Planet Data Depends On 5 Equations (Revised).pdf

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
I do this research
Gerges Francis Tawdrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
Planet Data Depends On 5 Equations (Revised)
The Author Authorized To Be Used By
Mr. Gerges Francis Tawdrous
A Student–Physics Department- Physics
& Mathematics Faculty –
Peoples' Friendship University of Russia
(RUDN University) – Moscow – Russia
Dr. Budochkina, Svetlana Aleksandrovna
Associate Professor (Mathematical Analysis
and Theory of Functions Department)
Peoples' Friendship University of Russia
(RUDN University) – Moscow – Russia
Phone +201022532292
E-Mail: mrwaheid@gmail.com
Curriculum Vitae http://vixra.org/abs/1902.0044
Phone +7 (495) 952-35-83
E-Mail: budochkina-sa@rudn.ru, sbudotchkina@yandex.ru
Website
http://web-local.rudn.ru/web-local/prep/rj/index.php?id=2944&p=19024
The Assumption Of S. Virgin Mary -Written in Cairo –Egypt –26th
June 2022
Abstract
Paper Argument
- Planet data means
- (Planet Mass, Diameter, Orbital Distance, Period, Inclination, Rotation Period And
Axial Tilt) (Planet Fact Sheet)
- Planet data be created based on exact equations as a plane creation process.
- The creation process uses exact equations to define this plane length, width, weight
and all other specifications and data
- The paper proves that based on 5 equations the planets data be created
- The paper disproves any random process, initial conditions or unknown historical
factors to be effective on the planet data creation.
- On the contrary,
- The paper proves (Based On Exact Equations The Planet Data Be Created)
- The equations are
- (1) Planet Orbital Distance Equation
- d2
= 4 d0 (d-d0)
- d = A Planet Orbital Distance
- d0= Its Direct Previous Neighbor Planet Orbital Distance
- (2) Planet Velocity Equation
- (V0
2
/ V2
) = 4 (1- (V2
/ V0
2
))
- V = Planet Velocity

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- (3) Planet orbital distance and velocity equation (Kepler law)
- (d1/ d2) = (v2/v1)2
- d = Planet Orbital Distance
- v = Planet Velocity
- (4) Planet Diameter Definition 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")
- I = Planet Orbital Inclination (of the planet whose diameter is "r")
- (5) Planet Velocity Is A Complementary One
- vt = 322 km
- 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
The Paper provides the equations and their tests
(Please scan the figure (
(Please scan the figure (ORCID
ORCID
ORCID
ORCID)

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1-Definitions And Refuted Theories
(1)
The Big Bang Theory Is Wrong –Because
It supposed the first energy was in a chaos form –in fact it was in a geometrical design
form and controlled by geometrical rules (The Principle Is Wrong)
(2)
Newton Theory Of The Sun Gravity Is Wrong – Because
One force should cause the planet creation and motion otherwise this planet will be
broken – by that – the force which caused the planet creation it causes its motion –
Not The Sun - (The Logic Is Wrong)
(3)
Planet Data Be Created Based On Exact Equations
As a plane or a rocket creation – you need exact equations to define the lengths,
widths and all dimensions and specifications – nothing be created by random
(This Concept Disproves The Initial Condition And Random Process Ideas)
(4) The Research Method
I Use Planet Data Analysis As A Method –
This method puts the planets data in comparison with the theory to test the theory
sufficiency.
For example – Planets order contradicts the gravitation equation, how to explain that?
(5)
One Geometrical Design
(i)
It's A Similar To The Chess Game On A Chess Board
The solar system motion is similar to chess battle motions – because
(ii)
The Motion of Planet (A) be in integration with the motion of planet (B) to create
together one geometrical effect of both planets motions.

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The eclipse phenomenon is a suitable example – Earth, the moon and the sun motions
be in harmony (To Produce The Eclipse Phenomenon)
(iii)
The planets creation data support this harmony of motions – for that reason – for
example - the moon diameter be (3475 km) to enable the system to create the total
solar eclipse
I want to say
The eclipse isn't a phenomenon be occurred by some random process – the eclipse is
an example for hundreds of similar phenomenon be created by integration of planets
motions –
We don't see this integration because we don't know the geometrical effect of this
motions integration result.
For example, we use nothing from the eclipse phenomenon, it be occurred but what
useful result we have from it?
(iv)
Why the planets motions integration is necessary?
My theory tells
The solar planets matters and distances be created from one energy and this energy be
provided by one light beam – that means – the light be consumed in the planets
matters and distances creation – how can we find another light beam if its energy be
consumed already?
My theory tells the solar system uses the planets motions energies total to produce the
sun rays (light beam) – This is a very important process because without it the cycle
will not be completed – the matter be created of light and the cycle will be complete
when the matter produces light –
That makes the integration of the planets motions is a necessary one
We don’t discuss these ideas here – we test and discuss the 5 equations

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The Paper Contents
1-Definitions And Refuted Theories (Page no. 2)
2- Planet Data Depends On 5 Equations (Tests and Discussions) (Page No. 5)
3- Planet Velocity Analysis (Page No. 39)
4- Planets Masses Rates Definition (Page No. 51)
Appendix No. 1 There's A Continuum In The Solar System

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2- Planet Data Depends On 5 Equations (Revised)
My 1st
Equation (Planet Orbital Distance Equation)
My 2nd
Equation (Planet Velocity Equation)
- V = A Planet Velocity
- V0= Its Neighbor Planet Velocity
My 3rd
Equation (Depend On Kepler Law Equation)
My 4th
Equation (Planet Rotation Period Equation)
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")
I = Planet Orbital Inclination (of the planet whose diameter is "r")
My 5th
Equation ( Planet Velocity Is A Complementary One)
v = Planet Velocity
t = another planet velocity be used as a period of time
Example
)
(
4 0
0
2
d
d
d
d −
=
)
1
(
4 2
0
2
2
2
0
V
V
V
V
−
=
2
1
2
2
1
)
(
v
v
d
d
=
I
r
s
v
v
=
=
2
1
km
t
v 322
=

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THE EQUATION TESTS

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Planet Orbital Distance Equation (My 1st
Equation)
- The equation depends on the planets order, for that , 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
- 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
(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) Ceres Motion
- (415)2
= 4 x 227.9 x (415-227.9)
)
(
4 0
0
2
d
d
d
d −
=

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- 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 be used with all planets sufficiently with max error (1%) except with
Earth (error 3%) and Neptune (error 4%)
(c)
We don't use the forma (d= 2d0) because it creates great errors with all planets data
And
The forma (d=2d0)2
causes max error (1%) with all ((Earth 3%) and (Neptune 4%)
But
The forma (d=2d0)3
causes NO Error at all
(d)
My 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

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(e)
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)
Means
These 27 data be created based on geometrical rules because we can conclude them
by mathematical calculations.
The Conclusion of Point No. (1)
The Planets Orbital distances, Periods and velocities Be Created Based On
Geometrical Rules
Means
Planet Motion Data Be Created Based on geometrical rules and can be concluded
theoretically without observation.

