Planet Diameter Be A Function In Its Rotation Period

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
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Gerges Francis Tawdrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
Planet Diameter Be A Function In Its Rotation Period
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
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The Assumption Of S. Virgin Mary -Written in Cairo –Egypt –2nd
August 2022
Abstract
Paper hypothesis
- Planet Diameter Be Created As A Function In Its Rotation Period
The hypothesis Explanation
- (1)
- Planet Data Be Created Based On Exact Equations –
- As a plane manufacture – the maker needs exact equations to define this plane
length, width, weight …etc otherwise this plane can't fly safely.
- The moving planet under the physical laws has to define its data based on exact
equations otherwise this planet can't move safely.
- I have discovered 5 equations can conclude Planets Data theoretically prove this
fact decisively.
- (2)
- Planet orbital distance be defined before this planet creation - because – the planets
motions leave an empty place for the new planet – by that- each planet orbital
distance be defined by the other planets orbital distances and motions trajectories.
- My first equation proves each planet orbital distance depends on its previous
neighbor planet orbital distance –
- d2
= 4d0 (d-d0) where d= planet orbital distance and d0= its neighbor orb. distance
- We test and discuss this equation in the current paper–

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- As a result –
- The planet found itself in an obligatory position be defined by other planets data
and the planet data itself has no role in its orbital distance definition.
- (3)
- Planet motion can be a dangerous one if this planet diameter isn't in harmony with
its orbital distance – means- the wrong diameter can cause the planet to be broken
(The mass can decrease planet velocity – but the wrong diameter can break it) –
for that reason – the planet has to create its diameter as a function in its orbital
distance to move safely.
- But
- If the function has only 2 variables (planet diameter and its orbital distance), If this
planet changes its orbital distance for any reason its diameter will be broken as a
result –
- The designer had to create the function between planet diameter and its orbital
distance as indirect function contains many other variables and by that if the planet
changes its orbital distance the other variables will be changed but the diameter
will be saved.
- As a result
- Planet diameter be created as a function in its rotation period and the rotation
period be a function in its velocity and the velocity be a function in its orbital
distance – by that – the function between planet diameter and its orbital distance
be created but contains also other variables to protect the diameter.
- My fourth equation proves this fact let's see it
- (4)
- Planet Diameter Definition Equation (My Fourth Equation)
- (v1/v2) = (s/r) =I
- v = Planet Velocity
- r= Planet Diameter

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- s= Planet Rotation Periods Number In Its Orbital Period
- I= Planet Orbital Inclination (a rate to inclination unit)
- v2, s, r and I be belonged to one planet and v1 be belonged to another planet
- we test, discuss and analyze this equation in this current paper.
- The equation tells each planet diameter be a function in (s) which is (the number of
its rotation periods in its orbital period)
- Shortly
- Planet diameter be created as a function in its rotation period –
- The planets data prove this fact
- But
- Why do planets data follow this equation? What's the geometrical effect on which
the equation depends?
Paper objective
- The equation depends on a rate of time which is (1 hour = 23.9 hours) - means
- Between 2 planets in the solar system this rate of time be created and this rate be
transported among the planets and based on this transported rate each planet be
forced to create its diameter as a function in its rotation period
- The paper analyzes this rate of time and discusses how it be created and what
geometrical effect it practices on the planets data to force them to create planet
diameter as a function in its rotation period.
- Please enjoy the reading (Please scan the figure (ORCID)
Gerges Francis Tawdrous +201022532292
Physics Department- Physics & Mathematics Faculty
Peoples' Friendship university of Russia – Moscow (2010-2013)
Curriculum Vitae https://www.academia.edu/s/b88b0ecb7c
E-mail mrwaheid@gmail.com, mrwaheid1@yahoo.com
gergesgerges@yandex.ru
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Paper Reference
Planet Motion Logic Disproves Newton Theory Of The Sun Gravity
https://www.academia.edu/s/500a64998b
or
https://app.box.com/s/gx8ys3g0snyngziwaezgkzceihvfy3cj

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Paper Contents
Subject Page No.
1-Introduction 6
2- Methodology 9
3- Planet Data Depends On Exact Equations 11
3-1 Preface 12
3-2 Planet Data Description 14
3-3 Planet Orbital Distance Equation 18
3-4 The Solar System Distances And Velocities Maps 22
4- Planet Diameter Definition Equation (My fourth Equation) 26
4-1 The Equation Test 27
4-2 The Equation Discussion 30
5- Planet Diameter Definition Equation Analysis 34
5-1 Preface 35
5-2 The Equation Effect Description 36
5-3 The Moon Effect Analysis 37
5-4 The Relative Motion Between The Moon And Pluto 40
5-5 The 3 Inner Planets effect on Pluto motion 49
5-6 The Moon And Pluto Motions Data Consistency 51
5-7 The Outer Planets Diameters Total Effect 54
5-8 The Moon Orbit Geometrical Structure 56
5-9 Why Does The Moon Apogee Orbital Radius =406000 Km? 58
5-10 Saturn Effect Analysis 64
5-11 The Moon And Saturn Motions Data Consistency 66
5-12 Planet 8 Days Cycle 70
5-13 Pluto Effect Analysis 76
5-14 Pluto And Neptune Data Consistency 78
5-15 Jupiter And The Moon Data Consistency 80
5-16 The Equation Units Analysis 81
5-17 Planet Diameter Analysis 83
Appendix no. (1) 84
References and Biography 94

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1- Introduction
- Why Does The Moon Orbital Period = The Moon Rotation Period =27.3 Days?
- Because Of The Tidal Locking!
- Can we be sure that (this answer is the correct one)?
- But
- Mercury day Period = 2 Mercury orbital periods = 3 Mercury rotation periods? Is
there any geometrical or physical reason behind?
- Venus day period = 2 Mercury rotation periods? Can there be a reason behind?
- Also
- (The moon day period /the moon orbital period) = (Venus rotation period / Venus
orbital period) – Why??
- (1)
- The basic difficulty in the solar system discussion is the vision of a single point -
because the solar system be similar to a machine of gears or one creature body and
each planet be a member in it – by that – we can understand nothing when we see
just one point of it – the integration is the secret behind the solar system motion
and without integration we can understand nothing of it
- I want to say
- The solar system data is similar to one picture be cut into pieces and we try to put
these pieces integrating each other to create the original picture – without the
integration we will misunderstand the solar system creation and motion facts
- I need to confirm on this meaning –we can understand by the data integration only-
the total solar eclipse gives us example – because – the sun, moon and Earth
diameters and distances be defined in comparison with the other players data to
cause the total solar eclipse.
- (2)
- The problem is known –it's the big bang theory – even if - it be refuted – its effect
still be found in physicists minds- because – the minds believe that planet data be

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created by random process, initial conditions and historical unknown factors –for
that no geometrical reason be found behind any planet data
- This wrong lesson teaches us (we should not try to explain how planet data be
created?) – the whole idea is wrong in its principle –because
- A flying plane under physical laws and its maker has to find exact equations to
define this plane length, width, weight…etc otherwise it can't fly safely
- The moving planet under physical laws has to define its data based on exact
equations otherwise it can't move safely -
- Here the gap be seen
- While each piece of data in the solar system be created based on exact equations
and geometrical rules the minds think the data be created by random and no need
to discover the geometrical reasons behind -
- The answer of the moon cycles periods equality can show this gap
- (3)
- The Moon Orbital Period = The Moon Rotation Period =27.3 Days – Not Because
Of Any Tidal Locking
- The fact is that,
- Planet diameter be created as a function in its rotation period – my fourth equation
proves that – the equation works from the Earth to Pluto only – why?– Let's see it..
- Planet Diameter Definition Equation (My Fourth Equation)
- (v1/v2) = (s/r) =I
- r= Planet Diameter
- s= Planet Rotation Periods Number In Its Orbital Period
- I= Planet Orbital Inclination (a rate to inclination unit)
- v2, s, r and I be belonged to one planet and v1 be belonged to another planet
- The rate (s) for the moon equal =1 and for that the moon be used as the base for
this equation –

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- By that,
- The moon cycles be used as the base of the equation and this equation effect
passes through all planets data and forces each planet to create its diameter as a
function in its rotation period-
- But
- The moon is a small planet and it's a strange to be used as a base for all planets
data – for that reason –
- Venus and Mercury motions support the moon motion cycles, that explains why
the cycles periods of the 3 planets show equality and rates frequently – shortly –
the three planets create one system based on which the moon orbital period be =
the moon rotation period =27.3 days, And based on this periods equality the
equation depends to define each planet diameter as a function in its rotation period.
- In this paper we test the equation, discuss it and discover the geometrical effect on
which the Equation depends.
- Let's refer to the paper contents in following….
- Point no.(3) proves that (Planet Data Depends On Exact Equations)
- (and test my first equation which defines planet orbital distance)
- Point no.(4) tests and discusses my fourth equation which proves planet diameter
be created as a function in its rotation period.
- Point no.(5) analyzes my fourth equation to discover the real geometrical effect on
which the equation depends- and discusses the 3 basic points of the equation (the
moon – Saturn – Pluto)
- Appendix no. (1)
- I have discovered 5 equations by which we can conclude planets data theoretically
– the paper discusses my first and fourth equations – and the rest 3eqautions tests
be put in the appendix no.(1) for reference.

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- 2- Methodology
- I use the planets data analysis to discover the solar system creation and motion
facts – The method is so useful.
- The method simply put the planets data in comparison with the theory and tries to
know if there's a consistency between both – let's use an example to explain how
this method works
- An Example
- The 3 planets (Mercury – Venus – Earth) give the interesting data! why?
- Because, the 3 planets be in order for their diameters, masses and orbital distances.
Can this order be found based on a geometrical rule? let's try to discover
- But
- Mars causes a question, because Mars causes to break this order!
- What hypothesis do we need to explain this interesting data?
- The hypothesis tells (Mars Original Position Was Between Mercury And Venus)
- If this is the original position of Mars the planets order will be
- (Mercury – Mars – Venus – Earth)
- The 4 planets be in order for their diameters, masses and orbital distances
- Can we prove this hypothesis? Yes
- Mars had migrated from its original orbital distance to its current one – and Mars
in its migration motion had collided with Venus and then with Earth and Mars
itself caused to create the Earth Moon!
- Giant-Impact hypothesis tells that, a planet in Mars Size had collided with the
Earth and caused the moon creation.
- Can Mars Itself do that? the theory tells No Hope
- But, Planets data analysis suggested that Mars had migrated from its original
orbital distance to its current one –

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- Let's move with this hypothesis for a while
- Suppose Mars was the second planet after Mercury and had migrated to its current
point (227.9 million km) and Mars had collided with Venus and then with Earth –
can this idea help Giant-impact hypothesis? for example can this idea answers
(Why Does Venus Have No Moon?)
- (a)
- Imagine Mars was the second planet after Mercury (84 mkm) and had migrated to
its current position (227.9 mkm), in its displacement, Mars was pushing by force
and had collided with Venus and pushed all debris with it in its motion direction
- Venus had found no debris around – for that Venus couldn't create its own moon-
- (b)
- Another question asks about (the origin of the lunar magma ocean!) Venus, The
Lunar Magma Ocean is came from Venus, it's a part of Venus found by the
collision between Mars and Venus but Mars had pushed all debris with it in its
motion direction and left Venus without debris
- Earth gravity is greater than Venus' and the debris lost some of their momentum
and by that the Earth could create its own moon where the moon rocks are
consisted of Venus, Earth and Mars debris
- The fact Mars has 2 moons is one more proof for this idea, because Mars with
small mass could attract 2 moons and Venus couldn't.
- The rest debris be attracted by Jupiter and consisted the asteroid belt
- Shortly
- The planets data analysis puts planet data in comparison with the theory explains
its motion to test if the theory be sufficient and to discover the geometrical rules
based on which this data be created.

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3- Planet Data Depends On Exact Equations
3-1 Preface
3-2 Planet Data Description
3-3 Planet Orbital Distance Equation
3-4 The Solar System Distances And Velocities Maps

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3-1 Preface
- Planet Data Be Created Based On Exact Equations
- Means,
- Planet mass, diameter, orbital distance, period, inclination, rotation period and
axial tilt – this data be created based on exact equations and mathematical
calculations and No single data be created by any random process.
- As a plane or rocket manufacture – the maker needs exact equations to define this
plane length, width, weight …etc and all its specifications and data otherwise this
plane can't fly
- The moving planet under the physical laws also has to define its data based on
exact equations otherwise this planet can't move safely.
- I have discovered 5 equations – By which we conclude Planets data theoretically
without observation – we discuss 2 of my 5 equations in this current paper. Which
are my first equation (defines each planet orbital distance) and my fourth equation
(defines each planet diameter).
- My fourth equation proves the following idea
- Planet diameter is created as a function in its rotation period and the rotation
period is a function in its velocity.
- In this paper we discuss my fourth equation and test it for all planets data – the
data proves the equation credibility sufficiently.
- But
- The current paper does more important job in this equation – the paper examines
the geometrical effect on which the equation depends
- Means,
- The planets data prove that each planet diameter be a function in its rotation period
and the rotation period be a function in its velocity – this fact be seen in the planets
data clearly – but why the data shows this fact? what geometrical effect be passed
through the planets motions and force each planet diameter to be created a function

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in its rotation and the rotation be a function in its velocity – we examine the real
geometrical effect on which the equation depends.
- By that – this current paper be dedicated for my fourth equation studying and
analysis where my first equation which defines each planet orbital distance be
discussed as a side subject.
- The discussion is divided into 3 points
- Point no. (3) discusses how planet data be created and tests my first equation
which defines each planet orbital distance
- Point no. (4) discusses my fourth equation and tests it by the planets data to prove
the equation credibility
- Point no. (5) discusses and analyzes the real geometrical effect which passes
through the planets and causes each planet diameter to be created as a function in
its rotation period –
- The geometrical effect in fact is a rate of time – where 1 hour of one planet motion
be = 23.9 hours of another planet motion– by that – this rate of time is the real
geometrical effect passes through the planets data and causes planet diameter to be
a function in its rotation period.
- Our discussion also answers old questions as (why doesn't planet use its orbital
velocity to define its rotation period?) and also (why does the moon rotation period
= the moon orbital period?) the answer isn't (for tidal locking) this idea isn't a fact
and wrong answer.
- The paper discussion analyzes a wide range of planets data.

