proportionality with Jupiter
orbital circumference 4900 mkm based on Jupiter circumference (449197
km).
o And
o Jupiter (13.1 km/s) moves during 10921 sec a distance = 142984 km =
Jupiter diameter.
o I don't need to prove that, this data be created based on geometrical system.
Because the data proves that strongly and clearly. I need to discover this
geometrical system based on which this data be created.
o Shortly
o Jupiter is the basic player found on other side and effects on the sun creation
process. Where this event is happening, Jupiter adopted the Earth moon as
its own son and uses its data as part of Jupiter own data. as a result the earth
moon (small planet) has some chance to effect on the sun creation data!
o Again
o Why We See The Sun Disc =The Moon Disc? Because
o (The sun diameter / the moon diameter) = (Earth orbital distance/ Earth
moon Distance)
o Why The Diameters Rate = The Distances Rate?
o Because all players are created in proportionality with 4900 mkm (Jupiter
orbital circumference).
o The data can't be created by any pure coincidence. The fact can be seen if
we try to answer the question (how the matter is created?)

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Equation No. (H)
4900 mkm = 0.384 mkm x 12756 km =0.406 mkm x 12104 km
- Where
- 0.384 mkm = the moon orbital distance
- 0.406 mkm = the moon orbital Radius At Apogee
- 12756 km = Earth Diameter
- 12104 km = Venus Diameter
- Equation no. (H) shows, the distance 4900 mkm controls the data around the
moon motion.
- Earth and Venus Diameters must be ancient and aren't changed. That tells the
value (4900 mkm) use the 2 planets diameters as references to define the moon
motion range. By that the moon orbital distance be =0.384 mkm depends on Earth
diameter and the moon apogee radius be 0.406 mkm depends on Venus diameter.
- The data shows this meaning but no geometrical mechanism be discovered to
answer how this data be created in proportionality with one another.
- More data can support the same meaning for example
- This data tells that (the apogee orbital radius 406000 km be defined by Venus),
more data supports this same meaning beefcake (406000 km = 3475 km x 116.75),
where 3475 km = The Moon Diameter and 116.75 days = Venus Day Period.
- The data gives the meaning but no geometrical mechanism be discovered.
- Shortly
- The distance 4900 mkm controls the sun creation process and the moon
orbital motion definition.
- The distance 4900 mkm uses Earth and Venus diameters as references to
define the moon orbital distance and its apogee radius.

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- Notice
- 1433 mkm =550.7 x 2.5986 mkm
- 2.598 mkm (the moon displacements total during 259.53 days),
- 550.7 mkm= Mars Jupiter Distance,
- 1433 mkm = Saturn Orbital Distance
- This data tells why the data considers the moon orbital circumference round the
sun be = 1433 mkm = Mars orbital circumference and Not be =940 mkm = Earth
orbital circumference.
- The data shows that, the Earth moon is the son of the inner planets. The moon isn't
belonged to the Earth alone where Venus effects greatly on the moon apogee
radius definition on one side where Mars still effect on the moon motion and
causes its orbital circumference around the sun be considered =1433 mkm = Mars
orbital circumference.
- The geometrical machine based on which this data be created has answers for the
question how this data be created.
Comments
- The distance 4900 mkm defines the moon diameter and its orbital motion range
- And
- The distance 4900 mkm controls the sun creation process and defines the sun
diameter.
- Because of that, the sun and moon data be created in proportionality as we have
discussed before.
- But
- Is the aforementioned data sufficient for this conclusion? Suppose (4900 mkm =
the sun diameter 1.392 mkm x 3475 km the moon diameter) can this data be a
sufficient for this heavy conclusion? Let's discuss the next data to answer this
question.

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Group No. (2)
Equation No. (J)
(2 x The Sun Mass) = Jupiter Mass x 2094
Where
2094 mkm = Jupiter Uranus Distance
- How to make the rate 2094 be = the distance 2094 mkm?
- Geometrically, this process depends on Jupiter motion. because Jupiter diameter
142984 km = (1 million km/7) where Mercury orbital inclination = 7 degrees
- That means, the interaction of motions between Jupiter and Mercury can produce
the distance 1 mkm based on geometrical methods.
- By that, the rate 2094 x 1 mkm = 2094 mkm (Jupiter Uranus Distance)
- Why do we need this data?
- I try to prove that, the sun is created by an effect of Jupiter motion. We have seen a
proportionality of data between the sun diameter and the distance 4900 mkm
(Jupiter orbital circumference) and also between the sun diameter and Jupiter
diameter. and this data shows a new proportionality between the sun mass and
Jupiter mass.
- Equation no. (J) shows that, Jupiter Uranus Distance has some role in the sun
creation (or perhaps Jupiter Uranus Motions Interaction does this role)
- Why does The Equation Use 2 Values Of The Sun Mass?
- We remember, Jupiter energy which is sent to Pluto used (2 solar days) where,
light supposed velocity (1.16 mkm/s) travels during one solar day a distance = the
solar planets orbital circumferences total, but Jupiter energy uses 2 solar days
which provides energy = 2 x the solar planets orbital circumferences total
- This is happened because of Neptune orbital circumference as we have discussed
before. So if the used energy be double value, and the matter is created of energy,
that caused the sun mass be used in double value.

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- Based on different data, Jupiter motion shows its role in the sun creation.
- Jupiter Uranus Distance shows specific role in this process.
- The difficulty is that, We can't discover the geometrical mechanism by which this
process is done or this data be created. the data tells what's happened but doesn't
tell how that's happened. By that the argument provides clear meanings but not
clear mechanism of motions.
- One more important we need to remember is the following
- 90000 mkm = 2872.5 mkm (Uranus orbital distance) x (97/3.1)
- Where
- C2
= 90000 mkm (if the period of time be = 1 second)
- 97 deg = 97.8 deg (Uranus Axial Tilt) – 0.8 deg (Uranus orbital inclination)
- 3.1 deg = Jupiter Axial Tilt
- We need this data because the sun rays be created from the value (C2
)
- The data shows that, the interaction between Uranus and Jupiter effects on the sun
creation process.
- This data supports the meaning we get from the analysis of the distance 2094 mkm
(Jupiter Uranus Distance)
- Notice
- The machine depends on layers, So, the rate (97/3.1) uses the value 97
- But, other rates can use different data for example
- (97.8 deg /3.1 deg) x 2872.5 mkm = 90560 mkm
- (where 90560 days = Pluto Orbital Period)
- Also
- (98.6 deg /3.1 deg) x 778.6 mkm x 2 = 49528 mkm
- Where
- 98.6 deg =97.8 deg (Uranus axial tilt) + 0.8 deg (Uranus orbital inclination)
- 778.6 mkm = Jupiter orbital distance and 49528 km = Neptune diameter.

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The Conclusions
- Jupiter orbital circumference 4900 mkm effects greatly on the machine motion
- The distance 4900 mkm defines the moon diameter and its motion range and by
that Jupiter motion effect on the moon motion be defined clearly
- Also
- The distance 4900 mkm defines the sun diameter 1.392 mkm, and that be done
through interaction with Jupiter diameter.
- These effects are found by the energy provided for the sun rays creation process
- This energy is seen through the interaction between Jupiter and Uranus motions.

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4-4 Mercury Jupiter Distance Analysis (720.7 mkm)
(1st
Point)
- In this point we try to prove that, the solar system distances be created in a
network form.
- We follow the light beam motion through the basic 3 planets (Mercury, Jupiter
and Pluto) and we try to prove that the distances be created in a network form
among these 3 planets.
- In following, let's remember the idea before to analyze the data….
(A)
o Light supposed velocity (1.16 mkm) be sent from Mercury based on a
period (4224 seconds), and during this period of time the light beam passes
a distance = 4900 mkm (Jupiter Orbital Circumference)
o Light motion be distinguished from the planet motion based on the rate (2π)
o means, the light moves (4900 mkm) and reach Jupiter (778.6 mkm)
o also
o 1 second of light motion = 1 hour of mercury motion and by that the period
4224 seconds creates Mercury day period (4222.6 hours) and causes it to
less than (4224 h) with the period (1.4 hours = 5040 seconds)
o Mercury moves during its day period a distance =720.7 mkm
o Shortly,
o Light motion passes a distance = 4900 mkm and
o Planet motion should pass a distance = 778.6 mkm
o But
o In fact, Mercury moves a distance =720.7 mkm
o And
o 778.6 mkm = 1.0725 x 720.7 mkm
o The rate (1.0725) is used frequently in the solar system distances and we
will void its discussion here.

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o As a result,
o Light moves (4900 mkm) and Mercury moves (720.7 mkm)
o Planet motion be considered created based on the light motion
(B)
o The light supposed velocity (1.16 mkm/s) travels from Mercury to Jupiter
and from Jupiter to Pluto.
o Let's suppose, The Light Causes The 3 Planets To Move Equal Distances
(720.7 mkm)
o Based on that, we analyze the distances among the 3 planets to test if these
distances be created as a network form and based on one geometrical design
as the theory claims.

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(2nd
Point) The Data Analysis And Discussion
I – Data
Group No. (A)
(1)
5848 mkm – 4900 mkm = 946.5 mkm
(2)
0.838 mkm/ days x 929 days = 778.6 mkm
(3)
0.838 mkm/ days x 5848 days =4900 mkm
(4)
(778.6 mkm/929 mkm) = (5040/4224) = (5848 mkm/4900 mkm) = 0.838
(5)
5848 mkm= 2π x 929 mkm
Group No. (B)
(6)
100733 mkm = 929 mkm x 108.2 mkm = 671 mkm x 149.6 mkm =
(7)
(929 mkm /629 mkm) =(940 mkm/636 mkm)
Group No. (C)
(8)
720.7 mkm = 0.406 mkm/day x 1775 days
(9)
(90560 days x 2) =1775 days x 102
(10)
90560 x 4 x 47.4 = 17.2 mkm.

