Another Force Effect On The Earth Moon Motion

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
Another Force Effect On The Earth Moon Motion
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 – 1st
November 2020
Abstract
Paper Hypothesis
- There's another force effects on the Earth moon motion
The Hypothesis Explanation
- The moon data shows that, there's one more force effects on the moon motion, the
hypothesis means that, the moon motion revolving around Earth is a result of
effect of 2 different forces – these 2 forces are (1) Earth gravity force (2) 2nd
force
effect on the moon motion
- This 2nd
force must be equal Earth gravity force, and it effects on the moon motion
orbit from the other side, and can be proved in the moon motion trajectory
- This 2nd
force has a great effect on the moon motion understanding, because
without it the moon orbit can't be created, but this force origin is a puzzle!…….
- The Earth mass = 4 planets (Mercury + Venus + Mars + the moon) masses total,
this data tells that, the 4 inner planets masses may produce a total gravity = Earth
gravity – but this conclusion can easily be disproved because of the great distances
between the 3 inner planets and the Earth moon
- The dilemma is that, the 2nd
force is a real one and its effect can be seen in the
moon motion, and the masses equality provided the required data to explain this
real force, but the distances disproves simply this proof…. How to solve that?
- The logical conclusion is that, the inner planets are attracted to each other by some
electric field which can provide such force spite of the long distances – i.e. the
gravity force between the inner planets doesn't depend on masses gravity but
charges attraction (Electric Field) and that can solve the distance question…
Paper Objective
- This paper objective is limited to prove the 2nd
force existence. The paper tries to
show this force effect as clear as possible on the moon motion trajectory, in
addition to refer to this 2nd
force origin.

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1- Another Force effects On The Moon Motion
1-1 The 2nd
Force Discovery 1-2 The 2nd
Force Proves & Origin
1-1 The 2nd
Force Discovery
I-Data
The Moon Orbital Triangle
Let's defines accurately this triangle data
AB = BC = 86000 km
EC =373000 km = Earth moon distance at total solar eclipse radius (error 1%)
EB = Perigee radius =363000 km
EF= The moon orbital distance =384000 km
ED = Apogee Radius =406000 km
Notice (I)
CD = 95930 km (The blue arrow, the value is defined by Pythagoras rule)
Notice (II)
The Point C is NOT The Moon, because the distance BC =86000 km and the distance
between perigee and apogee =43000 km, means this point is far from apogee with
43000 km, that's the point C is NOT the moon –
But (E) can be considered as Earth Point
And all other points can be considered really relative to Earth point, Just the Point C
causes this triangle NOT correspondent to the moon real data
Let's see the triangle angle
- Angle E = 13.33 degrees
- Angle A = 45 degrees
- Angle ADC = 116.3 degrees
- Angle ACD = 18.7 degrees
- Angle CDB = 63.7 degrees
- Angle DCB = 26.3 degrees
Let's start our analysis in following

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II- Discussion
- In The Figure
- The blue ball is the moon, and we see the Earth on one side, and on the other side
we see The Effect point of The Second Force,
- (The Effect point) in this figure, is seen in the moon orbital triangle in point (A)
- The moon orbital triangle simply proves the paper hypothesis, because the point
(A) is found out of the moon orbit (far from apogee radius with 43000 km), means,
the moon can never reach to this point (A), although this point (A) is one pillar of
the moon orbital triangle, i.e. the moon triangle data can't be created or defined
without the point (A)… how to explain that? it's simply another force effects on
the moon motion…
- The moon orbital triangle is a rich source of data, let's summarize some of it:
o The moon motion4 basic radiuses (Perigee =363000 km, Total solar eclipse
=377000 km, Orbital distance =384000 km, Apogee =406000km), these 4
radiuses can be defined based on each other by Pythagoras rule if one
dimension of the right triangle =86000 km (i.e. these 4 radiuses are defined
by this moon orbital triangle)
o The hypotenuse (CF) =88000 km, (the moon daily displacement=88000 km)
that means, the moon daily motion is defined also based on this orbital
triangle.
o From this orbital triangle the following question is raised, "If the moon
motion 4 basic radiuses are defined based on Pythagoras rule, so, Does the
moon use Pythagoras rule in its daily motion? and why?
o In answer this question, We have learnt that, The Moon indeed uses the
Pythagoras triangle in its daily motion, and for the question Why? the
answer is the following…
o The moon daily displacement is 88000 km, but the moon doesn't use this
displacement totally as daily displacement! Why Not? because during 29.53
days if the moon daily displacement =88000 km and can't be changed, the
total distance during 29.53 days = 2.59 mkm = 2π x 0.41 mkm (the moon