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3-2 Planet Data Description
- (1)
- As the preface stated (planet data be created based on exact equations), this
concept open a new chapter in the solar system geometry study –because – now we
no longer deceive ourselves by the wrong ideas tell planet data be create by
random process or initial conditions-these wrongs should be removed at end from
the physics book –
- In fact these idea be refuted simply and clearly – because – the planets move under
physical laws controls the solar system motions and under these laws no random
data can be found because in case it be found it will cause to break the moving
planet or stop its motion-
- Perfectly as a flying plane when we see it we consider that the maker had exact
equations for this plane dimensions and specifications otherwise this plane can't
fly.
- Shortly
- Planet Data Be Created Based On Exact Equations
- But, by what order is this happened? What data be created at first and what later?
What's the independent data and what's the depend? how to describe the planet
data creation? let's try to answer in following
- (2)
- Planet Orbital Distance Be Defined At First – Because – Planet orbital distance be
defined by its neighbor planet orbital distance – this idea be proved my first
equation we discuss here
- But even without my first equation the concept is correct because
- The new planet has to occupy the empty place – where the solar system isn't empty
and there are many moving planets, for that the new planet can't put itself in a
place of collision with the others- simply- the new planet has to occupy the
available place and this valuable place be defined by its neighbor.

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- My first equation proves this idea clearly –it tells
- d2
= 4d0 (d- d0) where d= planet orbital distance and d0 = its neighbor distance
we discuss and test this equation in the next point (no. 3-3)
- based on this vision
- The planet found itself in its position by force where this planet orbital distance
doesn't depend on any of this planet data neither mass nor diameter – shortly –
planet orbital distance be defined by the other planets effect on this planet motion
and this planet has nothing to do in its orbital distance definition -
- This situation is similar to a person in some crowded bus and the people pushed
him to the last place of the bus by force and he can do nothing to prevent it.
- (3)
- Planet Diameter Should Be A Function In Its Orbital Distance
- This sentence in a cornerstone in the planets motions analysis – where planet
diameter is the basic data and not its mass – because – the mass can decrease
planet velocity and may cause to create this planet orbital inclination – but the
wrong diameter can cause this planet to be broken.
- It's a simple idea – the mass can't cause to break the planet but the wrong diameter
can cause the planet to be broken through its motion – In this case the motion be so
dangerous for this planet because the planet moves against its internal structure.
- Imagine a truck be loaded by woods and the woods be out of this truck walls, now
when the truck tries to drive through (u-turn) inside a small street the woods be
broken – but the heavy weight can cause to decrease the velocity but not to break
the woods…
- Here we catch an important idea which help us to understand how planet data be
created – Planet diameter should be the second data be created after planet orbital
distance – planet diameter needs to be created based on exact and accurate
equation because the wrong diameter will cause the planet to be broken.
- Shortly

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- Planet Diameter Should Be A Function In Its Orbital Distance
- But
- If this function be consisted of 2 variables only (planet diameter and its orbital
distance) – if this planet changes its orbital distance for any reason this planet will
be broken.
- We don't know how Planet can change its orbital distance – but – if the function
contains only the 2 variables that will cause a continuous risk for the planet
because the change of its orbital distance be equal the planet death
- For example – I claim that –Mars original orbital distance was 84 million km and
Mars had migrated from it to its current position 227.9 million km –
- If the function has only the 2 variables Mars would be broken because the diameter
should be changed with the orbital distance change
- As a result
- The designer had created planet diameter as a function in its orbital distance but
the function contains more variables – and because of that –
- Planet diameter be created as a function in its rotation period and the rotation
period be a function in its velocity and the velocity be a function in its orbital
distance.
- That explains Kepler statement (Planet Orbit Defines Its Velocity) Why?
- Because
- Planet diameter be a function in its rotation and the rotation period be a function in
its velocity – And because the velocity is a function in its orbital distance - by that
the function between planet diameter and its orbital distance be created.
- This explanation shows the planet data creation direction and reason – by that -
we can analyze logically how this data be created.
- My fourth equation proves this fact which tells (Planet diameter be a function in its
rotation period and the rotation be a function in its velocity) -that shows the
importance of my fourth equation discussion.

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- (4)
- The previous analysis can help us to disprove Newton theory of the sun gravity
- Shortly
- By one force Planet be created and moving – because – if 2 forces caused this
planet creation and motion this planet will be broken by effect of 2 planets on it
- Means
- The Force Which Created The Planet It Causes Its Motion
- The sun created no planet and for that (No Planet Moves By The Sun Gravity)
- The point is
- Imagine a planet be created by any force and the sun forces this planet to revolve
around it by its gravity as Newton told – in this case – this planet will move
against its internal structure and this planet will be broken.
- It's Wrong Logic On Which Newton Theory Depended – the fact is that
- No Planet Moves By The Sun Gravity

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3-3 Planet Orbital Distance Equation
(My First Equation)
- d = A Planet Orbital Distance
- d0= Its Direct Previous Neighbor Planet Orbital Distance
- The equation depends on the planets order, for that , just 2 neighbor planets can be
used in this equation, means if (d is Venus distance, d0 be Mercury distance)
- The equation exceptions are,
- Earth depends on Mercury Not Venus – and Mars depends on Venus Not Earth
And Pluto depends on Uranus Not Neptune
- Note, we don't use the forma (d=2d0) instead we use the forma (d2
= 4d0 (d-d0))
because it uses the distances between the 2 planets and that decreases the errors
- Let's test the equation
(1) Venus Motion
- (108.2)2
= 4 x 57.9 x (50.3)
- d= 108.2 million km = Venus Orbital Distance
- d0= 57.9 million km = Mercury Orbital Distance
- 50.3 million km = The Distance Between Venus And Mercury
- Venus Depends On Mercury
(2) Earth Motion
- (149.6)2
= 4 x 57.9 x (149.6-57.9) (error 2.8%)
- d= 149.6 million km = Earth Orbital Distance
- d0= 57.9 million km = Mercury Orbital Distance
- Earth depends on Mercury and doesn't on Venus
(3) Mars Motion
- (227.9)2
= 4 x 108.2 x (227.9-108.2)
- d= 227.9 million km = Mars Orbital Distance
- d0= 108.2 million km = Venus Orbital Distance
- Mars depends on Venus and doesn't on Earth
)
(
4 0
0
2
d
d
d
d −
=

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(4) Ceres Motion
- (415)2
= 4 x 227.9 x (415-227.9)
- d= 415 million km = Ceres Orbital Distance
- d0= 227.9 million km = Mars Orbital Distance
- Ceres depends on Mars
(5) Jupiter Motion
- (778.6)2
= 4 x 415 x (778.6- 415)
- d= 778.6 million km = Jupiter Orbital Distance
- d0= 415 million km = Ceres Orbital Distance
- Jupiter depends on Ceres
(6) Saturn Motion
- (1433.5)2 = 4 x 778.6 x (1433.5- 778.6)
- d = 1433.5 million km = Saturn Orbital Distance
- d0 = 778.6 million km = Jupiter Orbital Distance
- Saturn depends on Jupiter
(7) Uranus Motion
- (2872.5)2
= 4 x 1433.5 x (2872.5- 1433.5)
- d= 2872.5 million km = Uranus Orbital Distance
- d0 = 1433.5 million km = Saturn Orbital Distance Uranus depends on Saturn
(8) Neptune Motion (error 4%)
- (4495.1)2
= 4 x 2872.5 x (4495.1- 2872.5)
- d= 4495.1 million km = Neptune Orbital Distance
- d0 = 2872.5 million km = Uranus Orbital Distance Neptune depends on Uranus
(9) Pluto Motion
- (5906)2
= 4 x 2872.5 x (5906- 2872.5)
- d= 5906 mkm = Pluto Orbital Distance
- d0 = 4495.1mkm = Neptune Orbital Distance Pluto depends on Uranus
- Notice the error is less 1% for all planets except (Earth 2.8%) and Neptune (4%)

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Discussion
(a)
The Equation tells each planet orbital distance depends on its previous neighbor
planet orbital distance – but – there are 3 exceptions which are –
Earth depends on Mercury not Venus –
Mars depends on Venus not Earth–
Pluto depends on Uranus not Neptune–
(b)
The equation shows – the distance from the sun to Pluto be distributed based on one
geometrical design – means – the distances be created together as one group in one a
network form – the distances be similar to the chess board distances- they are
distributed geometrically and in one network based on one design -
Shortly
No Single Distance be Created independently or individually – We deal with one
network of distances -
Means
By Using Mercury Orbital Distance (57.9 million km) (one data) We Can Conclude
All Planets Orbital Distances (9 Data) By Using Mathematical Calculations Only
(c)
Kepler stated (Planet orbit defines its velocity) – this concept is used in my third
equation (d1/d2) = (v2/v1)2
where d= Planet Orbital Distance and v = Planet Velocity
The concept tells– If we know a planet orbital distance, we can conclude its velocity
by mathematical calculations only
Shortly
By Using my 1st
equation (d2
= 4d0 (d-d0)) and kepler law and the One Data (Mercury
orbital distance = 57.9 million km) we can conclude by mathematical calculations
only All Planets Orbital Distances, Velocities And Periods (27 Data)
That proves the concept (Planet data be created based on exact equations)

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A Comment
The equation gives a complete different vision from the physics book – because it
tells planets data be created based on exact equations.
Kepler laws give us a beautiful description for the planets motions trajectories but
kepler never told that – (Planet distance depends on its neighbor distance)
And
Newton wrong theory tells (Planet motion depends on its mass) and by that (Planet
orbital distance depends on the sun and planet masses gravity)
Planets data show that (Planet orbital distance depends on its neighbor distance) and
by that – planets data disproves Newton theory of the sun gravity and his concept of
planet motion depends on its mass – this whole idea is an imaginary one -
That makes my first equation is a very new equation in concept where the physicists
believed that (Planet orbital distance should be defined by the sun gravity mass unless
the initial condition effected on it) – this whole idea is wrong-
The fact is that (Planet orbital distance depends on its neighbor distance) – and this
dependency caused to create the solar system distances in one Network form and as
one group of distances (as chess board distances)
Also my first equation solves the problem of Titius Bode law because it argues that
planet orbital distance depends on its neighbor planet orbital distance and not on the
numbers order.
Notice
Ceres Orbital Distance =414 Million Km And Its Orbital Period 1680 Days (its
velocity 17.9 km/s)

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3-4 The Solar System Distances And Velocities Maps
FIRST - The Map Of Distance
The distances from the sun to Pluto has 2 features which are
(1st
Feature)
Each Planet Orbital Distance Depends On Its Neighbor Orbital Distance – this feature
depends on my first equation (d2
= 4d0 (d-d0)) we have discussed before
(d= Planet Orbital Distance and d0 = Its Neighbor Planet Orbital Distance)
We have discussed it
(2nd
Feature)
The distances map depends on Jupiter and Pluto as 2 basic points of this map
(Proof)
Data
37100 million km – 4900 million km = 32200 million km
32200 million km x π = 100733 million km
Where
100733 million km = The planets orbital circumferences total
37100 million km = Pluto Orbital Circumference
4900 million km = Jupiter Orbital Circumference
Discussion
The data proves the idea - because
The 3 values (4900 , 37100 and 100733) depend on one another, any 2 values enable
us to conclude the third one theoretically
Means,
If we know Jupiter orbital circumference =4900 million km and The planets orbital
circumferences total = 100733 million km, We can conclude theoretically Pluto
orbital circumference =37100 million km – We do that by using the data and even
without my first equation (d2
= 4d0 (d-d0))

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The distances map tells –the map depends on 2 basic points (Jupiter and Pluto)
More analysis of Data can prove that Mercury point be used as an origin point for
these 2 planets (Jupiter and Pluto) –
Shortly - Mercury be the origin point based on which the 2 points (Jupiter and Pluto)
be created based on them the solar planets orbital circumferences total be created.
Notice
For the distances Map we need to notice that – the distance 4900 million km (Jupiter
orbital circumference) is the central distance in the solar system because
(i)
The inner planets orbital circumferences total be (Mercury 360 mkm + Venus 680
mkm + Earth 940 mkm + Mars 1433 mkm + 1433 mkm) = 4900 million km (1%)
The total = 4900 million km but the distance (1433 mkm) be used 2 times!
(ii)
Jupiter Orbital Circumference (4900 million km)
(iii)
Uranus needs 4900 days to pass a distance = Uranus Orbital Distance
Neptune needs 2 x 4900 days to pass a distance = Neptune Orbital Distance (2%)
Pluto needs 3 x 4900 days to pass a distance = Pluto Orbital Distance (1%)
(iv)
(10747 /9800) = (9800 /9007)
10747 days = Saturn Orbital Period
9007 million km = Saturn Orbital Circumference
9800 = 2 x 4900
The data tells, the value 4900 be used by all planets (in different units)
The distance for Jupiter (and the inner planets) be used as a period of time for Uranus,
Neptune and Pluto–and Saturn uses this value as a distance and as a period of time
Based on that - we conclude the distance 4900 million km is the central one in the
solar system

