Earth motion

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 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 –25th
November 2020
1- Abstract
Earth motion is one paper I have promised to write from years, just now I can do that!
Why have I promised but couldn't? ….the reason was the following…
- The moon orbital radius at total solar eclipse = 0.370 mkm, its circumference
= 2.32 mkm = 58.6 x 40080 km (Earth Circumference)
- Now let's imagine:
- If the moon orbital circumference (2.32 mkm) is a great gear and the Earth
circumference (40080 km) is a small gear …. Now let's these gears to rotate
- The great gear (2.32 mkm) needs to rotates (one time only), and that will
causes the small gear to rotate 58.6 times, means Earth circumference will
rotate around its axis 58.6 times but one rotation of Earth circumference means
one solar day, so Earth will rotate (58.6 solar days).
- But ….Earth moves per solar day a distance = 2.58 million km
- 2.58 mkm x 58.6 days = 149.6 million km (Earth Orbital Distance)
- Shortly….
- If the moon orbital circumference (a great gear) rotate around its axis one time,
that pushes Earth to rotate around its axis 58.6 times (58.6 days) and during this
period Earth moves a distance = Earth orbital distance.
- It's Our Old Puzzle, I have written almost in 2017 in my paper (the solar
- system geometry part no. 2) -
- The puzzle is clear and we have 2 options, (1st
) to consider all this data is
created by pure coincidence and I had refused this option (2nd
) to find the
geometrical reason or mechanism by which this data is created depending on
each other…. And I have found it….. that's why I write now this paper (Earth
motion), let's start our discussion to know how this data is created….

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2- Jupiter Effect On Earth Orbital Distance Definition
2-1 Preface 2-2 Jupiter Effect
2-3 Jupiter & Uranus motion interaction effect
2-1 Preface
I-Data
Equation No. 1
40080 seconds x 29.8 km/sec = 1.2 million km
- 40080 km (Earth Circumference) is used as a period of time,
- Equation no. (1) tells that, Earth velocity (29.8 km/second) needs a period =40080
seconds to pass a distance =1.2 million km – this value is so important because
Jupiter energy is found in light beams with supposed velocity =1.16 mkm/sec, and
that tells Earth motion is Jupiter energy source – so Earth during 40080 seconds
produce a distance =1.2 mkm, from which 0.04 mkm is saved for Earth and the
rest 1.16 mkm is sent to Jupiter in a light beam with supposed velocity 1.16
mkm/sec, and that shows Earth Motion is the solar group source of Energy.
- But
- On the other side – Equation no. 1 tells also – Earth needs to rotate around its axis
29.8 times to move (by rotation) a distance = 1.2 million km… means during 29.8
solar days Earth by rotation moves a distance =1.2 mkm….
- For 2.4 million km, Earth needs to rotate (29.8 times x 2 = 59.6 solar days)
- Why do we need this 2.4mkm? We know this value, let's remember it in following:
o The moon moves with Earth a typical motion to avoid the separation
between its and Earth motion… for that the moon moves per solar day 2.58
mkm with an angle 0.98562 degrees on the horizontal level – But
o Because of Lorentz length contraction phenomenon with a rate 1.0725, the
moon distance (2.58 mkm) is contracted to be (2.4 mkm)- and that created
the difference 0.17 mkm which the moon tries to cover by its additional
displacements per solar day. (i.e. (2.4 mkm) is passed by Earth rotation
during (59.6 solar days) and by the moon motion during 1 solar day!!

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Equation No. 2
59.6 solar days x 2.58 mkm (Earth velocity daily) = 153.4 million km
- We know, Earth orbital distance was 153 mkm and became 149.6 mkm (8211y)
- i.e. During 59.6 days Earth moves a distance = Earth Orbital Distance
- but the period 59.6 days is defined by (2.4 mkm)!
o How to understand that?
o Noteworthy, we have considered the value (2.4 mkm) to be created basically
by the length contraction phenomenon, i.e. it has no existence before,
because there's no motion distance = (2.4 mkm), because the moon moves
already (2.58 mkm) but the contraction caused this value to be (2.4 mkm).
o This consideration is incorrect, because the value (2.4 mkm) is found
basically in Earth motion… because it's needed to define the period 59.6
days which is required to define Earth Orbital Distance.
o That means…. The length contraction effect is used by a geometrical
machine, the contraction rate is defined geometrically and based on that, the
geometrical machine defines the velocity by which this contraction rate can
be produced….let's move in our analysis
- I want to say that,
- The value (2.4 mkm) of the moon orbital circumference is found before the moon
creation and is defined for the moon motion before its creation, the moon is a slave
created to do some job defined by many planets effects found before the moon is
created.
- The next question is
- Why the moon orbital motion (2.4 mkm) for one solar day after the
contraction effect is defined by Earth motion distance (by rotation) during
59.6 mkm?
- We know almost this question answer, let's try to discuss it in following….

