Paper Hypothesis -There's Another Similar Orbit For The Moon Motion The Hypothesis explanation - The triangle ACE shows that, a double value of the triangle perimeter is used for geometrical necessities in the moon orbital motion - i.e. - While the (ACE) triangle perimeter = 943817 km, the geometrical structure of the moon orbit uses the value 1887634 km (= 2 x 943817 km) as one of the basic motion values. - There's one more reason for this hypothesis, because the moon orbital motion space area contains (50%) of the Jupiter Whole Energy, that tells if the energy is transported to the moon orbit why just (50%) only? - I suggest that - Another similar orbit of the moon must be found which we can't see. This idea is supported by our previous argument, where the moon moves daily 2.58 mkm (as Earth) and because of the contraction this value became 2.41 mkm, So the tries to cover this difference (0.17 mkm) by its daily displacement (88000km) which is not enough and the moon needs to move another displacement (88000km) to cover the different distance, where this last displacement (88000km) we can't see, and based on that I suppose the moon must have another similar orbit through which the moon moves this additional displacement (88000km). - The paper tries to prove this fact Gerges Francis Tawdrous +201022532292

1. 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
The Moon Orbital Triangle Analysis (VII)
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 – 10th
November 2020
Abstract
Paper Hypothesis
- There's Another Similar Orbit For The Moon Motion
The Hypothesis explanation
- The triangle ACE shows that, a double value of the triangle perimeter is used for
geometrical necessities in the moon orbital motion
- i.e.
- While the (ACE) triangle perimeter = 943817 km, the geometrical structure of the
moon orbit uses the value 1887634 km (= 2 x 943817 km) as one of the basic
motion values.
- There's one more reason for this hypothesis, because the moon orbital motion
space area contains (50%) of the Jupiter Whole Energy, that tells if the energy is
transported to the moon orbit why just (50%) only?
- I suggest that
- Another similar orbit of the moon must be found which we can't see. This idea is
supported by our previous argument, where the moon moves daily 2.58 mkm (as
Earth) and because of the contraction this value became 2.41 mkm, So the tries to
cover this difference (0.17 mkm) by its daily displacement (88000km) which is not
enough and the moon needs to move another displacement (88000km) to cover the
different distance, where this last displacement (88000km) we can't see, and based
on that I suppose the moon must have another similar orbit through which the
moon moves this additional displacement (88000km).
- The paper tries to prove this fact

2. 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
2
1- Another Orbit Must Be Found For The Moon Motion
1-1 The Moon Orbital Triangle Perimeter
z
Figure No. (1) (my figure)
Please Note
(1) The blue dotted arrow creates a point (Z) between F & S
SZ = 7665 km ZF = 2414 km
CZS = 77.8 degrees CZF =102.195 degrees
(2) The Green arrow creates a point (Y) after the point D
DY = 2513.7 km DYA= 118.92 degrees

3. 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
3
Let's Review The Moon Orbital Triangle Data
- 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 = =47513.7 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) (BYC = 61.08 deg)
- (Uranus axial tilt = 97.8 degrees = FSC 0.6 degrees) and
- (CYA +0.6 deg = 119.5 deg where 119.5 x 0.99 = 118.3 deg (Neptune A. T)
- Angle under (E) = 13.33 degrees 1.1 degrees = 14.43 degrees

4. 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
4
The Triangles Perimeters Discussion
(1st
Point)
- The perimeter of the triangle ABC = 293620 km
- But
- 43 x 6792 km (Mars diameter) = 292056 km (error 0.5%)
- The previous data tells that, the moon orbit is created both planets (the moon and
Mars), where …
- The triangle perimeter is equivalent to the perigee apogee distance (=43000km),
i.e. Each 1000 km of this distance is equivalent to (6792 km = Mars diameter) of
the triangle (ABC) perimeter
i.e.
- We have here 2 cars moves with different rates, the moon moves with 1000 km but
mars moves with its diameter 6792 km and both finishes their motion on the same
rate (43).
(2nd
Point)
- The Perimeter Of The Triangle ACE = 943817 km
- And
- 943817 km x 2 = 1887634 km = (Approximately =1.88 mkm)
This triangle tells that "Another Orbit Must Be Found For The Moon Motion"
Equation No. 1
1887634 km = 86000 km x (Jupiter Mass/ Uranus Mass)
- 86000 km = The distance BC in the triangle (the perpendicular line) and based on
this value we calculate all the triangle distances and angles..
- Equation No. (1) tells that, there are 2 values of this triangle perimeter!
Because
- The 2 planets (Jupiter and Uranus) are the 2 players in the moon orbit where
Uranus is perpendicular on the moon orbit and create an angle 91.1 degrees with
The Moon Axial Tilt – and Jupiter causes the second for effects on the point (A),
means the 2 planets which are players in the moon orbit use their masses to create
this value (1887634 km) as a function of the triangle basic dimension (86000km)
- Equation No. (1) proves that, for some geometrical necessary, a double value of
the perimeter of this triangle (ACE) is used in the moon orbital geometrical
structure.

