Paper hypothesis The Moon Motion Basic Equation Is The Following θ Per Solar Day = θ Of The Previous Day + 0.985 degrees Where - The angle (θ) is the smallest one in Pythagoras triangle, that triangle is used by the moon motion to define its real displacement and height through its orbit. - The 0.985 degrees, is the moon original motion degrees which equals Earth motion degrees per solar day, confirming our conclusion that, the moon original motion is an identical motion to Earth motion, i.e. the moon moves 2.58 mkm per a solar day with an angle 0.985 degrees on the horizontal line… - Why the moon uses Pythagoras triangle in its motion? The moon motion daily distance =88000 km isn't used in its total as the moon real displacement through its orbit, Instead, the moon uses its motion distance (88000km) as the Pythagoras triangle hypotenuse, and then uses the triangle dimension (L = 88000km cos θ) as the real displacement. - The intelligent moon uses Pythagoras triangle to change its real displacement daily through its motion, that enables the moon to be nearer to the Earth. for example, Perigee orbit provides a max displacement = 77000 km and not greater, so spite the moon motion distance daily =88000km, the moon can move through perigee orbit by using Pythagoras triangle technique - Based on that, the angle (θ) defines (the real displacement L), and also the moon motion height above perigee radius 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 Motion Trajectory Analysis (VI)
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 – 28th
October 2020
Abstract
Paper hypothesis
The Moon Motion Basic Equation Is The Following
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
Where
- The angle (θ) is the smallest one in Pythagoras triangle, that triangle is used by the
moon motion to define its real displacement and height through its orbit.
- The 0.985 degrees, is the moon original motion degrees which equals Earth
motion degrees per solar day, confirming our conclusion that, the moon original
motion is an identical motion to Earth motion, i.e. the moon moves 2.58 mkm per
a solar day with an angle 0.985 degrees on the horizontal line…
- Why the moon uses Pythagoras triangle in its motion? The moon motion daily
distance =88000 km isn't used in its total as the moon real displacement through its
orbit, Instead, the moon uses its motion distance (88000km) as the Pythagoras
triangle hypotenuse, and then uses the triangle dimension (L = 88000km cos θ) as
the real displacement.
- The intelligent moon uses Pythagoras triangle to change its real displacement daily
through its motion, that enables the moon to be nearer to the Earth. for example,
Perigee orbit provides a max displacement = 77000 km and not greater, so spite
the moon motion distance daily =88000km, the moon can move through perigee
orbit by using Pythagoras triangle technique
- Based on that, the angle (θ) defines (the real displacement L), and also the moon
motion height above perigee radius, why? because with decrease the angle (θ)
the real displacement (L = 88000km cos θ) will be greater, means the moon has to
move into higher point to find a wider orbit to perform the greater displacement.
- This paper tries to test The Supposed Equation to define the angle (θ) daily and
based on it the moon real displacement and its motion height can be defined.

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- The Moon Motion Basic Equation Discussion
1-1 The Equation Test
The Moon Motion Basic Equation Is The Following
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
Where
- The angle (θ) is the smallest one in Pythagoras triangle, that triangle is used by the
moon motion to define its real displacement and height through its orbit.
How to test this equation?
- We will start with the moon in a apogee radius (actual data)
- And will add daily 0.985 degrees per solar day
- Till reach to perigee radius
- And test if the moon reach to this same perigee radius in the same time
- Let's uses several examples in following
Example No. 1
- On 31st
August 2009 the moon was on apogee (405269 km) ….and
- On 16th
September 2009 the moon reached perigee (364054 km)
- Can the equation predict these value? Let's see that
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
- What's θ when the moon be in apogee (405269 km)? 11.4 degrees
- Because
o 405269 km x 2π = 2.546 million km = 29.53 days x 86230 km
o (The real displacement (L) = 86230 km, and (L) = 88000 km Cos θ)
- 17 days x 0.985 degrees = 16.755 degrees
θ Per Solar Day = 11.4 degrees + 16.755 degrees =28.151 degrees
- L = 88000 km Cos θ = 88000 Cos (28.151)= 77590 km
- 77590 km x 29.53 days = 2.291 mkm = 2π x 364514 km
A Comment
(The registered 364054 km and the equation result is 364514 km, the difference
=0.12%)