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SECOND - The Map Of Velocity
The data shows that one map be found for the planets velocities – this map depends
on Jupiter and Uranus velocities – let's prove that in following –
Proof
Data
(i)
(2 x 100733 million km /197393 days) = (1.16/1.1318) = (0.6/0.5875)
Where
100733 million km = The Planets Orbital Circumferences Total
197393 days = The Planets Orbital Periods Total
1.16 million km/s = Light Supposed Velocity
0.6 million km/s = 2 x 0.3 million km/s (Light Velocity)
1.1318 million km/day = Jupiter Velocity Per A Solar Day
0.5875 million km/day = Uranus Velocity Per A Solar Day
(ii)
(1.16/0.6) = (47.4/24.1) = (35/17.9) = (13.1/6.8) Where
0.6 million km/s = 2 x 0.3 million km/s (Light Supposed Velocity)
47.4 km/s = Mercury Velocity 24.1 km/s = Mars Velocity
35 km/s = Venus Velocity 17.9 km/s = Ceres Velocity
13.1 km/s = Jupiter Velocity
6.8 km/s = Uranus Velocity
Discussion
the discussion supposes a light beam its velocity 1.16 million km per second be found
Data No. (i) shows the planets orbital circumferences and periods total depend on the
2 planets (Jupiter and Uranus) velocities in comparison with the 2 velocities of light
(the known velocity 300000 km/s and the supposed velocity 1160000 km/s)

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Data No. (ii) shows that, the rate of velocities between (Jupiter and Uranus) be used
also by (Mars and Mercury) and by (Venus and Ceres) – But
No any couple can be replaced in place of (Jupiter and Uranus) in Data no. (i) – that
shows, the planets orbital circumferences and periods be related to (Jupiter and
Uranus) velocities and not to any other couple of planets – they are the 2 basic players
in the design structure -
The data proves the idea tells
One Map of velocity be used for all planets velocities and in this map Jupiter and
Uranus are the 2 basic points (or 2 columns)
Notice
Light (300000 km/s) travels during 16330 sec a distance = 4900 million km
Light (1160000 km/s) travels during 4222.6 sec a distance = 4900 million km
Where
4900 million km = Jupiter Orbital Circumference (we have discussed before)
16330 hours = Mars Orbital Period
4222.6 hours = Mercury Day Period
Light motion uses 1 hour of planets cycles periods as one second of light motion
THIRD - Why Did The Designer Use 2 Maps For The Distance And Velocity?
Because the designer uses the phase between the distance map and velocity map to
define the planets diameters total– both Maps depend on Jupiter but the distance map
reaches to Pluto where the velocity map limits to Uranus – the phase between Pluto
and Uranus is found to define the planets diameters total 406000 km
This data will be useful in our analysis for my fourth equation (planets diameter
definition equation).

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4- Planet Diameter Definition Equation (My fourth Equation)
4-1 The Equation Test
4-2 The Equation Discussion

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4-1 The Equation Test
Planet Diameter Definition Equation (My Fourth Equation)
v = Planet Velocity
r= Planet Diameter
s= Planet Rotation Periods Number In Its Orbital Period
I= Planet Orbital Inclination (a rate to inclination unit)
v2, s, r and I be belonged to one planet and v1 be belonged to another planet
The planet (v1) be defined by test the minimum error
- Earth Equation uses Neptune velocity
- Mars Equation uses Pluto velocity
- Jupiter Equation uses the Earth moon velocity
- Saturn Equation uses Mars velocity
- Uranus Equation uses Neptune velocity (As Earth)
- Neptune Equation uses Saturn velocity
- Pluto Equation uses the Earth moon velocity (As Jupiter)
Notice / (The Equation Works From The Earth To Pluto Only)
The Equation Test
Earth equation (366.7/12756) = 5.4/ (29.8 x 2π) = 0.029
366.7 = Earth rotation periods number in Earth orbital period
12756 km = Earth diameter
29.8 km/s = Earth velocity
5.4 km/s = Neptune velocity
365.25 days = Earth orbital period (and Earth rotation period =23.9 hours)
I
r
s
v
v
=
=
2
1

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Mars equation (671/6792) = 4.7/ (24.1 x 2) =0.098 (error 1.2%)
671 = Mars rotation periods number in Mars orbital period
6792 km = Mars diameter
24.1 km/s = Mars velocity
4.7 km/s = Pluto velocity
687 days = Mars orbital period (and Mars rotation period =24.6 hours)
Jupiter equation (10500/142984) = 13.1/(27.78 x 2π) = 0.0734 (error 2.2%)
10500 = Jupiter rotation periods number in Jupiter orbital period
142984 km = Jupiter diameter
13.1 km/s = Jupiter velocity
27.78 km/s = The Earth Moon velocity
4331 days = Jupiter orbital period (and Jupiter rotation period =9.9 hours)
Saturn equation (24106 x2) /(120536) = 9.7/ 24.1 =0.4
24106 = Saturn rotation periods number in Saturn orbital period
120536 km = Saturn diameter
9.7 km/s = Saturn velocity
24.1 km/s = Mars velocity
10747 days = Saturn orbital period (and Saturn rotation period =10.7 hours)
(1/0.4) = 2.5 where 2.5 degrees = Saturn Orbital Inclination
Uranus equation (42683 / 51118) = 5.4/6.8 =0.8 (error 5%)
42683 = Uranus rotation periods number in Uranus orbital period
51118 km = Uranus diameter
6.8 km/s = Uranus velocity
30589 days = Uranus orbital period (and Uranus rotation period =17.2 hours)
0.8 degrees = Uranus Orbital Inclination

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Neptune equation (89143 /49528) = 9.7/ 5.4 =1.8
89143 = Neptune rotation periods number in Neptune orbital period
49528 km = Neptune diameter
9.7 km/s = Saturn velocity
59800 days = Neptune orbital period (and Neptune rotation period =16.1 hours)
1.8 degrees = Neptune Orbital Inclination
Pluto equation (14178 /2390) = 27.78/ 4.7 =5.9
14178 = Pluto rotation periods number in Pluto orbital period
23908 km = Pluto diameter
27.78 km/s = The Moon velocity
4.7 km/s = Pluto velocity
90560 days = Pluto orbital period (and Pluto rotation period =153.3 hours)

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4-2 The Equation Discussion
(1) The Equation Modifications
Many planets cause modifications for the equation – let's refer to them in following
(a)
Mars and Saturn use the number (2) which isn't found in the original equation
(b)
The Earth and Jupiter use the rate (2π) which isn't found in the original equation
(c)
Uranus equation causes a great error = 5%
All errors are less than (1%) except Jupiter (2.2%) and Mars (1.2%)
(d)
Pluto equation depends on the moon velocity – but – connected with Neptune
Because
(Pluto orbital period / Pluto rotation period) x 2π = (Neptune orbital period / Neptune
rotation period)
(e)
Jupiter and Saturn uses the rate (r/s) in place of the rate (s/r)
(f)
The orbital inclination rate (I) be produced in complex form – just with Saturn,
Uranus and Neptune the produced values refer to the planet orbital inclination clearly,
but with the others the produced values be complex – let's use an example
Example –Mars equation produces the value 0.098 – but
(1/0.098) = 10.2 = 2 x 5.1 (where the moon orbital inclination = 5.1 degrees)
Even if we accept this value – this isn't Mars orbital inclination– why??
7 deg (Mercury orbital inclination) = 5.1 degrees + 1.9 deg (Mars orbital inclination)
(5.1 degrees = The Moon Orbital Inclination)
Means - the definition isn't direct but with some complexity – that may be as result of
the moon motion effect on Mars motion.

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(2) The Equation Objective
The equation creates a function between planet diameter and (s) (its rotation periods
number in its orbital period) – then the rate (s) be a function in this planet velocity and
another equation shows that planet velocity be a function in its orbital distance - this
chain creates a function between planet diameter and orbital distance.
The equation shows that – planet diameter be created in harmony with its motion – as
in some canal water creates a vortex and for some reason the water lost its minerals
and salts around this vortex – with time some rock be created by the minerals and
salts and the rock be in a tube form through which the water moves – here the tube
dimensions be in harmony with this water motion because it be created by this water
motion effect.
(3) How Does The Equation Work?
The equation uses the moon and Pluto as 2 terminals of it – because
The Moon Orbital Period = The Moon Rotation Period =27.3 days, by that for the
moon the rate (s) be = 1
And
Pluto has 14177 rotation period in its orbital period and Pluto needs 14547 days to
pass a distance = 5906 million km = Pluto orbital distance.
By that the moon and Pluto each planet has 2 equal periods in its motion for that
reason the 2 planets be used as 2 terminals for the equation.
Notice
The moon orbital period = The moon rotation period =27.3 days not for any tidal
locking – the idea is wrong – the moon periods are equal because of my fourth
equation – where this equality of the 2 cycles be used as a base for the equation by
which the planets diameter be created as a function in its rotation period
This idea be discussed deeply in the next point no. (5)

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(4) Planet Rotation Period Analysis
According to kepler law (planet orbit defines its velocity), planet velocity be defined
by its orbital distance – by that –Planet should use its orbital velocity to define its
rotation period – and the rotation period by that will be a function in this planet
circumference -In this case Earth (29.8 km/s) would rotate around its axis in 22.4 min
only and not in 23.9 h
This is not the fact – Earth rotates around its axis in 23.9 hours –
Then we have to ask
Why doesn't planet use its orbital velocity to define its rotation period? because of
my fourth equation – the equation creates a function between planet diameter and its
rotation period and this is the reason which prevents planet to use its orbital velocity
to define its rotation period –
In fact Saturn is the only planet uses its orbital velocity to define its rots ton period
and Saturn does that because it works as a central point for my fourth equation as
we will see in the next point discussion (Point No. 5)
Shortly
My fourth equation refers to a geometrical effect starts from the moon and reaches to
Pluto, it passes through all planets data and forces each planet to create its diameter as
a function in its rotation period – or in more accurate words in the rate (s)
By that planet rotation period be defined almost before the planet diameter creation!
Can planet rotation period be defined before its diameter creation?
Let's use the moon as example -
The moon be created by collisions between planets – the collisions debris revolve
around the Earth and later these debris be collected together into the moon – we can
imagine that the rotation period can be defined before the diameter creation – because
the debris revolve around the Earth by some rotation period – and during the moon
diameter be growing some rotation period was used also.

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(5) Saturn Rotation Period Analysis
Saturn (9.7 km/s) moves during its rotation period (10.7 h) a distance = 373644 km
This distance equals approximately Saturn circumference (378675 km) (error 1.3%)
By that Saturn is the only planet uses its orbital velocity to define its rotation period
Jupiter (13.1km/s) also moves during its rotation period (9.9 h) a distance increased
4% than its circumference
Uranus (6.8 km/s) moves in its rotation period (17.2 h) a distance =2.6 its
Circumference
Neptune (5.4 km/s) moves in its rotation period (16.1 h) a distance = 2 its
Circumference
We can see that these distances be found by geometrical design – that tells we have an
effect passes through the planets and causes a geometrical effect through the planets
data
(6) Venus And Mercury Support The Moon Motion
The idea tells that – the moon motion be supported by Venus motion and Venus itself
be supported by Mercury motion – by that – the moon motion be supported by the 2
planets motions –
We need that because the moon cycles equality be used as the base of the equation
which is illogical because the moon is a small planet and the solar system can't
depend on it – but the moon cycles be supported by Venus and Mercury – by that the
equation base be supported strongly
We discuss that in details in the next point no. (5)
Notice
In the next point no. (5) we answer the questions
(Why Does The Equation Work From The Earth To Pluto Only?) and
(what's The Real Geometrical Effect Depend On Which The Equation Work?)

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5- Planet Diameter Definition Equation Analysis
5-1 Preface
5-2 The Equation Effect Description
5-3 The Moon Effect Analysis
5-4 The Relative Motion Between The Moon And Pluto
5-5 The 3 Inner Planets effect on Pluto motion
5-6 The Moon And Pluto Motions Data Consistency
5-7 The Outer Planets Diameters Total Effect
5-8 The Moon Orbit Geometrical Structure
5-9 Why Does The Moon Apogee Orbital Radius =406000 Km?
5-10 Saturn Effect Analysis
5-11 The Moon And Saturn Motions Data Consistency
5-12 Planet 8 Days Cycle
5-13 Pluto Effect Analysis
5-14 Pluto And Neptune Data Consistency
5-15 Jupiter And The Moon Data Consistency
5-16 The Equation Units Analysis
5-17 Planet Diameter Analysis

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5-1 Preface
What Do We Try To Do In This Point?
My fourth equation defines planet diameter as a function in its rotation period and the
rotation period as a function in its velocity – as we have discussed before –
The equation works from the moon to Pluto because the moon be used as the equation
base –
This story we have discussed based on the planets data – means – the data shows this
is a fact because the planets data follow the equation –
But
What's the real geometrical effect which be transported from the moon to Pluto
passing through all planets data and forces each planet diameter to be created as a
function in its rotation period according to the equation?
The planets data follow the equation – this is a fact - but why the planets data follow
the equation?
What's The Real Geometrical Effect on which the equation depends? And How This
Effect Be Transported From One Planet To Another? by what method the data can be
transported among the planets?
This point discusses this real geometrical effect by which the planets data follow the
equation -

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5-2 The Equation Effect Description
Let's summarize the idea in following….
The Equation works depending on three basic points which are (the moon – Saturn
and Pluto)
The moon and Pluto are the equation 2 terminals and Saturn be the central point of the
equation
Let's look at each point deeply in following
For The Moon – The Periods Are Equation Because
(The Moon Orbital Period = The Moon Rotation Period)
For Saturn – The Velocities Are Equal – Because
(Saturn Orbital Velocity = Saturn Rotational Velocity)
For Pluto – The Distances Are Equal – Because
(Pluto moves in its rotation period a distance = the moon
displacements total in its day period)
We examine these 3 points in details in our discussion to discover how the equation
work and why these planets show these features.