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apogee radius =0.406 mkm), and that means, if the moon uses only the
distance 88000 km as a daily displacement that means the moon will have to
move only through the apogee orbit (r=0.406 mkm) and can't be near to
Earth and never be nearer than the apogee radius
o The intelligent moon uses Pythagoras triangle, and uses its daily
displacement 88000 km as the right triangle hypotenuse, where the real
displacement (L = 88000 km Cosθ), where (θ) is the smallest angle in the
triangle – by this technique, the moon moves daily 88000 km but the real
displacement can be shorter, and by that the moon could reach to more near
orbit to The Earth
o For Example, the available displacement in perigee orbit (r=0.363 mkm) is
77000 km, so because the moon use Pythagoras triangle technique the moon
could move through Perigee orbit – otherwise the moon could never do
that.. please remember
o As we have discussed before (BC= L) is the real displacement, where (θ) is
changed with the moon orbit radius….! Why??
o We accepted that, Perigee orbit (r=0.363 mkm) supports only 77000 km as a
real daily displacement of the moon through the orbit – Any longer real
displacement needs more wide orbit to move through, and any wide orbit
needs the moon to move into higher point above perigee to find a wide orbit
o According to this triangle, when (θ) be smaller, (L) will be longer, which
needs more wide orbit means needs the moon to move into a higher point
above perigee …. That means …. The angle (θ) is a function of the moon
motion height above perigee radius…i.e. (θ) defines the real displacement
and the moon motion height above perigee radius
o Please remember (Gerges Equation For The Moon Orbital Motion) in
my previous papers
The Moon Motion Basic Equation
https://www.academia.edu/44396469/The_Moon_Motion_Basic_Equation
or
https://www.slideshare.net/Gergesfrancis/the-moon-motion-basic-equation
The Moon Motion Basic Equation (VI) (Test No.5)
https://www.academia.edu/44406776/The_Moon_Motion_Basic_Equation_IV_Test_No_3_

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- Still one question is left behind, let's ask it now, why the moon have to move daily
88000 km and needs to use Pythagoras triangle technique, why not change this
displacement (88000 km) itself daily?
- Let's answer …
- I suppose the moon moves daily a motion identical to the Earth motion, to save its
accompanying motion with Earth, that means, the moon daily moves 2.58 million
km with an angle =0.9856262 degrees on the horizontal line – (perfectly identical
to Earth motion)…
- If there's no Lorentz Length Contraction Phenomenon in the solar system, the
moon motion trajectory could be seen as a parallel line to Earth motion trajectory
around the sun… But
- Because of the length contraction effect (with rate 1.0725), the moon daily
distance 2.58 mkm is decreased to be 2.41 mkm, and that causes the moon motion
to be in risk to be separated from Earth motion as a result of the different distance
(0.17mkm)
- Based on that, the moon needs to move an additional distance = 0.17 mkm, for that
reason, the moon moves its daily displacement 88000 km under Earth Gravity
forces as additional distance, but it's not enough to cover the different distance,
- Because the 4 inner planets masses total =Earth mass (error 1%), I suppose that,
these 4 inner planets masses total produces a gravity force = Earth gravity force
and this force pushes the moon to move an additional displacement =88000 km,
means the moon move 2 equal additional displacements (88000 + 88000= 0.17mk)
- That's why 2nd
force must effect on the moon motion, because 2nd
additional
displacement (88000km) must be done to cover the different distance created by
the length contraction effect on the moon motion…
- The next question is, why we don't see this 2nd
displacement (88000 km)? we see
only the displacement (88000 km) done by Gravity force, and we don't see even
the moon original motion distance (2.41 mkm) because it's done in the Earth
orbital circumference displacement direction around the Sun, we see only the
moon displacement through its orbit on Earth which the moon does by Earth
gravity effect on it … the next figure can help this explanation
- In the figure, the moon displacement (88000km) is done in a closer trajectory to
Earth (the black Arrow), but the 2nd
displacement (88000 km), is done in the moon
motion original distance direction (2.41 mkm), So we don't see neither the original
motion (2.41 mkm) nor the 2nd
additional displacement (88000 km).