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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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My 2nd
Equation (Planet Velocity 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

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- Saturn Depends On Jupiter (The values 1.82 and 1.806 Error 1%)
(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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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)
- 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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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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My 4th
Equation ( Planet Rotation Period Equation)
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
Notice / (The Equation Works From The Earth To Pluto Only)
1- 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)
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)
I
r
s
v
v
=
=
2
1

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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
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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2- The Equation Analysis
(I)
The Equation elements are clear but just 2 elements aren't clear which are
(r) = Planet Diameter
The equation uses it as a period of time for example Earth diameter =12756 km the
equation uses it as 12756 seconds
(s) = The Rotation Periods Number In Planet Orbital Period
Example
Jupiter rotation period =9.9 hours and Jupiter orbital period =4331 solar days
4331 days x 24 hours = 9.9 hours x 10500
For that (s) for Jupiter be = 10500
The part (s/r) – how to understand its units?
If planet rotation period be =1 second
For our example
Jupiter rotation period =9.9 hours
If this period (9.9 hours) can be equal (1 second)
And the diameter (12756 km) be used in seconds (12756 seconds)
The value (s/r) will be (second /second) and will work correctly.
(II)
The second planet velocity (v1) can't be defined by a rule – we use the mini error
• Earth equation uses Neptune velocity
• Mars equation uses Pluto velocity
• Jupiter equation uses the Earth moon velocity (27.78 km/s)
• Saturn equation uses Mars velocity
• Uranus equation uses Neptune velocity
• Neptune equation uses Saturn velocity
• Pluto equation uses the Earth moon velocity (27.78 km/s)

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(III)
There Are Modifications In The Equation:
(a)
Mars and Saturn uses the number (2) which isn't found in the original equation
(b)
The Earth and Jupiter uses 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 clearly in 3 planets only (Saturn, Uranus
and Neptune) where the other planets results be complex values
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 complex may as result of the moon
motion effect on Mars motion.
Notice
Saturn equation produces 0.4 and we need to use (1/0.4) to reach to 2.5
2.5 degrees = Saturn orbital inclination

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3-Planet Rotation Period Analysis
(1)
Kepler stated (Planet Orbit Defines Its Velocity)
This law tells, if we know Planet orbital distance we can conclude its velocity
(2)
What's the planet rotation period?
The period the planet needs to rotate around its axis –
Planet velocity be defined by its orbital distance
Based on that
Planet rotation period should be a function in its circumference.
In other words
Planet rotation period should be the period planet needs to rotate around its axis by
using its velocity which is defined by its orbital distance
But
This is NOT the fact!
Only Saturn (9.7 km/s) moves during its rotation period (10.7 hours) a distance =
373644 km (= Saturn Circumference with error 1.3%)
Where
Jupiter (13.1km/s) moves during its rotation period (9.9 hours) a distance =466884 km
which is different from Jupiter Circumference (449197 km) with error (4%)
Not any planet uses its orbital velocity to define its rotation period!
That tells there's something not understood
Notice
Uranus (6.8 km/s) moves during its rotation period a distance = 2.6 x Uranus
Circumference
Neptune (4.7 km/s) moves during its rotation period a distance = 2 x Neptune
Circumference
My question is–why doesn't the planet use its velocity to define its rotation period?
Only Saturn does!!

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(3)
We notice - the physicist consider planet rotation period be created by random process
or initial condition – means – no geometrical reason be found behind.
But
The data tells– Planet rotation period is related to its circumference- the relationship is
not clear but it's the only one
Means
Planet rotation period is a function in this planet circumference only
Not a function in any other data – why?
Because
Planet orbital distance is defined based on its neighbor distance (by my first equation)
(d2
= 4d0 (d-d0) where d = planet orbital distance and d0 its neighbor distance)
And
planet velocity depends on its orbital distance (kepler law)
As a result
Planet rotation period has to depend on this planet circumference only
We have a question
Why should be found a relationship between planet diameter and rotation period?
(4)
Why Should Be Found A Relationship Between Planet Diameter And Rotation
Period?
Planet diameter should be created as a function in planet orbital distance – why?
Because
My first equation (d2
= 4d0 (d-d0) shows that
Each planet orbital distance depends on its neighbor planet orbital distance –means -
the distance from the sun to Pluto be seen as one piece of distance and be divided or
distributed based on one another and regardless any planet data – that tells – the

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planet position is defined before the planet data creation- this is a very important
notice we should analyze
Means
Before the planet creation its orbital distance was defined
Because the designer deals with the distance from the sun to Pluto as one piece of
distance and distributed it based on one geometrical design and in one network of
distances – that made the solar system distances to be similar to the chess board
distances– one group found depending on one another in a network form.
Shortly
The planets be forced for the their orbital distances – and then – these distances define
their velocities -
Let's suppose a planet diameter doesn't create a harmony with its orbital distance –
what will happen? This planet will be broken
Imagine you buy new 8 chairs for your table – can you do that without measurements
for the chairs to know if the place around the table be sufficient and can support these
chairs?
Planet diameter should be in harmony with its orbital distance and with its all motion
data to save this planet in its motion.
But
The planets had migrated – in My Mars Migration Theory – I had proved Mars had
migrated from its original orbital distance (84 million km), because Mars original
position was between Mercury and Venus, and later Mars had migrated to its current
position (227.9 million km)
Also – Pluto had migrated – in fact- Pluto was The moon of Mercury – and Pluto had
migrated with Mars Migration.
If the relationship between the planet diameter and orbital distance was a direct one –
the migrant planets may be destroyed as a result for their migration
Because of that,