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5-3 The Moon Effect Analysis
Why does the moon orbital period = the moon rotation period =27.3 days?
"Because Of The Tidal Locking" – A wrong answer!
Because
The moon cycles equality be found to be used as the base for the equation which
defines the planets diameters as we have discussed -
Here we have 2 answers for the same one question – I say my answer is the fact
For that –I have to provide facts prove my claim and disprove the other answer –
Shortly
The moon cycles periods equality depend on Venus and Mercury motions – by that –
the 3 planets motions cycles depend on one another –
Means - Venus and Mercury cycles be created to support the moon cycles periods
equality -Let's try to prove that in following
I- Data
(1)
1407.6 hours (Mercury Rotation Period) = 153.3 hours x 9.18
2802 hours (Venus Day Period) = 153.3 hours x 9.18 x 2
708.7 hours (The Moon Day Period) = 153.3 hours x (9.18/2)
(153.3 hours = Pluto Rotation Period)
Shortly
Venus day period = 2 Mercury rotation period = 4 the moon day period (error 1%)
Mercury day period = 2 Mercury orbital period = 3 Mercury rotation period
(2)
(90000 /4900) = 2 x 9.18
(3)
(29.53/27.3) = (243 /224.7) =1.0725
(4)
Mercury moves during its rotation period a distance = 243 million km (error 1%)

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II- Discussion
The idea tells – Venus and Mercury cycles supports the moon cycles and cause them
to use equal periods and as a result the moon orbital period = the moon rotation period
Data No. (1)
Data No. (1) tries to prove this idea – the equality of the moon 2 cycles periods should
be compared with Mercury and Venus cycles periods – because – many other equal
and rated periods be used for the 2 planets – it should be many pure coincidence if the
3 planets cycles aren't connected…
The data tells
Venus day period= 2 Mercury rotation period= 4 the moon day period (error 1%)
And
Mercury day period = 2 Mercury orbital period = 3 Mercury rotation period
Data no. (3) proves the cycles are connected – let's see it in following
Data no. (3)
(29.53/27.3) = (243 /224.7) =1.0725
Where
29.53 days = The Moon Day Period 27.3 days = The Moon Rotation Period
224.7 days =Venus Portal Period 243 days = Venus Rotation Period
The rate 1.0725 we have discussed frequently before because of its wide using in the
solar system
The data tells the cycles are connected
means, the equality and rates in the 3 planets cycles be mentioned for geometrical
necessity – that tells there's a geometrical reason caused Mercury day period to be = 2
Mercury orbital periods = 3 Mercury rotation periods.
Notice /
Data no. (5) connects Mercury motion with Venus rotation period –

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Mercury moves in its rotation period a distance =243 million km (error 1%) and if
each 1 million km = 1 day the period 243 days = Venus rotation period
The used rule is unknown but the data shows a connection.
Data no. (2)
(90000 /4900) = 2 x 9.18
Where
4900 million km = Jupiter Orbital Circumference
90000 million km = C2
for a period 1 second (c = light velocity)
This data tries to show that the rate 9.18 is found in the basic data of the solar system
motion- because Jupiter orbital circumference is the central distance in the solar
system and the value (C2
) refers to main energy –
I want to say – the rate 9.18 in the data no. (1) is found through the solar system main
data –means – the connection between Mercury, Venus and the moon cycles is a
deep connection in the solar system motion data and can form the solar system design
backbone.
Notice
The rate (9.18) is our main point of discussion, we will return to it for deep analysis
but here the discussion aimed only to prove that there's a connection between the 3
planets cycles periods and this connection can't be explained by the tidal locking idea
which explains the moon cycles periods equality – the correct explanation depends on
my equation – because the moon cycles periods equality causes the rate (s) to be equal
(= 1) and by that the moon motion be used as the base for the equation depends on
which all planets diameters be defined as functions in their rotation periods – here the
idea of support Mercury and Venus for the moon motion is a suitable idea because the
moon is a small planet and can't be a qualified base for the whole solar group but if
Venus and Mercury support the moon motion that creates a point of connection
between the 3 planets which can be used as the base of the equation controls all solar
planets. that also explains why the Equation works only from the Earth to Pluto.

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5-4 The Relative Motion Between The Moon And Pluto
Let's summarize the idea in following…
- (I)
- I add The velocities of (Mercury 47.4 + Venus 35+ the moon 29.8) = 112.2 km/s
(The moon velocity be considered 29.8 km/s = Earth velocity- because they aren't
separated in their revolutions around the sun)
- The 3 planets velocities total =112.2 km/s
- And
- Pluto velocity 4.7 km/s
- And
- We imagine that, the total velocity (112.2 km/s) moves relative to Pluto velocity
(4.7 km/s) where 112.2 km/s = 23.9 x 4.7 km/s
- This data means, 1 hour of the velocity (112.2 km/s) be = 23.9 hours of Pluto and
- As a result –
- We suppose that 1 hour of (Mercury, Venus or the moon) = 23.9 hours of Pluto
- Means
- The relative motion created a rate of time between the 2 planets – means-
- one hour of A Planet motion = 23.9 hours of Pluto motion
- This rate of time be created based on the total velocities (112.2 km/s)
- Notice
- We have studied the rate of time before and we have stated that the rate of time
depends on the planets orbital periods – for example
- Earth orbital period =365.25 days and Venus orbital period = 224.7 days by that
1 day of Venus = 1.62 days of Earth – this is the idea we have discussed.
- In our current discussion the rate of time be created depends on the velocities rate,
how can that be possible? Almost there's a connection between the planets orbital
periods and velocities and this connection depends on the geometrical design of
the planets orbital distances.

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- Shortly
- We will accept this idea for now –
- The idea tells – because the total velocity (112.2 km/s) =23.9 x 4.7 km/s that
creates a rate of time between a planet and Pluto – and by that –
- One hour of a planet = 23.9 hours of Pluto
- We can't define the planet because the velocity 112.2 km/sec isn't related to any
planet and we should search for the point on which the velocity be effective – in all
cases the rate be created between (One Planet And Pluto)
- (II)
- The previous idea neglected one point – which can be discovered by the question
- How the 3 planets velocities be added together? Which point moves by the
velocity (112.2 km/s) relative to Pluto velocity (4.7 km/s)?
- The answer of this question should be (Jupiter) – we prove that in (5-4-1)
The Conclusion
- My fourth equation defines each planet diameter as a function in its rotation period
and the rotation period be a function in its velocity.
- This is the idea we have learnt from the equation and planets data which follow the
equation perfectly- but
- The real geometrical effect which be started from the moon and reaches to Pluto is
the rate of time (1 =23.9) between 2 planets in the solar system – this rate of time
be found based on the relative motion between 2 points – and this rate of time
causes the planets diameters to be created as a function in their rotation periods –
by that – this rate of time passes through the planets data as a geometrical effect
creates the planets data based on its effect.
- The working 2 points which use this rate be (Jupiter and Pluto) as we will prove in
the next point -
- The working 2 points which created this rate be (The 3 planets velocities total
112.2 km/s in comparison with Pluto velocity 4.7 km/s)

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- 5-4-1 Planets Orbital Distances Distribution
- I- Data
- (1)
- Π x 37100 million km = 4900 million km x 23.9
- 100733 million km = 4900 million km (23.9 – Π)
- 100733 million km = Π ( 37100 million km - 4900 million km)
- (2)
- 100733 million km x 23.9 = 37100 million km x 20.5 Π
- 100733 million km = 4900 million km x 20.5
II- Discussion
- Data no. (1)
- Π x 37100 million km = 4900 million km x 23.9
- 100733 million km = 4900 million km (23.9 – Π)
- 100733 million km = Π ( 37100 million km - 4900 million km)
- Where
- 4900 million km = Jupiter Orbital Circumference
- 37100 million km = Pluto Orbital Circumference
- 100733 million km = The Planets Orbital Circumferences Total
- The 3 values (4900, 37100 and 100733) we have discussed before with the map of
distances where we have discovered that these 3 values depend on one another and
any 2 values can conclude the third one –
- Our current new data shows that – the 2 values depends on 4900 million km –
- Means, Jupiter orbital circumference (4900 million km) be created at first and
based on this distance the other 2 distances be created as functions in it – by that
37100 and 100733 be created depending on 4900 million km

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- The used factor is (23.9) this is the secret rate behind – if we know this rate and
Jupiter orbital circumference (4900 million km) we can conclude the 2 values
(37100 million km and 100733 million km)
- That tells the planets orbital distances distribution took into consideration the rate
(23.9) as a basic geometrical requirement in this distribution – why??
- Let's see the next data
- Data No. (2)
- 100733 million km x 23.9 = 37100 million km x 20.5 Π
- 100733 million km = 4900 million km x 20.5
- We have a new rate which is 20.5 what's this one?
- 2x 20.5 = 41 where (the planets orbital inclinations total = 41 degrees)
- That tells, the solar system distances be created based on (4900 million km Jupiter
orbital circumference) and the rate (23.9) which causes the planets orbital
inclination total to be = 41 degrees.
- The data leads to the following conclusions
- (1)
- Planet orbital distance depends on its neighbor planet orbital distance (be proved
by my first equation)
- (2)
- The planets orbital distances distribution depends on Jupiter orbital circumference
4900 million km
- (3)
- Pluto orbital distance and position depended on the rate (23.9) between Pluto and
Jupiter distances – by that – Pluto position be defined depending on Jupiter
position and the rate (23.9), That made this rate (23.9) as a basic one in the planets
distances distribution definition - and this rate be essential data.

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- (4)
- The rate (23.9) be effective on planets distances because this rate is found to
support the planets diameters definition -
- As we remember
- Planet diameter should be a function in its orbital distance to move safely and can't
be a direct function has only 2 variables (planet diameter and orbital distance) but
has to have many other variables to save this planet diameter in case of the planet
migration – by that - planet diameter be a function in rotation period and the
rotation be a function in the velocity then the velocity be a function in orbital
distance – this is the basic idea we deal with –
- Here the rate (23.9) is the rate (supposed) to be produced by planets relative
motions between the 3 planets and Pluto – and this rate of time (1 h =23.9 h) be
transported through the planets to cause each planet to define its diameter as a
function in its rotation period – this is the idea we have discussed –
- That explains why the planets distance distribution take into consideration this rate
(23.9) because the diameters should be created function in these distances and the
distances create configuration with the data to cause the function works softly
- Means
- The rate (23.9) is the feedback be sent from the planets diameters functions to the
distances to cause a general configuration and harmony of data.

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- 5-4-2 Jupiter Motion Relative To Pluto Motion
- I- Data
- (a)
- 103944 days x 0.406 million km = 37100 days x 1.1318 million km = 90560 days
x 0.466884 million km = 142984 seconds x 0.3 million km/s (error 2%)
- (b)
- (90560 /4331) = (103944 /4900)
- (c)
- (59800 /16.1) = (90560 /153.3) x 2π
II- Discussion
- Data no. (a)
- 103944 days x 0.406 million km = 37100 days x 1.1318 million km = 90560 days
x 0.466884 million km = 142984 seconds x 0.3 million km/s (error 2%)
- Where
- 103994 hours = 4331 days = Jupiter Orbital Period
- 90560 days = Pluto Orbital Period
- 37100 million km = Pluto Orbital Circumference
- 0.406 million km = Pluto Velocity Per Solar Day
- 0.466884 million km = Neptune Velocity Per Solar Day
- 1.13184 million km = Jupiter Velocity Per Solar Day
- Data no. (b)
- (90560 /4331) = (103944 /4900)
- 4900 million km = Jupiter Orbital Circumference
- 4331 days = Jupiter Orbital Period
- The previous data shows that – Jupiter motion uses 103944 in comparison with
90560 for Pluto motion – which refers to the rate (1 = 23.9)

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- Data no. (c)
- (59800 /16.1) = (90560 /153.3) x 2π
- Where
- 59800 days =Neptune Orbital Period
- 153.3 hours = Pluto Rotation Period
- 16.1 hours = Neptune Rotation Period
- This data tries to explain how Neptune motion be seen in data no. (a) where
Neptune moves during 90560 days a distance = Pluto motion distance during
103944 days – (where 90560 d =Pluto Orbital Period)
- This is happened because there's a deep interaction between Pluto and Neptune –
and in fact this interaction contains Uranus also – we have to examine this
interaction through the paper discussion.
- Notice no. (1)
- Light (300000 km/s) travels during 103944 seconds a distance = 32200 million km
(error 3%) where 32200 million km = Pluto orbital circumference 37100 – Jupiter
orbital circumference 4900
- I try to prove that a geometrical mechanism be found behind.
- Notice no. (2)
- We don't know yet if the rate of time (1=23.9) is a rate between the 3 planets and
Pluto or between Jupiter and Pluto – we know only this rate is found between the 4
planets and Pluto – till now it's not clear how it works
- And
- The geometrical concept behind the planets motions still needs to be discovered.
- But, why Jupiter is the point of the 3 planets motions effect? what connects Jupiter
with the 3 planets (Mercury- Venus and the moon) – we answer in following..

