Planet Diameter Equation Analysis.pdf

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 Equation Analysis
The Author Authorized To Be Used By
Mr. Gerges Francis Tawdrous
A Student–Physics Department- Physics
& Mathematics Faculty –
Peoples' Friendship University of Russia
(RUDN University) – Moscow – Russia
Dr. Budochkina, Svetlana Aleksandrovna
Associate Professor (Mathematical Analysis
and Theory of Functions Department)
Peoples' Friendship University of Russia
(RUDN University) – Moscow – Russia
Phone +201022532292
E-Mail: mrwaheid@gmail.com
Curriculum Vitae http://vixra.org/abs/1902.0044
Phone +7 (495) 952-35-83
E-Mail: budochkina-sa@rudn.ru, sbudotchkina@yandex.ru
Website
http://web-local.rudn.ru/web-local/prep/rj/index.php?id=2944&p=19024
The Assumption Of S. Virgin Mary -Written in Cairo –Egypt –16th
August 2022
Abstract
Paper hypothesis
Venus motion effect on the moon motion and causes the moon orbital period to be =
the moon rotation period =27.3 days
The hypothesis explanation..
- The hypothesis explanation can be summarized in 3 questions:
- (1) Can Planet Data Depend On Exaction Equations?
- (2) By What Equation Planet Diameter Can Be Defined?
- (3) How Can Planet Diameter Equation Change The Solar System Vision?
- Let's try to answer these questions in following…
(1st
Question)
Can Planet Data Depend On Exaction Equations?
- (1)
- Planet Data Be Created Based On Exact Equations –
- As a plane or rocket manufacture – the manufacturer needs exact equations to
define this plane length, width, weight and all specifications, otherwise this plane
can't fly safely.
- The moving planet under the physical laws has to define its diameter, mass, orbital
distance, period, inclination, rotation period, axial tilt …and all data based on
Exact Equations otherwise this planet can't move safely.

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- I have discovered 5 equations can conclude Planets Data theoretically without
observation which prove this fact decisively.
- Then we have to ask (How Planet Data Be Created In Order?)
- (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 this fact – because – it proves each planet orbital distance
depends on its neighbor planet orbital distance –
- d2
=4d0 (d-d0) where d= planet orbital distance and d0= its neighbor orbital distance
- Example (Venus orbital distance)2
= 4 (Mercury orbital distance) x50.3 million km
(Venus orbital distance=108.2 million km Mercury orbital distance =57.9 million km)
- This equation be tested and discussed in this paper – but its concept is a clear one –
it tells the planets leave an empty space for the new planet – for that reason each
planet orbital distance depends on its neighbor planet orbital distance.
- Logically the new planet can't disturb the current planets positions or motions
trajectories – by that –the orbital distance be defined by the neighbors positions –
- (3)
- The new planet has to revolve around the sun based on its orbital distance which
be defined obligatorily where no data of this planet be taken into consideration in
its orbital distance definition –neither mass nor diameter – instead – the distance
be defined based on the neighbor distance.
- But
- Planet diameter should be a function in its orbital distance – otherwise – this planet
will be broken through its motion –
- The function between planet diameter and its orbital distance is the necessary
requirement to cause the planet safe motion – almost – planet mass can't cause this
planet to be broken but it may decrease its velocity or creates orbital inclination –

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The planet geometrical motion form depends on Its Diameter – the wrong diameter
can cause this planet to be broken and destroyed.
- One more difficulty be found for the designer
- If the function contains only 2 variables which are planet diameter and its orbital
distance – in case this planet changes its orbital distance for any reason- this planet
will be broken also –
- As a result
- The designer had to create a function between planet diameter and its orbital
distance but also to make this function contains more variables – by that- if this
planet changes its orbital distance for any reason – the other variables will be
changed but the diameter will be saved -
- As a result
- The designer has created the planet diameter 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 3 variables (at least) which are this planet
rotation period, orbital period and velocity – in case of planet orbital distance
change these 3 variables will be changed as a result but the diameter will be saved.
- My fourth Equation proves this fact - let's see it in following…
(2nd
Question)
By What Equation Planet Diameter Can Be Defined?
- My 4th
Equation (Planet diameter Equation)
- (v1/ v2) = (s/r) =I
- v1 = planet velocity in second
- v2 = another planet velocity in second
- r = Planet Diameter of one planet of the 2
- s = The Planet Rotation Periods Number In Its Orbital Period
- (This value is belonged to the planet whose diameter is "r")

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- I = Planet Orbital Inclination (of the planet whose diameter is "r")
(means, 1.8 degrees be produced as the rate 1.8)
- 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 the rate (s) which is (the
number of its rotation periods in its orbital period)
- (for example Earth orbital period =365.25 days but (s) =366.7 rotation periods)
- The planets data follow this equation perfectly and prove that this equation shows
a fact in the solar planets data.
- The Equation works from the Earth to Pluto only – by that – the 3 first planets
(Mercury- Venus and the moon) aren't players in this equation – why??
- Let's try to answer this question ….
- The moon orbital period = the moon rotation period =27.3 days, for that reason the
rate (s) = 1
- As a result the moon motion be used as the base for this equation and because of
that the equation starts its work from the Earth and continues to Pluto.
- The planets data follow the equation as we show in the paper discussion –
- But, we have to ask,
- Can a small planet as the moon be used as the base for this equation?
- (4)
- Venus motion effect on the moon motion and causes the moon orbital period to be
= the moon rotation period – here we have a new fact disproves an old answer
- Because
- No tidal locking causes the moon orbital period be = the moon rotation period –
this idea is wrong totally – instead it's Venus motion effect on the moon motion -
let's prove that in following…
- (A)