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The geometrical design caused the planet diameter to depend on this planet motion
data through its rotation period – by that – the connection be through the velocity and
not through the orbital distance -
That shows the necessity of my fourth equation – it's the equation by which the planet
diameter be created –
Notice
The rule d = r x 1092
d = planet orbital distance and r = planet diameter
this rule proves the direct proportionality was found between planet diameter and
orbital distance as to enable the harmony of this planet motion –
but the planets migration caused to neglect this rule – by that some planets only use it
57.9 million km (Mercury orbital distance) = 4879 km (Mercury diameter) x 1092
149.6 million km (Earth orbital distance) = 12756 km (Earth diameter) x 1092
1433 million km (Saturn orbital distance) = 120536 km (Saturn diameter) x 1092
And even
84 million km (Mars ancient orbital distance) = 7070 km (Mars diameter) x 1092
(Mars diameter has changed by 4% from 7070 km to 6792 km as a result for its
migration and collisions)
Notice
Imagine water moves through a canal and in some point the water motion creates a
vortex – and – for some reason – the water has to loss its minerals and salts in this
vortex – later – the salts caused to create some rock in a tube form – through this rock
tube the water moves – and later the rock tube itself be released from its roots and be
moving with the water around the vortex while the water still moves through the tube
The rock tube dimensions be in full harmony with the water motion
Similar to this description, planet diameter be created depending on its motion

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4-The Equation Discussion
The Discussion Questions
I. Planet Orbit defines its velocity! Why?
II. Why Does The Equation Work From The Earth To Pluto Only?
III. How planet diameter be measured in seconds unit?
IV. How planet rotation period can be equal = 1 second?
V. Why do Jupiter and Saturn use the rate (r/s) in place of (s/r)?
(Question No. I) Planet Orbit defines its velocity! Why?
Kepler stated
(Planet orbit defines its velocity) means we can conclude planet velocity by
mathematical calculations if we know this planet orbital distance -
My first equation (d2
= 4d0 (d- d0)) enable us to conclude all planets orbital distances
by using Math. calculations and one data (Mercury orbital distance =57.9 million km)
Means,
By this one data (Mercury orbital distance =57.9 million km) and by using my first
equation and kepler laws we can conclude by math. Calculations only all planets
orbital distances, velocities and periods (27 data) – why?
Because
Planet (motion data) be created based on geometrical rules and exact equations
Simply, no random process or initial condition effect on planet data creation
Our equation tells a new meaning – that- based on planet velocity – planet rotation
period and diameter can be concluded by mathematical calculations
Means,
Planet creation data can be concluded theoretically – how that can be possible?
Because- planet creation data be created based on geometrical rules
Again

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Neither Random Process Nor Initial Conditions Effect On Planet Data Creation
Can this meaning be clear in our minds?
Planets positions be defined firstly and based on these positions the planets other data
be created based on geometrical rules! That tells planet orbital distance be defined
before planet creation!
This meaning – we have learnt from kepler law- kepler told – planet orbit defines its
velocity – that means – planet velocity doesn't depend on any data of this planet – it
depends only on its orbital distance –
Newton theory of the sun gravity is wrong simply – planet orbital distance depends on
its neighbor orbital distance (by my first equation) and planet velocity depends on its
orbital distance (by kepler law) and no one proof for Newton Concept (Planet Motion
Depends On Its Mass)
But
The meaning is still so strange –
It tells (Planet Position Defines Its Data!!)
Shortly
Planet orbit defines its velocity, and the velocity defines rotation period, diameter and
orbital inclination by our equation (my fourth equation)
That tells, planet position be defined at first and based on this position, planet data be
created in harmony with this position!
This description gives a reference to Bohr system of atom – where the orbit defines
particle energy level (and data) -
The information which we can confirm is that
Planet motion and creation data be created based on geometrical rules and exact
equations. Neither random process Nor initial condition effect on them.

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(Question No. II) Why Does The Equation Work From The Earth To Pluto Only?
Because
The Moon Orbital Period = The Moon Rotation Period
By that the element (s) in the equation be equal 1 and can't be used
Shortly
The equation tells we have 2 parts –
First part is the 3 planets (Mercury – Venus and the moon)
These 3 planets motions integrate together to cause one result which is (The Moon
Orbital Period = The Moon Rotation Period)
(I know, the physicist tells because of the tidal locking the moon periods are equal)
Second part, the rest planets from the Earth to Pluto
The rest planets depends on this fact (The Moon Orbital Period = The Moon
Rotation Period) to create a proportionality between their diameters and rotation
periods and by that the equation be working
This explain supposes the solar system moves depending on one geometrical design
(Explained in the paper definition – Point no. 1)
Notice
The equation works between 2 planets – The Earth Moon (Its Rotation Period = Its
Orbital Period) - and Pluto - The element (s) for Pluto = 14178 – and
Pluto needs 14178 solar days to pass a distance = Pluto orbital distance
That tells we have one system works between the 2 planets to support the equation.
Notice
(29.53 days /27.3 days) = (224.7 days /243 days) =1.0725
29.5 days = the moon day period and 27.3 days = the moon orbital period
224.7 days = Venus Orbital Period and 243 days = Venues Rotation Period
This data supports the idea

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(Question No. III) How Planet Diameter Be Measured In Seconds Unit?
The solar system motion uses (usually) planets and circumferences as periods of time
The following data supports this conclusion
- Jupiter (13.1 km/s) moves during 10921 seconds a distance = 142984 km = Jupiter
diameter-
- Where
- 10921 km = the Earth moon circumference
And
- 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
And
- Uranus (6.8 km/s) moves during 7510 seconds a distance =51118 km = Uranus
diameter-
- Where
- 7510 km = Pluto Circumference
And
- Venus (35 km/s) moves during 12104 seconds a distance =421056 km = Uranus
motion distance during Uranus rotation period -
- Where
- 12104 km = Venus diameter
- The data is a sample of many similar data – there's a geometrical reason to use
planets diameters and circumferences as periods of time.
- The equation uses planets diameters by a similar using in seconds units.

