- The document proposes the "Gerges equation for the moon orbital motion" which states that the angle θ per solar day equals the angle of the previous day plus 0.985 degrees. - It claims the moon uses a Pythagorean triangle concept in its motion, where the hypotenuse of 88,000 km represents the moon's daily displacement, and the real displacement is 88,000 km times the cosine of θ. - The equation is tested using actual moon distance data from January 2011, and it correctly predicts the moon's distance on the following day.

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 Basic Equation
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 – 29th
October 2020
Abstract
- I have discovered The Earth Moon Basic Equation
- I call it (Gerges equation for the moon orbital motion)
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)
The Equation Concept
- I claim, the moon original motion is an identical to The Earth one, i.e. the moon
moves daily a distance 2.58 mkm with an angle 0.985 deg. on the horizontal line.
- The moon motion daily rest angle (13.18 deg -0.985 deg= 12.195deg), it's created
by of the moon daily displacement (88000 km)
- The moon daily motion depends on the angle (0.985 deg) as Earth motion, each
day Earth adds (0.985 deg) to its revolution around the sun (till 360 degrees)
- But, Why the moon uses (The angle θ) in its motion? because
- The moon daily displacement (88000km) isn't used as a real displacement through
the moon orbit, instead, the moon uses this value (88000 km) to create another real
displacement suitable with the moon orbit radius… because of that, the moon uses
Pythagoras triangle in its motion. the moon displacement (88000km) will be used
as the right triangle hypotenuse, where the real displacement is (L =88000 cos θ).
- Why does the moon that? During 29.53 days, the available max displacement at
perigee orbit is only 77200 km (Because r=0.363mk x2π= 77200 x29.53 days)
means No greater displacement the moon can do through Perigee Orbit…
- Means, if the moon displacement (88000km) can't be decreased, the moon would
move only on the apogee orbit (r=0.406 mkm)..
- So, The moon by using Pythagoras triangle technique in its motion, enables itself
to move in more near orbit to the Earth.
- This paper tests my equation with many actual data of the moon motion.

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- Gerges Equation For The Moon Orbital Motion
1-1 The Moon Motion Description
1-2 The Pythagoras Triangle In The Moon Motion
1-3 Gerges Equation For The Moon Orbital Motion
1-4 The Equation Test
1-5 The Equation Test Results Discussion And Analysis
1-1 The Moon Motion Description
1. The moon moves per a solar day 2.58 mkm with an angle 0.985 degrees on the
horizontal line identically to Earth motion, because of that, the Earth and moon
aren't separated from each other during their motions.
2. If there's no (relativistic effects) in the solar system and specifically if there's no
lorentz length contraction effect on the moon motion distance, the moon would
move in a parallel line to Earth motion line
3. But, the length contraction effected on the moon motion distance (2.58 mkm)
with rate (1.0725) and caused the moon motion distance to be (2.41 mkm)
shorter than Earth motion distance (2.58 mkm) and put the moon in risk to be
separated from Earth
4. The moon has to move this additional distance (2.58 mkm -2.41 mkm =0.17
mk) to cover the different distance and save its accompanying with Earth…
5. Because of that, the moon moves its daily displacement (88000 km) under the
Earth mass gravity forces, but it's not enough to cover the distance
6. Because the 4 inner planets masses total (Mercury + Venus + Mars + moon) =
Earth Mass (error 1%) – I claim that, these 4 inner planets masses total
produce a gravity force = earth gravity force, which effect on the moon to push
it to move and additional distance (88000 km), so the different distance (0.17
mkm) is covered
7. but this claim has many questions, how these 4 inner planet can effect on the
moon motion spite of the great distances? I answered by claiming, the effect of
these 4 planets my be by some electric charges create some electric field –
which can effect spite the distance- the point we have discussed deeply before
8. But, there's another question, because we see only a motion of 88000 km, so
this additional one (88000 km) isn't seen by us! why?