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Equation No. A (Old Equation)
10921 km (the moon circumference) x 27.3 days = 0.3 mkm
- Equation no. (A) tells that, if the moon rotates around its axis one per solar day, so
the moon during its orbital period (27.3 days) will move a distance = 0.3 mkm =
light motion distance during 1 second (light known velocity 0.3 mkm/sec)
- We may remember the research 2nd
hypothesis, (Light Motion For 1 Second
Causes Planet Motion For 1 Solar Day)…
- Equation No (A) tells that, 27.3 days of the moon motion (rotation) is equivalent to
1 second of light motion! let's see one more equation ….
Equation No. B (Old Equation)
2.58 mkm passed by Earth motion during 1 solar day
2.58 mkm passed by the moon motion during 29.53 days (29.53 x 88000km)
2.58 mkm passed by Pluto during 153.3 hours
- We remember the rule (Equal Distances Are Used For Different Rate Of Time),
that means, 1 solar day of Earth motion = 29.53 solar days of the moon motion
and = 153.3 hours of Pluto motion.
- One more equation can support this conclusion
Equation No. C (Old Equation)
365.25 days x 29.53 days = 10747 days
- 365.25 (= Earth year), and 29.53 days (= the moon day period),
- Equation no. (C) supposes that, 1 solar day on the sun = 365.25 solar days on Earth
(different rates of time are found as relativistic effects by high velocity motion)
- Also equation (C) tells If 1 solar day of Earth = 29.53 solar days of the moon
- So
- 1 solar day on the sun = 10747 days on the moon, how can we be sure of that?
The period 10747 days = Saturn Orbital Period, and because it's a known cycle in
the solar system, we conclude that, this conclusion is correct because it produces
planets real data -
- And that supports the conclusion (1 day of Earth = 29.53 days of the moon)

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- Now we have 59.6 days = 2 x 29.8 days = 2.02x 29.53 days
- Almost 2 days of Earth = 59.6 days of the moon
- By 2 days this period (59.6 days) is produced and earth moves a distance = Earth
orbital distance…!
- How or why?
- The available answer now is that, Jupiter energy is sent to Pluto in light beams
form with supposed velocity =1.16 mkm/sec during 2 solar days – let's remember
this equation
o 1.16 mkm/sec x 86400 seconds x 2 = 200448 mkm, This is the energy sent
to Pluto and reflected to Neptune, from which Neptune consumed (14%) of
it and reflected the rest of energy toward the inner planets in 2 trajectories of
energy each one has 86400 mkm (because Space = Energy)
o So, Jupiter total energy depends on 2 solar days, that may help to explain
why earth orbital distance depends on 2 solar days also.
Equation No. 3
153.4 million km = 2.4 mkm x 63.7
- We know that, the moon at apogee radius (r= 0.406 mkm), has an orbital
circumference =2.55 mkm = 63.7 x 40080 km (Earth Circumference),
- That tells the moon motion range from perigee (r=0.363 mkm) to apogee (r=0.406
mkm), is defined by this value (2.4 mkm) which is found by Earth motion and not
defined by the moon motion…..!

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2-1 Jupiter Effect
Equation No. 4
778.6 mkm (Jupiter orbital distance) x 1.2 = 934 mkm = 720.7 x 1.3
- 928.2 mkm = Earth Jupiter Distance, where Earth and Jupiter be at 2 different
sides from the sun (928.2 mkm = 778.6 mkm +149.6 mkm)
- This equation is one reason to write this paper…
- Equation no. (4) tells us that, there's a geometrical reason to create the rate 1.2
- This equation tells that, the rate 1.2 is not an invented value, it's found basically
inside the solar system geometrical structure….
- The difference between two values (934 mkm and 928.2 mkm) can be found by
Earth displacement (from 153 mkm to 149.6 mkm).
Shortly
- Equation no. (4) (Part I) tells that, Earth Jupiter distance is defined based on (1.2)
and Earth orbital distance, means, 1.2 mkm (Earth motion by rotation during 29.8
days) is a basic player in definition of Earth and Jupiter orbital distances
- Equation no. (4) (Part II) tells that, 720.7 mkm (Mercury Jupiter distance) is a
player in the geometrical structure and produces 1.3 (Jupiter orbital inclination
=1.3 degrees)
- Please remember
o 778.6 mkm (Jupiter orbital distance) = 1.0725 x 720.7 mkm, where 720.7
mkm = Mercury Jupiter Distance, where 1.0725 is the contraction effect
o This equation tells us that, Jupiter orbital distance is equivalent to Mercury
Jupiter distance because the last is created from the first by contraction…
o That shows why both distances have as strong geometrical effect…
o And by these 2 distances Jupiter orbital inclination is created.