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- The moon daily displacement =88000 km and during 29.53 days (the moon day
period) the displacements total be = 2.598 million km = 2π x 413600 km
- The data tells us the moon orbital apogee radius should be 413600 km and also it
tells, because the moon daily displacement (88000 km) is so long, the moon should
revolve around the Earth through this apogee orbit its radius (413600km) only and
can't revolve around the Earth through any more near orbit…
- Not Facts
- The moon orbital apogee radius =406000 km only and the moon revolves around
the Earth through near orbits and can reach to perigee radius (363000 km).
- How Can The Moon Do That?
- (B)
- 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.
- 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 equation depends on this angle (θ) it tells θ1 = θ0 +1.7
- where (θ1) = today angle and (θ0) =yesterday angle and 1.7 degrees be used for
the moon daily motion in the equation
- (C)
- As a result for the moon using of the angle (θ), the moon orbital radiuses be
defined based on Pythagorean rule – because – the moon uses the angle (θ) in its

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motion – by that – Each point the moon passes shows this fact – and the radiuses
be defined based on one another by Pythagorean rule – let's prove that
- (363000 km)2
+ (86000 km)2
= (373000 km)2
- (373000 km)2
+ (86000 km)2
= (384000 km)2
- (384000 km)2
+ (86000 km)2
= (392000 km)2
- (392000 km)2
+ (86000 km)2
= (406000 km)2
(error 1%)
- Where
- 363000 km = The Moon Orbital Perigee Radius
- 373000 km = The Total Solar Eclipse Radius
- 384000 km = The Moon Orbital Distance
- The data shows, the moon orbital 4 basic radiuses be defined based on one another
by using Pythagorean rule.
- Now this type of motion is related to Venus motion – where – no other planet uses
this technique in its motion – the following data can prove this idea
- (i)
- (41.4 million km)2
+ (108.2 million km)2
= (115.8 million km)2
- (ii)
- (iii)
= (680 million km)2
- Where
- 41.4 million km = Venus Earth Distance
- 108.2 million km = Venus Orbital Distance
- 50.3 million km = Venus Mercury Distance
- 119.7 million km = Venus Mars Distance
- 115.8 million km = 2 x 57.9 million km Mercury Orbital Distance
- 670.4 million km = Venus Jupiter Distance

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- 680 million km = Venus Orbital Circumference
- I try to show that –Venus distances be defined based on Pythagorean rule as the
moon distances to the Earth – we should discuss why Venus needs to decrease its
distance by using this technique, but in all cases this is the moon motion behavior
and not the Earth – here we see a clear effect of Venus motion on the moon motion
–and we have a real reason to believe that Venus motion effect on the moon
motion to cause the moon orbital period = the moon rotation period = 27.3 days
- Where,
- (243/224.7) = (29.53/27.3) =1.0725
- Venus rotation period = 243 days and Venus orbital period =224.7 days
- The moon rotation period = 27.3 days and The moon day period = 29.53 days
- The rate 1.0725 we discuss deeply in this paper where great effects depend on it
- Notice
- 5.1 deg (the moon orbital inclination) =3.4 deg (Venus orbital inclination) +1.7
deg - But
- The moon motion equation tells θ1 = θ0 +1.7 where (θ1) = today angle and (θ0) =
yesterday angle – by that we see an effect of Venus motion on the moon motion
- Shortly
- Venus and Mercury support the moon motion and the 3 planets motions create one
system based on which the moon orbital period to be = the moon rotation period
and by that the rate (s) be =1 and the moon motion be used as the equation base.
- That tells the 2 periods equality depend on a strong point because it be supported
by 3 planets motions.
(3rd
Question)
How Can Planet Diameter Equation Change The Solar System Vision?
- Why all planets data follow this equation? The first equation concept is more clear
(d2
=4d0 (d-d0) where d=planet orbital distance and d0=its neighbor orbital distance)

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- because it tells the current planets positions and motions define the new planet
orbital distance by that each planet orbital distance depends on its neighbor
distance – the 1st
equation shows that the solar system as one building and the
planets be similar to stories in the same one building -but –
- The 4th
equation ((v1/v2) = (s/r) =I) tells something a very new
- It Tells Some Continuum Be Found In The Solar System
- What does mean " A Continuum"?
- Imagine a groundwater be found under 10 houses – it's a continuum – all houses
suffer from the same effects and features – it's the same one force moves through
all houses and effect on them –
- This meaning can be seen by the (4th
equation) clearly-because each planet defines
its diameter as a function in its rotation period- here we don't see a point depends
on a point as stories of one house but we see a force passes through all data and
nothing can stop it –
- I want to say, there's a geometrical effect be created by the moon motion and
passed through all planets to Pluto and forces each planet to define its diameter as
a function in its rotation period – this is a continuum- it's similar to the blood in a
creature body – it passes through all members does different jobs but nothing can
prevent it –this is the picture be understood by the (4th
equation) – but how to
understand it? how to create a continuum in the solar system?
- While we see the planets (matters) as source of energy and consider the space as a
sea separate the planets from one another (as the sea separate the ships from each
other) –the equation tells a new feature about the solar system – it tells
- From One Energy The Planets Matters And Distances Be Created
- This idea can be acceptable simply-because –the physics book accepts that matter
and space be created of energies – the equation tells one additional idea which is-
from The Same One Energy the planets diameters and their distances be created
that causes the continuum to be created in the solar system