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II- Discussion
Equation no. (1)
5848 mkm – 4900 mkm = 946.5 mkm
- Where
- This equation can be seen in another form as following:
- 5848 mkm (= 2π x 929 mkm) – 4900 mkm (=2π x 778.6 mkm) = (940 mkm =
2π x 149.6 mkm)
- Where
- 778.6 mkm = Jupiter Orbital Distance
- 149.6 mkm = Earth Orbital Distance
- 929 mkm = 788.6 mkm + 149.6 mkm (Earth Jupiter distances at 2 sides)
- (i.e. 929 mkm = Earth Jupiter distance while the 2planets be on 2 different sides
from the sun)
- The question we need answer is that
- Why (5848 mkm =2π x 929 mkm)?
- We should notice that the rate (2π) distinguish between the light motion and planet
motion.
- But we still can't catch the answer of the question (why 5848 mkm =2π x 929
mkm?)
- Let's return to Equation no. (1)

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Equation no. (1) (continued)
5848 mkm – 4900 mkm = 946.5 mkm
- Let's summarize the equation idea in following
(I)
- The basic data is that (5848 mkm = 2π x929 mkm) where 5848 mkm = Mercury
Pluto distance and (929 mkm) = Earth Jupiter distance when both be on 2 different
sides from the sun. why these 2 distances be connected with the rate (2π)? we still
search for an answer…
(II)
- The next information is a logical result where
- 5848 mkm – 4900 mkm = 946.5 mkm (where Earth orbital circumference =940
mkm) (error 0.6%)
- We need this information for 3 different branch of data discussion
- (1st
Branch)
- The distance 946.5 mkm accurately is equal the moon displacements total during
the period 10747 days, that means (88000 km x 10747 days = 946.5 mkm)
- Where
- 10747 days = Saturn orbital period
- Means,
- The moon displacement total be 940 mkm for a period = 10682 days (error 0.6%)
- The moon displacement total be 945.56 mkm for a period = 10747 days
- i.e.
- This distance (946.5 mkm) is mentioned for Saturn orbital period.

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- (2nd
Branch)
- Light supposed velocity needs 816 seconds to pass the distance =946.5 mkm.
- Accurately (5040 seconds – 4224 seconds = 816 second)
- But
- During 816 days Saturn (0.838 mkm/day) moves a distance = 684 mkm
- Where
- 680 mkm = Venus Orbital Circumference (error 0.6%)
- That means, the period (816 sec) be used by light velocity to pass Earth orbital
circumference (940 mkm). This same period (816 days) be used by Saturn motion
to pass Venus orbital circumference
- We remember that (1 second of light motion = 1 solar day of planet motion)
- That means, the accompaniment of Saturn motion for the data of light motion
between Mercury and Pluto was not just shadows for the motion. but even be
effective to create Venus orbital circumference while Earth orbital circumference
be created by light motion, Venus orbital circumference be created by Saturn
motion! we need more light to understand how can this be occurred
- Notice
- Saturn (0.838 mkm/day) moves during 929 days a distance = 778.6 mkm (Jupiter
orbital distance)
- Saturn (0.838 mkm/day) moves during 5848 days a distance = 4900 mkm (Jupiter
orbital circumference)
- (3rd
Branch)
- Light supposed velocity (1.16 mkm/s) travels during 800 seconds a distance = 929
mkm (Earth Jupiter Distance when both be on 2 different sides from the sun)
- But
- Saturn (0.838 mkm/ day) moves during 800 days a distance = 671 mkm (Venus
Jupiter Distance.

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- This data perfectly is similar to the previous one. means, Earth and Venus orbital
circumferences and their distances to Jupiter all these distances be created in one
process by an interaction of motion between light supposed velocity (1.16 mkm/s)
motion and Saturn motion
- While the light motion for 1 second be = planet motion for 1 soar day.
- Equation no. (6) can help our investigation
Equation no. (6)
100733 mkm = 929 mkm x 108.2 mkm = 671 mkm x 149.6 mkm
- Where
- 108.2 mkm = Venus orbital distance
- 149.6 mkm = Earth orbital distance
- 6714 mkm = Venus Jupiter Distance
- 929 mkm = Earth Jupiter Distance (The 2 Planets Be On The Sun 2 Sides)
- How to understand this equation?
- Light supposed velocity (1.16 mkm/s) travels during a solar day (86400 s) a
- Equation no. (6) tells that, from one source of Energy the 4 distances be created
- Specifically
- From the distance 100733 mkm, the 4 distances which are
- 108.2 mkm = Venus orbital distance
- 149.6 mkm = Earth orbital distance
- These 4 distances be created from the same one distance (100733 mkm)
- This information is supported perfectly the network theory ideas.

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- The discussion is a powerful proof for the distances distribution based on a
geometrical design. Simply Earth distance be created in proportionality with
Venus distance because they be created out of the same one source.
- Notice (1)
- The equation (5848 mkm = 2π x 929 mkm)
- Where
- 929 mkm = Earth Jupiter Distance (at 2 different sides from the sun)
- This equation provides one more reason for the proportionality of the data between
Earth motion and Pluto motion.
Notice (2)
- 929 mkm =2.574 mkm /day x 361 days
- (but the light needs 361 seconds to pass Venus orbital distance 108.2 mkm).
- I try to prove that, Venus an Earth orbital circumferences be created in
proportionality with the 2 planets distances to Jupiter (while Earth and Jupiter be
on different 2 side from the sun). these 4 distances be created in proportionality
with one another because all of them be created out of the distance (100733 mkm).
That also proves the concept that, the distance is energy and the branches distances
total = their parent distance.
- But
- The data analysis doesn't answer our question (why 5848 mkm = 2π x 929 mkm)?

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Why (5848 mkm =2π x 929 mkm)?
Equation no. (7)
(929 mkm /629 mkm) =(940 mkm/636 mkm)
- Where
- 929 mkm = Earth Jupiter Distance (The 2 Planets Be On The Sun 2 Sides)
-
- Equation no. (7) tells that, the data be used in proportionality to produce the
distance 636 mkm. And why this distance (636 mkm) be created?
- Jupiter moves during 636 days a distance = 720.7 mkm
- This is the information behind this data
- Shortly
- The distance (636 mkm) be created based on the distance (929 mkm) while Earth
orbital circumference (940mkm) be rated with Earth Jupiter distance (629 mkm).
- Why?
- Because
- Jupiter (13.1 km/s) moves during (4222.6 h) x (3600 s)= 200 mkm
- 629 mkm = Earth Jupiter Distance = π x 200 mkm
- That means,
- The distance 929 mkm is found to enable the distance (636 mkm) to be defined
based on (629 mkm)
- While
- Jupiter motion during Mercury day period (4222.6 hours) passes a distance
=200mkm = (629 mkm/π)
-
- Let's remember the question

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Why (5848 mkm =2π x 929 mkm)?
- Because
- The light suppose velocity (1.16 mkm/s) motion causes Mercury to move a
distance = 720.7 mkm duri0ng Mercury day period (4222.6 h) and the light motion
forces Jupiter to move during (Mercury day period 4222.6 h) a distance = 720.7
mkm.
- By that, the machine forces the gears to pass the required distance by motion
transportation among gears which causes the distance to be created based on one
geometrical design. Let's explain how this is done in following
o Light supposed velocity (1.16 mkm/s) passes a distance = 4900 mkm
(Jupiter orbital circumference) by using the period (4224 seconds)
o The period (4224 seconds) caused to create Mercury day period (4222.6 h)
and caused Mercury to move during it a distance = 720.7 mkm.
o Where
o 1 second of light motion = 1 hour of Mercury motion
o Light supposed velocity (1.16 mkm/s) caused Mercury day period to be (=
4222.6 h) and less than (4224 hours) with (1.4 hours = 5040 seconds) to
create a connection between Mercury and Pluto motions because light
supposed velocity during 5040 seconds a distance (= 5848 mkm Mercury
Pluto distance)
o That means, the period 5040 seconds be created with Mercury day period
and by that the distance (5848 mkm Mercury Pluto distance) be defined with
Mercury day period definition.
o i.e.
o Mercury day period creation caused to defined the distance (5848 mkm)
o The distance 5848 mkm = 2π x 929 mkm (Earth Jupiter distance on the sun
2 sides)
o That creates one data controls the 3 planets motions data

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o The distance 5848 mkm controls Mercury, Jupiter and Pluto motions.
o Shortly
o Light supposed velocity (1.16 mkm/s) creates the difference (5040 seconds)
between Mercury day period and the period (4224 hours).
o This different in time (5040 sec) creates a connection between the motion be
caused during Mercury day period for Mercury and the same period motion
for Jupiter and Pluto. the connection be in the period (5040 seconds).
o Shortly
o The distance 5848 mkm = 2π x 929 mkm because
o It connects Mercury motion for a distance (720.7 mkm) with Jupiter motion
for a distance (=720.7 mkm) with Pluto motion for a distance (=720.7
mkm).
o Now let's try to see this connection as deep as possible
- For Jupiter
o Jupiter moves during (4222.6 h) a distance =200 mkm
o But
o Earth Jupiter Distance 629 mkm = 200 mkm x π
o Equation no. (7) tells ((929 mkm /629 mkm) =(940mkm/636 mkm))
o By that, earth Jupiter distance (629 mkm) and at the sun 2 sides (929 mkm)
will be used in proportionality with earth orbital circumference (940 mkm)
to create the distance 636 mkm which will be used as a period of time,
where Jupiter moves during (636 days) a distance = 720.7 mkm
o Let's remember the question
o Why (5848 mkm = 2π x 929 mkm)? (the answer) To create a connection
between Mercury motion for a distance 720.7 mkm and Jupiter motion for
distance =720.7 mkm.