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1-2 The 2nd
Force Proves and Origin
I-Data
Equation No. (1)
(1.6 degrees x 0.985 degrees) /0.526 degrees = 3.02 degrees
Equation No. (2)
192.6 km = 63.7 km x 3.02
Equation No. (3)
733 mkm = 243 mkm x 3.02
Equation No. (4)
71 mkm = 23.45 mkm x 3.02
Equation No. (5)
Venus velocity daily x Mars velocity daily=Earth velocity daily x moon velocity daily
II-Discussion
Equation No. (1)
(1.6 degrees x 0.985 degrees) /0.526 degrees = 3.02 degrees
- We search here for the 2nd
force, if this claim is correct, how to prove this force
and what's its origin??
- 1.6 degrees = (360 degrees /224.7 days), where 224.7 days = Venus orbital period,
that means 1.6 degrees = Venus motion degrees per solar day
- 0.9856262 degrees = (360 degrees /365.25 days), Where 365.25 days = Earth
orbital period, that means 0.98562 degrees = Earth motion degrees per solar day.
- 0.526 degrees = (360 degrees /687 days), Where 687 days = Mars orbital period,
that means 0.526 degrees = Mars motion degrees per solar day.
- What's The Idea Behind This Equation?
- The hypothesis tells that, 2nd
force effects on the moon motion trajectory, and this
force is found because (Earth Mass = The 4 Inner Planets Masses Total),
regardless if the concept is a real or not, Any effect can be done on the Earth moon
motion by Venus and Mars (the most near planets), needs to be effective on Earth
motion- logically – neither Venus nor Mars can effect on the Earth moon without a
direct effect on Earth motion –
- Equation No. (1) tells that, Venus and Earth motions interaction (1.6x0.985=1.57),
this value shows the interaction between Earth and Venus motions in one side, and
this value (1.57 degrees2
/0.526 degree) =3.02 degrees

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- What does that mean?
- We have 3 gears work together, Venus and Earth motions interaction work in one
direction, and both in opposite direction to Mars motion effect, machine of 3 gears,
which causes to create the value 3.02 degrees
- Equation No. (1) tells that, there's some interaction between 3 planets motions and
this interaction causes to produce the value 3.02 degrees
- We know that 1 degree =1 mkm, so (3.02 degree =3.02 mkm but we know that
Venus motion per solar day =3.02 mkm/day)-
- Shortly, the value 3.02 degrees, shows the 3 planets motions interaction effect. But
how this value effect on the moon motion? We need to see more equations….
Equation No. (2)
192.6 km = 63.7 km x 3.02
- In the Triangle CFB (dimensions are 86000 km -18600 km -88000km) its
perimeter = 192600 km
- The angle (CDF) =63.7 degrees
- Equation No. (2), tells that the value 3.02 is seen in the basic triangle which
defines the moon motion daily (88000 km), and defines the sun pole declination
angle (63.7 degrees)…
- Note Please, the Point (D) is the first point on which the Point (A) can effect on
the moon motion- i.e. – if there's another force (or 2nd
force) effects on the moon
motion and this force effect point is found on the point (A), so this effect from the
point (A) on the moon orbital motion must be done on the point (D) because it's
the moon motion nearest point to the point (A).
- But it's another triangle …. (CDB), this is the most important triangle in the
moon motion, because (B defines perigee radius) and (D defines apogee radius) so,
the distance BD = the free space through which the moon can move =43000 km =
the distance from perigee to apogee – but how this triangle is related to the
previous one?! one old equation can help us

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Equation (No. A) (old Equation)
95930 km x (cos 23.45 degrees) = 88000 km (zero error)
- Let's remember (The Triangle CDB dimensions are 86000 -95930 km – 42500)
- Equation No. (A) tells that, the hypotenuse (CD) (95930 km) is defined accurately
based on the moon daily motion distance (88000 km) relative to Earth axial tilt
(23.45 degrees), where (CD x cos 23.45 =88000km)
- Remember Please, CF =88000 km
- That tells the angle (23.45 degrees = Earth axial tilt) is defined as relative to these
2 dimensions (95930 km and 88000 km),
- Also (CD Cos 23.45 =88000km), this equation form we know perfectly, because
the moon uses its daily displacement (88000 km) to create a real displacement
through its orbit based on this form of equation – as we have discussed before
where (L = 88000 km Cos θ) –
- That means, the moon daily motion (88000 km) causes to create the hypotenuse
95930 km and Earth axial tilt (because this value 88000 km is a fixed one and
defined by the difference in the daily motions between Earth 2.58 mkm and the
moon 2.41 mkm, i.e. This value 88000 km is the master value in this triangle)
i.e.
- Earth Axial Tilt =23.45 degrees because the moon daily displacement =88000 km
- Also the equation (L = 88000 km Cos θ) to define the moon real displacement
daily is an equation inherited from these triangles (CBF and CBD) interaction.
But
- If these 2 triangles cause to define Earth axial tilt 23.4 degrees, and the angle CDF
= 63.7 degrees = the sun pole declination, that means, some strong relationship is
found between Earth axial tilt and the sun declination, which may answer the
question why the sun circles the Earth during 365.25 days giving the same face
always…., that because their motions are interacted with each other based on their
basic angles which is not found between the sun and any other planet, making
Earth a very specific planet in the solar system.
- A question is still behind??
- How the value 3.02 degrees = 3.02 mkm can be equal 3.02 ( a rate without unit)?
- Because the moon orbital diameter = 1 million km, that means
- 3.02 mkm /1 mkm = 3.02