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- Shortly
- The solar planets matters and their distances be created of the same one trajectory
of energy – by that – the energy be not blockaded inside the planets matters but the
planets matters be geometrical points on this same one trajectory of energy
- For a simple description
- From one light beam energy the solar planets matters and their distances be created
– the light create the matter and space from the same stuff (energy) but
- By using different geometrical rules for creation, the matter and space be created
in different forms –this idea is a simple one also – because the oil and coal be
created of the same source but the creation methods caused their forms –that's the
same situation - the light created the matter and space from energy by using
different geometrical rules.
- That causes the continuum in the solar system – my fourth equation is one proof
and also mountains of data prove this fact (it tells, the solar system be created
based on a continuum of data-means- the data be transported through it).
- But
- What's the real geometrical effect based on which my 4th
equation depend?
- It's A Rate Of Time
- One hour of Mercury motion = 24 hours Of Pluto Motion
- This rate of time be created between Mercury and Pluto - where Mercury, Venus
and the moon motions be integrated to create the moon cycles periods equality as
the equation base - and Pluto be the end terminal of the equation – by that – we
have 2 terminals (the moon and Pluto) But Mercury motion be used behind the
moon motion – and the rate of time be between Mercury and Pluto – and this rate
of time be created and passed through all planets and forces them to create their
diameters as function in their rotation periods –
- But
- Where's this rate of time in the equation?

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- It's the rate (v1/v2) –the velocities rate creates the rate of time between the planets
– this idea can be acceptable simply in high velocity motions
- Also we have to ask
- Why can this rate of time force the planets data to follow the equation?
- The rate of time defines the rate of energy
- The river of water sends a great amount of water to the small canal but the canal
received only apart of this water and can't receive all amount -
- The rate of time between Mercury and Pluto is (1: 24)
- By that, Mercury energy will be divided by (24) and Pluto will receive energy =
(1/24) of Mercury energy –
- Now, we have a moving river and its outlet- the outlet can receive only 1/24 of the
total water- for that reason –the water in all passages be = (1/24) of the source
because the outlet controls the available water to be sent.
- That explains why Pluto data be found in all planets motions data analysis because
it's the range of available energy and no greater energy can be sent because the
outlet can't receive more that (1/24)
Paper conclusions
- (1)
- Newton Theory of the sun gravity is wrong because by one force the planet be
created and moving – otherwise this planet will be broken by effect of 2 forces on
it – means- the force which created the planet it causes its motion – for that reason
– No planet moves by the sun gravity –because the sun didn't create any planet
- (2)
- The big bang theory is a wrong one because the first energy wasn't in a chaos
form but was in a geometrical design form – as a result - planet data be created
based on exact equations –
- Shortly, the first energy never be in a chaos form and even if the first explosion be
done the energy be still in a geometrical design form

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Paper objective
- (I)
- The paper proves my fourth equation and analyzes it –in addition to analyze the
moon orbit design to prove that the orbit design be created as a result for my fourth
equation effect on the moon motion.
- (II)
- Venus motion effect on the moon motion and causes the moon orbital period be =
the moon rotation period
- (III)
- The solar system depends on a continuum which proves that- from one energy the
solar planets and their distances be created.
(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 Contents
Subject Page No.
1-Introduction 16
2- Methodology 20
Part No. One 22
A- Planet Data Depends On Exact Equations 23
A-1 Preface 24
A-2 Planet Orbital Distance Equation (My 1st Equation) 27
A-3 The Solar System Distances And Velocities Maps 31
A-4 Planet Diameter Equation (My 4th Equation) 36
A-5 My Three Rest Equations Tests 45
B- Venus Effect One The Moon Motion 53
B-1 Preface 54
B-2 Moon Orbital Motion Description 58
B-3 Venus Orbital Motion Description 61
B-4 The Proportionality Of The Moon And Venus Orbits Areas 62
B-5 The Cycles Equality 63
B-6 The Moon Orbital Motion More Analysis 64
B-7 The Moon Orbit Area Analysis 67
B-8 Venus Orbit Area Analysis 74
B-9 Venus And The Moon Data Consistency 78
C- Jupiter Effect On Venus And The Moon Motions 84
C-1 Preface 85
C-2 Planets Positions Description 90
C-3 Mercury and the Moon Motions for 30 million km 93
C-4 Planet 8 Days Cycle 95
C-5 The Main Idea 100
C-6 Jupiter And The Moon Data Consistency 104
C-7 Venus Be The Solar System Central Point 106
C-8 The outer planets diameters total analysis 108
D- Planet Diameter Equation Analysis 110
D-1 Preface 111
D-2 The Equation Effect Description 116
D-3 The Moon Effect Analysis 117
D-4 The Relative Motion Between The Moon And Pluto 120
D-4-1 Planets Orbital Distances Distribution 121
D-4-2 Jupiter Motion Relative To Pluto Motion 124
D-4-3 Jupiter And The 3 Planets Interaction 126
D-5 The 3 Inner Planets effect on Pluto motion 128
D-6 The Moon And Pluto Motions Data Consistency 130
D-7 The Outer Planets Diameters Total Effect 133
D-8 The Moon Orbit Geometrical Structure 135
D-9 Why Does The Moon Apogee Orbital Radius =406000 Km? 137
D-10 Saturn Effect Analysis 143