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(Question No. IV) How Planet Rotation Period Can Be Equal = 1 Second?
(Point no. 1)
It's a fact that planet rotation period be used as 1 second by light motion – planet data
proves this fact – as in following
(a)
Light (300000 km/s) travels during 366.7 seconds a distance = 110 million km
(108.2 million km = Venus Orbital distance error 1.8%)
(Earth orbital period has 366.7 Earth rotations periods)
(b)
Light (300000 km/s) travels during 671 seconds a distance = 201 million km
(201 million km = Jupiter Motion Distance during Mercury day period)
(Mars orbital period has 671 Mars rotations periods)
(c)
Light (300000 km/s) travels during 10500seconds a distance= 4950 million km x (2/π)
(4900 million km = Jupiter Orbital Circumference) (error 1%)
(Jupiter orbital period has 10500 Jupiter rotations periods)
(d)
Light (300000 km/s) travels during 24106seconds a distance=9210 million km x (π/4)
(9007 million km = Saturn Orbital Circumference) (error 2.2%)
(Saturn orbital period has 24106 Saturn rotations periods)
(e)
Light (300000 km/s) travels during 42683 seconds a distance=17680 million km x (3/
"1+π") (18048 million km = Uranus Orbital Circumference) (error 2%)
(Uranus orbital period has 24106 Uranus rotations periods)

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(f)
Light (300000 km/s) travels during 89143 seconds a distance=26743 million km
(26810 million km = Mars Neptune Orbital Circumference)
(Neptune orbital period has 89143 Neptune rotations periods)
(g)
Light (300000 km/s) travels during 14178 seconds a distance=26743 million km /(2π)
(26810 million km = Mars Neptune Orbital Circumference)
(Pluto orbital period has 14178 Pluto rotations periods)
Neptune and Pluto depend on the same distance because
(Pluto orbital period / Pluto rotation period) x 2π = (Neptune orbital period / Neptune
rotation period)
Notice
The data proves that – Planet rotation period be used usually as one second by light
motion –
Clearly the data proves this fact and shows that light motion be found behind planet
motion and causes to create planet data.
That also can explain how planet diameter and circumference can be used as periods
of time.

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(Point no. 2)
The Moon Motion Analysis
The moon motion shows that (A Planet Rotation Period Can Be = 1 Second)
If the moon rotates around its axis one time each solar day as the Earth does, where
the moon circumference =10921 km, by rotation the moon moves a distance equal its
circumference =10921 km
During the moon rotation period (27.3 days) the total distance will be
27.3 x 10921 km = 300000 km (error less than 1%)
Light (300000 km/s) travels during 1 second a distance = 300000 km
The 2 distances are equal – that may tell
The moon rotation period = 1 second of light motion
Notice
My theory of the solar system creation tells that
The planets matters and distances be created from one light beam
The produced matter isn't separated from its parent light beam – but moves with it a
unified motion by using different rates of time – based on that
The rate (1 second of light motion = 1 hour of planet motion) is a known one and used
in the solar system motion
Also the rate (1 second of light motion = 1 solar day of planet motion) is a known one
and used in the solar system motion
For example
Light (300000 km/s) travels during (16330 s) a distance = 4900 million km
(Mars orbital period = 16330 hours) -And
Light (1160000 km/s) travels during (4222.6 s) a distance = 4900 million km
(Mercury Day Period = 4222.6 hours)
Notice / light velocity (1160000 km/s) is my theory hypothesis.

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(Point no. 3)
Let's suppose each planet uses its orbital velocity to define its rotation period
Means,
Each planet uses its orbital velocity to pass a distance = its circumference and we will
calculate this period
Based on that
Mercury needs a period 324 seconds to pass a distance = Mercury Circumference
Venus needs a period 1087 seconds to pass a distance = Venus Circumference
Earth needs a period 1345 seconds to pass a distance = Earth Circumference
Earth needs a period 367 seconds to pass a distance = The Moon Circumference
Mars needs a period 886 seconds to pass a distance = Mars Circumference
Jupiter needs a period 34290 seconds to pass a distance = Jupiter Circumference
Saturn needs a period 39039 seconds to pass a distance = Saturn Circumference
Uranus needs a period 23617 seconds to pass a distance = Uranus Circumference
Neptune needs a period 28814 seconds to pass a distance = Neptune Circumference
Pluto needs a period 1600 seconds to pass a distance = Pluto Circumference
The total of periods = 131370 seconds
Light (300000 km/s) travels during 131370 seconds a distance = 39410 million km
39410 million km =1.0725 x 37100 million km (Pluto orbital Circumference) (error 1%)
That tells
The planets periods depend on the distance from the sun to Pluto – by that – the
planets diameters be rated to their distances to the sun through each planet motion
data - this is the concept for which the equation relationship be created.
Notice
The rate 1.0725 proves that (There's A Continuum In The Solar System)
We have discussed this rate before - And refer to it in appendix No. 1

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(Point no. 4)
10500 seconds x 47.4 = π2
x 50428 km (Uranus diameter 51118 km) (error 1.3%)
10500 seconds x 35 = 367500 km (366560 km = the outer planets diameters total)
10500 seconds x 29.8 = 312900 km (= 2 x Neptune Circumference)
10500 seconds x 27.78 = 291690 km (error 3%) (Light motion distance in 1 second)
10500 seconds x 24.1 = 2 x 40080 km (= Earth Circumference)
10500 seconds x 13.1 = 137550 km
10500 seconds x 9.7 = 2 x 51118 km (=Uranus diameter)
10500 seconds x 6.8 = 71400 km (=Jupiter Radius)
10500 seconds x 5.4 = 56700 km
10500 seconds x 4.7 = 49350 km (= Neptune Diameter)
It's interesting that the period 10500 s can be used by all planets velocities to defines
the most planets diameters!
Data Discussion
- The value (10500) is Jupiter data because Jupiter orbital period (4331 solar days)
has (10500) rotation periods (Jupiter rotation period = 9.9 hours)
- This value is used in seconds units (10500 seconds) by all planet velocities
- Logically, not all planets values can be used in place of (10500) because the most
diameters be created based on it – clearly – the geometrical design mentioned this
value (10500) – why??
- We remember Jupiter orbital circumference 4900 million km is the most important
distance in the solar system – and we have studied that – the solar system
geometrical design depended on it.
- Here also,
- The planets diameters definition depend almost on Jupiter value (10500)
- That may explain why Jupiter Diameter = 8 Planets Diameters Total
- All planets diameter = 2 Jupiter diameter + 1 Saturn diameter
- 2 Jupiter Circumferences - 2 Saturn Circumferences = 1 Jupiter Diameter
- (Error 1.3%) (Notice (56700 km = 4 x 14175))