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
9. the moon original motion (2.58 mkm) which became (2.41 mkm) isn't seen by
us, because it's a displacement on the Earth orbital circumference revolving the
sun, so this motion isn't seen by us because of its direction… I suppose that the
additional distance (88000 km) is done on the moon motion original trajectory
and by that we don't see this also… means we have 2 different directions of the
moon motion, one direction on its original trajectory and the other on the moon
orbit which is seen by us – as shown in the figure
10. The previous explanation tries to prove 2 claims
a. (1st
claim) The angle 0.985 degrees is The moon daily motion reason –
this angle is the motor which defines how the moon will move in the next
day
b. (2nd
claim) There's another force effect on the moon motion. this force I
have discovered in the Pythagoras triangle geometrical structure, and by
this triangle I have discovered (Gerges Equation for the moon orbital
motion). The Pythagoras Triangle is a very useful and trustee technique
used by the moon in its motion – and this triangle I have discovered from
years - i.e. the Pythagoras triangle is a very useful method for the moon
motion analysis because it gives us the proof for the second force effect
on the moon motion
Now let's discuss Pythagoras triangle which is used by the moon motion

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
1-2 The Pythagoras Triangle In The Moon Motion
- 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,
- This second force must be = Earth Gravity Force, so Earth force causes the moon
to move 88000 km and the second force causes the moon to move also 88000 km
- The Effect Point is found on a distance = 43000 km from apogee radius (r=0.406
mkm) and that means, this Effect Point at a distance 449197 km from The Earth
- In the triangle, Point (A) is (The Effect Point) which is found in the previous
figure – It refers to the other force effect on the moon motion
Please Take Care
- This triangle is a symbolic and not geometrical one, for example, the distance
EC=0.374 mkm (= Earth moon distance at total solar eclipse) and BE =0.363 mkm
(the moon perigee radius), and by Pythagoras rule (0.3632
+862
=3742
) but the
distance BC Doesn't = 86000 km, but equal 11000 km, means this triangle for
explanation and not for geometrical direct using, so please consider it in its using.
- Why do we need Pythagoras triangle?

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
Pythagoras triangle using
- How do we know that, the moon uses Pythagoras triangle in its motion?
- Because
- The moon motion 4 basic points (radiuses) are defined based on each other by
Pythagoras rule
- Specifically
o The perigee radius (0.363 mkm)
o The total solar eclipse radius (0.374 mkm)
o The moon orbital distance (0.384 mkm)
o The apogee radius (0.406 mkm)
- (0.374 mkm)2
= (0.363mkm)2
+(86000 km)2
- (0.384 mkm)2
= (0.374mkm)2
+(86000 km)2
- (0.3935 mkm)2
= (0.384 mkm)2
+(86000 km)2
- (0.403 mkm)2
= (0.3935 mkm)2
+(86000 km)2
(403 is very near to 406)
- More interesting data is found in this table,
- (Point A) is the most interesting feature of this triangle
- BE = Perigee radius = 0.363 mkm
- DE = Apogee radius = 0.406 mkm
- EC = total solar eclipse radius = 0.37 mkm
- EG = the orbital distance =0.384 mkm
- EA = 449197.2 km = Jupiter circumference
- EC = 374000 km = Saturn circumference
- AC = 120536 km = Saturn diameter
(The calculations error is around 1%)
The Question is
- If the moon uses Pythagoras triangle to define its motion 4 basic radiuses (perigee,
apogee, total solar eclipse and orbital distance), can that mean, the moon uses this
triangle in its motion usually? And by this question I have discovered that, the
moon uses Pythagoras triangle in its motion to decrease its real displacement
through its orbit to be less than 88000 km as suitable for the orbtial radius.

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
1-3 Gerges Equation For The Moon Orbital Motion
The Equation Concept
- The question is, Can the moon use Pythagoras triangle in its motion?
- In this right triangle
- The blue ball is the moon, the triangle hypotenuse =88000 km (the moon daily
displacement)
- The real displacement through the moon orbit (L) = 88000 Cos (θ)
- Why does the moon need to do that?
- 88000 km x 29.53 days = 2.589 mkm = 2π x0.41 mkm
- The previous calculation tells that, if the moon displacement is only 88000 km and
can't be less, so the moon during 29.53 days will pass a distance = its orbital
circumference at apogee radius (r= 0.406 mkm or even r= 0.41 mkm)
- i.e. the moon with this displacement (88000 km) can move only through the
apogee orbit – but
- when the moon displacement be (77200 km) the moon can during 29.53 days pass
a distance = its orbital circumference at perigee radius (r= 0.363 mkm)
- means, the Pythagoras triangle technique is found to enable the moon to be near to
Earth (showing the Earth great gravity effect on the moon motion)
- but on the other side, the moon moves its displacement (88000 km) to compete the
different distances between (2.58 mkm and 2.41 mkm) and by that to save its
accompanying with Earth motion – that shows the 2 forces which effect on the
moon to force it to use Pythagoras triangle in its motion
- How does this equation work?