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D-11 The Moon And Saturn Motions Data Consistency 145
D-12 Pluto Effect Analysis 150
D-13 Pluto And Neptune Data Consistency 153
D-14 Jupiter And The Moon Data Consistency (More Data) 155
D-15 The Equation Units Analysis 156
D-16 Planet Diameter Analysis 158
D-17 Questions And Answers 159
161
Part No. Two 175
3- One Geometrical Design Be Found Behind The Solar System 176
3-1 Preface 177
3-2 The One Geometrical Design Proves 178
3-3 The One Geometrical Design Reason 185
3-4 The Planet Motion Trajectory Analysis 191
3-5 The One Geometrical Design Discussion 193
3-6 The One Geometrical Design Result 200
4- The Solar System Geometrical Design Analysis 204
4-1 Preface 206
4-2 Newton Theory Of The Sun Gravity Is Wrong 208
4-3 The Solar System Is Created Based On One Geometrical Design 210
4-4 The One Geometrical Design Depends On One Light Beam Energy 211
4-5 The Solar System Be Created Out Of One Light Beam Its Velocity 1.16 Million
Km Per Second.
213
4-6 The Solar System Motion Depends On One Geometrical Design 216
4-7 THE DATA PROVE THE THEORY 218
4-8 The Solar System Geometrical Description 263
4-9 How The Matter Be Produced Periodically? 268
4-10 The Planets Unified General Motion (More Data) 270
4-11 Questions And Answers (Extending Discussion) 280
5- The Moon Orbital Apogee Radius Analysis 295
5-1 Preface 296
5-2 The Moon Orbital Motion Description (And Equation) 299
5-3 The Moon Orbital Apogee Radius Analysis 315
5-4 Can Uranus Motion Effect On The Moon Motion 321
5-5 The Moon Daily Displacement Analysis 325
5-6 The Moon And Mercury Motions Data Analysis 330
5-7 The Moon And Pluto Motions Data Consistency 333
5-8 Uranus Motion Effect On Pluto Motion 341
5-9 The Moon Orbital Apogee Radius Decreasing Details 344
5-10 The Moon Orbit Description
5-10-1 Preface
5-10-2 The Moon Orbital Triangle Description
5-10-3 The Moon Orbital Triangle Data Analysis
5-10-4 The Moon Orbital Triangle Major Points
348
349
350
360
361

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6- Jupiter Motion Effect On Earth And Venus Motions 397
6-1 Preface 398
6-2 Jupiter Motion Effect On Earth And Venus Motions 403
6-3 Jupiter Orbital Circumference Analysis 410
6-4 The Sun Rays Creation 414
6-5 Saturn Velocity Analysis 420
6-6 The Solar System Creation and Motion Theory 440
7- The Solar Planets Motions Use Different Rates Of Time 447
7-1 Preface 448
7-2 The Planets Motions Rates Of Time 450
8- The Solar Planets Rates Of Time Analysis 454
8-1 Preface 455
8-2 Venus Motion Rate of time 456
8-3 Earth Motion Rate of time 458
8-4 Mars Motion Rate of time 460
8-5 Jupiter Motion Rate of time 462
8-6 Saturn Motion Rate of time 464
8-7 Uranus Motion Rate of time 466
8-8 Neptune Motion Rate of time 468
8-9 Pluto Motion Rate of time 470
8-10 The Planets Orbital Distances Test 472
8-11 One Law Controls The Planets Orbital Periods And Distances 477
8-12 The General Discussion 478
9- The Planets Motions Rates Of Time Effect Analysis 481
9-1 Preface 482
9-2 Planets Motions Rates Of Time And Distances Data 483
9-3 The Data Equal Distances 487
9-4 The Data and the planets velocities. 495
9-5 The Data Distances And Rates Of Time Interaction 499
9-6 The Data General Discussion 509
9-7 Mars, Jupiter and Saturn Motions Analysis 511
9-8 Why Saturn And The Moon Use Equal Rates Of Time? 516
9-9 Why Mercury Use A Double Of Its Orbital Distance? 521
9-10 The Rate (4.61) be used between Pluto and the moon motion 522
10- Mars Migration Theory 526
10-1 Mars Migration Theory 527
10-2 Pluto Migration Theory 530
10-3 Planets Migration Theories Proves 532
10-4 Is There An Absent Planet In The Solar Group? 535
11-The Solar System Distances Be Created In A Network Form 538
11-1 Preface 539
11-2 The Continuum effect Through the Solar System Distances 541
11-3 The Solar System Distances Distribution 546

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11-4 The Solar System Distances Dependency On One Another 550
12- The Continuum Effect Proof 552
12-1 The Continuum Effect Proof 553
12-2 Saturn Motion Analysis 559
12-3 Planet Diameter Analysis 585
12-4 Why do the planets revolve around the sun if there's no sun gravity? 587
13- Planet Mass Effect On its Motion 588
13-1 Preface 589
13-2 Planet Mass effect on its Motion 591
13-3 Saturn and Earth Motions Interaction 602
13-4 Planets Velocities Proportionality 611
14- Saturn Motion Analysis 619
14-1 Preface 620
14-2 Saturn Diameter Analysis 622
14-3 Neptune Circumference Analysis 632
14-4 Neptune Day Period Analysis 641
14-5 Mercury Motion effect on Jupiter and Neptune Motions 650
14-6 Earth Motion Distance Daily Analysis 657
14-7 Uranus Day Period Analysis 661
14-8 The Inner Planets Motions Analysis 667
15- The Sun Age Description 675
15-1 Preface 676
15-2 The Sun Circles The Earth 677
15-3The Rate (1.0725) 678
15-4 The Sun Diameter Analysis 681
15-5 The Sun And Earth Motions Rate Of Time (1 day =365.25 days) 684
15-6 The Sun Rays Creation 592
16- Mercury Jupiter Distance Analysis 698
16-1 Mercury Jupiter Distance Analysis (720.7 mkm) 699
16-2 (Jupiter And Mercury Motions Analysis) 712
16-3 Jupiter Distances Analysis 716
Appendix No.1 The Solar System Equal Distances List 717
References and Biography 719