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Equation No. 5
2094 mkm = 1.2 mkm x 1737 km
- 2094 mkm = Jupiter Uranus Distance -
- 1.2 mkm = Earth motion distance (by rotation) during 29.8 days
- 1737 km = The Earth moon diameter
- Equation no. 5 is so beautiful equation…. Let's analyze its components
- 2094 mkm = Jupiter Uranus Distance ….. please remember
o A light beam with known velocity (0.3 mkm/sec) travels from Jupiter to
Uranus (2094 mkm) in a period 6939.75 seconds (and we know 6939.75
solar days = Metonic Cycle)
o 6939.75 seconds = 71x 97.8 seconds
o Where Uranus orbital distance 2872.5 mkm is created by this value (97.8s),
because (97.8 seconds x 0.3 mkm/sec) =29.33 mkm and then 29.33 mkm x
97.8 = 2872.5 mkm – shortly
o 2872.5 mkm = (97.8 seconds )2
x 0.3 mkm/sec (light known velocity)
- Uranus effects on the inner planets basically on the Earth moon orbit… we will
remember that in details in the next point....
- Shortly
- By light energy (6939.75 seconds), Uranus and Jupiter effects on the moon orbit
and that causes to create the moon based on calculations taken into consideration
the light motion – which is seen clearly in the moon circumference analysis and
discussion.
- Equation no. (5) tells clearly that, the distance (1.2 mkm) is a basic player in the
Earth moon creation.
- Please Note, the moon is created by planets collisions and these collisions debris
are attracted by Earth gravity to create its moon, spite of that, the moon diameter
and circumference are created based on light motion calculations which proves
that (Planet Motion Depends On Planet Motion)

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2-3 Jupiter & Uranus motion interaction effect
- Please let's remember
o Uranus axial tilt (97.8 deg) creates an angle (91.1 deg) with the moon axial
tilt (6.7 deg), but Uranus axial tilt is perpendicular on the moon orbit,
because Uranus creates a decline between the moon orbital triangle and the
horizontal level =1.1 degrees, by that the angle became 90 degrees
(perpendicularity)
o The angle 1.1 degrees Uranus created by stuffing consisted of 2 parts, the 1st
part is Jupiter and Saturn interaction which created an angle =0.6 deg and
seen in the orbital triangle by the green rectangle under the triangle, the 2nd
part, An angle (0.5 degrees) is consumed by the moon diameter (3475 km)
and means, Uranus axial tilt angle above the moon surface should be = 90.6
deg
o Because Uranus orbital inclination =0.8 degrees and vertically =90.8 deg, so
we tried to make the angle (90.6 deg) = to (90.8 deg)! how?
o We cut inside the moon diameter because it consumes (0.5 deg) and reach to
smaller moon diameter (1390 km) which consumes only (0.2 deg) so the
angle above this smaller diameter surface is 90.9 deg.
o At this paper end I insert the moon orbital triangle to be our discussion
reference
Uranus & Jupiter effect on the moon motion:
- They caused to create the moon diameter and its motion data relating to these 2
planets effect
- But one basic effect we need to refer here …. Which is
- To create one more orbit for the moon motion
- i.e.
- The Moon Motion Has To Orbits….

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The moon 2 Orbits
- Where's this 2nd
orbit of the moon motion,
- It's beyond the point (A) in the moon orbital triangle where the distance from
Earth to the point (A) = 449197 km
- The second orbit is found beyond this point far from Earth?
- How to know that?
- Because the lunar eclipse umbra length =1.392 million km (= the sun diameter)
- Let's see this value 1.392 mkm analysis
o 1.392 million km = 449197 km (Earth and the point (A) distance) + X
o X= 1.392 mkm – 449197 km =942803 km
o Where 942803 km =the moon orbital triangle perimeter
o i.e. 942803 km = 449197 km + 121620 km +373000km
- i.e.
- There's another moon orbital triangle beyond the point (A)
- That mean, the lunar eclipse umbra defines the 2nd
orbit or the moon motion.
- In next pages the moon orbital triangle

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Let's review The Earth Moon Orbital Triangle because we use it
Figure No. (1) (my figure)
Please Note
(1) SZ = 7665 km ZF = 2414 km
- CZS = 77.8 degrees CZF =102.195 degrees
(2) DY = 3475 km BCY = 28.39 degrees
(3) XB = 16203 km XCB = 10.67
- XCE = 66 degrees CX = 87513 km

