The Moon Motion Basic Equation (modified) (Revised) Abstract A Modification In The Moon Orbital Motion Equation The Old Form θ1 (Per Solar Day) = θ0 (of Previous Day) + 0.98562 degrees The Modified form θ1 (Per Solar Day) = θ0 ( of Previous Day) + 1.6 degrees - The Equation uses 1.5 degrees in place of the angle (0.98562 degrees) because the moon axial tilt (6.7 degrees) declines on the Earth Ecliptic line with 1.6 degrees. - This paper tests the new modification on the equation accuracy - The modified equation defines the moon position with an accuracy =96.12 % (where the error is (3.88 %)) The Equation Concept - The moon uses Pythagoras triangle to do the following job. While the moon moves a daily displacement =88000 km, The moon real displacement (L) through its orbit can be less than (88000 km) where (L =88000 Cos (θ)) - The moon needs to decrease its daily displacement (88000 km) through its orbit to enable the moon to revolve around Earth in more near orbits to Earth - The angle (θ) defines the moon real displacement and its distance to perigee radius (r=0.363 mkm). - For that reason the equation depends on the angle (θ). 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 Basic Equation (modified) (Revised)
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 – 16th
December 2020
Abstract
A Modification In
The Moon Orbital Motion Equation
The Old Form
θ1 (Per Solar Day) = θ0 (of Previous Day) + 0.98562 degrees
The Modified form
θ1 (Per Solar Day) = θ0 ( of Previous Day) + 1.6 degrees
- The Equation uses 1.5 degrees in place of the angle (0.98562 degrees) because the
moon axial tilt (6.7 degrees) declines on the Earth Ecliptic line with 1.6 degrees.
- This paper tests the new modification on the equation accuracy
- The modified equation defines the moon position with an accuracy =96.12 %
(where the error is (3.88 %))
The Equation Concept
- The moon uses Pythagoras triangle to do the following job. While the moon moves
a daily displacement =88000 km, The moon real displacement (L) through its orbit
can be less than (88000 km) where (L =88000 Cos (θ))
- The moon needs to decrease its daily displacement (88000 km) through its orbit to
enable the moon to revolve around Earth in more near orbits to Earth
- The angle (θ) defines the moon real displacement and its distance to perigee radius
(r=0.363 mkm).
- For that reason the equation depends on the angle (θ).

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- Introduction
- The moon moves daily (88000 km) on the right triangle hypotenuse (AC), but the
moon creates an angle (θ) between its motion direction and its orbital horizontal
level, by that the real displacement will be (L= 88000 km cos (θ)) through the
moon orbit, and by that, spite the moon moves 88000 km and the orbital horizontal
distance be less than (88000 km) and this is the objective for which the moon uses
Pythagoras triangle –
- As an example, if (θ) =28.63 degrees So the real displacement (L) = 77237 km, So,
if the moon real displacement daily be equal this value during 29.53 days the moon
will move a distance = 2.28 million km and this will be the moon orbital
circumference, BUT for what radius? 2.28 mkm = 2π x (0.363 mkm)
- The Moon Orbital Perigee Radius =0.363 mkm
- That means, the moon by a real displacement =77237 km can move around Earth
through the perigee orbit (radius =0.363 mkm), this is the useful result the moon
performs by using Pythagoras triangle,
- Now let's suppose the moon doesn't use Pythagoras triangle, what would happen?
- The moon daily displacement = 88000 km, during 29.53 days the moon moves a
distance = 2.598 mkm BUT for what radius? 2.598 mkm = 2π x (0.413 mkm)
- The Moon Orbital Apogee Radius =0.406 mkm
- So the moon will move along month revolving round Earth through its apogee
orbit (or even far from apogee orbit) because the total distance can't be passed
through any more near orbit around Earth…
- The data shows how Pythagoras triangle is so useful for the moon orbital motion.

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
The Angle θ
- The angle (θ) needs to pay some attention for its specific effect…let's summarize
the idea in following
o The angle (θ) changes the real displacement (L = 88000 cos (θ)), through the
moon orbit..
o We know that, when the real displacement (L) be shorter the moon can
move through near orbits to Earth and that means the moon will be near or
at Perigee radius (0.363 mkm)
o When the real displacement (L) be greater the moon has to move through
orbits far from Earth and by that the moon will be near or at to apogee orbit
o That means, the angle (θ) does change the real displacement (L) and also
change the distance between the moon to perigee or to apogee, shortly, the
angle defines the moon position (as a ship) between 2 river banks….
- Let's analyze the angle (θ) in some deep
o We know that (363000)2
+ (86000)2
= (373000)2
o In Pythagoras triangle with the previous dimensions, what's the angle (θ)
value? (θ) = 13.33 degrees
o Also (396800)2
+ (86000)2
= (406000)2
the angle (θ) = 12.229 degrees
o I have used (363000 km and 406000 km) because they are the perigee and
apogee radiuses between which the moon moves.
o The difference between angles = 1.1 degrees
i.e.,
The angle (1.1 deg.) controls the moon motion from perigee to apogee.

