Earth and Moon Motions Interaction

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
I do this research
Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
Earth and Moon Motions Interaction
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 –22nd
October 2020
Abstract
Paper Question
Why the moon orbital circumference =Earth Motion distance during 1solar day?
- This question was the first one I have faced in the moon orbital motion- may try to
solve it in following….
- Planet motion depends on light motion based on the hypothesis (1 second of light
motion causes Planet motion for 1 solar day), this hypothesis creates the 1st
rate of
time in the solar system which is (1 second of light = 1 day of planet)
- The 1st
rate of time dominates any planet motion – but – in the Earth Moon Orbit,
this rate is changed by Venus & Mars Motions Effect
- (86400 = 237 x 365), where 237 = (Venus rotation period/ Mars rotation period)
and 365 =(Earth orbital circumference /the moon orbital circumference)
- Venus & Mars Motions Interaction cause to remove the rate (237) from the
previous equation- and because of that (1 day on the sun = 365 days on Earth) and
that causes the moon orbital circumference = Earth motion distance daily
But
- The moon circumference shows also this same rate, as we have concluded that
from (10921 km the moon circumference x 86400 seconds = 940 mk) This
equation tells that, if Earth revolves around the sun a complete revolution in one
Solar Day Only, so the moon circumference in this case will equal the Earth
motion distance during 1 second…means the rate (1 is equivalent to 365) is seen
again in the moon circumference as we have seen it in the moon orbit..!
So
- (0.406 mkm (apogee radius) = 3475 km the moon diameter x 116.7), this equation
tells that, the moon diameter is created based on the moon orbital radius (at apogee
whose circumference =2.58 mkm= Earth motion daily) based on 116.7 where (Venus day
period = 116.7 days) means, the moon orbit rate of time which is created by Venus and
Mars effect, causes the moon diameter to be proportional to this same rate of time by
Venus effect- simply it's Venus whose create both effects (365 =116.7 x π)…. Can that
be real? the equation (19 degree= 3.02 deg x 2π) (1 deg =1 mkm) so, this equation tells
that Venus motion distance during 1 solar day (3.02mkm) causes the moon orbit to
regress 19 degrees per year (creating this same rate 1 to 365)… can this explanation help
to explain the moon orbital motion? let's discuss it.

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1- Introduction
The moon orbital motion has numerous of puzzles, let's show them in following:
(1)
- Because the Earth and moon move together always, it's necessary for the moon to
have a similar motion to Earth motion to avoid the separation –
- Specifically the moon has to move per solar day a distance =2.58 mkm and with
an angle (0.985 degrees) on the horizontal line perfectly as Earth does to avoid the
separation between Earth and Moon in their motions
- In fact these conditions are not performed by the moon motion, the moon daily
motion = 13.18 degrees and not 0.985 degree, also we can't catch how the moon
moves the distance 2.58 mkm because what we see only a displacement=88000 km
so we can't be sure how the moon does that,
- Note Please, the moon orbital circumference at apogee =2.58 mkm and if the
moon moves daily a distance =2.58 mkm, that means the moon has to revolve its
orbital circumference one time every solar day, but it doesn’t!
(2)
- The next hard question is the Metonic cycle, where the moon rotates Metonic
Cycle (19 years) which a cycle can't be observed in Earth or any other planet
motion- it's the moon unique cycle, and the question is, if Earth is the parent of the
moon how the follower can rotate this cycle which its parent doesn't?!
(3)
- The basic hard question is that, how to create a consistency between the moon data
and its real motion….. for example … the moon is created by the debris produced
from Earth collision with another planet (I claim it's Mars itself), so the moon
matter is created by this collision…Means the moon diameter is created from these
debris and with Earth gravity effect! But the previous data told us that the moon
diameter and circumference is created based on geometrical calculations! That's
cause some puzzle for us ….. because the equations ( 10921 km the moon
circumference x 86400 seconds = 940 mk) and (0.406 mkm (apogee radius) =
3475 km the moon diameter x 116.7), tell that the moon diameter is created
based on Geometrical Calculations… but how if the moon is created by some
collision between Earth and (Mars as my theory "Mars Migration" claims) (or
another planet as "Great-Impact Hypothesis" claims) – if the moon diameter is
created by collision and the moon diameter was some debris attracted by Earth
gravity to create the moon diameter, how these equations and similar hundreds are
created? we have to discuss deeply as possible the moon motion in following…
Paper Contents
2- Earth and Moon Motions Consistency
3- The Earth Moon and Jupiter Relationship
4- The Moon Orbital Triangle Data