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(Question No. V) Why do Jupiter and Saturn use the rate (r/s) in place of (s/r)?
- The solar system is one light beam – the light creates the planets matters and
distances
- In Saturn position there's some reflection of energy – the data is reflected on
Saturn both sides – that may explain why Saturn orbital distance = Saturn Uranus
distance – because the reflection causes to create equal energies and the distance is
energy –
- In following I show some of this reflected data to prove the idea
- The idea is that,
- 4900 million km is used a distance for Jupiter and a period for Uranus – just light
motion can use the values as distances or periods equivalently – and even Saturn
itself use it as a period and as a distance – the data be in following:
- 4900 million km = Jupiter Orbital Circumference
- And
- 4900 solar days Uranus needs to move a distance = Uranus Orbital Distance
- Also
- 2 x 4900 solar days Neptune needs to move a distance = Neptune Orbital Distance
- 3 x 4900 solar days Pluto needs to move a distance = Pluto Orbital Distance
- For Saturn
- (9007 /9800) = (9800 /10747) where
- 9007 million km = Saturn Orbital Distance
- 10747 solar days = Saturn orbital Period
- 9800 million km = 2 x 4900 million km
- I try to prove that, the value 4900 is the basic value in the solar system
- We have a clear reference for one geometrical design be found behind.

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(My 5th
Equation) Planet Velocity Is A Complementary One
v = Planet Velocity
t = another planet velocity be used as a period of time
Example
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 35km/s the error 5%)
km
t
v 322
=

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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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1.16 million km/s = Light Supposed Velocity
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
Conclusions
- Planet Velocity Be Created Complementary For Another Planet Velocity
- The 2 planets velocities be controlled by one distance (1.16 million km)
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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3- Planet Velocity Analysis
3-1- Preface
3-2- Planet Velocity In Comparison with Light velocity
3-3- Planet Velocity Effect On Its Motion
3-1- Preface
- This point tries to prove that (Planet Velocity is a Complex Value)
- That means, planet velocity be defined by some complex mathematical design,
- Here, at first we have to disprove Newton Theory which tells (Planet motion
depends on its mass) and (planet moves by the sun gravity effect on it)
- These ideas are so wrong and nothing of them be related to the facts
- Planet velocity is a very good point of discussion to disprove Newton ideas
frequently and proves that a complex geometrical design be found behind –
- In fact
- While Newton theory of the sun gravity is lived for more than four centuries- it
should be so interesting discussion which disproves it – but in fact – the discussion
provides more interesting idea than Newton theory refutation – the discussion tells
us–A Planned Geometrical Design Be Found Before The Solar System Creation-
- Planet velocity discussion can do this role perfectly – and we can through planet
velocity analysis prove that Newton theory is wrong – and more important to
discover that A Planned Design Was Found Before The Solar System Creation
- Let's try to make our discussion so fruitful and useful
- In following let's write each rule and data we know about planet velocity,
- Notice
- Planet velocity definition is a complex process because planet velocity be defined
as a function in light velocity – this idea we try to discuss and prove in our
discussion.
- Shortly, let's ask…

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- How Be Planet Velocity Defined?
- (Item No. 1)
- Kepler Stated (Planet Orbit Defines Its Velocity)
- (Item No. 2)
- Planet velocity be decreased with its orbital distance increasing, by that we have a
proportionality between planet velocity and its distance – we notice No similar
proportionality be found between planet velocity and its mass because the
velocities be decreased gradually but the masses can't support this graduation.
- (Item No. 3)
- My (1st
Equation) d2
= 4d0 ( d- d0)
- d = Planet Orbital Distance and d0= Its neighbor Planet Orbital Distance
- The equation depends on the planets order, means, just 2 neighbor planets can be
used in this equation -
- The equation shows, each planet orbital distance be defined based on its neighbor
planet orbital distance
- My (2nd
Equation)
- (V0)2
/(V)2
= 4 x (1- (V)2
/(V0)2
- V = Planet Velocity and V0= Its Previous neighbor Planet Velocity
- The equation also depends on the planets order, means, just 2 neighbor planets can
be used in this equation -
- The equation shows, each planet velocity be defined based on its neighbor planet
velocity
- My (3rd
Equation) (depend on Kepler Laws)
- (d1/d2) = (v2/v1)2
- This equation doesn't depend on the planets order – means- any 2 planets can be
used in this equation –

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- We have noticed that before
- My first equation tells, 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.
- That tells (Planet Velocity Be Defined By Some Complex Process) – also because
of my equations no. (4 and 5)
- My (4th
Equation)
- (v1/v2) = (s/r) = I
- v = Planet velocity and r= a diameter of one of the 2 planets
- s= the planet rotation periods number in its orbital period
- (the value "s" is belonged to the planet whose diameter is "r")
- I= Planet orbital inclination (of the planet whose diameter is "r")
- The equation tells, planet diameter and rotation period be created as a function in
the rate (v1/v2)
- My (5th
Equation)
- (vt) = 322
- The equation also some more complex process behind planet velocity definition

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- Item No. 4)
- Why does the moon rotate Metonic Cycle? This cycle be continuous for 19 years
but Earth doesn't move this Cycle, why the moon does it?
- The fact is that, Uranus orbital distance 2872 million km = 19.2 Earth orbital
distance 149.6 million km
- If the 2 planets velocities be equal, while Uranus revolves around the Sun one
complete revolution Earth would revolve around the sun 19 revolutions (19 years)
- The number (19 years) be so attractive to suppose that (the moon may move
Metonic Cycle under Effect of Uranus motion),
- Of course the gravitation equation refuses this idea and also the 2 planets velocities
aren't equal – by that – the hypothesis is wrong in its principle
- But the number (19 years) give a great attraction to search behind – we have
discussed that in previous paper
- The point is, we have a reference tells the planets velocities should be equal!!
- (Item No. 5)
- Planets velocities shows strange features let's refer to one of them in following…
- (6.8)2
= 47.4 (error 2.5%)
- (5.4)2
= 29.8 (error 2.1%)
- (9.7)2
= 2 x 47.4 (error 0.7%)
- (4.7)2
= π x 6.8 (error 3.4%)
- (24.1)2
= 4π x (6.8)2
(No error)
- (13.1)2
= 171.6 = 2π x 27.78 (1.7%)
- (17.9)2
= 322.22
- 6.8 km/s = Uranus Velocity 47.4 km/s = Mercury Velocity
- 5.4 km/s = Neptune Velocity 29.8 km/s = Earth Velocity
- 9.7 km/s = Saturn Velocity 17.9 km/s = Ceres Velocity
- 4.7km/s = Pluto Velocity 24.1 km/s = Mars Velocity
- The data shows so strange feature… why these values be rated as such??