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1- Introduction
If The Moon Orbital Period Doesn't Equal The Moon Rotation Period – Can The
Moon Orbit Design Be Changed As A Result?
- Let's try to answer in following …
- The moon orbital period = the moon rotation period =27.3 days because my fourth
equation uses the moon motion as the Equation base (as I claim) –let's remember it
- My 4th
Equation (Planet diameter Equation)
- (v1/ v2) = (s/r) =I
- This equation defines planet diameter as a function in its rotation period – or
accurately in the rate (s)
- For the moon (s) be =1 and because of that the moon motion be used as the base of
the equation and the equation works from the Earth to Pluto only
- Can the moon orbit design be effected also by this equation? I have 3 groups of
data can be used as answer for this question – let's see them
- (Data No. 1)
- The moon orbital apogee radius =406000 km =the planets diameters total
- The moon orbital perigee radius =363000 km =the outer planets diameters total
(error 1%)
- The distance between the perigee and apogee be =43000 km

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- We notice that the moon moves through this distance (43000km) and if we remove
the moon diameter from this distance – the space be 39500 km=the inner planets
diameters total = Earth Circumference (error 1%)
- The data tells, the planets diameters values be taken into consideration in the moon
orbit designing process – we have no explanation except my fourth equation effect
on the moon orbit design.
- (Data No. 2)
- The outer planets diameters total =366556 km = the perigee radius 363000 km +
the moon diameter 3475 km
- But
- 366556 km = 3475 km (the moon diameter) x (655.7 /2π) (error 1%)
- And
- 366556 km = 2390 km (Pluto diameter) x 153.3
- Where
- 655.7 hours = the moon rotation period
- 153.3 hours = Pluto rotation period
- This data also can be explained only by my fourth equation because it connects
between planets diameters and their rotation periods – as the equation does
- Also
- The moon and Pluto are the equation 2 terminals planets
- (Data No. 3)
- (1)
- The moon orbit area = 103944 million km2
- (2)
- 103944 million km2
=2.598 million km x 40000 km

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Where
- 2.598 million km = The moon displacements total in 29.53 days
- 2.41 million km = The moon orbital circumference at orbital distance (384000km)
- 2.28 million km = The moon orbital circumference at perigee radius (363000 km)
- Also 43000 km and 40000 km we have discussed before
- The data tells the moon orbit area defines the moon orbital radiuses in its motion –
here we deal with a geometrical design depends on the orbit area and defines the
radiuses lengths-
- In the paper discussion we should prove that the moon orbit area is connected with
Venus orbital area and Jupiter motion data –
- We remember that Venus motion effect on the moon motion and causes the moon
orbital period to be =the moon rotation period – and by this equality – the moon
motion be used as my fourth equation base – Venus supports the moon motion by
its effect on it –
- We discuss this data in details in the paper - but the data shows that – some
geometrical reason be found behind the moon orbit design and this geometrical
reason depends on a function between (planet diameter and its rotation period)
because these 2 values only be used frequently in the moon design data.
- Based on this analysis we conclude simply
- If the moon orbital period doesn't equal the moon rotation period, the moon orbit
design would be changed as a result
- Notice
- (We disprove the tidal locking idea as a reason for the moon periods equality)
- Let's see the paper contents in following…

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- The Paper Contents
- The paper is divided into 2 parts
- Paper Part No. One
- This part analyzes and proves planet diameter equation and also proves that the
moon orbital period be = the moon rotation period by effect of Venus motion on
the moon motion
- Paper Part No. Two
- This part proves that, the solar system be built on one design because the planets
matters and their distances be created from the same one energy
- For the details please review the contents table

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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.
- Notice
- Mars Migration Theory be discussed and proved in point no. (10) of this paper

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Paper Part No. One
(Pages from 22 to 174)
This paper part analyzes and proves Planet Diameter Equation
and also proves the paper hypothesis tells
(The Moon Orbital Period Be = The Moon Rotation Period By Effect Of Venus
Motion On The Moon Motion)

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A- Planet Data Depends On Exact Equations
A-1 Preface
A-2 Planet Orbital Distance Equation (My 1st
Equation)
A-3 The Solar System Distances And Velocities Maps
A-4 Planet Diameter Equation (My 4th
Equation)
A-5 My Three Rest Equations Tests

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A-1 Preface
- What proves for the concept (Planet Data Be Created Based On Exact Equations)?
- Let's answer in following
- (1)
- I have discovered 5 Equations Prove the concept and enable to conclude the
planets data theoretically without observation –
- But the concept is proved even without my 5 equations
- (2)
- As a plane or rocket manufacture – the manufacturer needs exact equations to
define this plane length, width, weight and all specifications, otherwise this plane
can't fly safely.
- The moving planet under the physical laws has to define its diameter, mass, orbital
distance, period, inclination, rotation period, axial tilt …and all data based on
Exact Equations otherwise this planet can't move safely and will be broken by its
motion.
- Simply, the safe motion is a proof that this planet data be created based on exact
equations –
- Let's refer to my 5 equations –all equations be tested in this point – where and the
paper discusses and analyzes the first and fourth equations basically
- My 1st
Equation (Planet Orbital Distance Equation)
- d2
= 4d0 (d-d0)
- d = A Planet Orbital Distance
- d0= Its Direct Previous Neighbor Planet Orbital Distance
- My 2nd
Equation (Planet Velocity Equation)
- (V0
2
/V2
) = 4 (1 – (V2
/ V0
2
)
- V = A Planet Velocity
- V0= Its Neighbor Planet Velocity

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- My 3rd
Equation (Depend On Kepler Law Equation)
- (d1/d2)=(v2/v1)2
- v = Planet Velocity
- My 4th
Equation (Planet Rotation Period Equation)
- (v1/ v2) = (s/r) =I
My 5th
Equation ( Planet Velocity Is A Complementary One)
vt =322 km
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
- This point discussion be divided into the following points
- A-2 Planet Orbital Distance Equation (My 1st
Equation)
- This equation defines each planet orbital distance depends on its neighbor distance
We test the equation with all planets data and discusses its concept.