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- 5-4-3 Jupiter And The 3 Planets Interaction
- I- Data
- (1)
- Mercury moves during its day period a distance = 720.7 million km = Jupiter
Mercury Distance
- (2)
- Venus moves during its orbital period a distance = 680 million km
- (Jupiter Venus Distance =670.4 million km "error 1.4%")
- (3)
- Earth moves during its orbital period a distance = 940 million km
- (Jupiter Earth Distance = 929 million km "error 1.2%")
- Notice
- Jupiter Earth Distance be = 929 million when the 2 planets be on 2 different sides
from the sun.
- Why do the three planets move during their cycles distances = their distances to
Jupiter? This is connected with the total velocities 112.2 km/sec - but how?
- I try to shows a geometrical mechanism be found behind this data which we don’t
know and by that the data be seen for us as puzzles while a geometrical effect be
found behind.
- (4)
- The inner planets orbital circumferences total be (Mercury 360 mkm + Venus 680
mkm + Earth 940 mkm + Mars 1433 mkm + 1433 mkm) = 4900 million km (1%)
- The total = 4900 million km but the distance (1433 mkm) be used 2 times!
- This data we have seen before – it's another data connects the 3 planets with
Jupiter – I try to show we have a reason to suppose that the total velocity (112.2
km/s) works on Jupiter point because Pluto uses 23.9 with Jupiter and the 3 planets
create deep connections with Jupiter.

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- (5)
- 778.6 million km (Jupiter orbital distance) =1.16 million km /s x 670 seconds
- 778.6 million km (Jupiter Mercury distance) =1.16 million km /s x 629 seconds
- 5906 million km (Pluto orbital distance) =1.16 million km/s x 5127seconds
- Where
- 670 million km = Jupiter Venus Distance
- 629 million km = Jupiter Earth Distance
- 5127 million km = Jupiter Pluto Distance
- We know that Jupiter distances depend on 1.16 where we have studied that before
in details – but the 3 inner planets and Pluto distances to Jupiter be the most clear
distances in Jupiter data depend on the (1.16)
- I use different data to show that a connection point must be found on Jupiter
between the 3 inner planets from one side and Pluto from the other side.
- (6)
- 1.1318 million km = 112.2 km/s x 5040 seconds x 2
- Where
- 1.1318 million km = Jupiter Motion Distance Per A Solar Day
- 112.2 km /s = The 3 planets velocities total (47.4 +35+ 29.8)
- 5040 seconds = The Period Mercury needs to make its day =4224 hours
- This more data shows a connection between the 3 planets and Jupiter
II- Discussion
- I put different types of planets data to show that the connection between the 3
inner planets and Jupiter can't be a simple one but it's a deep connection and based
on it huge amount of data be created – which shows its effect on the planets
creation and motion data.
- The difficulty is that - many used geometrical rules be unknown for that we can
catch clearly the geometrical machine behind the data.

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5-5 The 3 Inner Planets effect on Pluto motion
- I- Data
- (1)
- 90560 days = 4222.6 hours x 9.18 x 2.33
- (2)
- ((2x 153.3 x 3600)/ (5040)) =23.9 x 9.18
- (3)
- 90560 =23.9 x 346.6 x 10.9
- (4)
- (406000 km / 88000 km) = (708.7 h/ 153.3 h) = 4.61 =(9.18/2)
II- Discussion
- Data no. (1)
- 90560 days = 4222.6 hours x 9.18 x 2.33
- Where
- 90560 days =Pluto Orbital Period
- 4222.6 hours = Mercury Day Period
- 2.33 = ??
- Mercury velocity (47.4 km/s) = Pluto velocity (4.7 km/s) x 10.08 –because of that
- The rate 2.33 be found between (one day of Pluto and one hour of Mercury)
because 2.33 x 10.08 = 24
- The rate 2.33 be used because we use Pluto orbital period (90560 days) in
comparison with Mercury Day Period (4222.6 hours)
- The data tries to show that the rate (9.18) is the basic one between the 3 inner
planets and Pluto
- Notice /
- we have studied this rate (9.18) in the beginning of this discussion and we still use
it because it's the basic one between the 3 planets and Pluto

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- Data no. (2)
- ((2x 153.3 x 3600)/ (5040)) =23.9 x 9.18
- Where
- 153.3 hours = Pluto rotation period = Pluto day period
- 5040 seconds = a period be required by Mercury day period to be =4224 hours
- The data shows that, the same rate (9.18) be used between different data of the
same planets – that tells a geometrical mechanism be found behind this rate and
effect on these planets data.
- Data no. (3)
- 90560 =23.9 x 346.6 x 10.9
- Where
- 346.6 days = The Nodal Year
- 10.9 =??
- The moon orbital apogee radius should be 413600 km but it decreased and be only
406000 km where 406000 km = 413600 km x cos (10.9 degrees)
- It's a complex data – later we will see more shared data be used by the moon and
Pluto supports this one – and also we should discuss how the moon orbital apogee
radius be 406000 km – our current data here only aims to show that the rate (23.9)
be found between Pluto orbital period and the nodal year.
- Data no. (4)
- (406000 km / 88000 km) = (708.7 h/ 153.3 h) = 4.61 =(9.18/2)
- 406000 km = Pluto motion distance during solar day
- 88000 km = The Moon Displacement during solar day
- 708.7 hours = the moon day period
- 153.3 hours = Pluto day period =Pluto rotation period
- This data be discussed in the next point no.(5-6)

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5-6 The Moon And Pluto Motions Data Consistency
I- Data
- (A)
- 5906 mkm (Pluto orbital distance) = 940 mkm (Earth orbital circumference) x 6.3
- 153.3 hours (Pluto day period) =24hours (Earth Day Period) x6.3 (error 1.4%)
- 90560 days (Pluto orbital period) = 1461 days (Earth Cycle) x 6.3 x π2
- (B)
- Pluto (4.7 km/s) moves during a solar day = 406000 km = apogee radius
- Pluto (4.7 km/s) moves during Pluto day 153.3 h = 2.5986 km = the moon
displacements Total during 29.53 days
- Earth moves during a solar day =2.574 mkm is different with 2.598 mkm by 1%
- (C)
- 406000 km (Pluto motion daily) / (88000 km the moon displacement) = 4.61
- (708.7h the moon day period /153.3h Pluto day period) = 4.61
- (D)
- Pluto day be created as a function in the moon cycles – the data proves that -
- Tan (12.19) x 708.7 hours = 153.3 hours (708.7 h = the moon day period)
- Tan (13.17) x 655.7 hours = 153.3 hours (655.7 h = the moon rotation period)
- (10.96 deg /1.7 deg) = (153.3 h /24 h) (error 1%)
- 13.177 degrees = The Moon Daily Motion Degrees
- 12.19 degrees = 13.177 degrees – 0.9856 degrees (Earth motion daily degrees)
- (E)
- Pluto (4.7 km/s) moves during 88000 seconds a distance 413600 km
- Pluto (4.7 km/s) moves during 10.7 h (Saturn day period) a distance 181800 km
- (F)
- (708.7 h /10.7 h)= (655.7 h/ 9.9 h)= (224.7/3.4)= (2 x 153.32
h)/708.7 h = 66.2
- 10.7 h = Saturn Day Period
- 9.9 h = Jupiter Day Period
- 153.3 h = Pluto Day Period

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- 224.7 days = Venus Orbital Period
- 3.4 degrees = Venus Orbital Inclination
- Notice No. 1
- 17.4 degrees = The Inner Planets Orbital Inclinations Total
- 17.2 degrees = Pluto Orbital Inclination (error 1%)
- 23.6 degrees = The Outer Planets Orbital Inclinations Total
- 23.4 degrees = Earth Axial Tilts (error 1%)
- 17.4 degrees = 5.1 deg (the Moo Orbital Inclination) x 3.4 deg (Venus Orbital
Inclination)
- Notice No. 2
- Pluto (4.7 km/s) moves during a solar day a distance =406000 km
- Pluto (4.7 km/s) moves during a Pluto day period a distance = 2.598 mkm
- Pluto (4.7 km/s) moves during 88000 seconds = 413600 km
- Pluto (4.7 km/s) moves during 10.7 h (Saturn day period) = 181800 km
- All distances be used in the moon orbital motion data
II- Discussion
- The data gives a sense for the idea – and tells – some concrete base be found under
it – because – it's not one data be in consistency between the 2 planets (the moon
and Earth on one side and Pluto on the other side) but almost all planets data be in
consistency…
- In fact - we can conclude all Pluto data based on the moon and Earth data – if
there's no a geometrical reason behind we would lose any logical thinking here –
- The data shows the fact clearly and tells that based on a geometrical reason the
data of (the Earth and moon) on one side be connected with Pluto data on the other
side.

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A Comment Of The Previous Discussion (No.1)
- The previous 6 points of the discussion tried to show that- A Geometrical Effect be
created on the moon and passes through the planets data – where this effect causes
each planet diameter to be created as a function in its rotation period as the
equation proves clearly.
- This geometrical effect be the rate of time (1 hour = 23.9 hours) which be found
between 2 points in the solar system (almost between Jupiter and Pluto) as the data
shows.
- But, the fact is that,
- I can't catch clearly this geometrical effect – and the data shows its effect but its
nature isn't discovered yet also we don't know how the data be transported through
the planets – the only available solution depends on light motion which we have
discussed in the previous paper –please review the paper references.
- I want to say
- In the next (3) points we will see one more effect of this rate of time (or this
geometrical effect which be found behind this rate of time) – because –
- The moon orbit geometrical structure depends on this effect – by that – we will see
many puzzled changes be found in the moon orbit where no available explanation
for these changes except by this rate of time geometrical effect – we still don't
know how this effect can cause these changes but no any source of effect be found
in the moon orbit except this one and we have to suppose that it’s the reason –
- Shortly
- The effects in the data which we have discussed in the previous 6 points will be
seen in geometrical effects and works be found inside the moon orbit where the
moon orbital motion depends on these effects which tells us that the equation uses
the moon as its base and forced the moon to create its orbit in a specific
geometrical design to be suitable for this equation – let's prove that in following..

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5-7 The Outer Planets Diameters Total Effect
I-Data
(1)
366556 km = 2390 km x 153.3 = 3475 km x (655.7/2π)
(2)
Saturn (9.7 km/s) moves during its rotation period a distance =373644 km
(3)
366556 km = 15327 x 23.9
II- Discussion
Data No. (1)
366556 km = 2390 km x 153.3 = 3475 km x (655.7/2π)
Where
366556 km = The Outer Planets Diameters Total
2390 km = Pluto Diameter
3475 km = The Moon Diameter
153.3 hours = Pluto rotation period
655.7 hours = the moon rotation period
Simply the planet diameter relative to The Outer Planets Diameters Total produces a
rate can be used as this planet rotation period – of course a geometrical mechanism be
found behind this data – we just don't know how the geometrical rules work -
I want to say – the rate of time (1 =23.9) is a reference to this geometrical effect
which passes through the planets data and causes their diameters be function in their
rotation periods – The rate of time is just an indicator for this geometrical effect –
because it controls a wide range of planets data but the geometrical rules and effects
are unknown for us.
Data No. (2)
Saturn (9.7 km/s) moves during its rotation period a distance =373644 km

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The data shows Saturn also has a connection with this distance 366556 km because
the distance 373644 km is different from 366556 km with (2%) while the distance
373644 km is different from (378675 km Saturn Circumference) with (1.3%)
There's a geometrical machine behind this data because Saturn is the third player in
the equation with the moon and Pluto – by that - Saturn also has a connection with
the outer planets diameters total 366556 km – but the used geometrical rule2 is
unknown.
Data No. (3)
366556 km = 15327 x 23.9
Where
366556 km = The Outer Planets Diameters Total
15327 km = Mercury Circumference
The data shows the rate (23.9) one more time proves that Mercury is a player in the
equation…
I wish we can extend our thinking because the rates of the diameters be used as
periods of time (rotation periods) and this using refers to unknown geometrical effect
for us-
I want to say- the geometrical machine works and produces its results and we can't
catch these results because we don't know the used geometrical rules - for that – we
see puzzled data which be created based on a geometrical machine
For example - the simple question –
Why planet rotation period can be defined as a rate between its diameter and the outer
planets diameters total? what's the great geometrical effect of this outer planets
diameters total?
Notice
We have to keep in mind this number 366556 km because it's a main distance in the
moon orbit geometrical structure.