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
2- The Modified Equation Test
The Modified form
θ1 (Per Solar Day) = θ0 ( of Previous Day) + 1.6 degrees
2-1 The Equation Test (6-30 June 2020)
I-Data
Date The Earth Moon Distance
6-6-2020 369418
7-6-2020 373729
8-6-2020 378917
9-6-2020 384534
10-6-2020 390096
11-6-2020 395156
12-6-2020 399345
13-6-2020 402395
14-6-2020 404153
15-6-2020 404574
16-6-2020 403718
17-6-2020 401733
18-6-2020 398840
19-6-2020 395303
20-6-2020 391409
21-6-2020 387432
22-6-2020 383607
23-6-2020 380110
24-6-2020 377044
25-6-2020 374451
26-6-2020 372338
27-6-2020 370703
28-6-2020 369571
29-6-2020 369005
30-6-2020 369106
Data Reference is
https://www.timeanddate.com/moon/egypt/cairo?month=1
Note Please
- In the ascending motion from perigee to apogee the angle (1.6 deg) will be minus
- In the descending motion from apogee to perigee the angle (1.6 deg) will be
positive.

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
II- The Equation Test
θ1 (Per Solar Day) = θ0 ( of Previous Day) + 1.6 degrees
1st
The Ascending Motion
(1) 6 –June -2020
Jun 6 369418 km Jun 7 373729km (The Next Day)
On 6/6/2020 the moon Earth distance was 369418 km
So we need to know where the moon will be on the next day (7/6/2020)?
Let's try to do that….
- The radius =369418 km so C (orbital circumference) = 369418 x 2π=2.321 mkm
- The moon moves it during 29.53 days ( L0= 2.321 mkm/29.53) = 78602.1 km
- L0= 78602.1 km but L0 = 88000 km Cos θ0
- θ0= 26.721 degrees
- θ1= θ0- 1.6 = 26.721 – 1.6 = 25.121 degrees
- L1 = 88000 km (Cos 25.121 degrees) = 79676.3 km
- During 29.53 days = L1 x29.53 = 79611 km x 29.53 = 2.352 mkm
- 2.352 mkm = 2π x 374466.6 km
- i.e. The moon on next day (7/6/2020) should be on 374466.6 km
The registered value on 7/6/2020 (373729 km)
The difference = 373729 km – 374466.6 km = - 737.6 km

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
(2) 7 –June -2020
Jun 7 373729 km Jun 8 378917km (The Next Day)
On 7/6/2020 the moon Earth distance was 373729 km
So we need to know where the moon will be on the next day (8/6/2020)?
Let's try to do that….
- The radius =373729 km so C (orbital circumference) = 373729 x 2π=2.348 mkm
- The moon moves it during 29.53 days ( L0= 2.348 mkm/29.53) = 79519.4 km
- L0= 79519.4 km but L0 = 88000 km Cos θ0
- θ0= 25.36 degrees
- θ1= θ0- 1.6 = 25.36 – 1.6 = 23.76 degrees
- L1 = 88000 km (Cos 23.76 degrees) = 80541.2 km
- During 29.53 days = L1 x29.53 = 80541.2 km x 29.53 = 2.378 mkm
- 2.378 mkm = 2π x 378531.3 km
- i.e. The moon on next day (8/6/2020) should be on 378531.3 km
The registered value on 8/6/2020 (378917 km)
The difference = 378917 km –378531.3 km = 385.7 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
(3) 8 – June -2020
Jun 8 378917 km Jun 9 384534km (The Next Day)
On 8/6/2020 the moon Earth distance was 378917 km
So we need to know where the moon will be on the next day (9/6/2020)?
Let's try to do that….
- The radius =378917 km so C (orbital circumference) = 378917 x 2π=2.3808 mkm
- The moon moves it during 29.53 days ( L0= 2.3808 mkm/29.53) = 80623.3 km
- L0= 80623.3 km but L0 = 88000 km Cos θ0
- θ0= 23.627 degrees
- θ1= θ0- 1.6 = 23. 627 – 1.6 = 22.027 degrees
- L1 = 88000 km (Cos 22.027 degrees) = 81576.6 km
- During 29.53 days = L1 x29.53 = 81576.6 km x 29.53 = 2.408 mkm
- 2.408 mkm = 2π x 383397.4 km
- i.e. The moon on next day (9/6/2020) should be on 383397.4 km
The registered value on 9/6/2020 (384534 km)
The difference = 384534 km – 383397.4 km = 1136.6 km