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- A-3 The Solar System Distances And Velocities Maps
- This point shows that the solar system has one map for the distances and one map
for the velocities where the phase between the 2 maps be found to cause the
planets diameters to be created as function in their orbital distances.
- A-4 Planet Diameter Equation (My 4th
Equation)
- My fourth equation proves shows that, planet diameter be created as a function in
its rotation period – we test the equation with all planets data and discuss shortly it
concept
- But
- This equation (my fourth equation) be analyzed in details in the point No. (D) -
the equation analysis be put in independent point because of its wide and deep
discussions and arguments.
- A-5 My Three Rest Equations Tests
- In this point we test the rest 3 equations to be used as a reference.

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A-2 Planet Orbital Distance Equation
(My First Equation)
- 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 distance 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.
And also it tells (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 – also disproves the gravitation equation.
No initial conditions effect on any planet data –because planet data be created based
on exact equations and mathematical calculations.
That makes my first equation is a very new equation in concept where the physicists
believe 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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A-3 The Solar System Distances And Velocities Maps
The Idea Summary
Based on many sources of knowledge we have to consider the solar system as one
machine of gears – or one creature body- or one building – and the planets be similar
to members in this same creature body-
My first equation proves that – the distance between the sun and Pluto be designed as
one piece of distance – and my fourth equation proves that a continuum be found in
the solar system and that causes the data transportation and integration among the
planets.
I want to say-
Mountains of proves show that- the solar system is One Geometrical Design and
can't be considered as separated planets revolving around the sun.
As a result –
I provide 2 maps in this point – one map for the solar system distance which deals
with the distance from the sun to Pluto as one piece of distance and the planets be
distributed in it based on one geometrical design.
Another map for the velocities which consider the planets velocities total as the basic
value which be distributed for the planets based on one geometrical design.
The maps show and prove the data treatment based on the data total and not based on
individuals data – that shows – the integrating motion is a planned job for the solar
planets –
The map of distance depends on Jupiter and Pluto but the map of velocity depends on
Jupiter and Uranus – here Jupiter is used in both maps but while the distance map
reaches to Pluto the velocity map stops at Uranus why??
Because the phase between the 2 maps be used to define the planets diameters as
function in their orbital distances –let's discuss these 2 maps in following…

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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))
The distances map tells –the map depends on 2 basic points (Jupiter and Pluto)

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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 – We should discuss the reason in the paper discussion

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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
1.16 million km/s = Light Supposed Velocity
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
Notice, The discussion supposes a light beam its velocity 1.16 million km/s be found
– we prove this fact Later

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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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A-4 Planet Diameter Equation (My 4th
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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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 around (1%) except Jupiter (2.2%)
(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 motions be used as 2 terminals for the equation.
But
We still need to discover the consistency of these 2 planets motions – because – they
don't create comparable sense of motion – the data should be analyzed to know why
Pluto 14177 rotation periods and 14547 days be used as comparable with the moon
cycles periods (27.3 days) and how this is done? For example we may be forced to
suppose that (Pluto rotation period 153.3 hours be used equal to one solar day 24
hours?) (as a rate of time be used in the planet motion)

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Notice
The moon orbital period = The moon rotation period =27.3 days not for any tidal
locking – this idea is wrong and will be disproved through the discussion…
The moon periods equality be produced by Venus motion effect on the moon motion -
where Venus and Mercury supports the moon motion to cause its orbital period be =
its rotation period to make (s) to be =1
By that the moon motion be used as the equation base
(4) The Equation basic points
The Equation depends on 3 basic points
The moon where (The Moon Rotation Period = The Moon Orbital Period)
Saturn where (Saturn Rotation Velocity = Saturn Orbital Velocity)
Pluto where (Pluto Rotation Distance = Pluto Orbital Distance)
Pluto distances equality should be discussed in details in point No.(D-12)
And we should know why the Equation depends on these 3 planets and creates this
strange feature –
Shortly
We have to ask why this feature is necessary for the equation to be working
(5) 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

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rotation period and this is the reason which prevents planet to use its orbital velocity
to define its rotation period –
Saturn is the only planet uses its orbital velocity to be equal its rotational velocity –
Saturn motion be distinguish among the planets – but it's part of the geometrical effect
of the equation on the planets motions – we should notice that
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
We analyze Saturn motion through the equation analysis in point No. (D-10)
(6) The Equation Geometrical Effect
Let's ask
Why does planets data follow this equation? What's this geometrical effect which be
started from the moon and reaches to Pluto and passes through all planets forces each
planet to create its diameter as a function in its rotation period (or in the rate (s))?