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3-2- Planet Velocity In Comparison with Light velocity
I - Data
The data supposes that – a light beam its velocity 1.16 million km per second be
found and effective in the solar system motion –
Means, we compare planet velocity with 2 velocities of light – the light known
velocity 300000 km/s and the light supposed velocity 1.16 million km/sec.
300000 km = 3600 km x 83.33
1160000 km = 3600 km x 322.2
Column No. 1 Column No. 2
Planet 322.22 83.333
Mercury 47.4 6.8 1.76 = (π)0.5
Venus 35 9.2 2.38
Earth 29.8 10.8 2.8
The moon 27.78 11.6 3
Mars 24.1 13.37 3.4578
Ceres 17.92 17.92 4.65
Jupiter 13.1 24.6 6.361 = 2π
Saturn 9.7 33.21 8.6
Uranus 6.8 47.4 12.25 = 4π
Neptune 5.4 59.67 15.432
Pluto 4.7 68.56 17.73

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II- Discussion
- Let's summarize the idea in following
- (1)
- We have 2 light velocities effective on the solar system motion (my hypothesis)
one light velocity is the known one (300000 km/s) and the other is the supposed
one (1160000 km/s) – The 2 light velocities effect on planet motion –Why?
because the planet itself is A Point on this light beam – means –
- This light beam creates one trajectory of energy and the planet is a point on this
trajectory by that (all planet data) be defined by (light motion) – means
- We see the moon and considers the moon diameter is 3475 km – we see that by
our eyes– but – the fact is that – A light beam moves behind and creates this
diameter 3475 km by this light motion –
- Here we try to analyze how the 2 light velocities effect on planet motion trajectory
- (2)
- (The Table Column No. 1)
- The data in column No. (1) compares between planet velocity and Light supposed
velocity (1.16 million km/sec)
- The planets velocities are complementary clearly – as a result
- 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
- Venus (35 km/s) moves during 2 x 4.7 hours a distance = 1.16 million km
- Pluto (4.7 km/s) moves during 2 x 35 hours a distance = 1.16 million km
- (error 2%)
- Earth (29.8 km/s) moves during 2 x 5.4 hours a distance = 1.16 million km
- Neptune (5.4 km/s) moves during 2 x 29.8 hours a distance = 1.16 million km

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- Mars (24.1 km/s) moves during 13.1 hours a distance = 1.16 million km
- Jupiter (13.1 km/s) moves during 24.1 hours a distance = 1.16 million km
- (Error 2%)
- (In fact Jupiter moves during 24.6 hours a distance 1.16 mkm)
- Venus (35 km/s) moves during 9.7 hours a distance = 1.16 million km
- Saturn (9.7 km/s) moves during 35 hours a distance = 1.16 million km
(error 5%)
- Why be planets velocities complementary one another based on this distance 1.16
million km?
- Because the planets velocities be created as a function in the light supposed
velocity (1.16 million km per second)
- We have to accept this answer as a hypothesis till perform our analysis
- In all cases we should find the real geometrical reason for which the planets
velocities be created complementary one another based on this velocity or distance
per a second (1.16 million km)
- (3)
- We have 2 light velocities (0.3 million km/s) and (1.16 million km/sec) how these
2 velocities be in harmony with each other?
- (1.16/0.3) x 2π = 24.3
- Where
- 24.3 refers to (24 hours = the solar day) (error 1%)
- 1.16 and 0.3 are the 2 light beam velocities (300000 km/s and 1160000km/s)
- This calculation tells how the matter be created – let's summarize it:
- The matter creation process INPUT was the light beam with velocity 1.16 million
km per second - and the creation process OUTPUT was (matter + space+ light
beam with velocity 300000 km/sec)
- The light (300000 km/sec) be created as a side product in the matter creation

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- But
- What's this rate (2π)??
- This rate (2π) is the reason of all circular motions in the universe – while – the
light moves in a straight lines – and the born matter should move with its parent
light beam in straight lines – the production of the light known velocity (300000
km/s) caused the straight line trajectory of motion to be in circular trajectory and
that causes the planets motions trajectories to be in elliptical forms –
- We should notice that – the matter creation is a process done for one time – and
the light known velocity (300000 km/s) be produced as a side product also for one
time only – after the creation – the light (300000 km/s) travels and the matter be
created forever – because
- The matter can't return again into its original light beam form because a part of the
energy be traveled already which is the light known velocity – means –
- The input was a light beam its velocity 1.16 million km per second (Energy =E)
and the output was (matter + space + light known velocity) (Energy = E)
- But
- The light known velocity (300000 km/s) be traveled and disappeared by that the
total energy be less than (E) and by that the energy in the matter be a prisoner in
this matter forever
- The light known velocity (300000 km/s) traveled and caused to decrease the total
energy (E) and also caused the motion trajectories to be in circular or elliptical
forms in place of the straight line forms.
- Let's try to see this picture – after the matter creation –
- A light beam will be seen traveling to the end of the universe and the produced
matter will move in circular trajectories forever – but – in darkness – no more light
be found – because – the source (1.16 million km per second) be used to produce
(matter+ space + light known velocity) and the light known velocity is traveling
now and moving to the end of the universe where the matter be created by the rest

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energy in movable case and the trajectory be a circular one which will not change
again by that the matter will revolve in the circular trajectory forever in darkness.
- The motion is available but no light be available!!
- (3) (The Table Column No. 2)
- The column no.2 deals with light known velocity (300000 km/s)
- We see clearly that,
- 1 second of light Motion be = 2π hours of Jupiter Motion,
- 1 second of light Motion be = 4π hours of Uranus Motion,
- But
- Jupiter (13.1 km/s) moves in (24.6 hours) a distance 1.16 million km
(Light supposed velocity for 1 second)
- And
- Uranus (6.8 km/s) moves in (24.6 hours) a distance 0.6 million km
(Light known velocity for 2 seconds)
Please review my paper
A Planned Design Be Found Before The Solar System Creation
https://www.academia.edu/s/037dae1264
or
https://www.academia.edu/79019056/A_Planned_Design_Be_Found_Before_The_Solar_System_Creation
or
https://app.box.com/s/hazuljgpj4z4dsrv6pny5akv1blxzokm
or
https://app.box.com/s/drf36zvpskhk8wze6ojuhhb3a0mf8gr0