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2- Earth and Moon Motions Consistency
2-1 The Earth and Moon Motion Degrees Analysis
2-2 The Moon Motion Basic Difficulty
2-1 The Earth and Moon Motion Degrees Analysis
I-Data
(A)
- Earth moves during 29.53 solar days a value 29.2 degrees (29.2 = 29.53 x 0.985)
- The moon moves during 29.53 solar days 389.2 degrees (13.18 x 29.53 days =
389.2 = 29.2 +360)
(B)
- Earth moves during 1 solar day a distance = Moon motion distance during
29.53 solar days = Pluto Motion during its day period (153.3 hour)
II-Discussion
Data No. (A)
- Earth and Moon move the same value of degrees (29.2 degrees) during the same
period of time (29.53 solar days)
- The previous data shows the motions consistency reason – and this data shows
clearly why the moon is belonged to Earth perfectly, because the moon is in full
consistency with Earth motion- but how the moon does that- spite of the different
effects practiced on it during its motion? this question is the tool by which we may
understand how the moon can rotate Metonic Cycle during its revolution around
Earth…
- Means, if the moon rotate Metonic Cycle during 19 years and the moon saves its
accompanying with earth motion during this period, that means, Metonic Cycle
motion must be seen in the previous data! But where is it?!
Data No. (B)
- Because "Earth moves during 1 solar day a distance = moon motion distance
during 29.53 days"= Pluto Motion during its day period (153.3 hour), We
concluded that 1 day on Earth = 29.53 days on the moon, this conclusion is so
important because it gives an explanation for the following equation…
- (10921 km "the moon circumference) x 27.3 days = 0.3 mkm) this equation tells
that, if the moon rotates around its axis one time per solar day as earth does, the
moon will pass a distance = light motion for 1 second, that tells us the next
conclusion…."1 second of light motion is equivalent to 27.3 days of moon motion"
but we know that 29.53 days of the moon = 1 day on Earth, also we know that

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29.53 = 1.0725 x 27.3 which tells that, 29.53 days is equivalent to 27.3 days, based
on that 1 day on Earth = 27.3 days on the moon = 1 second of light motion … this
simply the basic rate of time in the solar system motion (1 sec is = 86400 sec) this
rate depends on the hypothesis (Light Motion For 1 Second Causes Planet
Motion For 1 Solar Day)
- Why 29.53 days on the moon is equivalent to 153.3 hours on Pluto? Why this is
useful at all?
The Earth Moon And Pluto Rotation Periods Interaction:
a. 655.7 h. (the moon rotation period) – 153.3 h. (Pluto rotation period) = 502.4 hours
b. 655.7 h. (the moon rotation period) x 153.3 h. (Pluto rotation period) = 100518
Comments
- Equation no. (a) tells that, the difference =502.4 hours, but we know that, 500
seconds are required for the light known velocity (0.3 mkm/sec) to pass from the
sun to Earth… we also know that, the basic connection between Pluto and Earth is
this 500 seconds because during 500 seconds light suppose velocity (1.16 mkm/s)
travels a distance =580 mkm = Pluto motion distance during 1433 days …. This is
an old discussion in which we have proved that the number 500 seconds is the
basic connection between Earth and Pluto, that also explains why Pluto during
365.25 days moves a distance = 149.6 mkm (Earth orbital distance) and that
explain why earth velocity = 2π x Pluto velocity…
- Equation no. (a) tells that, this 502.4 hours is used as 500 seconds by light motion
to create earth orbital distance, what geometrical mechanism is used here, we will
left this question behind….
- Equation no. (b), light supposed velocity 1.16 mkm/sec travels during 1 solar day
a distance = 100224 mkm, and the distance passed by light can be seen by us as
time period (x =ct), and because of that the result 100518 hours2
can be seen as
100518 mkm2
which is very near to the value 100224 mkm (error 0.2%),
- Equation no. (b), tells that both the Earth moon and Pluto motions depends on
light motion (supposed velocity 1.16 mkm/sec) ! this conclusion can be supported
by the next 2 old equations
Equation no. (c), 27.3 days x 43000 km = 1.16 mkm, this equation tells that,
if the Earth moon is a point (without diameter) and move from perigee to
apogee one time per a solar day, so the moon will pass during its orbital period
27.3 days (= its rotation periods) a distance = 1.16 mkm = light motion distance
during 1 second.
Equation no.(d), 7510 km (Pluto circumference) x 153.3 hours = 1.16 mkm,
this equation tells that, if Pluto rotates around its axis one time per an hour, so
Pluto will pass during its day period a distance = 1.16 mkm = light supposed
velocity (1.16 mkm/sec) motion for 1 second (153.3 h = Pluto rotation period)
27.3 days of the moon = 153.3 h. of Pluto = 1 day on Earth = 1 s. of light motion