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We have answered this question before
It's a rate of time (One Hour Of Mercury Motion Be =24 Hours Of Pluto Motion)
On this rate of time the equation depends
How can the rate of time effect on planets data?
Because
The rate of time controls the amount of passed energy– here we see another vision for
the solar system
The planets matters and their distances be created from the same one energy and the
energy passes through all planets and causes their motions – by that - the energy isn't
blockade inside the planet body in mass form but the energy be in mass form and in
space form and be as a river water moves from a point to another
The planets matters be similar to geometrical points – as when the water creates a
rock tube of its salts and minerals and the water passes through this tube – the tube
walls be made of the water itself but the tube has a geometrical form distinguish from
the water – and still the tube can't continue in life without the water because it’s the
source of this tube salts and minerals –
As a result, The Solar System Be Similar To One Trajectory Of Energy –
Here, the rate of time be so powerful because it controls the passed amount of energy-
as the child hands can't hold all sweets be given by his mother hand- the rate of time
(1/24) makes Pluto to receive only (1/24) of the energy be sent from Mercury-
But
Because Pluto is the river outlet and no other place to put more energy in it – that
makes the energy which passes from Mercury to Pluto = (1/24) of Mercury energy
Here we get one more result – Pluto Data Controls The Solar System –
Because no other outlet be available for that all planets have to take into consideration
Pluto data because the energy which passes through each planet be defined by the rate
of time between Mercury and Pluto – here we see how the equation works and why all
planets data follow it.

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(7) The Equation Deep Analysis
My fourth Equation be analyzed in details in point No. (D)
I put the equation analysis in independent point because it's a wide analysis – where
many questions be left behind
Above all we need to know why Jupiter and Saturn use the rate (r/s) in place of (s/r)?
we discuss in point No. (D-17)
Also - the rate (s/r) itself be puzzled because
s= planet rotation periods number in its orbital period – for Earth it be 366.7
r= planet diameter –for Earth be 12756 km
how to define the units (s/r)??
because 366.7 be (366.7 Earth rotation periods) but (Earth rotation period =23.9 h)
how to create a consistency between these 2 values (s/r)?
We have to consider 23.9 hours to be equal =1 second
And
12756 km to be used as 12756 seconds
This is the way to create a harmony of the data –but how to use these units?!
How planet diameter be used as a period of time by the rate (1km = 1 second)?
And
How the planet rotation period (23.9 hours) be used as (one second)?!
We answer these questions in the Equation analysis Point NO. (D)
Here to prove the equation analysis will be interesting – I put some using of planets
diameters as periods of time – to show the analysis will provide discoveries …
Data
Jupiter (13.1 km/s) moves in 10921 seconds a distance= 142984 km= Jupiter diameter
Uranus (6.8 km/s) moves in 7510 seconds a distance= 51118 km= Uranus diameter
Pluto (4.7 km/s) moves in 10921 seconds a distance= 51118 km= Uranus diameter
Pluto (4.7 km/s) moves in 51118 seconds a distance= 2 x120536 km= Saturn diameter
Pluto (4.7 km/s) moves in 2 x120536 s a distance= (Jupiter motion distance daily)
(10921 km =the moon circumference and 7510 km =Pluto circumference)

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A-5 My Three Rest Equations Tests
In following I provide my rest 3 equations and their tests with planets data to be used
as reference for our discussion.
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 respectively.
- 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 an imaginary one.

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My 3rd
Equation (Depends on Kepler Law)
- v = Planet Velocity
- 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
- 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
- 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.

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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 point is that the distance (1.16 million km) controls all planets motions –we
should discuss this distance later – because
The data analysis shows that a light beam its velocity be = 1.16 million km per second
be found and effect on the planets motions data – that causes the velocities depend on
this distance as a reference – We need to discuss that Part No. 2 of this paper.

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B- Venus Effect One The Moon Motion
B-1 Preface
B-2 Moon Orbital Motion Description
B-3 Venus Orbital Motion Description
B-4 The Proportionality Of The Moon And Venus Orbits Areas
B-5 The Cycles Equality
B-6 The Moon Orbital Motion More Analysis
B-7 The Moon Orbit Area Analysis
B-8 Venus Orbit Area Analysis
B-9 Venus And The Moon Data Consistency

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B-1 Preface
Can Venus Effect On The Moon Motion?
The idea tells (Venus causes the moon rotation period be = the moon orbital period)
This idea disproves (the tidal locking as a reason)–That may change the physics book!
Is There Any Proof For Venus Effect On The Moon Motion and Its Cycles?
Let's try to answer in following…
- (A)
revolve around the Earth through this apogee orbit its radius (413600 km) only and
- Not Facts
- (B)

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- The moon orbital motion depends on this angle (θ) it tells θ1 = θ0 +1.7
- where (θ1) = today angle and (θ0) =yesterday angle and 1.7 degrees be used for
the moon daily motion in the equation
- (C)
- (363000 km)2 + (86000 km)2 = (373000 km)2
- (373000 km)2 + (86000 km)2 = (384000 km)2
- (384000 km)2 + (86000 km)2 = (392000 km)2
- (392000 km)2 + (86000 km)2 = (406000 km)2 (error 1%)
- Where
- (D)
- Now this type of motion is related to Venus motion – where – no other planet uses
this technique in its motion –
- The following data can prove this idea
- (i)
- (41.4 million km)2 + (108.2 million km)2 = (115.8 million km)2
- (ii)