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For Pluto
Equation no. (8)
720.7 mkm = 0.406 mkm/day x 1775 days
- Where
- 0.406 mkm = Pluto motion distance during a solar day
- For what we search here?
- For a connection between the distance 5848 mkm and Pluto motion for a distance
720.7 mkm!
- Let's move step by step
- Pluto (0.406 mkm/day) moves during a period 1775 days a distance 720.7 mkm
- Now we have the required distance (720.7 mkm) and the period is 1775 days
- Now
- What's the relationship between this period (1775 days) and the distance 5848
mkm?
- Let's see the next equation….
Equation no. (9)
(90560 days x 2) =1775 days x 102
- Where
- 90560 days = Pluto Orbital Period
- 1775 days = the period Pluto needs to pass the distance 720.7 mkm
- 102 = ??
- Mercury Pluto Distance 5848 mkm = 101 x 57.9 mkm (Mercury orbital distance)
- Pluto Orbital Distance 5906 mkm = 102 x 57.9 mkm (Mercury orbital distance)
- The rate 102 be between Pluto Orbital Distance / Mercury Orbital Distance
- The data also can be considered referring to the distance 5848 mkm with error
(1%)
- The tells the period 1775 days be defined in proportionality with Pluto orbital
period (90560 days) based on the distance 720.7 mkm

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Equation no. (10)
90560 x 4 x 47.4 = 17.2 mkm.
- Where
- 90560 days = Pluto orbital period (here this period be used in second units)
- Mercury (47.4 km/s) moves during (4 x 90560 seconds) distance =17.2 mkm
- Pluto orbital inclination (17.2 degrees) be created based on this distance 17.2
mkm.

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Earth Jupiter Distance Analysis
- The previous analysis tries to show that, the distance between Earth and Jupiter
across the sun (929 mkm = 778.6 mkm +149.6 mkm) is defined as a basic point for
Mercury Pluto Distance (5848 mkm) where
- 5848 mkm = 929 mkm x 2π
- Let's analyze this distance in following
- The figure shows the sun and the 3 planets (Mercury – Venus – Earth)
- The 3 planets be drawn on 2 sides of the sun. and Jupiter is the planet (J)
- Now let's start our analysis
- The distance between Earth and Jupiter (on one side) = 778.6 -149.6 = 629 mkm
- The distance between Earth and Jupiter (on 2 sides) = 778.6 +149.6 = 929 mkm
- The distance between Earth and Mercury (on 2 sides) =57.9 +149.6 = 207.5 mkm
- The distance between Earth and Venus (on 2 sides) =108.2 +149.6 = 257.8 mkm
-
- From equation no. (6)
- 100733 mkm = 929 mkm x 108.2 mkm = 671 mkm x 149.6 mkm
- We know that
- (940 mkm /680 mkm) = (929 mkm/671 mkm)
- Where
- 929 mkm = Earth Jupiter Distance (on the sun 2 sides)
- 671 mkm = Venus Jupiter Distance ( on one side)
- The data tells

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- The distance 929 mkm be used geometrically in wide range. Where such using
isn't available for the other distances…
- Why the distance across the sun is important geometrically? Let's see one example
to test these distances geometrical effect on the solar system motion.
Example no. (1)
Equation no. (A)
Sin (23.6 deg) x 257.8 mkm = 103.4 mkm
- S= The Sun
- V = Venus
- E= The Earth
- ES = Earth Orbital Distance = 149.6 mkm
- SV = Venus Orbital Distance = 108.2 mkm
- EV = 103.4 mkm
- Notice
- The distance between Venus and Earth = 41.4 mkm and NOT 103.4 mkm. So how
this triangle be built?!
- let's move step by step:
o Earth orbital distance 149.6 mkm + Venus orbital distance 108.2 mkm =
257.8 mkm (This distance be found when Earth and Venus be on 2 different
sides from the sun with an angle 180 degrees between the 3 players)
o 257.8 mkm x sin (23.6 deg) = 103.4 mkm (=EV distance In The Triangle)
o 23.45 deg = Earth axial tilt (has a difference with 23.6 deg be 1%)

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o In our triangle (SEV) the angles be defined geometrically as following
o The Angle (S) = 43.6 degrees
o The Angle (V) = 90 degrees
o The Angle (E) = 46.4 degrees (= 2 x 23.2 deg = Earth Axial Tilt error 1%)
o Now let's analyze this triangle in following
- Is this triangle (SEV) be an imaginary one? because the distance between Venus
and earth be 103.4 mkm and the real distance = 41.4 mkm?
- We see that, the distance between the Earth and Venus can be even = 257.8 mkm
when the 2 planets be on 2 different sides from the sun. So if the distance can be =
257.8 mkm, that means, it can be also 103.4 mkm. That means this triangle can't
be an imaginary one.
- Why this triangle is useful?
- Because of the angle (E) = 2 x 23.2 degrees (Earth Axial Tilt) (error 1%)
- That tells, Earth axial tilt is produced by some interaction be found among Earth,
Venus and The Sun Data
- Notice
- Why Venus angle =90 degrees? The triangle data is in proportionality with the
planets motions real data. but why the angle is 90 degrees?
- What Useful Results We Get From This Triangle? 2 Basic Results
- (1st
Result)
- The triangle tells that, some geometrical design be used to produce the data of
Earth and Venus orbital distances in addition to the sun diameter. also Earth axial
tilt be defined in this interaction. The triangle tells some geometrical design be
found under this data. because of that this triangle is found as a part of this
geometrical design.

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- (2nd
Result)
- The Sun Be Perpendicular On Venus Planet. Specially, the angle between the sun
position and Venus planet position is (90 degrees). By that the sun must be on
vertical axis (z-axis) while Venus moves on (x-y plain).
- This description of the sun position needs a deep analysis to know how this
position be defined…
Notice
41.6 mkm = 0.384 mkm x 108.2 mkm (Venus Orbital Distance)
- 41.4 mkm = Venus Earth Distance (different from 41.6 mkm with 0.4%)
- 0.384 mkm = The Earth Moon Orbital Distance
- The data tells, The Earth moon orbital distance also be defined accordingly
- The paper provides an interesting approach to explain the solar planets motions.
- While the solar planets motions description inherited from Newton Theory of the
sun mass gravity tells that each planet motion is independent from the other
planets motions. As a result each planet orbital distance be defined independently,
The suggested description tells that, each planet motion takes into consideration
the other planets motions.
- We here deal with a building found depending on one another. based on this
vision, because Venus orbital distance =108.2 mkm, that causes Earth orbital
distance to be 149.6 mkm
- And because of the 2 planets orbital distances be (108.2 mkm and 149.6 mkm) the
Earth moon orbital distance be (0.384 mkm).
- The vision is completely different from what Newton told us. while Newton
supposed the space is limitless and can contain hundreds of planets moving with
effect on each other. the fact is the 9 planets motions effect on one another and
forces each planet to define each motion in proportionality with the other planets.

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The vision is different because Newton vision of (NO Effect) created separated
trajectories of motions. And based on this vision we see the planets as cars move in
track independently from one another (even the real cars can't be independent
because the track is the same one and all have to move in harmony otherwise they
will collide). The suggested description tells that, the motions be built as a building
and each motion depend on the other motions and by that they need one
geometrical design controls and manages all of them.

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Discussion
- This triangle I put here because it shows the distance between the earth and Venus
can be =257.8 mkm by put the 2 planets on 2 different sides from the sun.
- This behavior perfectly is similar to Earth Jupiter distance (929 mkm) while the 2
planets be on different 2 sides from the sun.
- I try to show that, it's part of the distances distribution system to use the planets
positions across the sun.
Notice
- The distance between Earth and Mercury across the sun be = 149.6 mkm + 57.9
mkm =207.5 mkm
- Pluto (0.406 mkm/day) moves during 511 days a distance = 207.5 mkm
- Where
- 511 degrees = The Solar Planets Axial Tilts Total
- A Summary
- The distances between Mercury and Pluto be distributed based on geometrical
design to create a proportionality between the 3 planets motions for periods (720.7
mkm).
- The distance distribution be performed by light motion effect on te planets
distances creation.

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4-5 Saturn Motion data analysis.
I-Data
Group No. (I)
(1)
5848 mkm – 4900 mkm = 946.5 mkm
(2)
0.838 mkm/ days x 929 days = 778.6 mkm
(3)
0.838 mkm/ days x 5848 days =4900 mkm
(4)
(778.6 mkm/929 mkm) = (5040/4224) = (5848 mkm/4900 mkm) = 0.838
Group No. (II)
(5)
816 sec x 1.16 mkm/s = 940 mkm
(6)
800 sec x 1.16 mkm/s = 929 mkm
(7)
816 days x 0.838 mkm/day = 680 mkm
And
(8)
929 mkm x 108.2 mkm = 671 mkm x 149.6 mkm = 100733 mkm
Group No. (III)
(a)
100733 mkm = 0.838 mkm/ days x 120536 days
(b)
(30589 days /27.32 days) = (940 mkm /0.838 mkm)

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II- Discussion
- What Do We Try To Do In This Discussion?
- One observation I have observed frequently in the previous discussion, let's try to
summarize it in following
- The light motion (as we have discussed in different data) always be accompanying
with Saturn motion!
- How to understand that?
- The light motions data can be produced (always) by Saturn motion data!
- We have some accompanying motion done by Saturn for the light motion!
- The data can explain this meaning in more better form
- Example No. (1)
- But
- Saturn (0.838 mkm / day) moves during 120536 days a distance = 100733 mkm
- Where
- 120536 km = Saturn Diameter
- In the data the value 120536 km be used as a period of time (120536 solar days)
- We have seen that, Planet diameter or circumference be used as periods of time.
Based on that Saturn diameter isn't exceptional but follows the same.
- As usual the planet diameter or circumference be used in (seconds unit) and satrun
isn't exceptional because
- Light known velocity (0.3 mkm/s) travels during (120536 sec) a distance = 2π x
5757 mkm (Earth Pluto distance )
- And because the light uses the period (120536 in seconds units), the planet motion
uses this period as (120536 days) that because (1 second of light motion be
equivalent to 1 solar day of planet motion)