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
How Does This Equation?
- We accept that, the real displacement (L =88000 cos θ), when (L) be smaller the
moon can move through near orbit to Earth and when (L) greater the moon has to
move through more far orbit from Earth –
But
- As it's clear, the angle (θ) can change the real displacement (L),
i.e.
- The angle (θ) can't be greater than 30 degrees because the moon can't move lower
than perigee radius (r=0.363 mkm)
- So the angle (θ) is ranged from (1 to 30 degrees), and
o When the angle (θ) be smaller the real displacement (L) be greater and the
moon will move through higher orbit,
o When the angle (θ) be greater the real displacement (L) be smaller and the
moon will move through lower orbit,
i.e.
o the angle (θ) defines the moon real displacement through its orbit and the
moon motion height above Perigee orbit
o Shortly, the angle (θ) defines the moon motion properties…
o The equation add the angle (θ) of previous day to the daily motion angle
(0.985 degrees) to define the moon motion today …
Gerges Equation For The Moon Orbital Motion
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
- Let's test this equation in following to show if it works suitably

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
1-4 The Equation Test With Actual Data
Part (I)
We use here the moon distances to Earth during 1-9 January 2011 (the data is selected
randomly)
On 1/1/2011 the moon Earth distance was 381393 km
So we need to know where the moon will be on the next day (2/1/2011)?
Let's try to do that….
- The radius =381393 km so C (orbital circumference) = 381393 x 2π=2.396 mkm
- The moon moves it during 29.53 days ( L0= 2.396 mkm/29.53) = 81150 km
- L0= 81150 km but L0 = 88000 km Cos θ0
- θ0= 22.756 degrees
- The moon move from Perigee to Apogee so we will discount (NOT add) the angle
- θ1
=
θ0
–
0.985 degrees = 22.756 – 0.985 = 21.77 degrees
- L1 = 88000 km (Cos 21.77) = 81723.6 km
- During 29.53 days = L1 x29.53 = 81723.4 km x 29.53 = 2.4132989 mkm
- 2.41329 mkm = 2π x 384088.4 km
- i.e. The moon on next day (2/1/2011) should be on 384088.4 km
The registered value on /2/1/2011 (385186 km)
The difference = 385186 km -384088.4 = 1097.4 km
We will use this same calculation day after day to see how the equation can describe
the moon motion ….

9. 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
9
On 2/1/2011 the moon Earth distance was 385186 km
So we need to know where the moon will be on the next day (3/1/2011)?
Let's try to do that….
- The radius =385186 km so C (orbital circumference) = 385186 x 2π=2.4211 mkm
- The moon moves it during 29.53 days ( L0= 2.4211 mkm/29.53) = 81990 km
- L0= 81990 km but L0 = 88000 km Cos θ0
- θ0= 21.29789 degrees
- The moon move from Perigee to Apogee so we will discount (NOT add) the angle
- θ1
=
θ0
–
0.985 degrees = 21.29789 – 0.985 =20.312272 degrees
- L1 = 88000 km (Cos 20.312272) = 82527.68 km
- During 29.53 days = L1 x29.53 = 82527.68 km x 29.53 = 2.43704 mkm
- 2.43704 mkm = 2π x 387711 km
- i.e. The moon on next day (3/1/2011) should be on 387711 km
The registered value on /2/1/2011 (388673 km)
The difference = 388673 km -387711 km = 962 km

10. 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
10
On 3/1/2011 the moon Earth distance was 388673 km
So we need to know where the moon will be on the next day (4/1/2011)?
Let's try to do that….
- The radius =388673 km, so C (orbital circumference) = 388673 x 2π=2.4421 mkm
- The moon moves it during 29.53 days ( L0= 2.4421 mkm/29.53) = 82699 km
- L0= 82699 km but L0 = 88000 km Cos θ0
- θ0= 19.98828 degrees
- The moon move from Perigee to Apogee so we will discount (NOT add) the angle
- θ1
=
θ0
–
0.985 degrees = 19.98828 – 0.985 =19.0026 degrees
- L1 = 88000 km (Cos 19.0026) = 83204.3 km
- During 29.53 days = L1 x29.53 = 83204.3 km x 29.53 = 2.457 mkm
- 2.457 mkm = 2π x 391047.5 km
- i.e. The moon on next day (3/1/2011) should be on 391047.5 km
The registered value on /2/1/2011 (392285 km)
The difference = 392285 km -391047.5 km = 1237.5 km