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
Example No. 2
- On 19-3-2009 the moon was on apogee (404302 km) ….and
- On 2-4- 2009 the moon reached perigee (370014 km)
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
- What's θ when the moon be in apogee (404302 km)? 12 degrees
o 404302 km x 2π = 2.5413 million km = 29.53 days x 86059 km
o (The real displacement (L) = 86059 km, and (L) = 88000 km Cos θ)
- 15 days x 0.985 degrees = 14.77 degrees
θ Per Solar Day = 12 degrees + 14.77 degrees =26.77 degrees
- L = 88000 km Cos θ = 88000 Cos (26.77)= 78568 km
- 78568 km x 29.53 days = 2.32 mkm = 2π x 369108 km
A Comment
(The registered 370014 km and the equation result is 369108 km, the difference
=0.24%)
Example No. 3
- On 31-1-2008 the moon was on apogee (404532 km) ….and
- On 14-1- 2008 the moon reached perigee (370216 km)
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
- What's θ when the moon be in apogee (404532 km)? 11.9 degrees
o 404532 km x 2π = 2.54277 million km = 29.53 days x 86108 km
o (The real displacement (L) = 86108 km, and (L) = 88000 km Cos θ)
- 15 days x 0.985 degrees = 14.784 degrees
θ Per Solar Day = 11.9 degrees + 14.784 degrees =26.684 degrees
- L = 88000 km Cos θ = 88000 Cos (26.684)= 78627 km
- 78627 km x 29.53 days = 2.321 mkm = 2π x 369388 km
A Comment
(The registered 370216 km and the equation result is 369388 km, the difference
=0.22%)

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
Example No. 4
- On 27-5-2007 the moon was on apogee (405458 km) ….and
- On 12-6- 2007 the moon reached perigee (363778 km)
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
- What's θ when the moon be in apogee (405458 km)? 11.3 degrees
o 405458 km x 2π = 2.5485 million km = 29.53 days x 86305 km
o (The real displacement (L) = 86305 km, and (L) = 88000 km Cos θ)
- 17 days x 0.985 degrees = 16.755 degrees
θ Per Solar Day = 11.3 degrees + 16.755 degrees =28.055 degrees
- L = 88000 km Cos θ = 88000 Cos (28.055)= 77659 km
- 77659 km x 29.53 days = 2.293 mkm = 2π x 364841.7 km
A Comment
(The registered 363778 km and the equation result is 364814.7 km, the difference
=0.28%)
Example No. 5
- On 6-12-2007 the moon was on apogee (406235 km) ….and
- On 22-12- 2007 the moon reached perigee (360817 km)
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
- What's θ when the moon be in apogee (406235 km)? 10.7 degrees
o 406235 km x 2π = 2.5534 million km = 29.53 days x 86470 km
o (The real displacement (L) = 86470 km, and (L) = 88000 km Cos θ)
- 17 days x 0.985 degrees = 16.755 degrees
θ Per Solar Day = 10.7 degrees + 16.755 degrees =27.45 degrees
- L = 88000 km Cos θ = 88000 Cos (27.45)= 78090 km
- 78090 km x 29.53 days = 2.306 mkm = 2π x 366841.7 km
(There's a great difference between the equation result and the registered value, we
have to discuss that in the Equation discussion)
A Comment
(The registered 360817 km and the equation result is 366841.7 km, the difference
is some how great and the equation here is almost inaccurate)

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
Example No. 6
- On 9-4-2006 the moon was on apogee (405551 km) ….and
- On 25-4- 2006 the moon reached perigee (363737 km)
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
- What's θ when the moon be in apogee (405551 km)? 11.31 degrees
o 405551 km x 2π = 2.548 million km = 29.53 days x 86290 km
o (The real displacement (L) = 86470 km, and (L) = 88000 km Cos θ)
- 17 days x 0.985 degrees = 16.755 degrees
θ Per Solar Day = 11.31 degrees + 16.755 degrees =28.067 degrees
- L = 88000 km Cos θ = 88000 Cos (28.067)= 77650 km
- 77650 km x 29.53 days = 2.293 mkm = 2π x 364798 km
A Comment
(The registered 363737 km and the equation result is 364798 km, the difference is
0.29%)
Example No. 7
- On 11- 8- 2004 the moon was on apogee (405291 km) ….and
- On 27-8- 2006 the moon reached perigee (365106 km)
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
- What's θ when the moon be in apogee (405291 km)? 11.38 degrees
o 405291 km x 2π = 2.5475 million km = 29.53 days x 86270 km
o (The real displacement (L) = 86270 km, and (L) = 88000 km Cos θ)
- 17 days x 0.985 degrees = 16.755 degrees
θ Per Solar Day = 11.38 degrees + 16.755 degrees =28.135 degrees
- L = 88000 km Cos θ = 88000 Cos (28.135)= 77600 km
- 77600 km x 29.53 days = 2.291 mkm = 2π x 364566 km
(The registered 365106 km and the result is 364566 km, the difference is 0.14%)
Example No. 8
- On 23- 11- 2001 the moon was on apogee (404396 km) ….and
- On 6-12- 2001 the moon reached perigee (370114 km)
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
- What's θ when the moon be in apogee (404396 km)? 12 degrees
o 405291 km x 2π = 2.5419 million km = 29.53 days x 86079 km
o (The real displacement (L) = 86079 km, and (L) = 88000 km Cos θ)
- 14 days x 0.985 degrees = 13.8 degrees
θ Per Solar Day = 12 degrees + 13.8 degrees =25.8 degrees
- L = 88000 km Cos θ = 88000 Cos (25.8)= 79228 km
- 79228 km x 29.53 days = 2.33 mkm = 2π x 372213 km
(The registered 370114 km and the result is 372213 km, the difference is 0.5%)