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- I want to say
- Saturn motion data be found simply in the parallel line to all light motions data
through the distance from Mercury to Pluto. we can't catch why Saturn is the
player on the other side. What I can do, is to prove this claim and show that Saturn
data be found frequently in parallel with light motion data through the distance
between Mercury and Pluto…
- Let's try to do that in following..
- Notice, we import the data from Jupiter discussion with their numbers. And the
new data be used in another sequence.
- Equation no. (1)
- 5848 mkm – 4900 mkm = 946.5 mkm
- Where
- 946.5 mkm = Earth orbital circumference (= 940 mkm error 0.7%)
- Equation no.(1) we have discussed deeply before
- What we need her is to review the value 946.5 mkm, that because
- The Moon Displacement Daily = 88000 km
- Earth Orbital Circumference = 940 mkm
- If the moon passes the distance (940 mkm) by using its daily displacement 88000
km that needs a period = (10747 days) (error 0.6%)
- Where
- 10747 days = Saturn Orbital Period
- And we know that,
- Saturn orbital period causes a deep interaction between the earth moon and Saturn
motions. by that the data of equation no. (1) refers clearly to Saturn orbital period
where we don't understand why Saturn be used in this data which defend the
distance between Mercury and Pluto.

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- Equation no. (2)
- 0.838 mkm/ days x 929 days = 778.6 mkm
- Saturn (0.838 mkm/day) moves during a period (929 days) a distance = 778.6
mkm (Jupiter orbital distance)
- And
- Equation no. (3)
- 0.838 mkm/ days x 5848 days = 4900 mkm
- Saturn (0.838 mkm/day) moves during a period (5848 days) a distance =4900
mkm (Jupiter Orbital Circumference).
- The data simply proves the claim. In each equation Saturn motion data be used in
parallel. We simply don't know why?
Equation no. (4)
(778.6 mkm/929 mkm) = (5040s/4224s) = (5848 mkm/4900 mkm) = 0.838
- Where
- 778.6 mkm = Jupiter orbital distance
- 929 mkm = Earth Jupiter Distance when be both on 2 sides from the sun.
- 4900 mkm =Jupiter orbital circumference
- 5040 seconds = a required period for Mercury day period to be =176 solar days
- 4224 seconds = the period through which the light beam supposed velocity (1.16
mkm/s) travels during it’s a distance 4900 mkm
- Equation no. (4) tells something very interesting…
- Saturn motion distance during a solar day (0.838 mkm / day) is equivalent to the
rate (0.838)
- The rate (0.838) be used as a rate between the 2 basic periods (5040 s and 4224 s)
- This rate (0.838) = Saturn Motion Distance During A Solar Day

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- Here we can see a feature of the motion interaction which causes Saturn data to be
in parallel to the light motion data, which is
- Saturn motion distance depends on the solar day (0.838 mkm/day) and
- The light motion distance depends on the solar day
- The unification of the period of time may create a unification for motions data.
- But, In fact
- It's a basic question to find the rate between the 2 basic periods of light supposed
velocity (1.16 mkm/s) motion which are (5040 s and 4224 s) , to find this rate
(0.838) be equivalent to Saturn velocity during a solar day (0.838 mkm/day)
Equation no. (5)
816 sec x 1.16 mkm/s = 940 mkm
Equation no. (6)
800 sec x 1.16 mkm/s = 929 mkm
- Equations no. (5 and 6) tell that,
mkm = Earth Orbital Circumference
- And
mkm = Earth Jupiter Distance while the 2 planets be on the sun 2 sides.
- Now Earth distances be defined clearly based on the motion of light supposed
velocity (1.16 mkm/s) - BUT
Equation no. (7)
And
- Equation no. (7) tells that
- Saturn (0.838 mkm/day) moves during 816 days a distance = 680 mkm = Venus
orbital circumference

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- And
- Saturn (0.838 mkm/day) moves during 800 days a distance = 671 mkm = Venus
Jupiter Distance (A Direct Distance)
- We know that,
- Light motion for 1 second is equivalent to planet motion for 1 solar day
- And by that Saturn uses the same period in days units, in place of seconds units by
the light motion.
- Venus distances be defined based on Saturn motion while Earth distances be
defined based on light motion by using the same periods of time.
- The data proves the point of view that, Saturn motion data be found in parallel
with light motion data.
- The frequent using of Saturn motion data trough light motion data creates a
question needs explanation…
Equation no. (8)
929 mkm x 108.2 mkm = 671 mkm x 149.6 mkm = 100733 mkm
- Equation no. (8) we have discussed before.
- Light supposed velocity (1.16 mkm/s) travels during a solar day (86400s) a
distance =100733 mkm.
- And
- 929 mkm = Earth Jupiter Distance (the 2 planets on the sun 2 sides)
- Equation no. (8) shows that, Earth and Venus distances be created in comparison
with one another, because the 4 distance be controlled by the original distance
100733 mkm.
- But

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- The equations no. (5,6 and 7) shows that, Earth distances be created based on light
supposed velocity motion but Venus distances be created based on Saturn motion
by using the same periods of time based on different rate of time (1 second of light
motion = 1 solar day of planet motion).
- That means,
- Saturn motion isn't some additional motion in the data
- On the contrary
- Saturn motion be effective as much as the light motion itself and the 2 results of
the light and Saturn motions be used in complementary wit one another creating
one general result as a final result (100733 mkm).
- We need to understand why Saturn motion be effective on the data?
- Why specifically Saturn motion?
Group No. (III)
(a)
100733 mkm = 0.838 mkm/ days x 120536 days
(b)
(30589 days /27.32 days) = (940 mkm /0.838 mkm)

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- Saturn Motion More Analysis
- Light velocity 300000 km/s
- While Jupiter distances to the inner planets and Pluto shows clearly that they are
created based on light beam its velocity = 1.16 mkm/s,
- Saturn data shows frequently a connection with light know velocity (0.3mkm/s)
- Let's refer to this data in following
- The moon (2.4 mkm) moves during 10747 days a distance = 25920 mkm
- Saturn (0.838 mkm) moves during 30589 days a distance = 25920 mkm
- Where
- Light known velocity (0.3 mkm/) travels 25920 mkm during a solar day (86400 s)
- Also
- (Saturn diameter 120536 km/ 9.7) = (300000 km /24.1)
- Where
- 300000 km = light motion distance during 1 second
- 9.7 km /s = Saturn velocity
- 24.1 km /s = Mars velocity
- Also
- (2 x 120536 km Saturn diameter) /24.1 = (300000 km /29.8)
- 24.1 km /s = Mars velocity
- 29.8 km/s = Earth velocity
- In different data Saturn shows, it has a connection with light known velocity (0.3
mkm/s).

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- Notice (1)
- There are 3 reasons to connect Saturn orbital period with the earth moon motion
which are
(1) The Moon Displacement (88000 km) can pass earth orbital circumference (940
mkm) in a period (=10747 days =Saturn orbital period)
(2) The moon moves daily a distance (2.4 mkm) and during 10747 days the total
distance = 25920 mkm = light known velocity (0.3 mkm/s) motion distance
during solar day.
(3) The period 10747 days = 365.25 days x 29.53 days
Notice (2)
- Saturn (0.838 mkm/day) moves during the period (30589 days= Uranus orbital
period) a distance =25920 mkm (error 1%)
- This data creates a connection between (Uranus orbital period 30589 days) and
Saturn velocity (0.838 mkm/day) and with the earth moon motion data
- But
Equation no. (b)
(30589 days /27.32 days) = (940 mkm /0.838 mkm)
- Where
- 30589 days = Uranus Orbital Period
- 27.3 days = The moon Orbital Period
- 940 mkm = Earth orbital Circumference
- 0.838 mkm/day = Saturn Velocity Per A Solar Day
- Equation no. (b) tells that, more connection be found between Uranus orbital
period and Saturn motion velocity with the moon motion data.

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4-6 The Solar System Distances As A Network (More Data)
I-Data
(1)
100733 mkm = 5848 mkm x 17.2
(2)
(100733 mkm x 2) = 4900 mkm x 41
(3)
(100733 mkm = 18048 mkm x 5.6
(4)
100733 mkm = 4345.5 mkm x 23.18
(5)
100733 mkm = 360 mkm x 278.4
(6)
100733 mkm = 940 mkm x 107.2
II-Discussion
- Where
- 100733 mkm = The Solar Planets Orbital Circumferences Total
- The data tries to show that, the distance 100733 mkm controls many planets
distances and causes to create the planets orbital inclination or axial tilts a result
for this control.
- Let's summarize the equations in following
Equation no. (1)
100733 mkm = 5848 mkm x 17.2
- Where
- 17.2 degrees = Pluto Orbital Inclination

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- Based on out discussion we remember that the distance 5848 mkm is the most
important distance in Pluto data. and based on it Pluto orbital inclination (17.2
deg) be created.
Equation no. (2)
(100733 mkm x 2) = 4900 mkm x 41
- Where
- 2 x 100733 mkm = The Solar System Total Energy
- 41 degrees = the solar planets orbital inclinations total
- We should remember the equation
- 2 x 86400 s x 1.16 mkm/s = 2 x 100733 mkm = (37100 mkm – 4900 mkm ) x2π =
28255 mkm + 2 x 86400 mkm.
- This equation explains the solar system total energy (2 x 100733 mkm0 which is
used in out equation no. (2)
- Equation no. (2) tells that the solar planets orbital inclinations total (41 deg) be
defined based the solar system total energy.
Equation no. (3)
100733 mkm = 18048 mkm x 5.6
- Where
- 18048 mkm = Uranus Orbital Circumference
- 5.6 deg =5.1deg (the moon orbital inclination)+ 0.5 deg the moon diameter angle)
- Equation no. (3) refers to a deep relationship between Uranus and the earth moon.