11. 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
11
On 4/1/2011 the moon Earth distance was 392285 km
So we need to know where the moon will be on the next day (5/1/2011)?
Let's try to do that….
- The radius =392258 km, so C (orbital circumference) = 392258 x 2π=2.4648 mkm
- The moon moves it during 29.53 days ( L0= 2.4648 mkm/29.53) = 83467.6 km
- L0= 83467.6 km but L0 = 88000 km Cos θ0
- θ0= 18.46887 degrees
- The moon move from Perigee to Apogee so we will discount (NOT add) the angle
- θ1
=
θ0
–
0.985 degrees = 18.46887 – 0.985 =17.48324 degrees
- L1 = 88000 km (Cos 17.48324) = 83934.8 km
- During 29.53 days = L1 x29.53 = 83934.8 km x 29.53 = 2.47859 mkm
- 2.47859 mkm = 2π x 394480 km
- i.e. The moon on next day (3/1/2011) should be on 394480 km
The registered value on /2/1/2011 (395847 km)
The difference = 395847 km -394480 km = 1366.3 km

12. 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
12
On 5/1/2011 the moon Earth distance was 395847 km
So we need to know where the moon will be on the next day (6/1/2011)?
Let's try to do that….
- The radius =395847 km, so C (orbital circumference) = 395847 x 2π=2.4871 mkm
- The moon moves it during 29.53 days ( L0= 2.48718 mkm/29.53) = 84225.5 km
- L0= 84225.5 km but L0 = 88000 km Cos θ0
- θ0= 16.841805 degrees
- The moon move from Perigee to Apogee so we will discount (NOT add) the angle
- θ1
=
θ0
–
0.985 degrees = 16.841805 – 0.985 =15.85617 degrees
- L1 = 88000 km (Cos 15.85617) = 84651.64 km
- During 29.53 days = L1 x29.53 = 84651.64 km x 29.53 = 2.4997632 mkm
- 2.4997632 mkm = 2π x 397849.66 km
- i.e. The moon on next day (3/1/2011) should be on 397849.66 km
The registered value on /2/1/2011 (399137 km)
The difference = 399137 km -397849.66 km = 1287.34 km

13. 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
13
On 6/1/2011 the moon Earth distance was 399137 km
So we need to know where the moon will be on the next day (7/1/2011)?
Let's try to do that….
- The radius =399137 km, so C (orbital circumference) = 399137 x 2π=2.5078 mkm
- The moon moves it during 29.53 days ( L0= 2.5078 mkm/29.53) = 84925.5 km
- L0= 84925.5 km but L0 = 88000 km Cos θ0
- θ0= 15.18979 degrees
- The moon move from Perigee to Apogee so we will discount (NOT add) the angle
- θ1
=
θ0
–
0.985 degrees = 15.18979 – 0.985 =14.204172 degrees
- L1 = 88000 km (Cos 14.204172) = 85309.6 km
- During 29.53 days = L1 x29.53 = 85309.6 km x 29.53 = 2.519193 mkm
- 2.519193 mkm = 2π x 400942 km
- i.e. The moon on next day (3/1/2011) should be on 400942 km
The registered value on /2/1/2011 (401905 km)
The difference = 401905 km -400942 km = 963 km

14. 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
14
On 7/1/2011 the moon Earth distance was 401905 km
So we need to know where the moon will be on the next day (7/1/2011)?
Let's try to do that….
- The radius =401905 km, so C (orbital circumference) = 401905 x 2π=2.5252 mkm
- The moon moves it during 29.53 days ( L0= 2.5252 mkm/29.53) = 85514.5 km
- L0= 85514.5 km but L0 = 88000 km Cos θ0
- θ0= 13.649010 degrees
- The moon move from Perigee to Apogee so we will discount (NOT add) the angle
- θ1
=
θ0
–
0.985 degrees = 13.649010 – 0.985 =12.664274 degrees
- L1 = 88000 km (Cos 12.664274) = 85859.08 km
- During 29.53 days = L1 x29.53 = 85859.08 km x 29.53 = 2.535418 mkm
- 2.535418 mkm = 2π x 403524.4 km
- i.e. The moon on next day (3/1/2011) should be on 403524.4 km
The registered value on /2/1/2011 (403900 km)
The difference = 403900 km -403524.4 km = 375.6 km