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3-3- Planet Velocity Effect On Its Motion
- We don't prove here – if a light beam its velocity 1.16 million km /sec be found
really in the solar system or not - also we don't prove if light motion effect on
planets motions or data –
- We here suppose just there are 2 velocities of light which are (the known one 0.3
million km/sec and the supposed one 1.16 million km/sec)
- It's just a hypothesis –
- And
- We try to conclude if these 2 velocities can effect on planet motion - so, let's try to
discover that in following
- (Item No.1)
- (1.16 million km /0.3 million km) = 3.866
- And
- 3.866 x 2π = 24.3
- The value (24.3 hours) is different from (the solar day 24 hours) by (1%) and
different from (24.6 hours = Mars rotation period) by (1%)
- Shortly
- If 2 light beams effect on planet motion, the solar day will be created as (24 hours)
also (Mars rotation period 24.6 h can be created as a result)
- But what's this (2π)? how to explain it??
- (Item No.2)
- 1 second of light motion = 2π hours of Jupiter motion
- It's easy to prove
- Light travels a distance = 300000 km in one second
- Jupiter (13.1 km/s) moves a distance = 300000 km in 6.28 hours (=2π hours)
(error 1.2%)

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- Also
- Uranus (6.8 km/s) moves a distance = 300000 km in 12.56 hours (=4π hours)
(error 2.5%)
- Why the rate (2π or 4π Be Produced Here?)
- It's a geometrical process – as we have discussed - let's remember it here
- 3.866 x 2π = 24.3
- Because
- 3.866 be produced as a rate between the 2 light velocities (1.16 and 0.3)
- And
- 24 hours = the solar day which is the basic period of time
- The rate between these 2 values be = 2π
- The rate (2π) be produced as a geometrical necessary product.
- Can this analysis be useful for our understanding for the solar system motion?
- (Item No.3)
- My theory claims that
- The solar planets and their distances be created out of one light beam its velocity =
1.16 million km /sec
- In details
- The planets matters and distances be created out of one light beam whose velocity
1.16 million km/sec.
- The Matter is created out of light but the matter doesn't separate its light parent but
moves together one unified general motion by using different rates of time – that
may explain – why 1 second of light motion = 2π hours of Jupiter motion –
because light and matter move together by using different rates of time
- Now what's the sun?
- The sun is the planets motions result and not reason – that means -
- The sun rays energy isn't provided by any nuclear interaction inside the sun-
instead – the planets motions energies be accumulated and by using different rate

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of time for the sun motion – this accumulated energy be using by the sun to create
its rays – that explains – why planets motions distances total in 1461 solar days =
light motion distance in one solar day because the sun uses the rate 1461 days of
the planets = 1 solar days of the sun motion
- Here we have one more shock, the sun doesn't cause the planets motions instead
the sun itself be created by the planets motions – that tells more data – the sun be
as a cycle because it's created based on the planets motions – that means – the sun
existence depends on a cycle
- That tells the sun rays (300000 km/s) be created later after the planets creation and
motion

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4- Planets Masses Rates Definition
I –Data
(1) (0.642/0.33) = 1.94
(2) (0.073 /0.0146) = 5.1
(3) (4.87/5.97) = 0.81
(4) (4.87 /0.073) = 66.7 = (708.7 /10.7) = (655.7/9.9)
(5) (1898 / 568) = 3.34
(6) (568/4.87) = 116.6
(7) (0.33 /0.0131) = 49
(8) (568/102) = (0.073 /0.0131) = 5.56 = (29.8/5.4)
(9) (102 /5.97) = 17.1
(10) (1898 /102) = (5.97/0.33) = 18.6 (error 2.8%)
No. (11) (The Planets Masses Total 2666.7) / (Uranus mass 86.8) = 30.772
No. (12) (The inner planets masses total 11.8) /(Venus mass 4.87) = 2.44
Where
- 0.33 x 1024
kg = Mercury Mass
- 4.87 x 1024
kg = Venus Mass
- 5.97 x 1024
kg = Earth Mass
- 0.0.73 x 1024
kg = The Earth Moon Mass
- 0.642 x 1024
kg = Mars Mass
- 1898 x 1024
kg = Jupiter Mass
- 568 x 1024
kg = Saturn Mass

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- 86.8 x 1024
kg = Uranus Mass
- 102 x 1024
kg = Neptune Mass
- 0.0131 1024
kg = Pluto Mass
- 47.4 km/s = Mercury velocity
- 24.1 km/s = Mars velocity
- The rate between the masses be so near to (planet orbital inclination)
- For example
- (Mars Mass / Mercury Mass) = 1.94
- While Mars orbital inclination = 1.9 degrees
- 7 deg (Mercury orbital inclination) = 1.9 deg+ 5.1 deg (the moon orb. Inclination)
- Shortly
- The masses rates can be used as planets orbital inclinations (rates)
- But the inclinations configuration between the planets caused these rates to be
effected by some change as a result.
- Also some specific rates – for example – the moon orbital inclination 5.1 degrees
be produced by the masses rate between the moon and Pluto – for myself – I don't
find this data is a strange one – I know that - Pluto motion is the most planet
motion effect on the Earth moon motion and vice versa – I know by these planets
data analysis but the geometrical rules are still absent by that I can't support my
answer with geometrical analysis –
- The rate 116.6 is the most important ever because of Venus day period (116.75
days) – also I can't prove how Saturn motion effect on Venus cause to produce this
rate 116.6 and Venus day 116.75 days – I can provide only data – such as
- (Venus circumference)2
= Saturn Orbital Distance (error 1%) and
- Venus Circumference x π = Saturn diameter (error 1%)
- Shortly, More Geometrical Rules Cover The Direct Relationship