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Equation no. (4)
100733 mkm = 4345.5 mkm x 23.18
- Where
- 23.4 degrees = Earth Axial Tilt (error 1%)
- 4345.5 mkm = Earth Neptune Distance
- Equation no. (4) tells, the distance between Neptune and the Earth is the most
important distance for Earth motion that because the energy is sent from Neptune
to Earth as we have discussed before.
- Based on this distance earth axial tilt be created showing the effect of the distances
distribution on the Earth motion data.
Equation no. (5)
100733 mkm = 360 mkm x 278.4
- Where
- 360 mkm = Mercury Orbital Circumference
- 278.4 degrees = The outer planets axial tilts total
- Mercury motion be considered as the master planet motion. that because the light
supposed velocity (1.16 mkm/s) motion started from Mercury
- For that mercury orbital circumference (360 mkm) defines The outer planets axial
tilts total (278.4 degrees).
Equation no. (6)
100733 mkm = 940 mkm x 107.2
- Where
- 107.2 degrees = 90 degrees + 17.2 degrees (Pluto orbital inclination)
- But

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- 17.4 degrees = The Inner Planets Orbital Inclinations Total
- (17.4 deg x 0.99 =17.2 deg)
- And
- 23.6 degrees = The outer Planets Orbital Inclinations Total
- (23.6 deg x0.99 = 23.4 deg earth axial tilt)
- Where
- 23.6 degrees + 17.2 degrees = 23.4 degrees +17.4 degrees
The data explains why Earth orbital circumference define Pluto orbital inclination
(17.2 degrees). That's happened because of the dep interaction between earth and
Pluto motions.

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5- The Sun Age Historical Proves
5-1 The Sun Circles The Earth
5-2 The Rate (1.0725)
5-3 The Sun Diameter Analysis
5-4 The Sun Rays Analysis

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5-1 The Sun Circle 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.

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5-2 The Rate 1.0725
I- Data
(A)
Why These Distances Are Equal?
(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
= Mercury Jupiter Distance x 2
(2)
Mercury Neptune Distance = Saturn Pluto Distance
Jupiter Pluto Distance = Uranus Neptune Circumference
Earth Neptune Distance = Mercury Saturn Circumference (error 0.5%)
(3)
Jupiter Mercury Distance = 2 Mercury Orbital Circumference
Jupiter Venus Distance = Venus Orbital Circumference (error 1.5%)
Jupiter Earth Distance = Earth Orbital Circumference (error 1.2%)
(Earth and Jupiter at 2 different sides from the sun)
(4)
Jupiter Mercury Distance = Mars Orbital Distance x π (error 0.6%)
Jupiter Uranus Distance = Venus Jupiter Circumference (error 0.8%)
Pluto Orbital Distance = Earth Orbital Circumference x 2π

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(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
=
=
=

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II – Discussion
- The previous data shows that
- 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)
- Why?
- The solar system motion analysis lead us to the conclusion that, the machine works
by equal distances feature (let's remember it in following)
- An Example
- Earth moves during its day period (24 h) a distance = Pluto motion distance during
its day period (153.3 h) (error 1%).
- We have observed this equality of distances.
- The periods of time are defined because they are defined by planets motions.
means, these periods of time aren't common periods but defined periods,
depending on planets motions.
- Means, these 2 distances be equal by some mentioned process. Because they
depend on 2 defined periods of time (depend on 2 planets motions).
- And,
- We have discovered that, this method is used by all solar planets motions. That
means, The equal distances method is used (thousands of times) in the solar
system motion.
- Because of this heavy using (50%) of all distances in the solar system to be equal
one another.
- That explains the data group No. (A) and makes it understandable.
- Means,
- 50% of all distances in the solar system are equal because of the equal distance
method Using.
- (The Equal Distances Method is discussed in details in point no.8)

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- Now
- The point needs explanation is the data group no. (B) where the distances are NOT
equal But Rated with 1.0725 (Why?)
- Why 40% of distances are rated with the same rate (1.0725)?
- 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
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.

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5-3 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

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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.
- This idea we have discussed before in the paper methodology, where the
(supposed) diameter 12430 km be defined geometrically and not created actually
- The Sun Diameter Can Have A Similar Discussion.
- 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)

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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

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5-4 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

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(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.

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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.

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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.

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- 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)
Notice
The planets unified general motion supports the proof with powerful arguments which
makes the proof more understandable and trustee.

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The Theory Proves Discussion
Part No. (II)
- This paper part discusses the planets motions nature to prove that,
- (1st
) The Planets Move One Unified General Motion
- Which Disproves Newton Concept Of The Planet Motion Independency
- (2nd
) The Inner Planets Motions depend on a source of motion different from
the outer planets motions where the 2 motions can't be provided by the same
one reason.
- Which disproves Newton Theory of the sun mass gravity
The Part Contents
6- The Solar Planets Move One Unified General Motion
7- The Solar Planets Motions Depend On Different Rates of Time
8-The Solar Planets Motions Analysis

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6- The Solar Planets Move One Unified General Motion
6-1 The Planets Unified General Motion Description
6-2 The Solar Planets Motions Are Complementary One Another
6-3 The Solar Planets Move An Unified Motion (A Team Motion)

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6-1 The Planets Unified General Motion Description
- I suppose that (The Planets Motions Create One Unified General Motion)
- This hypothesis tells the planets motions be similar to Train Carriages Motion
- Also It tells the planets motions be similar to Chess Pieces Motions
- The hypothesis 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
- The hypothesis gives a different vision from the planets classical description.
because its tells, (The Solar Planet Data Be Created Based On Geometrical
Calculations And Rules). The classical description tells planets data be created by
random process and without geometrical reasons behind.

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6-2 The 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

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(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

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(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)

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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.

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- Notice
- The previous analysis of motions prove the interactions of motions among the
three planets (Mercury, Mars and Jupiter). These interactions we have referred to
them during our analysis for Jupiter orbital circumference (4900 mkm).
- The 2 parts of discussions can provide a clear vision for how the planet motion be
done. Because the planets motions data help us greatly and protect us from any
imaginary vision. Because of that, the motions interaction which we have
concluded based on the planets data analysis concerning Jupiter orbital
circumference be proved here clearly by the 3 planets motions analysis
- It's simply a fact
- If the geometrical machine based on which this data be created be discovered the
reasons will be known to answer how or why these motions have interactions.
- But even before this machine discovery the data shows the motions interactions in
both sides. In the planets data and in the planets motions.

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6-3 The Planets Move One 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

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(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

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- 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.

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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

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- 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)

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7- The Solar Planets Motions Depend On Different Rates of Time
7-1 Preface
7-2 Earth And Pluto Rate Of Time
7-3 Mercury Motion Rate Of Time
7-4 Saturn Motion Rate Of Time
7-5 Uranus And Jupiter Rate Of Time

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7-1 Preface
- The different rates of time is the logical result for light motion accompanying
planet motion.
- The matter is created out of light but be not separated from its light parent, on the
contrary, they move together one unified motion. but they use different rates of
time for their motions.
- The basic rate of time is (1 second of light motion = 1 solar day of planet motion)
- The solar planets divides this great rate (1s =86400 s) among their motions and
work as a great clock.
- The rate of time between Earth and the sun (1 s = 365.25 s) is produced through
this process.
- But
- How to prove that? for example If
- 1 hour of Mercury motion = 9.18 hours of Pluto motion, can we prove that?
- The rate of time is one data of the planets motion data, because of that we can
discover the rates of time by the planets motions data analysis.
Notice
- Mercury data analysis refers to the rate 1 hour of Mercury motion = 1 second of
light motion and
- 1 hour of Mercury motion = 24 hours of Jupiter motion
- The combination of these 2 rates produce the basic rate (1s =86400 s) but the
planets depend on one another to operate the great clock.

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7-2 Earth And Pluto Rate Of Time
I- Data
Group no. (1) (Earth and Pluto Rate Of Time)
(A)
- 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%).
(B)
- (Earth velocity / Pluto velocity) = (Pluto Day period/ earth day period) = 6.37 (0.8%)
(C)
- 1 hour of Earth motion= 29.53 hours of the moon motion (explained in discussion)
(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
(4)
2 x 153.3 h = 4.61 x 66.8 But
(708.7 h /10.7 h) = (655.7 h /9.9 h) =66.23 (error 1%).
(5)
413600 km cos (10.96 deg) = 406000 km
(6)
(10.96 deg /1.7 deg) = (153.3 h /24 h) (error 1%)

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II- Discussion
- The Rate Of Time Depends On Planets Velocities Rate (A hypothesis)
- But, this rule sometimes be violated and in this case we can use the simple rule
tells (Equal Distances Causes Different Rates Of Time)
- 1 hour of Earth motion = 6.37 hours of Pluto motion
- This rate is concluded simply from Earth and Pluto velocities
Because
- Earth velocity (29.8 km/s) = Pluto velocity (4.7 km/s) x 6.37
- Also
- Pluto Day Period (153.3 h) = Earth Day Period (24 h) x 6.37
- The rate (6.37) which creates the rate of time between Pluto and Earth Motions,
effects greatly on the 2 planets motions data. As we have seen.
- Let's summarize the data idea in following:
- There's a proportionality between Pluto Data on one side and Earth and its moon
data on the other side. No Pure Coincidence can be claimed, because the
connection among the three planets be defined based on their positions to one
another. the question be (How This Data Be Created?)