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
1-2 The Equation Discussion
- The previous examples shows the equation works suitably…
- Any way, the previous examples can't be a proof for this equation, why?
- Because they are selective data and not selected randomly
- I have selected specific data from the moon perigee and apogee tables, this specific
data works suitably with the Equation while many other data can't works totally, as
we have seen in examples No. 5, the equation result is inaccurate completely…
o Let's explain this specific data
o I used a long period between perigee and apogee (15 days or more), but
when the period is 13 days or less always the equation can't work
o Also I used 2 near values for perigee and apogee
o Almost when the values approximately be (apogee 404000 km and perigee
368000, with a time period between them around 15 days, the equation
works suitably… means, if the difference between perigee and apogee is
more, so we need longer time… that's how this equation works…
- The next question is Why the Equation works with some data and doesn't wok
with the others? to answer let's remember some features of the angle (θ)…
The Angle (θ) Features
- We know that, the angle (θ) can't be greater than (29) because the moon perigee
radius = 0.363 mkm, and (θ) = (29 degrees) the real displacement will be 77000km
which needs an orbital radius = 0.361742 mkm which is lower already than
Perigee radius, means the greater (θ) angle, needs the moon to move lower than
Perigee radius which isn't usual in the moon motion…
- The Basic reason is that, because each degree between the range (1 degree to 29
degrees) of the angle (θ) causes a different increase in the real displacement,
means, not each angle causes the same increased distance in the real displacement,
on the contrary, each angle causes difference increased distance… let's test that
with real data to see as clear as possible
(1)
o for the angle (θ) = 10 degrees, the real displacement will be 86663 km
o for the angle (θ) = 11 degrees, the real displacement will be 86383 km
(The increased distance = 280 km)
(2)
o for the angle (θ) = 17 degrees, the real displacement will be 84154.8 km
o for the angle (θ) = 18 degrees, the real displacement will be 83693 km
(The increased distance = 461.8 km)
(3)
o for the angle (θ) = 25 degrees, the real displacement will be 79755 km
o for the angle (θ) = 26 degrees, the real displacement will be 79093.8 km
(The increased distance = 661.2 km)

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
- The problem simply is that, not each degree causes the same increased distance,
but in the previous (8) examples, the calculations consider that each degree causes
an equal increased distance… because of that the equation result can't be equal the
registered value…
- Simply we need to move with the moon with each (0.985 degree) day by day, and
this following may show clearly that, the moon motion is defined clearly by this
same equation
The Moon Motion Basic Equation Is The Following
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
- So we have to analyze the moon motion day by day based on this same equation to
prove this claim
- The equation simply shows the moon motion orbital trajectory, and this equation is
almost trustee and hope to test it with the moon motion day by day to show clearly
its creditability
- So in the next paper we have to test the moon motion day by day by this same
equation
- Also, we may answer the question, why the moon needs to change its real
displacement through its orbit without change in its daily motion? the question
tells us that, there are 2 forces effect on the moon, the 1st
one forces the moon to
move daily this 88000 km without change and the 2nd
one force the moon to move
as near as possible to Earth and by that the moon found no better technique than to
use Pythagoras triangle to create a real displacement different from its daily
motion…. that means, Pythagoras triangle is a geometrical necessity in the moon
motion and by this triangle may the 2 forces be more clear… let's try to do these
tasks in the next 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
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
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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:
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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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