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- Notice
- (102 /5.97) = 17.1
- We have seen that – the couple Earth and Neptune velocities be used with my
fourth and fifth equations - both equations use the same couple
- Now
- 102 x 1024
kg = Neptune Mass
- 5.97 x 1024
kg = Earth Mass
- 17.2 degrees = Pluto orbital inclination
- 17.4 degrees = The Inner Planets Orbital Inclinations Total
- 23.6 degrees = The outer Planets Orbital Inclinations Total
- 23.4 degrees = Earth axial tilt
- That tells we have a system (17.2 and 17.4 error 1%) and (23.6 and 23.4 error 1%)
- It's one system create a distribution for planets
- The value 17.2 depends on (Earth and Neptune masses) why??
- Earth Circumference = The 5 Inner Planets Diameters Total
- Earth Mass = The 4 Inner Planets Diameters Total
- But
- Neptune Circumference = The outer Planets Diameters Total (without Neptune)
- Again we deal with one system – the 2 points (Earth and Neptune) be the 2 basics
for this job.
- Now let's take a look on each data of the masses group of data

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More Discussion
(1) (0.642/0.33) = 1.94
(Mars mass / Mercury mass) = 1.94 but Mars orbital inclination = 1.9 deg – and
7 deg (Mercury orbital inclination) = 1.9 deg + 5.1 deg (the moon orbital inclination)
The data shows an interaction between the 3 planets orbital inclinations
(2) (0.073 /0.0146) = 5.1
(Notice 0.0146 = Pluto old registered mass but the new is 0.0131)
5.1 deg (the moon orbital inclination)
17.2 deg (Pluto orbital inclination) = 7 deg + 2 x5.1 deg (the moon orbital inclination)
(3) (4.87/5.97) = 0.81
Venus Mass = 80% of Earth Mass (0.8 deg = Uranus orbital inclination).
(4) (4.87 /0.073) = 66.7 = (708.7 /10.7) = (655.7/9.9)
(5) (1898 / 568) = 3.34
Jupiter Mass = 3.34 x Saturn mass (3.4 deg = Venus orbital inclination)
What's the relationship between Jupiter and Saturn with Venus? Data no. (4) answers
708.7 h = The Moon Day Period
655.7 h = The Moon Rotation Period
10.7 h = Saturn Rotation Period
9.9 h = Jupiter Rotation Period
Some complex geometrical interaction caused to produce Venus orbital inclination
3.4 deg by effect of Jupiter and Saturn masses .
Notice
3.4 deg (Venus orb. Inc) + 7 deg (Mercury orb. Inc)= 2 x5.1 deg (the moon orb. Inc)

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(6) (568/4.87) = 116.6
Saturn mass = 116.6 Venus Mass
We have seen this rate before and I have claimed that – Venus day period (116.75
days) be produced as a result for this rate
Also
26.6 deg = Saturn axial tilt but 26.6 deg + 90 deg = 116.6 degrees
The data some many geometrical interactions – the point is that – Neither the
geometrical rules be known Not their geometrical effects by that we learnt nothing
form the data.
(7) (0.33 /0.0131) = 49
Mercury Mass = 49 Pluto Mass but (49 =72
) and
Mercury Orbital Inclination = 7 degrees
(8) (568/102) = (0.073 /0.0131) = 5.56 = (29.8/5.4)
So interesting data
Saturn Mass = 5.56 Neptune Mass and
The moon mass = 5.56 Pluto Mass
And
Earth velocity = 5.56 Neptune velocity –
We remember that the couple (Earth and Neptune) be used in my (4th
) and (5th
)
equations – means – the 2 equations use this same couple which isn't repeated for any
other couple
5.6 degrees = 0.5 deg + 5.1 deg (The Moon Orbital Inclination)
And
0.5 degrees = The Moon Angular Diameter
(9) (102 /5.97) = 17.1
We have studied this data

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(10) (1898 /102) = (5.97/0.33) = 18.6 (error 2.8%)
(Jupiter Mass / Neptune Mass) = (Earth Mass / Mercury Mass)
No. (11) (The Planets Masses Total 2666.7) / (Uranus mass 86.8) = 30.772
31 =π3
No. (12) (The inner planets masses total 11.8) /(Venus mass 4.87) = 2.44
2.5 deg = Saturn Orbital Inclination
I wish the data help to prove my theory that
The Solar System Creation And Motion Depends On One Geometrical Design

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Appendix No. 1
There's A Continuum In The Solar System
The continuum effect proves (From One Energy The Solar Planets Matters And
Their Distances Be Created)
The continuum is a force effected on a wide range of data – the data types be different
but the effect is the same! Let's see an example
Example / (the rate 1.0725)
This rate (1.0725) effects on a wide range of data in the solar system – the data be
different in types
(i) 40% Of All Distances In The Solar System Be Effected By This Rate 1.0725
(ii) 50% of Planets axial tilts also be rated by it
(iii) many cycle periods be rated also by it and many other data
Let's refer to some data in following
Jupiter Distances
778.6 mkm (Jupiter orbital distance) = 1.0725 x 720.7 mkm (Jupiter Mercury distance)
720.7 mkm (Jupiter Mercury distance) = 1.0725 x 671 mkm (Jupiter Venus distance)
671 mkm (Jupiter Venus distance) = 1.0725 x 629 mkm (Jupiter Earth distance)
629 mkm (Jupiter Earth distance) = (1.0725)2
x 550 mkm (Jupiter Mars distance)
Planets Axial Tilts
(28.3/26.7) = (26.7/25.2) = (25.2/23.4) = (122.5/113.4) = ( 97.8 /91.3) = 1.0725
28.3 degrees = Neptune Axial Tilt 26.7 degrees = Saturn Axial Tilt
25.2 degrees = Mars Axial Tilt 23.4 degrees = Earth Axial Tilt
113.4 degrees= 23.4 degrees +90 degrees
122.5 degrees= Pluto Axial Tilt 97.8 degrees = Uranus Axial Tilt
91.3 degrees = 90 degrees + 1.3 degrees (Jupiter Orbital Inclination)
The rate (1.0725) also be used between cycles periods for example

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Cycles Periods
(29.53 days /27.3 days) = (224.7 days /243 days) =1.0725
29.5 days = the moon day period and 27.3 days = the moon orbital period
224.7 days = Venus Orbital Period and 243 days = Venues Rotation Period
We can explain the continuum effect only by the hypothesis (from One Energy the
planets matters and distances be created) -Because
The continuum was in the original one energy, that explains how it effects so widely
on data.

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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
ORCID https://orcid.org/0000-0002-1041-7147
Facebook https://www.facebook.com/gergis.tawadrous/
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Curriculum Vitae http://vixra.org/abs/1902.0044
Please scan the figure (ORCID)

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