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5-8 The Moon Orbit Geometrical Structure
I- Data
(1)
366556 km = The Outer Planets Diameters Total = 3475 km +363080 km
40080 km = The Inner Planets diameters total = the Earth Circumference
406000 km = The Planets Diameters Total
(Notice/ Pluto Moves During A Solar Day A Distance = 406000 km)
(2)
17.4 degrees = The Inner Planets Orbital Inclinations Total
17.2 degrees = Pluto Orbital Inclination (error 1%)
23.6 degrees = The Outer Planets Orbital Inclinations Total
23.4 degrees = Earth Axial Tilts (error 1%)
17.4 degrees = 5.1 deg (the Moo Orbital Inclination) x 3.4 deg (Venus Orbital
Inclination)
17.2 degrees = 2 x 5.1 deg (the Moo Orbital Inclination) +7 deg.
7 degrees = 5.1 deg (the Moo Orbital Inclination) + 1.9 deg (Mars. Orb. Inclin)
II- Discussion
Data No. (1)
This data tells – if the moon be on its perigee radius (363000 km), the distance from
the Earth to the moon includes the moon diameter be = 366556 km= the outer
diameters total – in this case the rest distance after the moon diameter to its apogee
radius be (406000 km – 366556 km = 39500 km)
The distance (39500 km) be different with 1% from the inner planets diameters total =
40080 km = the Earth Circumference
That tells the moon orbit is divided into 3 parts according to the planets diameters
total – the apogee radius refer to all planets diameters total and perigee radius refers to
the outer planets diameters total and the distance between them refers to the inner
planets diameters total – of course there's a geometrical reason behind

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The point is that,
The data (366556 / the moon diameter 3475 km) = (655.7 /2π)
(655.7 hours = the moon rotation period)
This data refers to some geometrical machine which effects on the moon orbital
radiuses (perigee and apogee)
It's so important result because the moon could change its orbital radiuses if its
rotation period be changed or its diameter be changed – we here very near to the
meaning be suggested from the equation – but how the geometrical effect be done?
What force caused this effect to be real and effect on the moon orbital radiuses?
Data No. (2)
The data tells that
Pluto orbital inclination (17.2 degrees), the moon orbital inclination (5.1 degrees) and
the inner planets orbital inclination total (17.4 degrees) – these values be created by
one machine of data –
We can't see separated values or data from one another – we see one stream of data be
created by one force for one process and for that the data be in harmony with one
another
These are summarized data –I have tried to show that the moon orbit geometrical
structure be created based on this geometrical effect which be found behind the
suggested rate of time we have discussed as a reason for the equation –
Shortly
We deal with a great machine and it effects greatly on the planets motions and data -
In the next point (No. 5-9) we examine the moon motion in more details to show that
more data of the moon motion be in harmony with this geometrical effect which is
covered by the suggested rate of time.

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5-9 Why Does The Moon Apogee Orbital Radius =406000 Km?
(1)
The Planets Diameters Total = 406000 km
Pluto moves in a solar day a distance = 406000 km
The moon orbital apogee radius = 406000km
(2)
Earth moves during its day period (24 hours) a distance = 2.574 million km
Pluto moves during its day period (153.3 hours) a distance = 2.59 million km
The Earth moon displacements total during 29.53 days = 2.59 million km
The 3 distances are equal (error 1%)
Why the 3 planets move equal distances in their days periods?!
(3)
The Moon Orbital Motion
The moon daily displacement =88000 km and during 29.53 days the displacements
total be = 2.598 million km = 2π x 413600 km
The data tells the moon orbital apogee radius should be 413600 km and
The moon daily displacement (88000 km) is long, because of that, the moon should be
prisoner in the orbit with radius (= 413600 km) and the moon can't revolve around the
Earth through any more near orbit!
Not facts – The moon orbital apogee radius =406000 km and the moon revolves
around the Earth through near orbits and can reach to perigee radius (363000 km).
How Can The Moon Do That?
The intelligent moon creates an angle (θ) between its motion direction and its orbit
horizontal level by that the real displacement (L) through the orbit be less than (88000
km) because it be (L = 88000 km cos θ), as a result the total displacements be less
than (2.598 million km) and that makes the moon orbital apogee radius to be
decreased from 413600 km to 406000 km.

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We should pay attention to the angle (θ), because this angle controls the moon motion
features – where- with the angle (θ) increasing the real displacement (L) be shorter
and the moon can revolve around the Earth through more near orbits – but –with the
angle (θ) deceasing the real displacement (L) be longer and that pushes the moon far
from the Earth to more far orbits.
The moon orbital motion depends on this angle (θ) it tells θ1 = θ0 +1.7
where (θ1) = today angle and (θ0) =yesterday angle
Now, one more question be raised, why the moon apogee radius be 406000 km? why
not shorter if the moon uses this technique which enable the moon to decrease its
orbital apogee radius as possible? Why specifically the radius 406000 km be chosen?
Because 406000 km = The Planets Diameters Total
We still be connected with my fourth equation – and the geometrical effect based on
which the planets diameters be created as functions in their rotation periods
Shortly
The orbital apogee radius 406000 km be defined by effect of this equation on the
moon motion
(4)
Why Does The Moon Move Daily A Displacement = 88000 km?
- The moon moves with The Earth and by its Earth velocity daily, this fact we know
because the Earth and the moon move together and don't separate each other
through the motions course revolving around the sun.
- Means,
- The moon moves per solar day a distance = Earth motion distance per solar
day = 2.574 million km.
- But
- The moon distance (2.574 mkm) be contracted by the rate 1.0725 and for that this
distance 2.574 mkm be 2.4 mkm

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- Now the moon difficulties be started, because, the difference 176000 km will
cause the moon to be separated from the Earth in their motions course
- For that reason the moon moves a displacement =88000 km (50%) depending on
the Earth gravity –
- We notice that, we don't see the moon motion for the distance 2.4 mkm neither the
original one 2.574 mkm, we see only the moon displacement 88000 km in the
Earth sky –
- The question we need to solve is that, why the moon doesn't separate from the
Earth if the different distance be 176000 km and the moon moves only 88000 km?
how the rest (88000 km) be adjusted?
- This story is a complex one and we need to move step by step to catch the idea
behind
- Firstly, how the rest distance (88000 km) be adjusted? This question answer be
provided by the generous Mercury, because Mercury is the basic helper behind the
moon motion - any way – Mercury uses very strange language for us – where the
moon displacement be 88000 km Mercury sees it as (88 days = Mercury orbital
period) and while the required distance is 176000 km, (Mercury day period =176
days approximately) – means – Mercury is our hope now –
- And
- This rate (1.0725) we have discussed frequently before –it's the basic rate in the
solar system and 40% of all distances in the solar system be rated by it and around
50% of all planets axial tilts be rated by it and also this is the rate between the
moon and Venus cycles (29.53 /27.3) = (243/224.7) =1.0725
- This rate (1.0725) is the reason to decrease the moon distance from 2.574 million
km to 2.4 million km.
- The data tells a deep geometrical machine be found between the 3 planets
(mercury, Venus and the moon) and by that not only the 2 planets cycles support
the moon cycles but even the 3 planets almost integrate their motions into one
unified motion.

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(5)
- Metonic Cycle
- Why does the rotate Metonic Cycle (19 years)?
- Because of Uranus motion effect on the moon motion – we know that because
- Uranus Orbital Distance =19.2 Earth Orbital Distance - means
- If Uranus and Earth velocities be equal, while Uranus revolves around the sun one
revolution Earth would revolve 19 revolutions (19 years)
- That's why I have suggested Uranus effect is the reason of the moon Metonic
Cycle
- In fact Uranus has 3 effects which are
- (a)
- Uranus motion effect on the Earth moon to rotate Metonic Cycle –and by this
effect - Uranus cause the moon orbital inclination to be 5.1 degrees and this
inclination causes to decrease the moon orbital apogee radius from 413600 km to
406000 km
- (b)
- Uranus effects on Pluto motion and causes Pluto rotation period to be (153.3 h)
- We know that because Uranus rotation period =17.2 hours and Pluto orbital
inclination be =17.2 degrees – the geometrical rule is unknown but Uranus is the
reason.
- Also Uranus uses Pluto rotation period in (Planet 8 days cycle) –we discuss it with
Saturn effect analysis (Point no.5-10)
- (c)
- Uranus axial tilt effects on all planets axial tilts and prevent the overturning motion
of any planet around the sun.
- This effect shows why Earth and Uranus use Neptune velocity in their Equations
(in my fourth Equation) – because Uranus guides all planets motions directions
and Uranus is one column of the velocity map 2 columns – where Earth is the
central point of the solar system motion

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- Notice (1)
- (153.3/ 53.9) = (1.16/0.406)
- Where
- 153.3 hours = Pluto Rotation Periods
- 53.9 hours = The 4 Outer Planets Rotations Periods Total
- 406000 km = Pluto motion distance per solar day
- 1160000 km = light supposed velocity
- Notice (2)
- I wish the (new) type of motion which be done by Planets be seen by us – because
- Uranus effect on Pluto motion to increase its rotation period (and day period) to be
(153.3 hours) but Pluto moves during this period (153.3 h) a distance= 2.95 million
km = The moon displacements total during its day period (29.53 days) = Earth
motion distance during its day period (24 h) (error 1%)
- Here Uranus caused Pluto rotation period to be (153.3 h) for a geometrical reason
by which the planets move equal distances – but we understand nothing! Because
we don't realize any produced effect by the motions equal distances - Because we
don't know the produced geometrical effect we don't understand the whole process.
- It's a type of motion - the planet does- (which be unknown for us)
- We see only planet revolution around the sun but planet does many different other
motions which we don't observe nor understand because their geometrical effects
are unknown for us –

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A Comment Of The Previous Discussion (No.
A Comment Of The Previous Discussion (No.2
2
2
2)
)
)
)
- The discussion of the previous points (5-7, 5-8 and 5-9) give us 2 clear information
which are
- (1)
- The used geometrical rules to create the moon orbit are unknown
- (2)
- The rate of time (1 =23.9) coves a great geometrical effect – and the moon orbit be
created in comparison with the data be provided by this rate of time
- The moon orbit should be a proof for this rate existence and geometrical effect on
the solar system creation.
- But
- As long as we don't know the effect of the diameters rate on the rotation periods
and how to distinguish between the distance and planets diameters we can't reach
to a clear meaning in this discussion
- The curiosity here comes form Lorentz –
- Because Lorentz equation for length contraction effect doesn't distinguish between
particle length and a distance – both can be contracted by high velocity motion
- Here also
- The diameters values be used as distances and we can't catch the geometrical
reason behind.

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5-10 Saturn Effect Analysis
- Let's remember the equation basic points
- The equation depends on 3 points (The Moon –Saturn –Pluto)
- For The Moon
- The periods are equal because
- The moon orbital period = the moon rotation period =27.3 days
- For Saturn
- The velocities are equal because
- Saturn Orbital Velocity = Saturn Rotational Velocity
- Saturn is the an unique planet in the solar system which uses its orbital velocity to
be = its rotational velocity – and because of that –
- Saturn (9.7 km/s) moves during its rotation period (10.7 hours) a distance =
373644 km = Saturn circumference 378675 km (error 1.3%)
- For Pluto
- We supposed the distances be equal –we should discuss that in Pluto discussion.
- But, even if, this vision is a correct one and the periods be equal on the moon and
the velocities be equal on Saturn and the distances be equal on Pluto – how does
the Equation work? why do these equalities effect on the equation work?
- I try to show we still need to deepen our discussion for better understanding
- Now, let's ask, what's Saturn effect on the equation? Let's see this data
- Saturn moves in its rotation period a distance = its circumference (error 1.3%)
- Jupiter moves in its rotation period a distance = its circumference (error 4%)
- Uranus moves in its rotation period a distance = 2.6 x its circumference
- Neptune moves in its rotation period a distance = 2 x its circumference

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- These 4 planets data refer to some geometrical effect behind – this data can’t be
created by any random process – a geometrical effect be found behind and caused
this data.
- This data discussion can be more clear with (Planet 8 Days Cycle) we discuss in
point no. (5-12)

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5-11 The Moon And Saturn Motions Data Consistency
Let's summarize the idea in following…
Because The Moon Motion Be The Equation Base And Saturn Motion Be The Central
Point In This Equation. The 2 Planets Motions Data be in deep harmony and even
they use identical data- let's refer to them in following
(1)
The moon orbital radius in the total solar eclipse be = 373644 km = Saturn motion
distance during its rotation period (10.7 h) where (Saturn Circumference =378675 km
-error 1.3%)
That makes Saturn a goal for all proportionality of dimensions based on w1hich the
total solar eclipse be created – as example
(a)
(The sun diameter /the moon diameter) =(Earth orbital distance /Earth moon distance)
Based on this equality we see the sun = the moon disc and that enable the total solar
eclipse to be occurred – Earth moon distance here refers to Saturn data
(b)
Earth orbital distance 149.6 million km = The sun diameter 1.392 million km x 109
The sun diameter 1.392 million km = the Earth diameter x 12756 km x 109
The Earth moon distance (373644 km) = the moon diameter 3475 km x 109
The sun diameter / the Earth moon distance = Earth diameter / the moon diameter
(2)
Saturn orbital period =10747 days - and
The moon displacements total during 10747 days = 940 million km = Earth orbital
circumference
By that the 2 planets use the period 10747 days.