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Equation no. (C)
1 hour of Earth motion = 29.53 hours of the moon motion.
- Because
- The moon displacements total during the moon day period (29.53 days) = Earth
motion distance during a solar day Period. (error 1%)
- Because of that
- 1 Day Of Earth Motion =29.53 Days Of The Moon Motion.
- We can see that, the machine works with consistency, for example
o 1 hour of Earth motion =6.37 hours of Pluto motion
o 1 hour of Pluto motion = 4.61 hours of the moon motion
o And
o 1 hour of Earth motion = 6.37 x 4.61 = 29.53 hours of the moon motion
o The rates of time are created based on each planet motion data and be in full
harmony as we see. That tells we deal with a geometrical mechanism works
consistency…
- Let's see the rate of time of Pluto and the moon motions

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Discussion (continued)
Equation no. (1)
(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
- 708.7 hours = The Moon Day Period
- Based on the rule (Equal distances creates different rates of time).
- 1 hour of Pluto motion = 4.61 hours of the moon motion.
- This rate of time be concluded because Pluto moves during its day period (153.3 h)
a distance = the moon displacements total during its day period (29.53 days). By
that 708.7 hours of the moon motion = 153.3 hours of Pluto motion. And So, the
rate of time be concluded (1 h of Pluto motion =4.61 h of the moon motion)
- Again the rates of time causes (2 Planets All Data) to be in proportionality with
one another. in following equations we see that Pluto and the moon motions data
be in full proportionality which is very interesting data.
Equation no. (2)
Tan (12.2) x 708.7 hours = 153.3 hours
- Where
- The angle 12.2 degrees = 13.177 deg – 0.98562 deg
- 0.98562 degrees = The Earth Motion Degrees Per A Solar Day
- Equation no. (2) 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).

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Equation no. (3)
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
- Equation no. (3) shows, Pluto rotation period is created depending on the moon
rotation period based on the angle (13.177 deg).
- The data of the 3 equations are strong and prove the idea clearly, because it's A
Directed Data. Means, this data be created based on geometrical mechanism be
found between Pluto and the moon motions. For that reason, the equations use (all)
data of the 2 planets. That shows the data is directed and can't be created by any
random process. It's data be created based on geometrical reasons and rules.
- 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.
Equation no. (4)
2 x 153.3 h = 4.61 x 66.8 But
(708.7 h /10.7 h) = (655.7 h /9.9 h) =66.23 (error 1%).
- 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
- The rate (4.61) we have defined, where (1 hour of Pluto motion =4.6 hours of the
moon motion).
- Equation no.(4) tells that this rate (4.61) is known by Jupiter and Saturn motions.
- That tells, the rate (4.61) is transported among the solar planets and that explains
how this rate be created between the moon motion and Pluto motion.

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Equation no. (5)
413600 km cos (10.96 deg) = 406000 km
- Where
- 406000 km = Pluto Motion Distance During A Solar Day
- Equation no. (5) shows that, the motions interaction between the moon and Pluto is
a deep interaction. And effects on the planet motion basic data. Because
- The moon displacements total during 29.53 days =2598693 km = 2π x 413600 km
- The distance (413600 km) should be the moon apogee orbital radius if the moon
apogee orbital circumference = the moon displacements total during 29.53 days.
- This is the fact which should be performed
- But
- The moon apogee orbital circumference be (2550973 km) less than the moon
displacements total during (29.53 days) with 2%. This fact arises the question
Why? the moon apogee circumference should be 2598693 km (surely). But for
some geometrical miracle this distance became =2550973 km!
- The radius 413600 km be changed into 406000 km and this change be done based
on the equation no. (5). The angle (10.96) caused this change. We have discussed
this angle before deeply because it's the main angle in the moon orbit geometrical
structure (we will review a short notice about it later). The angle (10.96 deg)
causes the radius 413600 km to be 406000 km – why?
Equation no. (6)
(10.96 deg /1.7 deg) = (153.3 h /24 h) (error 1%)
- The angle 10.96 deg be created based on the rate between Pluto Day Period
(153.3h) and Earth Day Period (24 h). this rate be used between the angle (10.96)
which causes to decrease the radius from 413600 km to 406000 km, and the angle
1.7 deg which is the moon daily motion angle in the moon orbital equation (θ1= θ0
+ 1.7 deg). (The moon orbital equation be in my previous paper)

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Notice (1)
- The moon displacements total 2598693 km should be the moon apogee orbital
circumference, by that, the moon apogee orbital radius should be =413600 km
- But
- It's not true, the moon apogee orbital radius =406000 km
- Why? because of the angle (10.96 degrees)
- Let's remember this angle in following:
o The moon orbital perigee radius =363000 km
o The moon orbital apogee radius = 406000 km
o If we draw these 2 radiuses in 2 circles as in the figure
o The tangent DB = 181843 km, also
o The perigee orbital circumference = 2.28 mkm
o (181843 km /2.28 mkm) = 0.08
o 137 degrees x 0.08 = (10.96 degrees)
o As we remember that 137 degrees = 95.1 deg x 1.44 deg where
o 95.1 deg =90 deg + 5.1 deg (the moon orbital inclination)
o 1.44 deg = the moon orbit regression degrees per month.
o We have supposed that the angle 137 degrees is created based on the moon
orbital inclination (degrees) and the moon orbit regression angle monthly.
And this angle (137 deg) is the basic angle in the moon orbital geometrical
structure (hypothesis).
o The calculations of the tangent DB = 181843 km aimed to explain one basic
idea which is, the moon orbit is built geometrically to create the rate (0.08)
by which the moon orbital inclination and regression will create the angle
(10.96 deg).
o Let's explain that more clearly

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o The moon orbit inclination and regression controls the moon orbital motion.
this is a hypothesis. Now, the moon orbital motion creates the moon orbit
based on a geometrical design, the moon orbital motion uses the moon orbit
geometrical design to create the angle (10.96 deg) from the basic one (137
deg). by that the original angle is (137 deg) which is created by the moon
inclination and regression unification, and the born angle is (10.96 deg)
which is the motor of the machine.
o What we need to refer to is that, the moon orbit geometrical design is used
in this process. That's why the distance 413600 be decreased and became
406000 km. By that, the main player is the angle (10.96 deg).
- This explanation answers the question,
- Why Pluto Motion Distance During A Solar Day = The Moon Apogee Orbital
Radius=406000 km?

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Notice (2)
- The previous data still have another reading let's summarize it in following:
o Pluto day period is a very long day period in comparison with the outer
planets days periods, Why?
o We notice also that, Pluto Moves During Its Day Period A Distance =
Earth Motion Distance During Earth Day Period. That means, Pluto Day
Period be extending to perform this requirement.
o i.e.
o Pluto Day Period be extending to be 153.3 h to enable Pluto to move a
distance = Earth motion distance during Earth day period.
o Because of that,
o Pluto day period (and rotation period) be created as functions in the Earth
moon day and rotation periods as we have seen in the data discussion.
o i.e.
o The interaction between Pluto motion on one side with the Earth and its
moon motions on the other side caused the following results:
o (1st
Result) the moon apogee orbital radius be decreased from 413600 km to
be 406000 km
o (2nd
Result) Pluto Day Period be extending to be =153.3 days.
Notice (3)
- Uranus effect is the main force behind the changes for the 2 planets data.
- Uranus effect on Pluto is proved by Planet 8 days Cycle where Uranus uses Pluto
day period in its motion.
- Uranus effect on the moon is proved by the fact, Uranus orbital circumference =
19.2 Earth orbital circumference. The moon moves a cycle called Metonic cycle
and extends for 19 years. Metonic Cycle is caused almost by Uranus effect on the
moon motion.

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7-3 Mercury Motion Rate Of Time
I- Data
(1)
112.2 km/s = 4.7km/s x 23.9
(2)
1407.6 hours =153.3 hours x 9.18
(3)
90560 days = 4222.6 hours x 9.18 x 2.33 (error 2%)
(4)
(2 x 153.3 h x 3600 s) /5040 s =23.9 x 9.18
II- Discussion
Equation (1)
112.2 km/s = 4.7 km/s x 23.9
- Where
- 112.2 km/s = The 3 Planets Velocities Total
- 4.7 km/s = Pluto Velocity
- Let's suppose, the 3 planets velocities (Mercury, Venus and Earth) be unified in
one velocity =112.2 km/s, So this velocity works as one gear and Pluto velocity
works as another gear, move relative to one another.
- By this supposed motion of 4 planets, the rate (1 hour = 23.9 hours) be created and
that causes 1 hours of a planet motion be = 23.9 hours of another planet motion.
- This process has a great effect on the solar planets rates of time because if the
planets motions rates of time depend only on the planets velocities that makes the
max rate is (10.08). Means, 1 hour of a planet motion be = 10.08 hours of another
planet motion. And this is a very narrow range of the rates of time. The 3 planets
velocities total creates an extension for this process and enables the system to
create the rate (1h = 23.9 h).

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Equation (2)
1407.6 hours =153.3 hours x 9.18
- Where
- 1407.6 hours = Mercury Rotation Period
- 153.3 days = Pluto Day Period.
- (9.18) = The rate of time between Mercury and Pluto motions.
- We conclude that mercury and Pluto motions rates of time is (9.18) based on this
equation. Means
- 1 hour of Mercury Motion = (9.18) hours of Pluto Motion.
- By this same method we conclude the rate of time between Pluto and the moon
motions. It was the rate between the moon day period and Pluto day period. But
here the rate (9.18) is between Mercury Rotation Period and Pluto Day Period
(where Pluto day period =153.3 h and Pluto rotation period = -153.3 h)
Equation (3)
90560 days = 4222.6 hours x 9.18 x 2.33 (error 2%)
- Where
- 90560 days = Pluto Orbital Period
- 4222.6 hours = Mercury day period
- 2.33 = ??
- Mercury velocity (47.4 km/s) = 10.08 x Pluto velocity (4.7km/s). if the planets
rates of time depend on their velocities. That means 1 hour of Mercury motion =
10.08 hours of Pluto motion. and in this case 24 hours of Pluto motion = 2.37
hours of Mercury Motion.
- That because the equation uses (Pluto orbital period 90560 solar days) in
comparison with 4222.6 hours (Mercury day period). The equation proves that the
rate (9.18) is the used one between the 2 planets.