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
(4) 9 – June -2020
Jun 9 384534 km Jun 10 390096km (The Next Day)
On 9/6/2020 the moon Earth distance was 384534 km
So we need to know where the moon will be on the next day (10/6/2020)?
Let's try to do that….
- The radius =384534 km so C (orbital circumference) = 384534 x 2π=2.416 mkm
- The moon moves it during 29.53 days ( L0= 2.416 mkm/29.53) = 81818.4 km
- L0= 81818.4 km but L0 = 88000 km Cos θ0
- θ0= 21.60 degrees
- θ1= θ0- 1.6 = 21. 60 -1.6 = 20 degrees
- L1 = 88000 km (Cos 20 degrees) = 82692.9 km
- During 29.53 days = L1 x29.53 = 82692.9 km x 29.53 = 2.4419 mkm
- 2.4419 mkm = 2π x 388644 km
- i.e. The moon on next day (10/6/2020) should be on 388644 km
The registered value on 10/6/2020 (390096 km)
The difference = 390096 km – 388644 = 1452 km

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
(5) 10 – June -2020
Jun 10 390096 km Jun 11 395156km (The Next Day)
On 10/6/2020 the moon Earth distance was 390096 km
So we need to know where the moon will be on the next day (11/6/2020)?
Let's try to do that….
- The radius =390096 km so C (orbital circumference) = 390096 x 2π=2.451 mkm
- The moon moves it during 29.53 days ( L0= 2.451 mkm/29.53) = 83002 km
- L0= 83002 km but L0 = 88000 km Cos θ0
- θ0= 19.4033 degrees
- θ1= θ0- 1.6 = 19.4033 – 1.6 = 17.8033 degrees
- L1 = 88000 km (Cos 17.8033 degrees) = 83785.8 km
- During 29.53 days = L1 x29.53 = 83785.8 km x 29.53 = 2.474 mkm
- 2.474 mkm = 2π x 393780.5 km
- i.e. The moon on next day (11/6/2020) should be on 393780.5 km
The registered value on 11/6/2020 (395156 km)
The difference = 395156 km –393780.5 km = 1375 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
(5) 11 – June -2020
Jun 11 395156 km Jun 12 399345 km (The Next Day)
On 11/6/2020 the moon Earth distance was 395156 km
So we need to know where the moon will be on the next day (12/6/2020)?
Let's try to do that….
- The radius =395156 km so C (orbital circumference) = 395156 x 2π=2.482 mkm
- The moon moves it during 29.53 days ( L0= 2.482 mkm/29.53) = 84078.5 km
- L0= 84078.5 km but L0 = 88000 km Cos θ0
- θ0= 17.169 degrees
- θ1= θ0- 1.6 = 17.169 – 1.6 = 15.569 degrees
- L1 = 88000 km (Cos 15.569 degrees) = 84771 km
- During 29.53 days = L1 x29.53 = 84771 km x 29.53 = 2.503 mkm
- 2.503 mkm = 2π x 398411 km
- i.e. The moon on next day (12/6/2020) should be on 398411 km
The registered value on 12/6/2020 (399345 km)
The difference = 399345 km – 398411 km = 934 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
(6) 12 – June -2020
Jun 12 399345 km Jun 13 402395 km (The Next Day)
On 12/6/2020 the moon Earth distance was 399345 km
So we need to know where the moon will be on the next day (13/6/2020)?
Let's try to do that….
- The radius =399345 km so C (orbital circumference) = 399345 x 2π=2.509 mkm
- The moon moves it during 29.53 days ( L0= 2.509 mkm/29.53) = 84970 km
- L0= 84970 km but L0 = 88000 km Cos θ0
- θ0= 15.0794 degrees
- θ1= θ0-1.6 = 15.0794 – 1.6 = 13.4794 degrees
- L1 = 88000 km (Cos 13.4794 degrees) = 85575.9 km
- During 29.53 days = L1 x29.53 = 85575.9 km x 29.53 = 2.527 mkm
- 2.527 mkm = 2π x 402193.6 km
- i.e. The moon on next day (13/6/2020) should be on 402193.6 km
The registered value on 13/6/2020 (402395 km)
The difference = 402395 km –402193.6 km = 202 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
(7) 13 – June -2020
Jun 13 402395 km Jun 14 404153 km (The Next Day)
On 13/6/2020 the moon Earth distance was 402395 km
So we need to know where the moon will be on the next day (14/6/2020)?
Let's try to do that….
- The radius =402395 km so C (orbital circumference) = 402395 x 2π=2.528 mkm
- The moon moves it during 29.53 days ( L0= 2.528 mkm/29.53) = 85618.8 km
- L0= 85618.8 km but L0 = 88000 km Cos θ0
- θ0= 13.3592 degrees
- θ1= θ0- 1.6 = 13.3592 – 1.6 = 11.7592 degrees
- L1 = 88000 km (Cos 11.7592 degrees) = 86153.1 km
- During 29.53 days = L1 x29.53 = 86153.1 km x 29.53 = 2.5441 mkm
- 2.5441 mkm = 2π x 404906.3 km
- i.e. The moon on next day (14/6/2020) should be on 404906.3 km
The registered value on 14/6/2020 (404153 km)
The difference = 404153 km – 404906.3 km = (-) 753.4 km
Note Please
- This day is the last day in the moon ascending motion from Perigee to apogee, on
the next day the moon will stay on apogee for one more day (404574 km) on (15th
- 6-2020), and then on (16th
-6-2020) will start its descending motion (403718
km). So we will start the descending motion analysis in next page, and because it's
a descending motion we will add the value 0.985 degrees to the angle (θ)