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- (50.3 million km)2 + (108.2 million km)2 = (119.7 million km)2
- (iii)
- (670.4 million km)2 + (119.7 million km)2 = (680 million km)2
- Where
- 41.4 million km = Venus Earth Distance
- 50.3 million km = Venus Mercury Distance
- 119.7 million km = Venus Mars Distance
- 115.8 million km = 2 x 57.9 million km Mercury Orbital Distance
- 670.4 million km = Venus Jupiter Distance
- 680 million km = Venus Orbital Circumference
- I try to show that –Venus distances be defined based on Pythagorean rule as the
moon distances to the Earth – we should discuss why Venus needs to decrease its
distance by using this technique, but in all cases this is the moon motion behavior
and not the Earth – here we see a clear effect of Venus motion on the moon motion
–and we have a reason to believe that Venus motion effect on the moon motion to
cause the moon orbital period = the moon rotation period = 27.3 days
- Venus motion effect on the moon motion has a theoretical explanation we study
with my fourth equation analysis –
- Let's summarize it in following..
- My fourth equation shows a function be found between planet diameter and its
rotation period or the rate (s) where (s = planet rotation periods number in its
orbital period)
- For The Moon Motion (s = 1)
- By that the equation uses the moon motion as its base because of that the equation
works from the Earth to Pluto only –
- Where the moon is a small planet – logically it can't be used as a base for an
equation effects on all planets data –for that reason – Venus motion supports the

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moon motion and by this support the moon, Venus and Mercury motions can be
seen as one system of motions based on which the moon orbital period be = the
moon rotation period and the equation depends on this equality of the periods.
- That create a strong base of the equation and its geometrical effect which forces all
planets (from the Earth to Pluto) to create their diameters as function in their
rotations periods.
- This point discusses Venus Motion effect on the moon motion and prove many
basic data be produced as a result of this effect.

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B-2 Moon Orbital Motion Description
- Why does the moon orbital apogee radius =406000 km?
revolve around the Earth through this apogee orbit its radius (413600km) only and
- Not Facts
- The moon orbital motion depends on this angle (θ) it tells θ1 = θ0 +1.7
- where (θ1) = today angle and (θ0) =yesterday angle
- 1.7 degrees be used as the moon daily motion degrees for the equation

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- 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
- This result be produced by my fourth equation which proves a function be found
between planet diameter and its rotation period – where the equation uses the
moon motion as its base – we discuss that in point no. (D)
- Notice
- (363000 km)2
+ (86000 km)2
= (373000 km)2
- (373000 km)2
+ (86000 km)2
= (384000 km)2
- (384000 km)2
+ (86000 km)2
= (392000 km)2
- (392000 km)2
+ (86000 km)2
= (406000 km)2
(error 1%)
- Where
- This conclusion is so effective in the moon orbital motion explanation because it
shows accurate details in the moon motion – we conclude simply that – this angle
(θ) using is a feature be related to the moon motion – and not a general feature for

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all planets – also this angle (θ) using is a proved behavior by the moon motion data
because without this using the moon apogee radius should be =413600 km and the
moon would be prisoner in this orbit and revolves around the Earth only through
this orbit.
- The point is that, this using of the angle (θ) is a behavior of Venus motion and the
moon uses this same behavior as a result for Venus motion effect on the moon
motion –
- In following let's prove that –Venus motion depends on Pythagorean rule also.

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B-3 Venus Orbital Motion Description
I - Data
(1)
(41.4 million km)2
(2)
(50.3 million km)2
(3)
(670.4 million km)2
= (680 million km)2
Where
41.4 million km = Venus Earth Distance
108.2 million km = Venus Orbital Distance
50.3 million km = Venus Mercury Distance
119.7 million km = Venus Mars Distance
115.8 million km = 2 x 57.9 million km Mercury Orbital Distance
II - Discussion
The data proves that, Venus distances to Mercury, Earth, Mars and Jupiter be defined
based on one another by using Pythagorean rule – we here have identical behavior of
motion be done by 2 planets (Venus and the moon) and we have to consider that a
deep connection must be found between them –
The point is that - we know why the moon uses Pythagorean rule – because the moon
tries to decrease its daily displacement length and tries to revolve around the Earth
through more near orbits that this very far one its radius (413600 km) and also the
moon decreased its apogee from this far one to (406000 km) by using this intelligent
technique – here we have a geometrical reason behind the angle (θ) using by the moon
in its orbital motion – but why Venus uses a similar behavior?
Shortly, why does Venus use Pythagorean rule in its orbital motion and as a result the
rule controls the distances between Venus and the other planets? We should answer
this question in point no. (B-9)

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B-4 The Proportionality Of The Moon And Venus Orbits Areas
I- Data
(1)
The Moon Orbital Area From Perigee (363000 km) To Apogee (406000 km) Be
Equal = 103944 Million km2
This area is calculated for an area between 2 concentric circles small radius =363000
km and great radius =406000 km
(2)
Venus Orbital Area (the distance between the sun and Venus) be = 36780 x 1012
km
(3)
36780 x 1012
km = 103944 million km x 0.354 million km
II-Discussion
The moon perigee radius = 363000 km and is different from 354000 km with (2.5%)
The data tells the moon and Venus orbits areas are in proportionality and the moon
defines its perigee radius (363000km) by an effect of Venus motion - in fact Jupiter
also effects to define the moon perigee radius but we discuss that later.
Notice
Venus effect on the moon motion be seen basically in the moon orbital distance
384000 km –we have to prove that by different data – but the definition of the
distance 354000 km is found in comparison with 384000 km basically based on the
following data
(243/224.7) = (29.53/27.3) =(0.384/0.354) = 1.0725
243 days = Venus Rotation Period 224.7 days = Venus Orbital Period
29.53 days = The Moon Day Period 27.3 days = The Moon Orbital Period
We discuss this data in the next point (B-9)

Planet Diameter Equation Analysis.pdf

Planet Diameter Equation Analysis.pdf

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