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Equation No. (3)
733 mkm = 243 mkm x 3.02
(733 x 0.99 = 727 mkm)
- 243 mkm = the distance mercury moves during its rotation period (58.6 days)
But
- 243 mkm = the distance Marts moves during Venus day period (116.7 days)
But
- 243 days = Venus rotation period
On the other side
- Venus during 365 days moves a distance = 2π x 175.94 mkm (175.94 days =
Mercury day period)
How to understand that??
- If the 4 inner planets masses create a gravity force equal Earth gravity, and by this
force the 4 inner planets pushes the moon to move its additional displacement
(88000km) we can't catch this force…..!! what we can do, is to discover the
interaction between the inner planets motions to create this effect on the moon
motion…
- Venus and Mercury motions have some secret between each other, how can
discover it?? Mercury moves a distance which Venus sees as a period of time, and
on the other side, Venus moves a distance Mercury sees as a period of time…
that's clearly seen by the previous data…. But this relationship isn't closed for
them only…!
- Mars also moves a distance (based on Venus day period) =243 mkm but Venus
sees this distance as a period of time =243 days (Venus rotation period)
- the interaction is just seen clearly –
- the interaction between Mercury and Mars Motions must be seen in Venus rotation
period (243 days) which is seen as 243 mkm
- The Conclusion/ Mercury & Mars Motions Interaction is seen in Venus rotation
period 243 days (or 243 mkm)
- Equation No. (3) 733 mkm =243 mkm x 3.02, but we know that, 720 mkm =
Mercury Jupiter Distance = 2 Mercury orbital circumference, also as we know
that, Mercury moves during its day period a distance =720 mkm = Mercury Jupiter
distance!
- Why this is useful? let's move step by step…. 243 mkm shows the interaction of
Mercury and Mars motions, and 3.02 shows the interaction between Venus, Earth
and Mars Motions – both factors are used to define (Mercury Jupiter Distance)!
Why?? because Jupiter is the inner planets parent, we know that perfectly, so if the
inner planets needs to practice their total effect, they have to do that through their
parent (Jupiter) – because Jupiter is the inner planets source of energy.

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- But, is this true, Jupiter the inner planets parent? Let's remember the answer ..
- Mercury moves during its day period a distance = Jupiter Mercury distance
- Venus moves during its orbital period a distance = Venus Mercury distance
- Earth moves during its orbital period a distance = Earth Mercury distance
(Error 1.4%, Note Earth & Jupiter should be at the sun 2 different sides)
i.e.
- The 3 inner planets define their orbital circumference to be equal their distances to
Jupiter – that because – Jupiter is their source of energy- this is a deep discussion
please review it (Jupiter Effect On The Inner Planets)
- Still a question behind, because 733 mkm doesn't equal 720 mkm but there's an
error 2%??
- We know that, the rate 99% is found because of a geometrical necessity, so for 2
values (M & N) when we use 99% of both, we find the total error is 2%
- Equation No. (3) tells that, the rate 3.02 expresses the 3 inner planets motions
interaction (Mercury, Venus and Mars) effect on the Earth Moon motion, where
this effect is practiced by Jupiter effect directly on the Earth moon because Jupiter
is the inner planets parent
- Notice No. (1)
- Mercury orbital period =88 days, but if 1 day = 1000 km ( in place of 1 mkm), so
these 88 days will be seen at the moon motion as 88000 km (the Moon daily
displacement) and that tells us, Not only Mercury, Venus and Mars motions causes
this using of the distances to be periods of time, but also the moon motion shows
this same feature, proving that, this feature is transported to the moon motion from
these 3 inner planets – which proves the paper hypothesis that (the 3 inner planets
practices force on the moon motion trajectory which is the 2nd
force we have seen
in the moon orbital triangle)
- Notice No. (2)
- (Jupiter axial tilt =3.1 degrees = 3.02 degrees + 0.1 degree) but we know that,
Mars orbital inclination was 1.8 degree before its migration and after it became 1.9
degrees, this is the required difference 0.1 degrees…,
Mars orbital inclination was 1.8 deg because Earth diameter =1.8x
Mars ancient diameter (7070 km) but
Mars orbital inclination became 1.9 deg because Earth diameter =1.9x
Mars current diameter (6792 km)
Please review, my paper (Giant-Impact Hypothesis Supports "Mars
Immigration Theory")