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Equation (4)
(2 x 153.3 h x 3600 s) /5040 s =23.9 x 9.18
- Where
- 5040 seconds are required for Mercury day period to be 176 solar days
- 153.3 h = Pluto Day Period.
- 23.9 = the rate between the 3 planets velocities total (112.2 km/s) and Pluto
velocity (4.7 km/s).
- This equation also proves that the rate of time between Mercury and Pluto is (9.18)
- The data proves that, 1 hour or Mercury motion = (9.18) of Pluto motion.
Notice
- Different data of Mercury and Pluto Motions shows that the rate of time between
them is (9.18)
- Because this rate is used in different data in the 2 planets motions data, we
conclude that it's a fact and real one.
- I wish the proves show its power clearly. Because we deal with geometrical data.
We find for example a rate (9.18) between Mercury and Pluto motions data and
this rate be used between them frequently. That happens also with Earth (6.37) and
with the moon (4.61).
- The point is that, the rate of time has a massive effect on planets motion but we
still need to discover this effect clearly.
- It's noticeable to take into consideration the method by which the rate of time
effect on Planet Motion.
Notice
- If Mercury and Pluto motions rate of time be defined based on their velocities
only, The rate will be 10.08 (means 1 h of Mercury motion = 10.08 h of Pluto
motion). the rate (9.18) effect only on Pluto and Mercury motions but no other
planets. For example (Mercury 1 hour =46.1 h of The Earth moon motion), in this
case (46.1 = 10.08 x 4.61) , the data doesn’t use (9.18) but (10.08).

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7-4 Saturn Motion Rate Of Time
I- Data
(1)
Saturn moves daily 838080 km = the Earth moon displacement 88000 km x 9.5
(2)
Saturn diameter 120536 km = Earth diameter 12756 km x 9.5
(3)
Saturn orbital distance 1433.5 mkm = Earth orbital distance 149.6 mkm x 9.5
(4)
Earth axial tilt 23.4 deg = Saturn orbital inclination 2.5 deg x 9.5 (error 1.7%)
(5)
Saturn Orbital Period 10747 days = 365.25 days x 29.53 days
II-Discussion
- Equation no.(1) shows that, the rate of time be produced as a rate between Saturn
motion distance per a solar day and the moon displacement per a solar day (9.5)
and this rate controls different data between Saturn on one side and Earth with it
moon on the other side.
- This feature is similar perfectly to Pluto data proportionality with Earth data.
- And also the data between Pluto and Mercury.
- The data tells that, the rate of time has a massive effect on the planets motions
data. This feature which we see frequently in different planets data.
- The data (1, 2 ,3 and 4) shows that the rate of time (9.5) creates a proportionality
between Saturn and Earth motions data.
- Let's discuss Equation No. (5) in following
Equation no. (5)
Saturn Orbital Period 10747 days = 365.25 days x 29.53 days
- We know that,
- 1 day of Earth Motion = 29.53 days of the moon motion
- And also we know

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- 1 day of the sun motion = 365.25 days of Earth Motion
- Based on that,
- 1 day of the sun motion = 365.25 x 29.53 days of The moon Motion, means
- 1 Day Of The Sun Motion = 10747 Days Of The Moon Motion
- Where
- Saturn Orbital Period =10747 days
- How this data be created?
- More interesting data be in our waiting
- The Earth orbital circumference =940 mkm but the moon daily displacement be
=88000 km, that means
- If the moon passes Earth orbital circumference (940 mkm) by using it daily
displacement (88000 km) the moon needs a period = 10747 days (error 0.6%)
- This data tells, the period 10747 days be defined regardless Saturn motion. or in
better words, the period 10747 days be defined before Saturn Creation.
- The data forces us to guess that, the Earth moon must be created before Saturn
creation. and as a result for the interaction with the moon motion Saturn uses the
period be defined by the moon motion. (also Saturn orbital distance =1433 mkm =
Mars orbital circumference. That tells may some interaction of motions between
Saturn and Mars caused Saturn to use mars orbital circumference as its orbital
distance. And if we put this data beside each other we may conclude that Saturn is
created after Mars migration and the moon creation).
- One more data can help our investigation
- The moon daily motion distance = Earth daily motion distance =2.574 mkm
(because they are not separated from one another)
- The moon distance be effected by the rate 1.0725 and because of that (2.574 mkm
=1.0725 x 2.4 mkm) the moon motion distance daily be =2.4 mkm
- During 10747 days the moon moves (2.4 mkm) a total distance =25920 mkm

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7-5 Uranus And Jupiter Rate Of Time
1 hour of Mercury motion = 24 hours of Jupiter motion = 24 hours of Uranus motion
(For Jupiter)
Old Equation No. (A)
4780 mkm = 1.1318 mkm per solar day x 4222.6 days
- Where
- 1.1318 mkm Per Solar Day = Jupiter Motion Distance Per A Solar Day
- 4222.6 hours = Mercury Day Period
- The distance 4780 mkm is different from 4846 mkm with (1.38%)
- 4846 mkm = The Inner Planets Orbital Circumferences Total
- We have seen this equation before in our analysis for Jupiter orbital circumference
(4900 mkm).
- Equation no.(A) tells,
- 1 Hour Of Mercury Motion= 1 Solar Day Of Jupiter Motion.
- This rate equal Uranus rate also let's see it in following
-
(For Uranus)
Notice
- Mercury Uranus Distance = 2815.2 mkm
- Mercury rotation period = 1407.6 hours
- 2 x Mercury rotation periods = 2815.2 hours
- If 1 mkm = 1 hour, So the distance 2815.2 mkm be = 2 Mercury rotation periods
And
- Mercury (47.4 km/s) moves during its rotation period (1407.6 h) a distance = 243
mkm (error 1%)
- 243 solar days = Venus rotation period
- If each 1 mkm be =1 solar days, the distance 240 mkm be = 243 days (error 1%)

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Based on that
- Mercury as a player uses (1 mkm = 1 hour) and (1 mkm = 1 day =24 hours)
- That creates the rate (1 hour = 24 hours)
- But between whom? Almost
- Between Mercury and Uranus, means
- 1 hour of Mercury Motion =24 hours of Uranus Motion.
Notice
- 1 h of Pluto motion=2.62 h of Jupiter motion =2.62 h of Uranus motion.
- The rate 2.62 is so important for Jupiter and Uranus because
- Uranus Circumference 160592 km x 2.62 = 421056 km (Uranus Motion
Distance During Its Day Period)
- Jupiter Circumference 449197 km x 2.62 = 1160000 km (Jupiter Motion
Distance During Mars Rotation Period) (error 1.4%)
- This data we have discussed before.

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8- The Solar Planets Motions Analysis
8-1 Preface
8- 2 The Equal Distances Method
8-3 The Inner And Outer Planets Motions Be Distinguished

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8-1 Preface
- The paper claims that,
- The solar planets move a type of motion be different from the motion type Newton
had imagined based on The Sun Mass Gravity Concept.
- The previous discussion proves that:
1. The Planets Move One Unified General Motion, Disproving Newton Concept
Of Planet Independent Motion
2. The Planets Motions Depend On Different Rates Of Time.
- These Features Contradicts Newton Theory Of The Sun Mass Gravity.
- But
- There are more interesting features which prove that, the solar planets motions be
different in nature from the motion Newton had mentioned by The Sun Mass
Gravity.
- Among these features we study 2 basic features which are
o The Equal Distances Method
o The Distinguish between The inner and outer Planets Motions.
- Let's start our discussion in following…

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8- 2 The 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%)
(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.

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- 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).

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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

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Let's summarize the equal distances using range in following:
(1)
The data in appendix no. (1) shows that
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

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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

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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 use by light and planet motions
or by planet and planet motions.
- That makes this method so useful and suitable for the solar planets motions

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8-3 The Inner And Outer Planets Motions Be Distinguished
8-3-1 The Inner Planets Motion
I- Data
(1)
Earth Mass = The Masses total of (Mercury + Venus + Mars)
(2)
Earth Circumference = The 5 inner planets diameters total
(Means the diameters total of (Mercury + Venus + The Moon + Earth + Mars))
(3)
2 x Earth Circumferences = 2 Mars Circumferences + Venus Circumference
(4)
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. (1) 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.

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- 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.
Equation no. (2)
Earth Circumference = The 5 inner planets diameters total
(Means the diameters total of (Mercury + Venus + The Moon + Earth + Mars))
Equation no. (3)
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.

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- The previous 2 equations no. (2 and 3) 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.
Equation no. (4)
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
- We know the distance can't be charged unless under specific conditions, but I try
to show geometrical effects are found. It's not necessary for this effect to be a
direct one, but it can be found through interactions.
- 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
- We have discovered that the planets motions can't be independent from the other
planets motions on the contrary the planets motions be similar to chess board
motions where each motion be calculated geometrically. This features (Chess

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Board Motion) we have discussed in the planets unified general motion. and I want
to say that, 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)
A Conclusion
- The electric field idea aims to create a source of motion for the inner planets. By
that the moving planets are the inner planets. That tells the inner planets motions
depend on the inner planets electric field. But the outer planets motions depend on
the motion transportation from the inner planets motions to the outer planets
motions.
- By that we have 2 different types of motions but both originated from the same
one source by 2 different methods and that enable the planets motions to be
integrated in one unified general motion
- If the 2 types of motions be proved that disprove Newton Theory of The sun mass
gravity.

