The Solar System Distances Network Theory

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
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Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
The Solar System Distances Network Theory
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 –21st
August 2021
Abstract
Paper hypothesis
- The Solar System Distances Be Created In A Network Form
The hypothesis explanation
- The paper hypothesis tells that,
- The solar system be created by energy provided by a light beam its velocity =1.16
mkm/s
- The light motion velocity creates the solar planets matters and their distances
- The solar system distances creation be done by one light beam motion energy and
caused the distances to be created in a network form
- That means,
- The solar system distances be created and distributed based on one geometrical
design. Where no distance be created independent from the other distances
- Based on this vision
- The distances among the solar planets be similar to the railways but the planets be
similar to the trains.
- As a result no planet can move freely in space, on the contrary, the planets move
obligatory motions on their railways
Please scan the figure (ORCID)

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1-Introdcution
- This paper provides The Solar System Distances Network Theory
- The Theory tells that,
- The solar distances be created in a network form. Where the distances be created
depending on one another based on one geometrical design.
- This paper discussion should be considered as a part of the previous paper
discussion.
Please review the previous paper.
Light Velocity Disproves Newton Theory Of The Sun Mass Gravity (Revised)
https://www.academia.edu/50922874/Light_Velocity_Disproves_Newton_Theory_Of_The_Sun_Mass_Gravity_Revised_
https://www.slideshare.net/Gergesfrancis/light-velocity-disproves-newton-theory-of-the-sun-mass-gravity-revised
https://app.box.com/s/47fwd0gshir636xt0i3wpso8lvvl8vnv/file/848070482722

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2-The Solar System Distances Network Theory
2-1 The Theory Basic ideas
2-2 Jupiter Motion Analysis Proves The Network Theory
2-3 Saturn Motion Analysis Proves The Network Theory
2-4 The Solar System Distances As A Network (Examples for discussion)

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2-1 The Theory Basic ideas
(I) The Distances Description
- The Solar System Distances Be Created Based On One Geometrical Design.
- That makes the distances similar to the railways and the planets be similar to the
trains.
- Also that makes the distances to be created in a network form.
(II) The Distances Creation
- (Distance = Energy)
- The theory supposes, the distance be created by a light motion. that means, the
space is created by light motion energy where the light provides the required
energy for these distances creation.
- The light motion energy creates the space and matter, by that, the space is created
in the same process by which the matter be created. means, no matter without
space and no space without matter.
- The light motion provides the required energy for matter and its distance creation.
- The theory supposes that, the solar system be created out of one light beam its
velocity =1.16 mkm/sec. and by that the planets matters and their distances be
created by this light beam energy. and that makes all planets data be created
complementary with one another and alls distances be created in a network form.
- Because (Distance = Energy), and the distances are created based on one
geometrical design, so the distances total be equal the original distance which
transports their energy (this idea will be discussed in more detail later)
(III) Can The Distances Network Form Effect On Planet Motion Description?
- The planet can't move freely in space, on the contrary, the distance be created and
defined by light motion energy. and that forces the planet to move in a defined
obligatory trajectory of motion. the planet has to follow the light motion trajectory
because the light energy created the distance through which the planet moves.

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- Now
- Light motion is distinguish from the planet motion through the same distance,
because light motion (for first) time will create the distance but planet motion
doesn't need energy because the distance be created already.
(IV) What's The Useful Result Of The Distances As A Network Form?
- The distances network form be useful because we can conclude any distance value
based on the other distances. Where, the distances can't be independent distances
from one another, but the distances be created based on one another, which enable
us to predict any distance value and also can help us to explain why any distance
has its value…
(V) What Does Prove The Distances Network Theory?
- 50% Of All Distances in the solar system be equal one another (Appendix no. 1
provides a list for these distances)
- 40 % Of All Distances in the solar system be rated with one another based on the
same one rate (1.0725) (Appendix no. 1 provides a list for these distances also)
- The data analysis be provide in this paper discussion proves that many distances be
created based on geometrical design and in proportionality with other distances.
(VI) Why The Distances Be In A Network Form?
- While we are so confused to explain the solar planets order. The fact is that, the
solar planets order is a typical to the interference frame be created by light
coherence in the double slits experiment (Young Experiment).
- the frame which produced bright fringes and dark fringes. If we accept that the
bright fringes be the planets and the dark fringes to be the distances (space) by that
the planets order be a typical to the interference frame.