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Equation No. (4)
71 mkm = 23.45 mkm x 3.02
- 23.45 degrees = Earth Axial Tilt
- We know that, the Length contraction greatest rate is 71, means 71 mkm will be
contracted to be 1 mkm, where the moon orbital diameter =1 mkm
- That tells us how the 1 day of the sun = 365.25 days of Earth motion? because 71
mkm is contracted to be 1 mkm and transferred into 1 day and used by the sun
- 71 mkm which is not contracted, will be multiply with 5.1 degrees ( the moon
orbital inclination) to produce 365.25 mkm which is used by Earth as 365.25 days
- Equation No. (4) supports the claim that, Earth Axial Tilt Is Created As A
Result Of This 2nd
Force Effect On The Moon Orbital Motion.
- This fact which we have discovered in the moon orbital triangle analysis
Equation No. (5)
Venus velocity daily x Mars velocity daily=Earth velocity daily x moon velocity
daily = Mercury velocity daily x Ceres velocity daily =2π mkm2
- This equation we have seen frequently…
- Equation no. (5) tells that, the inner planets velocities are created depends on each
other, the 6 planets velocities are found by one function,
- That tells us, the interaction between the planets motions is a fact proved by this
equation only – simply each planet defines its velocity to be complementary to
another planet….
- Venus and Mars motions interaction effect on the moon motion is seen clearly by
the equation, because Earth and moon velocities are defined based on Venus and
Mars velocities – the data simply proves the paper hypothesis

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More Data in this triangle
The Moon Orbital Triangle
Please Note
The angle FCB = 12.195 degrees
Where
13.18 degrees (the moon daily motion) - 0.985 degrees (Earth motion) =12.195 deg
0.985 degrees = Earth motion degrees daily = The moon motion degrees daily
- Because the moon moves (originally) an identical motion to Earth motion
- The angle (12.195 deg) is found in the triangle whose hypotenuse =88000 km and
that tells us, the basic concept is correct and the angle (0.985 degrees) is the moon
original motion angle,
- That explains the moon motion basic equation definition….
Gerges Equation For The Moon Orbital Motion
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
- The angle θ is the smallest one in a Pythagoras triangle
- 0.985 degrees = (360 degrees /365.25 days)
- Where this angle 0.985 degrees is the daily motion reason for the Earth and the
moon, I add this angle (0.985 deg) each day to the angle (θ) of the previous day to
define the current day angle (θ) and by which the real displacement (L) and the
moon motion height above perigee radius is defined
- This equation I have discussed and tested in many papers please review
The Moon Motion Basic Equation
https://www.academia.edu/s/46a3441e4a
or
https://www.academia.edu/44396469/The_Moon_Motion_Basic_Equation
or
https://www.slideshare.net/Gergesfrancis/the-moon-motion-basic-equation

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References
The Moon Motion Trajectory Analysis (II)
https://www.academia.edu/44368860/The_Moon_Motion_Trajectory_Analysis_II_
or
https://www.slideshare.net/Gergesfrancis/the-moon-motion-trajectory-analysis-ii
Light Motion Features Are Discovered in Planet Motion
https://www.slideshare.net/Gergesfrancis/light-motion-features-are-discovered-in-planet-motion
or
https://www.academia.edu/44286772/Light_Motion_Features_Are_Discovered_in_Planet_Motion
Can Different Rates Of Time Be Found In The Solar System Motion?(II)
https://www.academia.edu/44334645/Can_Different_Rates_Of_Time_Be_Found_In_The_Solar_System_Motion_II_
Does Particle Data Depend on Its Motion? (Lorentz Transformations Analysis)
https://vixra.org/abs/1912.0134
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
Gerges Francis Tawdrous +201022532292
Curriculum Vitae http://vixra.org/abs/1902.0044
E-mail mrwaheid@gmail.com
Linkedln https://eg.linkedin.com/in/gerges-francis-86a351a1
Facebook https://www.facebook.com
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ORCID https://orcid.org/0000-0002-1041-7147
Quora https://www.quora.com/profile/Gerges-F-Tawdrous
Google https://scholar.google.com/citations?user=2Y4ZdTUAAAAJ&hl=en
Academia https://rudn.academia.edu/GergesTawadrous
List of publications http://vixra.org/author/gerges_francis_tawdrous

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