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8-3-2 The Outer Planets Motion
The Outer Planets Motions Are Distinguish By 2 Features
- The outer planets motions be distinguished from the inner planets motions by 2
basic features which are
- (1st
Feature) Planet 8 days Cycle doesn't use any inner planet motion.
- (2nd
Feature) The outer planet moves during its day period a distance equal to, or
rated with its circumferences which is a feature doesn't be used by the inner planets
motions.
- Planet 8 days cycle we have studied before
- So,
- Let's discuss in following feature no. (2)

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(2nd
) Planet moves during its day period a distance = Its Circumference
The outer planet moves during its day period a distance equal to, or rated with its
circumferences
I-Data
(1)
Saturn moves During Its Day Period A Distance=Its Circumference (error 1.3%)
(2)
Jupiter moves During Its Day Period A Distance =Its Circumference (error 4%)
(3)
Uranus moves During Its Day Period A Distance =260% of Its Circumference
(4)
Neptune moves During Its Day Period A Distance =200% of Its Circumference
(5)
Pluto moves During Its Day Period A Distance =345.6 x Its Circumference

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II-Discussion
- We should notice that, Jupiter (4%) causes Planet 8 days Cycle
- And in Planet 8 days cycle we have seen that, a different distance = Jupiter
diameter be defined between Jupiter motion distance and its circumference which
creates planet 8 days cycle for Jupiter motion.
- I want to say that, the planet circumference is a player in this planet motion. and
because of that, the (4 great planets circumferences) have deviations from their
circumferences because of Planet 8 days cycle.
- Why does the outer planets be different from the inner planet?
- The outer planets move by observing their circumferences but the inner planets
moves observing their orbital circumferences. That may be happened because of
the electric field controls the inner planets motions and causes their motions be
controlled by their orbital circumferences. But the outer planets depend on the
motion transportation from the inner planets, and by that the outer planet almost
uses its circumference as a measurement for its motion.
Saturn Motion Analysis
- We analyze Saturn motion as an example for the 4 planets motions, in following
- Saturn (9.7 km/s) moves during its day period (10.7 hours) a distance = 373644 km
= 98.67 % of Saturn Circumference (378675 km).
- The difference between the distance (373644 km) and Saturn Circumference =
5031 km. (= 1.3%)
- Why is this interesting?
- Because if a planet moves during its day period a distance = its circumference that
will create a rate between this planet orbital period and day period
- Means
- (Planet Orbital Period/ Day Period) = (Planet Orbital Circumference / Planet
Circumference)
- For that reason, we interest in planet motion distance during its day period.

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- Saturn motion performs this rate
- But
- It shows that, the rate in Saturn data be changed because of geometrical necessities
which I claim they are produced as a result for (planet 8 days cycle). Let's
summarize the idea in following
o The inner planets move under their electric field because of that their
motions be defined by this field effect, but
o The outer planets receive the motion by transportation of motion. because of
that, the outer planets have no a guide in their motion, they use their
circumference as a guide and measurement.
o If this rule be working, the outer planet would move during its day period a
distance = its circumference - But
o Because planet 8 days cycles which causes the motion transportation among
the outer planets, that creates an interaction which changes the rule and
caused the outer planets to move during their days periods distances (NOT)
equal their Circumferences, but (rated with their circumferences)
o Jupiter (8 day cycle) shows clearly that, the planet circumference is a player
in the cycle and proves the interaction which is found among the outer
planets as a result for planet 8 days cycle.
o This explanation details disproves clearly Newton Description because the
motion (here) is transported from one planet to another clearly and with
strong proves which disproves the sun masses gravity concept
o The point is that, the interaction among the outer planets motions cause the
change in the rule and creates planet 8 days cycle. But. The rule still works
spite the deviation.
o I want to say that,

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o Saturn moves During Its Day Period A Distance=Its Circumference
(error 1.3%), But the machine feels that no error in Saturn motion.
because this error is found as a result for the configuration with Planet
8 days cycle
By this same explanation all outer planets motions be controlled. The errors in the
planets motions from the rule (Outer Planet moves during its day period a distance =
its circumference) still works sufficiently, the errors be found as a result for planet 8
days cycle. (hypothesis)

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Appendix No.1
The Solar System Equal Distances List
(A)
Why These Distances Are Equal?
(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
= Mercury Jupiter Distance x 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
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.

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(B)
Why These Distances Are NOT Equal?
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 %)
(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
=
=
=

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Appendix No.2
Planet Orbital Distance Equation
Gerges Equation For Planet Orbital Distance
d2
= 4do (d – do)
Where
d= A Planet Orbital Distance
do= The Previous Planet Orbital Distance
Earth depends on Mercury
Mars depends on Venus
Pluto depends on Uranus
Discussion
Why Does A Planet Orbital Distance Depend On Its Neighbor Planet?
- Because
- The Solar Planets Motions Depend On Other Solar Planets Motions And Not On
The Sun Mass Gravity
- Also
- Many Solar Planets Motions Depend On Their Direct Neighbor Solar Planets
Motions

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The Equation Application
Gerges Equation For Planet Orbital Distance
d2
= 4do (d – do)
d= A Planet Orbital Distance
do= The Previous Planet Orbital Distance
- (227.9 mkm)2
= 4 (108.2 mkm) (119.7 mkm)
- Where
- 227.9 mkm = Mars Orbital Distance
- 119.7 mkm = Venus Mars Distance
- (108.2 mkm)2
= 4 (57.9 mkm) (50.3 mkm)
- Where
- 50.3 mkm = Mercury Venus Distance
- (149.6 mkm)2
= 4 (57.9 mkm) (91.7 mkm) (error 2.5%)
- Where
- 91.7 mkm = Mercury Earth Distance
- (Earth depends on Mercury in place of its direct neighbor Venus)
- (227.9 mkm)2
= 4 (108.2 mkm) (119.7 mkm)
- Where
- 119.7 mkm = Venus Mars Distance
- (Mars depends on Venus in place of its direct neighbor The Earth )
- (415 mkm)2
= 4 (227.9 mkm) (187.1 mkm)
- Where
- 415 mkm = Ceres Orbital Distance
- 187.1 mkm = Mars Ceres Distance
- (778.6 mkm)2
= 4 (415 mkm) (363.6 mkm)
- Where
- 415 mkm = Ceres Orbital Distance
- 363.6 mkm = Ceres Jupiter Distance
- (1433.5 mkm)2
= 4 (778.6 mkm) (655 mkm)
- Where

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- 1433.5 mkm = Saturn Orbital Distance
- 655 mkm = Saturn Jupiter Distance
- (2872.5 mkm)2
= 4 (1433.5 mkm) (1439 mkm)
- Where
- 2872.5 mkm = Uranus Orbital Distance
- 1433.5 mkm = Saturn Orbital Distance
- 1439 mkm = Saturn Uranus Distance
- (4495.1 mkm)2
= 4 (2872.5 mkm) (1622.6 mkm) (error 4%)
- Where
- 4495.1 mkm = Neptune Orbital Distance
- 1622.6 mkm = Uranus Neptune Distance
- (5906 mkm)2
= 4 (2872.5 mkm) (3033.5 mkm)
- Where
- 5906 mkm = Pluto Orbital Distance
- 3033.5 mkm = Uranus Pluto Distance
(Pluto depends on Uranus in place of its direct neighbor Neptune)
Notice
- 3 Planets Violate The Equation, (Why?)
- For example
- Pluto depends on Uranus and not on Neptune why?
Because of interaction of motion be found between Pluto and Uranus, which we have
proved by different data (for example by Planet 8 days Cycle)

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Appendix No.3 Planet Gravity Equation
Gerges Equation For Planet Gravity
I- Data
Table No. 1
Planet Equation Resulted Registered Error
The Sun (Earth/Sun diameters rate)2
(Sun Mass/
Earth Mass)* earth gravity
= (1/109)2
* 333000*9.8= 274
274 274 0
Moon (3.66)2
x (0.073 / 5.97) x 9.8 1.6 m/sec2
1.6 m/sec2
0
Mercury (2.61)2
x (0.23/5.97) x 9.8 3.7 m/sec2
3.7 m/sec2
0
Venus (1.0538)2
x 4.87 /5.97) x 9.8 8.9 m/sec2
8.9 m/sec2
0
Earth 9.8 9.8 9.8 0
Mars (1.878)2
x (0.642 / 5.97) x 9.8 3.7 m/sec2 3.7 m/sec2 0
Jupiter (0.0892)2
x (1898 / 5.97) x 9.8 24.7 m/sec2 23.1 m/sec2 6.9%
Saturn (0.10582)2
x (568/5.97) x 9.8 10.44 m/sec2 9 m/sec2 16%
Uranus (0.25)2
x (86.8/5.97) x 9.8 8.9 8.7 2.3%
Neptune (0.2575)2
x (102/ 5.97) x 9.8 11.1 m/sec2
11 m/sec2
0.9%
Pluto (5.337)2
(0.0131 / 5.97) x 9.8 0.6125 m/sec2
0.6 m/sec2
2%
II- Discussion
1. We can see that, the Equation is working sufficiently with all planets and even the sun
follows this same equation, Except With Jupiter And Saturn There Are great
Errors (6.9% and 16% respectively…. Why?) (because of the rate 1.0725 effect
which is used here in form 7.1)
2. we need to observe and analyze deeply The Sun Equation in row No.1 of the
previous table, because this equation supports my claim that the sun is
originated from the same source from which the solar group is originated,
otherwise why the sun herself follows the same equation? That may support my
claim that, the sun rays is created depends on solar planets motions
accumulation….
gravity
Earth
)
Mass
Earth
The
Mass
Planet
The
(
)
Diameter
Planet
The
Diameter
Earth
(
gravity
planet
The 2
×
×
=

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References and Biography
(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_
Light Motion Features Are Discovered in Planet Motion
https://www.slideshare.net/Gergesfrancis/light-motion-features-are-discovered-in-planet-motion
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
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The Solar System Distances Network Theory (Revised)

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