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- More Data
- (3)
- 10747 days x 2.4 mkm/ day = 25920 million km
- (4)
- 17.2 x 3600 x 0.838 mkm/ day = 2 x 25920 million km
- 17.2 x 3600 x 88000 km/day = 2 x 2723 million km
- (5)
- The values 10.7 and 10.9
- (6)
Discussion
- The data proves that– A deep harmony of the motion data be found between Saturn
and the moon – let's examine the data in following…
- Data No. 3
- It's interesting data - because – light 300000 km/s moves during a solar day
(86400 s) a distance =25920 million km
- The moon moves equal distance during Saturn orbital period – here we have more
than one interesting data – because
- The moon displacement per solar day be =88000 km and during Saturn orbital
period 10747 days the displacements total be 940 million km (Earth Orbital
Circumference).
- But – we know that – the moon moves daily a distance = Earth motion distance
daily because they aren't separated from one another and by that we know the
moon moves per solar day 2.574 million km where the rate (1.0725) effects on this
distance and contracted it to be (2.4 million km) by this last distance per solar day

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the moon moves during Saturn orbital period (10747 days) a distance = Light
motion distance in solar day.
- It's of course interesting data -
- Data No. 4
- 17.2 x 3600 x 0.838 mkm/ day = 2 x 25920 million km
- 17.2 x 3600 x 88000 km/day = 2 x 2723 million km
- This data uses Uranus day period (17.2 x 3600 = 61920 seconds),
- If the 2 planets (Saturn and the moon) use this value in days units (61920 days) in
this case
- Saturn will pass a distance = 2 x 25920 million km and
- The Moon will pass a distance = 2 x 2723 million km and
- Where
- 25920 million km = light (300000 km/s) motion distance during a solar day
- 2723 million km = Uranus Earth Distance
- Notice
- This data is a complex one because it tells during Uranus day period (61920
seconds) Saturn moves a distance = 600000 km
- But Uranus moves equal distance (600000 km) in a period =24.6 hours = Mars
rotation period -But
- Jupiter moves in 24.6 hours (Mars rotation period) a distance = 1.16 million km =
light supposed motion distance in one second.
- I can't catch the geometrical machine behind but this data be created based on one
another.
- Data No. 5
- The values 10.7 and 10.9
- The period 10.7 hours = Saturn oration period = Saturn day period

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- But, What's 10.9??
- The moon orbital apogee radius should be 413600 km (because the moon
displacements total during 29.53 days be 88000 km x 29.53 = 2.59 mkm= 2π x
413600 km)- as we have discussed before.
- Where 413600 km cos (10.9) = 406000 km.
- By that this 2 values (10.7 and 10.9) are different with 2% which tells they be rated
to each other – it may be the reason to decrease the moon orbital apogee radius
with 2%.
- Data No. 6
- This data tells, Saturn moves the distance 100733 million km in a period = 120536
days where
- 100733 million km = The Planets Orbital Circumferences Total
- 120536 km = Saturn Diameter

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5-12 Planet 8 Days Cycle
Planet 8 days cycle is a cycle I have discovered found between the 4 planets (Jupiter,
Saturn, Uranus and Neptune) where Uranus motion uses Pluto day period (= Pluto
rotation period)
The cycle depends almost on 4% found as a difference between Jupiter motion
distance during its rotation period and Jupiter circumference – the cycle describe the
data behavior without explanation for this behavior reason –
The cycle proves that – the 4planets moves as one machine of gears where their
motions be integrated with one another.
(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) (96%) + 17687 km (4%)
- 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

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- (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
- 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)

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- 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
- Pluto during its day period (153.3 hours) moves a distance = the Earth moon
displacements total during 29.53 days (the moon day period) = Earth motion
distance during earth day (24 hours) (error 1%) which shows Uranus caused Pluto
day period to be =153.3 hours for a geometrical necessity and reason.
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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A Comment Of The Previous Discussion (No.3
3
3
3)
)
)
)
- The discussion of the previous points (5-10, 5-11 and 5-12) shows that
- The equality of Saturn orbital velocity and rotational velocity is a result depends
on a geometrical machine – where a continuous geometrical effect be seen on the
outer planets data-
- Means
- A geometrical effect be passed through the planets data and be seen by the equality
of Saturn orbital velocity and rotational velocity – this continuous geometrical
effect can be seen in (Planet 8 days cycle)
- This vision supports the equation concept tells (The equation depends on a
geometrical effect be passed through Planets data and forces each planet to create
its diameter as a function in its rotation period)
- In our discussion we have examined Saturn motion and have discovered this
geometrical effect based on which Saturn orbital velocity be = Saturn rotational
velocity.
- That may explain why the outer planets diameters total be a player in Pluto and the
moon rotation periods definition as rates to their diameters – here we deal with a
great geometrical machine contains the outer planets –one feature of this machine
we have discovered which is (Planet 8 days cycles)
- This discussion should be added to Neptune and Pluto data consistency (point
no.5-14) which shows more features for this same machine behind the outer
planets data.

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5-13 Pluto Effect Analysis
As we have discussed –My fourth equation depends on 3 points (the moon –Saturn
and Pluto)
For The Moon – The Periods Are Equation Because
(The Moon Orbital Period = The Moon Rotation Period =27.3 days)
For Saturn – The Velocities Are Equal – Because
(Saturn Orbital Velocity = Saturn Rotational Velocity)
For Pluto – The Distances Are Equal – Because
what are the equal distances? Let's see the data
I-Data
(1)
The 3 distances are equal (error 1%)
Why the 3 planets move equal distances in their days periods?!
(2)
2.59 million km = 346.6 x 7510 km
(3)
Pluto orbital circumference = 7510 km x 4.94 million km
(4)
Pluto (4.7 km/s) moves during 88000 seconds a distance =413600 km
II-Discussion
What's the distances equality of Pluto? in the moon and Saturn the data related to the
orbital and rotation – at the distances that will be – Pluto circumference and its orbital
circumference – where the rate between both = 4.94 million

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But
Pluto moves during its rotation period a distance = 2.59 million km =346.6 x 7510 km
(Pluto Circumference)
346.6 days = The Nodal Years
That may tell Pluto motion effects on the moon orbit and causes its regression yearly
and the nodal year creation
The basic secret is found in the first data
Data No. (1)
The question is (What's The Result Of These Distances Equality?)
What's the geometrical effect we produce if 2 planets move equal distances in their
rotation periods
Here's the secret which can add a new book of geometry to the physics library.
The point is that
The 3 planets velocities total be 112.2 km/s = 4.7 km/s (Pluto velocity) x 23.9
By that one hour of a planet = 23.9 hours of Pluto
But
153.3 hours /23.9 hours = 2π (error 2%)
But
5906 million km (Pluto Orbital Distance) = 940 million km (Earth Orbital
circumference) x 2π
That shows the data be created based on this rate of time – and by that – this rate of
time covers a geometrical effect passes through the planets data to effect on it and
create it by this effect.
Here the motions distances equality shows a great significance

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5-14 Pluto And Neptune Data Consistency
I- Data
(1)
(90560 /153.3) x 2π = (59800/ 16.1)
(2)
Uranus Orbital Distance = 2 x Saturn Orbital Distance
Neptune Orbital Distance = π x Saturn Orbital Distance
Pluto Orbital Distance = (π+1) x Saturn Orbital Distance
(3)
Uranus needs 4900 days to pass a distance = Uranus Orbital Distance
Neptune needs 2 x 4900 days to pass a distance = Neptune Orbital Distance
Pluto needs 3 x 4900 days to pass a distance = Pluto Orbital Distance
Max error 2%
(4)
406000 km = Pluto velocity (4.7 km/s) x 86400 seconds
406000 km = Uranus velocity (6.8 km/s) x 59800 seconds
(5)
Neptune (5.4 km/s) moves during 155597 seconds a distance = 2 x 421056 km
II- Discussion
The data tries to prove that a deep interaction be found between Uranus, Neptune and
Pluto motions data. let's refer to each data in details
Data No. (1)
(90560 /153.3) x 2π = (59800/ 16.1)
Where
90560 days = Pluto Orbital Period
59800 days = Neptune Orbital Period
153.3 hours = Pluto rotation period
16.1 hours = Neptune rotation period

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Data no. (4)
406000 km = Pluto velocity (4.7 km/s) x 86400 seconds
406000 km = Uranus velocity (6.8 km/s) x 59800 seconds
406000km= the moon orbital apogee radius = the planets diameters total = Pluto
motion distance in a solar day period.
Also
Uranus (6.8 km/s) moves during 59800 sec a distance =406000 km
Where
59800 days = Neptune orbital period
Data no. (5)
Neptune (5.4 km/s) moves during 155597 seconds a distance = 2 x 421056 km
Where
421056 km = Uranus Motion Distance During Uranus Day Period
155597 km = Neptune Circumference

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5-15 Jupiter And The Moon Data Consistency
I- Data
(1)
Jupiter (13.1km/s) moves during 10921 s a distance = 142984 km = Jupiter diameter
Where the moon circumference =10921 km (where 1 km= 1 sec)
(2)
4900 million km = Jupiter Circumference (449197 km) x 10921 km
4900 million km = The Sun Diameter x The Moon Diameter
(4900 million km = Jupiter Orbital Circumference)
And
(Jupiter diameter x π2
= the sun diameter – error 1.4%)
(3)
Jupiter orbital distance 778.6 million km = Earth orbital distance 149.6 million km x 5.2
(5.1 degrees = The Moon Orbital Inclination)
(4)
The moon orbital apogee circumference should be 2.59 million but the fact it be
2550973km the difference = 47720km= Jupiter motion during one hour.
The data tries to prove that a deep interaction and consistency be found between the
moon and Jupiter motion data.

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5-16 The Equation Units Analysis
Let's review the Equation in following
Planet Diameter Definition Equation (My Fourth Equation)
v = Planet Velocity
r= Planet Diameter
s= Planet Rotation Periods Number In Its Orbital Period
I= Planet Orbital Inclination (a rate to inclination unit)
v2, s, r and I be belonged to one planet and v1 be belonged to another planet
The planet (v1) be defined by test the minimum error
- Earth Equation uses Neptune velocity
- Mars Equation uses Pluto velocity
- Jupiter Equation uses the Earth moon velocity
- Saturn Equation uses Mars velocity
- Uranus Equation uses Neptune velocity (As Earth)
- Neptune Equation uses Saturn velocity
- Pluto Equation uses the Earth moon velocity (As Jupiter)
A Question
Are The Equation Units Be In Harmony?!
Not because
(s/r) a rate can't be understandable where
s= planet rotation period number in its orbital period
r= uses km units
for example for Jupiter s =10500 and its units be 10500 Jupiter rotation periods
(Jupiter rotation period =9.9 hours) but
412984 km = Jupiter diameter - how to create a harmony for these units?
Planet rotation period should be considered = one second
I
r
s
v
v
=
=
2
1

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Jupiter rotation period 9.9 hours and we have to consider this 9.9 hours = 1 second
And
Planet diameter value be used in second units
Based on that
(s) for Jupiter =10500 Jupiter rotation period will be used as 10500 seconds
And
142984 km will be used as 142984 seconds
By that the rate (s/r) can find a harmony for its units.
Example – Jupiter diameter be 142984 km be used as 142984 seconds
How this data can be used by these units?!
Here we have 2 questions
How planet diameter can be used as a period of time?
This question is answered in point no. (5-17)
And
How planet rotation period can be used as (1 second)?
This question be answered in my previous paper which discusses light motion effect
on planet motion – please review it
Planet Motion Logic Disproves Newton Theory Of The Sun Gravity
https://www.academia.edu/83621274/Planet_Motion_Logic_Disproves_Newton_Theory_Of_The_Sun_Gravity
or
https://app.box.com/s/oap9ssvbaf9ikqcyuqy13wl2xpze2lp3

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5-17 Planet Diameter Analysis
I- Data
(1)
- Jupiter (13.1 km/s) moves during 10921 seconds a distance = 142984 km = Jupiter
diameter- (10921 km = the Earth moon circumference)
(2)
- Pluto (4.7 km/s) moves during (10921 s) a distance = 51118 km = Uranus diameter
- Pluto (4.7 km/s) moves during (51118 s) a distance = 2 x 120536 km = Saturn
diameter.
- Pluto (4.7 km/s) moves during (2 x 120536 s) a distance = 1.13184 million km =
Jupiter motion distance per a solar day
(3)
- Uranus (6.8 km/s) moves during 7510 seconds a distance =51118 km = Uranus
diameter- (7510 km = Pluto Circumference)
(4)
- Venus (35 km/s) moves during 12104 seconds a distance =421056 km = Uranus
motion distance during Uranus rotation period -
- (12104 km = Venus diameter)
- Also
- Venus (3.024 million km per solar day) moves during 12104 days a distance = 2 x
18048 million km = Uranus Orbital Circumference) (error 1.4%)
II- Discussion
- The data tells –planet diameter and circumference can be used as period of time –
- It's not clear how or why – but the data shows the facts
- This feature may be found because planet diameter be created as a function in its
rotation period and the designer needed to create a measurement to diameter
definition – in all cases the type of motion is unknown.