5. 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
5
Equation No. 2
1887634 km = 97.8 degrees 19300 km
- But (366556 km (the outer planets diameter total) = 19 x 19300 km
- 97.8 degrees = Uranus Axial Tilt
- 19 degrees the moon orbit regresses during a year.. but why we use here the outer
planets diameters total? because the perigee radius 363000 km + 3475 km (the
moon diameter) = the outer planets diameters total…
- i.e. if the moon stand on the perigee radius and we measure the distance to Earth
including the moon diameter we will measure the outer planets diameters total –
it's Not pure coincidence … because the apogee radius 406000 km = All solar
planets diameters total, it's simply one feature of the moon orbit geometrical
structure
- so this value (366556 km) = 19 degrees x 19300 km, i.e. each degree is equivalent
to 19300 km….
i.e.
- the moon orbit regresses 19 degrees each year, and each degree of these (19 deg) is
equivalent to a distance =19300 km if we use the outer planets diameters total as
its medium – so the degrees are equivalent to some (km) by this rate
- In our Equation No (2), Uranus axial tilt is used in place of 19 degrees and based
on that we will use the triangle double perimeter (1887634 km)… I try to show
that, this value (1887634 km) is used for real purposes and that proves its existence
- Let's ask why Uranus Axial Tilt (97.8 degrees) is used in place of (19 degrees)?
Because 97.8 deg =19 deg x 5.1 degrees (the moon orbital inclination =5.1 deg),
means these 3 values consist together one machine, i.e. Uranus axial tilt (97.8 deg)
effected to produce these 2 values (19 deg and 5.1 deg). Simply Uranus axial tilt is
their parent – and by that- if the value 19 degrees produces 19300 km for each 1
degree by using the outer planets diameter total as medium. So this same machine
produces 19300 km for each degree of Uranus axial tilt by using the perimeter
double value (1887634 km) as medium..
Equation No. 3
1887634 km = π3
x 60891 km (Saturn radius) (error 1%)
- 31 = (Uranus Axial Tilt / Jupiter Axial Tilt)
- Saturn diameter = AC (The Triangle ACE)
i.e.
- We can replace this (Saturn radius) with (1/2 of AC)
- Again
- Equation No. (3) tells that, the 2 players planets have interaction and produces (π3
)
which based on which Saturn radius (or 0.5 of AC) is produced, the equation tells,
there's a geometrical necessary to use a double value of the triangle perimeter.

6. 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
6
Equation No. 4
1887634 km = 2π x 300000 km
- Why should use double values of this triangle perimeter (ACE)??
- Based on what this hypothesis should be suggested??
Let's remember
- The Research 1st
hypothesis (Planet Motion Depends On Light Motion)
- The Motion Rule (A Light Moves Through A Circle Circumference And A
Planet Moves Along Its Radius)
- Because of that, We have found frequently the rate (2π) in Jupiter data analysis,
- Equation No. (4) tells that, we have 2 motions here, the first motion is a planet
motion which moves 300000 km and the second motion must be a light motion to
pass the distance (300000 km x 2π = 1887634 km), that tells us the used distance
(1887634 km) is just real one and found for a geometrical necessary…
- But
- Does the moon move 300000 km?!
- 10921 km (the moon circumference) x 27.3 days =300000 km, this old equation
told us that…
o If the moon rotates around its axis one rotation per a solar day (as Earth
does), so the moon will travel during its orbital period a distance =300000
km = light motion for 1 second (light known velocity 0.3 mkm/sec)
o The moon doesn't rotate around its axis (actually) once per a solar day, But
the light motion considers this motion as real.
Equation No. 5
The area of the moon motion distance
(from perigee to apogee 43000 km) = 103883 mkm =
50% of Jupiter Total Energy (error 3.5%)
- The Moon Orbit Energy =50% Of The Total Energy
So
- Where's The Other 50% Of The Energy
i.e.
- If the moon orbital triangle perimeter (ACE) = 943817 km and the used value is
(1887634 km) – that means we see only the half of fact
- The other 50% must be found in the moon orbit but we don't see it

7. 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
7
Equation No. 6
1887634 km = 366556 km (the outer planets diameters total) x 5.1 (error1%)
(5.1 degrees = the moon orbital inclination)
1887634 km = 6792 km (Mars diameter) x 278.4
(278.4 degrees = The Solar Planets Axial Tilt Total)
1887634 km = 15327.8 km (Mercury circumference) x 122.5
(122.5 degrees = Pluto Axial Tilt)
1887634 km = 142984 km (Jupiter diameter) x 13.1 (error 0.8%)
(13.1 = Jupiter diameter /the moon circumference)
Many data proves this value (1887634 km) is a real one and in using in the moon data
Equation No. 7
1887634 km = 1.88 mkm (Mars orbital inclination =1.9 deg)
1 mkm = 1 deg, so this value (1.88 mkm is equivalent to 1.9 deg) (error 1%)
(We know Mars orbital inclination 1.9 deg is used as a rate for 75% of all inner
planets distances, in addition to use as a rate between Earth and Mars orbital periods)
Equation No. 8
2070 mkm = 550.7 mkm (Mars Jupiter distance) x2 x 1.88
550.7 mkm (Mars Jupiter Distance) x 2 = Mars Jupiter orbital diameter
2094 mkm = Jupiter Uranus Distance (with 2070 mkm error 1%)
- Equation No. (8) tells, again, the 2 basic players in the moon orbit have interaction
between each other (this time in their distance) based on this same rate 1.88
Equation No. 9
940 mkm (Earth orbital circumference) = 1.88 x 500 mkm
- Light known velocity (0.3mkm/sec) needs 500 second to reach Earth from the sun,
the period (500 s) is used as a distance value (500 mkm) and the rate 1.88 is found
in a basic relationship between the light and earth motions…
Equation No. 10
3033.5 mkm (Uranus Pluto distance) = 1.88 x 1622.6 mkm (Uranus Neptune distance)
- The rate 1.88 is used by Uranus in 2 relationships connected Uranus with Neptune
and Pluto –
Equation No. 11
5127.4 mkm (Jupiter Pluto distance) = 1.88 x 2737 mkm
(2737 mkm the distance from Uranus to point 135.5 mkm far from the sun), the
distance 2737 mkm we have discussed in the previous paper where the light travels
through this distance in opposite Pluto motion through 43000 km…these motions
have a great effect on the moon orbit (please review the previous paper).

8. 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
8
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
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