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2-2 The Moon Motion Basic Difficulty
I - Data
Equation No. (2-1)
25920 mkm = 17.75 mkm x 1461 days
Equation No. (2-2)
25920 mkm = 2.41 mkm x 10747 days
II – Discussion
Equation No. (2-1)
25920 mkm = 17.75 mkm x 1461 days
Where
25920 mkm = a distance passed by light known velocity (0.3mkm/sec) during a solar
day (86400 seconds)
17.75 mkm = the distance passed by all planets motions total during a solar day
1461 days = Earth Cycle (365+365+365+366 =1461 days)
But
- The value 17.75 mkm = the planets motions distances total per a solar day, this
value include the Earth moon motion distance which is considered equal to Earth
own motion to avoid the separation between them, so the moon motion distance =
2.58 mkm
- Equation (2-1) tells that, light motion distance during 1 solar day is equal all
planets motions distances total during 1461 days
- We need Equation no. (2-1) to confirm that, the Earth moon motion distance
per a solar day = Earth motion distance per solar day = 2.58 mkm
Equation No. (2-2)
25920 mkm = 2.41 mkm x 10747 days
- Equation (2-2) tells that, the Earth Moon motion distance during 10747 days can
be equal 25920 mkm, if Earth moon motion distance per a solar day = 2.41 mkm!
- This equation contradicts directly the previous one!
But
- 10747 days = Saturn orbital period and it’s created depending on a defined cycle
And
- Because, 1 day on the sun = 365 days on Earth and 1 day on Earth = 29.53 days on
the moon, so 1 day on the sun = 10747 days on the moon
- i.e.
- The value 10747 days = Saturn orbital period, and also created based on the rate of
time we have discovered and proved, this rate of time is proved by different
equations in this paper and in many previous paper we can't just remove all of that,
simply the moon motion distance per a solar day (really) must be = 2.41 mkm
- i.e. The Moon Has To Have 2 Different Velocities In The Solar Day!!

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- This is not a complex question…. Let's solve it in following…
- The moon motion distance per a solar day =2.58 mkm …. But
- Because of the length contraction effect we see this distance under the contraction
rate (1.0725) to be = 2.41 mkm,
i.e.
- The moon moves really 2.58 mkm but we it as 2.41 mkm because of the length
contraction effect….
- This explanation will have a serious question relating to Saturn Orbital Period, let's
give this question some consideration in following…..
o If the moon moves 2.58 mkm per solar day, so the moon will pass the
distance 25920 mkm during a period = 10046.5 days,
But
o Because of the contraction, the moon motion distance become 2.41 mkm,
and because of that the period became 10747 days (=Saturn orbital period)
And
o (10046.5 days /2.23 days) = 4505 days (where 2.23 d =29.53-27.3)
o (10747 days /2.23 days) = 4819.2 days (= 4900 error 1.6%)
o 4500 mkm = 2π x 720 mkm (Mercury Jupiter Distance)
o 4900 mkm = 2π x 778.6 mkm (Jupiter Orbital Distance)
o The equation tells that, the moon motion distance contraction is created by
the contraction found between Mercury and Jupiter (as we know that
778.6mkm "Jupiter orbital distance" =1.0725 x 720 mkm "Mercury Jupiter
distance"….this contraction which causes Jupiter orbital period 4331 days
to be = 2π x 687 days Mars orbital period, this same contraction defines
Saturn orbital period 10747 days, that tells us why Saturn orbital period =
2.5 x Jupiter orbital period (where 2.5 deg = Saturn orbital inclination)
o Please remember (Equal distances are used for different rates of time),
9010 mkm (Saturn orbital circumference)= 2π x Mars orbital circumference,
so 10747 days (Saturn orb. Period) = 2π x 687 days x 2.5 (where 687 days
= Mars orbital period),
o The data tells that, the moon distance contraction is caused by Mercury &
Jupiter Distances contraction, that proves my claim (one light beam creates
all the planets and their orbital distances)
Regardless all that,
Where Is The Main Difficulty In The Moon Orbital Motion
Or
Why the moon daily displacement =88000 km?
Let's try to answer in following ….