15. 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
15
On 8/1/2011 the moon Earth distance was 403900 km
So we need to know where the moon will be on the next day (9/1/2011)?
Let's try to do that….
- The radius =403900 km, so C (orbital circumference) = 403900 x 2π=2.5377 mkm
- The moon moves it during 29.53 days ( L0= 2.5377 mkm/29.53) = 85939 km
- L0= 85939 km but L0 = 88000 km Cos θ0
- θ0= 12.424731 degrees
- The moon move from Perigee to Apogee so we will discount (NOT add) the angle
- θ1
=
θ0
–
0.985 degrees = 12.424731 – 0.985 =11.4391052 degrees
- L1 = 88000 km (Cos 11.4391052) = 86251.91 km
- During 29.53 days = L1 x29.53 = 86251.91 km x 29.53 = 2.547 mkm
- 2.547 mkm = 2π x 405370 km
- i.e. The moon on next day (3/1/2011) should be on 405370 km
The registered value on /2/1/2011 (404896 km)
The difference = 404896 km -405370 km = - 474 km
Please Note
(1)
The moon still between the heights (403900-403275) ascending and descending
during 4 days
(2)
When the moon descending from (403275 km) in its motion to Earth, each degree
causes a difference around (3000 km) which is 3 folds of the previous results, so
the calculations of the moon motion toward Earth are removed because it's just
similar to the previous calculations but with greater error - through the data list
in the references the reader can perform these calculations easily
Now let's discuss the results in following

16. 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
16
The Results Discussion And Analysis
During 9 days, the moon ascends its orbit from the radius (381939 km) to its apogee
radius (404896 km)
During this 9 days, the equation defines the moon next day point with error around
(1000 km) and this value is not great because the distance between perigee and
apogee = 43000 km or more and this error is considered as 2.5%
That tells us
The Equation Is A Correct One And Trustee
But
Many questions still we have – let's ask them in following
(1)
- Why we discount the angle (θ) and not added? Because the logical concept tells
that, the moon moves per day (0.985 degrees) and this angle is the reason of the
moon motion – because of that, this angle should be added – but in the moon
motion from perigee to apogee we have to discount the angle and not addition it –
this process is related to Pythagoras triangle analysis we have to return to it,
because its geometrical meaning can help us to know the angle sign if positive or
negative
- That means, more analysis we need to do for Pythagoras triangle
(2)
- Why the moon descending motion from apogee to perigee cause sop great error.
Where our previous examples error is around (2.5%), the moon descending motion
causes error with the equation with around (7%), that may tell there's one more
factor still effect on the moon motion toward Earth but not effect on the moon
motion far from Earth – this also we need to discuss in next papers..
The moon data reference is
http://jgiesen.de/moondistance/index.htm

17. 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
17
Part (II)
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%)
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%)

18. 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
18
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%)
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%)

19. 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
19
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)
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%)

20. 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
20
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%)
Note Please
- The previous examples shows the equation works suitably…
- But theses (8) examples 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…
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…

21. 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
21
- 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)
- 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…
We use here the moon distances to Earth during 1-9 January 2011

22. 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
22
1-5 The Equation Test Results Discussion And Analysis
We have 3 groups of proves for this equation which are
(1st
Group)
The moon motion 4 basic points (perigee – total solar eclipse – orbital distance –
apogee) – these 4 points are defined based on each other by Pythagoras rule with
using the dimension 86000 km – (error around 1%)
(2nd
Group)
The equation works suitably with the period from 1 to 9 January 2011 (random data)
and the error was around (2.5%) which makes the equation correct and trustee
(3rd
Group)
A selective group of data (8 examples), which shows that the equation works suitably
Discussion
- The previous test results tell lead us to many important conclusions
1. The Equation Is A Correct And Trustee One
2. There's another factor effect on the moon motion, Specially in the moon
motion descending toward Earth, but this factor doesn't effect almost on the
moon ascending motion far from the Earth –
3. The moon stayed on apogee radius for around 4 days, that because the real
displacement difference based on each degree be so small when the angle
(θ) be less than 10 degrees – that's why the moon could be many days on
(apogee radius). That explain How the moon after 9 January stay on apogee
radius for long time.
Please Note
- Because the moon motion 4 basic points are defined based on Pythagoras triangle,
an because Pythagoras triangle is built depending on the Point (A) (which is the
effect point) – that proves my claim- there's another force effect on the moon
motion in addition to Earth force – because the moon motion basic points are
defined based on a point can't be defined by Earth force – please review this
triangle deeply

23. 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
23
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
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