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- Where the greatest fringe be in the middle (Jupiter), and the fringes be decreased
on both sides regularly (Mars is exceptional because of its migration). But still
Uranus is exceptional because Neptune should be in its position.
(VI) Why The Distances Be In A Network Form?
- The solar system motion depends on the planets motions interaction and not on the
sun mass gravity because the sun herself be created after all planets creation and
motion.
- The planets motions interaction depends on planets motions for equal distances in
defined periods of time. (the equal distances method will be discussed later)
- Because the planets motions depend on the equal distances method, because of that
- 50% of all distances be equal one another
- 40% of all distances be rated with one another with the rate (1.0725)
- The planets orbital and internal distances be created in a network form as a result
for the planets motions interaction and specially because of the planets motions
dependency on The Equal Distances Method.
- Notice
- The equal distances method be discussed deeply in the previous paper

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

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

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

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

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

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

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- This data perfectly is similar to the previous one. means, Earth and Venus orbital
circumferences and their distances to Jupiter all these distances be created in one
process by an interaction of motion between light supposed velocity (1.16 mkm/s)
motion and Saturn motion
- While the light motion for 1 second be = planet motion for 1 soar day.
- Equation no. (6) can help our investigation
Equation no. (6)
100733 mkm = 929 mkm x 108.2 mkm = 671 mkm x 149.6 mkm
- Where
- 108.2 mkm = Venus orbital distance
- 149.6 mkm = Earth orbital distance
- 6714 mkm = Venus Jupiter Distance
- 929 mkm = Earth Jupiter Distance (The 2 Planets Be On The Sun 2 Sides)
- How to understand this equation?
- Light supposed velocity (1.16 mkm/s) travels during a solar day (86400 s) a
distance = 10733 mkm
- Equation no. (6) tells that, from one source of Energy the 4 distances be created
- Specifically
- From the distance 100733 mkm, the 4 distances which are
- 108.2 mkm = Venus orbital distance
- 149.6 mkm = Earth orbital distance
- 929 mkm = Earth Jupiter Distance
- These 4 distances be created from the same one distance (100733 mkm)
- This information is supported perfectly the network theory ideas.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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References and Biography
(Preprint No. 1) Mars Migration Theory
https://www.academia.edu/49051037/_Preprint_No_1_Mars_Migration_Theory
(Preprint No. 2) The Moon Orbital Motion Equation
https://www.academia.edu/49051029/_Preprint_No_2_The_Moon_Orbital_Motion_Equation
The Solar Planets Motions Description (Revised)
https://www.academia.edu/49464125/The_Solar_Planets_Motions_Description_Revised_
The Moon Orbital Motion Geometry (II)
https://www.academia.edu/45181646/The_Moon_Orbital_Motion_Geometry_II_
Light Motion Features Are Discovered in Planet Motion
https://www.slideshare.net/Gergesfrancis/light-motion-features-are-discovered-in-planet-motion
Dr. Budochkina, Svetlana Aleksandrovna
Associate professor - Candidate of physico-mathematical sciences (2005)
http://www.mathnet.ru/eng/person22119
List of publications on Google Scholar
List of publications on ZentralBlatt
https://mathscinet.ams.org/mathscinet/MRAuthorID/757317
http://elibrary.ru/author_items.asp?spin=6087-3245
http://orcid.org/0000-0003-3447-0425
http://www.researcherid.com/rid/G-7453-2014
http://www.scopus.com/authid/detail.url?authorId=6507007003
https://www.researchgate.net/profile/Svetlana_Budochkina
Full list of
publications:
http://web-local.rudn.ru/web-
local/prep/rj/index.php?id=2944&p=15209
Mr. Gerges Francis Tawdrous +201022532292
Physics Department- Physics & Mathematics Faculty
Curriculum Vitae http://vixra.org/abs/1902.0044
E-mail mrwaheid@gmail.com
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Also https://www.slideshare.net/Gergesfrancis?utm_campaign=profiletracking&utm_medium=sssite&utm_source=ssslideview
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The Solar System Distances Network Theory

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