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The Moon Displacement Per A Solar Day = 88000 Km
- The Moon Real Motion Distance Per A Solar Day = 2.58 mkm
- Because of the length contraction effect
- We see this distance 2.58 mkm under the contraction rate 1.0725 to be 2.41 mkm
- If the moon moves 2.41 mkm per a solar day, that will to separate the moon from
the Earth in their motions…!
- The moon has to find this difference 2.58 mkm- 2.41 mkm =0.17 mkm
- Now the question is raised clearly….
- Is Lorentz length contraction real phenomenon or illusion of measurement?!
This is the question I have fought to find its answer ….and let's write its answer..
- Physics is the science of measurements, and if Lorentz length contraction effect is
proved by empirical results (as happened really), so what's measured is the real
one – means – if a particle length measurement shows the contraction under high
velocity motion, so this particle own length is contracted, I have one more reason
to support this opinion …. Which is
o If the particle data is in consistency with measurements while this particle
moves with Zero velocity relative to me, and if this particle data got illusion
of measurements when this particle moves by high velocity relative to me,
that means (I'm the universe reference point) – to remove our effect on the
experimental results we have to accept that the length contraction causes a
real contraction for the particle own length...
o Based on this conclusion
- The moon distance 2.58 mkm which is contracted by Lorentz length contraction
effect rate (1.0725) and became 2.41 mkm, so the moon real motion is 2.41 mkm
- But how to understand that?!
- The equation 25920 mkm = 17.75 mkm x 1461 days tells that, the moon moves a
distance = 2.58 mkm, and also the moon doesn't separate from earth motion which
proves that the moon moves per a solar day a distance = 2.58 mkm and not 2.41
mkm!!
2.41 mkm (the moon contracted motion distance) = 0.17 mkm = 2.58 mkm
i.e.
- The moon does its displacement per a solar day (88000 km) to cover this
difference (0.17 mkm) but the moon displacement is not to cover it! so how
the moon solves this problem

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More Discussion (The Moon Displacement Per A Solar Day = 88000 Km)
- The moon motion still provides 2 puzzles which are
- (1st
Puzzle) if the moon moves per a solar day why it doesn't revolve its orbital
circumference one revolution every solar day?
- (2nd
Puzzle), the moon motion displacement is 88000 km and not 170000 km and
how this displacement can cover this distance?
- Let's try with the 2nd
puzzle at first,
o 116.75 days (Venus Day Period) /58.66 days (Mercury rotation period) =2
o If the moon uses this rate, simply 88000 km x 2 = 170000 km
o So the required distance (0.17 mkm) is produced by the moon displacement
per solar day (88000 km) multiply with (2)
o We can notice simply that, Venus Day Period (116.7 days) is the basic toll
effects on the moon motion and its orbit creation, that because the rate of
time (1 to 365) which is created in the Moon orbit by Venus and Mars
Motions effect creates the Earth orbital period (365.25 days= π x116.7 days)
showing how Venus day period is effective on Earth and Moon Motions…
- Now, Let's try with the 1st
puzzle,
- This puzzle we should solve in the next paper because it gives us 3 questions
which are:
o The moon motion distance 2.58 mkm is a hidden one?
o The moon metonic motion is a vertical and hidden one?
o How the moon daily distances (88000 km) uses this rate 2 (=116.7/58.66)
geometrically? Or by what geometrical mechanism this is occurred?

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References
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
session
https://www.academia.edu/s/69edd1d0ea
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_
Matter Origin and Creation
https://www.academia.edu/43214351/Matter_Origin_and_Creation
Is Saturn The Last Planet Created In The Solar System? (II)
https://www.academia.edu/43197587/Is_Saturn_The_Last_Planet_Created_In_The_Solar_System_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
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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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