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Appendix no. (1) (My 3 Equations)
I have discovered 5 equations by which we can conclude planets data theoretically –
the paper discusses my first and fourth equations and in this appendix I provide the
tests of my rest 3 equations.
Planet Velocity Equation (My 2nd
Equation)
- V = A Planet Velocity
- V0= Its Neighbor Planet Velocity
- The equation depends on the planets order, means, just 2 neighbor planets can be
used in this equation, So if (d is Venus distance, d0 be Mercury distance)
- The equation exceptions are, Earth depends on Mercury Not Venus – and Mars
depends on Venus Not Earth And Pluto depends on Uranus Not Neptune.
- The equation system is very similar to my first equation system (Planet Orbital
Distance Equation)
(1) Venus Velocity
- (V0)2
/ (V)2
= 1.834
- (V)2
/ (V0)2
= 0.5452
- 4 (1- 0.5452) = 1.819
- (V0) = 47.4 km /s = Mercury Velocity
- (V) = 35 km /s = Venus Velocity
- Venus Depends On Mercury (The values 1.834 and 1.819 error 1%)
(2) Earth Velocity
- (V0)2
/ (V)2
= 2.53
- (V)2
/ (V0)2
= 0.39525
- 4 (1- 0.3952) = 2.4189
- (V0) = 47.4 km /s = Mercury Velocity
- (V) = 29.8 km /s = Venus Velocity
- Earth Depends On Mercury (The values 2.418 and 2.53 error 4 %)
)
1
(
4 2
0
2
2
2
0
V
V
V
V
−
=

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(3) Mars Velocity
- (V0)2
/ (V)2
= 2.1091
- (V)2
/ (V0)2
= 0.47413
- 4 (1- 0.47413) = 2.1034
- (V0) = 35 km /s = Venus Velocity
- (V) = 24.1 km /s = Mars Velocity
- Mars Depends On Venus (The values 2.109 and 2.103 No Error)
(4) Ceres Velocity
- (V0)2
/ (V)2
= 1.812
- (V)2
/ (V0)2
= 0.55166
- 4 (1- 0.55166) = 1.793
- (V0) = 24.1 km /s = Mars Velocity
- (V) = 17.9 km /s = Ceres Velocity
- Ceres Depends On Mars (The values 1.81 and 1.79 Error 1%)
(5) Jupiter Velocity
- (V0)2
/ (V)2
= 1.867
- (V)2
/ (V0)2
= 0.53559
- 4 (1- 0.53559) = 1.857
- (V0) = 17.9 km /s = Ceres Velocity
- (V) = 13.1 km /s = Jupiter Velocity
- Jupiter Depends On Ceres (The values 1.867 and 1.857 NO Error)
(6) Saturn Velocity
- (V0)2
/ (V)2
= 1.8238
- (V)2
/ (V0)2
= 0.548278
- 4 (1- 0.548278) = 1.80688
- (V0) = 13.1 km /s = Jupiter Velocity
- (V) = 9.7 km /s = Saturn Velocity
- Saturn Depends On Jupiter (The values 1.82 and 1.806 Error 1%)

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(7) Uranus Velocity
- (V0)2
/ (V)2
= 2.034818
- (V)2
/ (V0)2
= 0.49144
- 4 (1- 0.49144) = 2.0342
- (V0) = 9.7 km /s = Saturn Velocity
- (V) = 6.8 km /s = Uranus Velocity
- Uranus Depends On Saturn (The values 2.034 and 2.0342 NO Error)
(8) Neptune Velocity
- (V0)2
/ (V)2
= 1.5857
- (V)2
/ (V0)2
= 0.63062
- 4 (1- 0.63062) = 1.477
- (V0) = 6.8 km /s = Uranus Velocity
- (V) = 5.4 km /s = Neptune Velocity
- Neptune Depends On Uranus (The values 1.585 and 1.477 Error 7%)
(9) Pluto Velocity
- (V0)2
/ (V)2
= 2.093
- (V)2
/ (V0)2
= 0.4777
- 4 (1- 0.4777) = 2.089
- (V0) = 6.8 km /s = Uranus Velocity
- (V) = 4.7 km /s = Pluto Velocity
- Pluto Depends On Uranus (The values 2.093 and 2.089 NO Error)
- Notice
- The equation errors are (Neptune 7%), and (Earth 4%) but all other planets errors
are less than 1% -
- Ceres Orbital Distance =414 Million Km And Its Orbital Period 1680 Days (its
velocity 17.9 km/s)

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- The Equation Discussion
- My first and second equations depend on planets order –means- Just 2 neighbors
planets can be used in these 2 equations – the 2 equations behave typically and the
errors are similar also
- Shortly
- Each Planet orbital distance (and velocity) depends on its previous neighbor data –
- But
- Earth depends on Mercury Not Venus
- Mars depends on Venus Not Earth
- Pluto depends on Uranus Not Neptune
- All planets calculations errors are around 1% except
- Earth (4%) and Neptune (7%)
- (The great errors be because of the square value –the real error is only 3% and 4%)
- Simply we can conclude that, the planets orbital distances and velocities depend on
their neighbors orbital distances and velocities in order.
- Notice
- Newton Concept (Planet motion depends on its mass) lost its 2 components,
Neither Planet orbital distance nor its velocity depend on its mass – by that no
proof for Newton concept at all – the idea is imaginary one.

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My 3rd
Equation (Depends on Kepler Law)
- d = A Planet Orbital Distance
- Kepler Law stated (Planet Orbit Defines Its Velocity), this equation depends on
this concept
- The equation doesn't depend on the planets order – Any 2 planets can be used
- Example No. (1)
- (108.2 mkm /57.9 mkm) = (47.4 /35)2
(error 1.8%)
- Where
- 108.2 million km = Venus Orbital Distance
- 57.9 million km = Mercury Orbital Distance
- 35 km/s = Venus Velocity
- 47.4 km/s = Mercury Velocity
- Example No. (2)
- (149.6 mkm /57.9 mkm) = (47.4/29.8)2
(error 2%)
- Where
- 149.6 million km = Earth Orbital Distance
- 57.9 million km = Mercury Orbital Distance
- 29.8 km/s = Earth Velocity
- 47.4 km/s = Mercury Velocity
- Example No. (3)
- (149.6 mkm /108.2 mkm) = (35/29.8)2
(No error)
- Where
- 149.6 million km = Earth Orbital Distance
- 108.2 million km = Venus Orbital Distance
- 29.8 km/s = Earth Velocity
- 35 km/s = Venus Velocity
2
1
2
2
1
)
(
v
v
d
d
=

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- The Equation Discussion
- My first and second equations depend on the planets order –but this third equation
doesn't depend on the order
- Means,
- Any 2 Planets can be used in it
- For that I have provided just 3 examples – and all other planets be similar
- The errors be in range (2%)
- Notice
- My second equation is the logical one between the first and this third equation –
let's explain that in details
- My first equation tells that, Planet orbital distance depends on its neighbor planet
orbital distance – by that – the equation depends on the planets order-
- But
- Kepler stated (Planet Orbit Defines Its Velocity)- that means – if we know planet
orbital distance we can conclude its velocity theoretically –based on that my third
equation be created -
- But the third equation doesn't depend on the planets order – any 2 planets can be
used in this equation – how can that be done? If the distances be created based on
one another how this third equation be free from the planets order?
- Because the velocities be distributed based on the rule by which the distances be
distributed – that be clear in my second equation – one rule be used for both data
(distance and velocity) distribution – and as a result – the distribution be similar
and kepler could create its law and the third equation be free from the planets
order.
- Notice
- Planet velocity is so effective player on its data creation and the rate (v1/v2) be so
useful and effective in different using as we will discuss later.

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Planet Velocity Is A Complementary One (My 5th
Equation)
v = Planet Velocity
t = another planet velocity be used as a period of time
Example
Mercury (47.4 km/s) moves during 6.8 seconds a distance = 322 km but
Uranus (6.8 km/s) moves during 47.4 seconds a distance = 322 km
By that, Planet velocity be used as a period of time for the distance 322 km - Why??
Details Data
(1)
Mercury (47.4 km/s) moves during 6.8 hours a distance = 1160000 km
Uranus (6.8 km/s) moves during 47.4 hours a distance = 1160000 km
(2)
Mars (24.1 km/s) moves during 13.1 hours a distance = 1160000 km
Jupiter (13.1 km/s) moves during 24.1 hours a distance = 1160000 km
(error 2%)
(3)
Earth (29.8 km/s) moves during 2 x 5.4 hours a distance = 1160000 km
Neptune (5.4 km/s) moves during 2 x 29.8 hours a distance = 1160000 km
(4)
Venus (35 km/s) moves during 2 x 4.7 hours a distance = 1160000 km
Pluto (4.7 km/s) moves during 2 x 35 hours a distance = 1160000 km
(error 2%)
Shortly
The distance 1160000 km be used as a reference to create planets velocity based on
one another. (why?)
Notice
Saturn (9.7 km/s) moves during 33.2 hours a distance = 1160000 km
(between 33.2 and Venus velocity 35 km/s the error 5%)
km
t
v 322
=

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The Equation Discussion
Data Analysis
The equation tells
Mercury velocity be complementary with Uranus Velocity
Venus velocity be complementary with Pluto Velocity
Earth velocity be complementary with Neptune Velocity
Mars velocity be complementary with Jupiter Velocity
But
Saturn velocity be complementary with Venus Velocity (with great error 5%)
Why? How Does Each Planet Choose Its Mate?
We notice that
The couple (Earth and Neptune) be used in my (5th
equation) and my (4th
equation)
The same couple be used in both equations –
A Question
Why does Mercury choose Uranus to be its mate? Jupiter choose Mars why?!
The answer
- Mercury (47.4 km/s) moves during 6.8 hours a distance = 1.16 million km
- Uranus (6.8 km/s) moves during 47.4 hours a distance = 1.16 million km
Why Mercury and Uranus? (as example)
This is a result of the geometrical distribution of the planet velocities – the next data
explains that
(1)
(2 x 100733 million km /197393 days) = (1.16/1.1318) = (0.6/0.5875)
Where
100733 million km = The Planets Orbital Circumferences Total
197393 days = The Planets Orbital Periods Total

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0.6 million km/s = 2 x 0.3 million km/s (Light Supposed Velocity)
1.1318 million km/day = Jupiter Velocity Per A Solar Day
0.5875 million km/day = Uranus Velocity Per A Solar Day
(2)
(13.1/6.8) = (47.4 /24.1) (error 2%)
Where
13.1 km/s = Jupiter Velocity
6.8 km/s = Uranus Velocity
47.4 km/s = Mercury Velocity
24.1 km/s = Mars Velocity
The data shows that, the solar system geometrical design creates a frame for its
creation and motion based on defined points by defines velocities – means- it's
necessary for Mercury to choose Uranus and Jupiter to choose Mars because based on
the geometrical design be created
(We have discussed this data in point no. 3 and will be discussed in more details in the
next points)
Conclusions
- Planet Velocity Be Created Complementary For Another Planet Velocity
- The 2 planets velocities be controlled by one distance (1.16 million km)
Notice
Saturn has no mate – I use Venus velocity but it causes an error 5 % -
Let's explain this data in details
Saturn (9.7 km/s) travels during 119587 seconds a distance 1.16 million km
The period 119587 s = 33.2 hours
Venus velocity =35 km/s for that reason I use Venus velocity for Saturn
Also a very deep connection be found between Venus and Saturn which supports the
idea – but
In fact Saturn depends on its own diameter (120536 km) where (1km= 1 second)

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The error be less (1%)
The difficulty is that,
Planet diameter and circumference be used as a period for usually in the solar system
data – by that – Saturn using of its diameter is belonged to a different category of data
The next examples can help our discussion
- Jupiter (13.1 km/s) moves during 10921 seconds a distance =142984 km (Jupiter
diameter) (10921km = the moon circumference)
- And
- Pluto (4.7 km/s) moves during 10921 seconds a distance =51118 km (Uranus
diameter)
- Pluto (4.7 km/s) moves during 51118 seconds a distance = 2 x 120536 km (Saturn
diameter) (where 51118km= Uranus diameter)
- Pluto (4.7 km/s) moves during 2 x 120536 seconds a distance = 1.13184 mkm
(=Jupiter motion distance during a solar day)
- And
- Uranus (6.8 km/s) moves during 7510 seconds a distance =51118 km (Uranus
diameter) (7510 km = Pluto circumference)
Simply it's a new geometrical rule we don't know and based on this rule Saturn uses
its own diameter as its mate.
The data shows a geometrical rule be used but it's unknown for us
Notice
The mates selection process needs more analysis – for example
The couple (Earth and Neptune) be used in my (5th
equation) and my (4th
equation)
Why?? may be because
(5.4)2
=29.8 (error 1%)
5.4 km/s = Neptune velocity 29.8 km/s = Earth velocity
That tells, planet velocity is a complex value we need to analyze it more deeply – let's
do that in the following point.

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References and Biography
The Solar System Be Created By A Light Beam Its Velocity 1.16 mkm/Sec (Revised)
https://www.academia.edu/65143114/The_Solar_System_Be_Created_By_A_Light_Beam_Its_Velocity_1_16_mkm_Sec_Revised_
(Preprint No. 1) Mars Migration Theory
https://www.academia.edu/49051037/_Preprint_No_1_Mars_Migration_Theory
(Preprint No. 2) The Moon Orbital Motion Equation
https://www.academia.edu/49051029/_Preprint_No_2_The_Moon_Orbital_Motion_Equation
The Moon Orbital Motion Geometry (II)
https://www.academia.edu/45181646/The_Moon_Orbital_Motion_Geometry_II_
Dr. Budochkina, Svetlana Aleksandrovna
Associate professor - Candidate of physico-mathematical sciences (2005)
http://www.mathnet.ru/eng/person22119
List of publications on Google Scholar
List of publications on ZentralBlatt
https://mathscinet.ams.org/mathscinet/MRAuthorID/757317
http://elibrary.ru/author_items.asp?spin=6087-3245
http://orcid.org/0000-0003-3447-0425
http://www.researcherid.com/rid/G-7453-2014
http://www.scopus.com/authid/detail.url?authorId=6507007003
https://www.researchgate.net/profile/Svetlana_Budochkina
Full list of
publications:
http://web-local.rudn.ru/web-
local/prep/rj/index.php?id=2944&p=15209
Mr. Gerges Francis Tawdrous +201022532292
Peoples' Friendship university of Russia – Moscow (2010-2013)
E-mail mrwaheid1@yahoo.com gergesgerges@yandex.ru
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Planet Diameter Be A Function In Its Rotation Period

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