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Let's Review The Moon Orbital Triangle Data
(1st
Point)
- The figure I brought from internet to use in the Explanation -
- We have supposed that the inner circle is Perigee orbit and the
outer circle is apogee orbit – and we have calculated the tangent
AB = 181843 km
- AB = 363686 km (= perigee radius approximately)
- Perigee radius r =0.363 mkm Apogee radius r =0.406 mkm
- Based on that, the triangle (ODB) is a specific Pythagoras triangle (1, 2 and 51/2
)
- i.e. the triangle (ODB) angles are 26.564 degrees and 63.435 degrees
NOTE
- for these 2 angles (26.564 deg and 63.435 deg) we have searched, because these 2
angles will correct many data in the orbital triangle.
(2nd
Point) The Moon Orbital Triangle Data Correction
- EB = Perigee radius = 363000 km
- ED = Apogee radius = 406000 km
- EA= (Jupiter Circumference) =449197 km
- AC = (Saturn diameter) =121620 km (error 1%)
- ES = total solar eclipse radius = 373000 km (error 1%)
(EC = 373000 km = Earth moon distance at T. Solar eclipse, BUT point C is NOT
the moon position in T. solar eclipse, because the distance BC= 86000 km but the
distance between perigee point and total solar eclipse point = 11000 km)
- BS= (the moon Circumference) =10921 km
- BZ = 18586 km BF =21000 km
- BD = DA = 43000 km (BY =46475 km)
- BA = BC = 86000 km
- CS = = 86690 km
- CZ= (the moon daily displacement) =88000 km
- CF= 88526.8 km CD =96150.9 km
THE ANGLES
- The angle between the black and red lines (under E) = 1.1 degrees
- (E) = 13.33 degrees (C)= 121.67 degrees (A) = 45 degrees
- (ECB) = 76.67 degrees (BCA) = 45 degrees
- (BCS = 7.23 deg) (BCZ = 12.195 deg) (BCF = 13.72 deg) (BCD = 26.564 deg)
(ACD = 18.435 deg)
- (BSC = 82.7 deg) (BZC = 77.8 deg) (BFC = 76.82 deg) (BDC = 63.434 deg)
- (CSA =97.23 deg) (CZA =102.195 deg) (CFA= 103.7 deg) (CDA = 116.564 deg)
- (CYA = 118.92 deg) (
- (Uranus axial tilt = 97.8 degrees = FSC 0.6 degrees)
- Angle under (E) = 13.33 degrees 1.1 degrees = 14.43 degrees

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The Moon Orbital Motion
- Please remember why we need the moon orbital triangle….
o The moon daily displacement =88000 km but the moon doesn’t use it as its
real displacement but instead the moon uses Pythagoras triangle to define its
real displacement
o Based on that
o The moon uses the right triangle dimension (L= 88000 km Cos θ) where
this (L) is the moon real displacement through its orbit daily
o The angle (θ) is the smallest angle in the right triangle, and it effects on the
moon real displacement and its height in motion above perigee radius!
o Why?
o Because the displacement 88000 km during 29.53 days is a great distance
can be provided only by the apogee orbit whose radius (r=0.406 mkm) so if
the moon uses only 88000 km as a real displacement daily, the moon would
move only through apogee radius
o But
o Because the moon uses real displacement technique (L= 88000 km Cos θ)
so the moon has the ability to move through lower orbits with the Earth, and
based on that, when the angle θ be smaller the real displacement be greater
and needs more wide orbit to be performed which force the moon to move
in high orbits above perigee radius (r=0.363 mkm).
o Then based on that I have suggested the moon motion equation which is
Gerges Equation For The Moon Orbital Motion
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
o Then by more analysis, we have discovered that, a 2nd
force effects on the
moon orbital motion and this force effects on the point (A) in the moon
orbital triangle – where this point is an essential part of the triangle while it's
far from apogee radius with 43000 km
o The 2nd
force is a result of interaction gravity forces between the sun, Earth
and Jupiter on 2 points (Earth and its moon), and because of this interaction
Jupiter causes some gravity force (10% of Earth gravity force) to be effected
on the point (A) and causes the moon motion to apogee radius.
o Then Uranus axial tilt perpendicularity effects analysis gives us the
suggestion that another orbit must be found for the moon motion and this
orbit is found under the first one as described in the following figure.

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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
Researcherid https://publons.com/researcher/3510834/gerges-tawadrous/
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

Earth motion

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