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
1st
The Descending Motion
(8) 15 – June -2020
Jun 15 404574 km Jun 16 403718 km (The Next Day)
On 15/6/2020 the moon Earth distance was 404574 km
So we need to know where the moon will be on the next day (16/6/2020)?
Let's try to do that….
- The radius =404574 km so C (orbital circumference) = 404574 x 2π=2.542 mkm
- The moon moves it during 29.53 days ( L0= 2.542 mkm/29.53) = 86082.4 km
- L0= 86082.4 km but L0 = 88000 km Cos θ0
- θ0= 11.983 degrees
- θ1= θ0+ 1.6 = 11.983 +1.6 = 13.583 degrees
- L1 = 88000 km (Cos 13.583 degrees) = 85538.7 km
- During 29.53 days = L1 x29.53 = 85538.7 km x 29.53 = 2.5259 mkm
- 2.5259 mkm = 2π x 402018.6 km
- i.e. The moon on next day (16/6/2020) should be on 402018.6 km
The registered value on 16/6/2020 (403718 km)
The difference = 403718 km – 402018.6 km = 1700 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
(9) 16 – June -2020
Jun 16 403718 km Jun 17 401733 km (The Next Day)
On 16/6/2020 the moon Earth distance was 403718 km
So we need to know where the moon will be on the next day (17/6/2020)?
Let's try to do that….
- The radius =403718 km so C (orbital circumference) = 403718 x 2π=2.536 mkm
- The moon moves it during 29.53 days ( L0= 2.536 mkm/29.53) = 85900 km
- L0= 85900 km but L0 = 88000 km Cos θ0
- θ0= 12.5413 degrees
- θ1= θ0+ 1.6 = 12.5413 +1.6 = 14.1413 degrees
- L1 = 88000 km (Cos 14.1413 degrees) = 85333.2 km
- During 29.53 days = L1 x29.53 = 85333.2 km x 29.53 = 2.519 mkm
- 2.519 mkm = 2π x 401053.1 km
- i.e. The moon on next day (17/6/2020) should be on 401053.1 km
The registered value on 17/6/2020 (401733 km)
The difference = 401733 km – 401053.1 km = 680 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
(10) 17 – June -2020
Jun 17 401733 km Jun 18 398840 km (The Next Day)
On 17/6/2020 the moon Earth distance was 401733 km
So we need to know where the moon will be on the next day (18/6/2020)?
Let's try to do that….
- The radius =401733 km so C (orbital circumference) = 401733 x 2π=2.524 mkm
- The moon moves it during 29.53 days ( L0= 2.524 mkm/29.53) = 85478 km
- L0= 85478 km but L0 = 88000 km Cos θ0
- θ0= 13.75 degrees
- θ1= θ0+ 1.5 = 13.75 +1.6 = 15.35 degrees
- L1 = 88000 km (Cos 15.35 degrees) = 84860.7 km
- During 29.53 days = L1 x29.53 = 84860.7 km x 29.53 = 2.506 mkm
- 2.506 mkm = 2π x 398832.4 km
- i.e. The moon on next day (18/6/2020) should be on 398832.4 km
The registered value on 18/6/2020 (398840 km)
The difference = 398840 km –398832.4 km = 8 km

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
(11) 18 – June -2020
Jun 18 398840 km Jun 19 395303 km (The Next Day)
On 18/6/2020 the moon Earth distance was 398840 km
So we need to know where the moon will be on the next day (19/6/2020)?
Let's try to do that….
- The radius =398840 km so C (orbital circumference) = 398840 x 2π=2.505 mkm
- The moon moves it during 29.53 days ( L0= 2.505 mkm/29.53) = 84862.36 km
- L0= 84862.36 km but L0 = 88000 km Cos θ0
- θ0= 15.346 degrees
- θ1= θ0+ 1.6 = 15.346 +1.6 = 16.946 degrees
- L1 = 88000 km (Cos 16.946 degrees) = 84179 km
- During 29.53 days = L1 x29.53 = 84179 km x 29.53 = 2.4858 mkm
- 2.4858 mkm = 2π x 395628.4 km
- i.e. The moon on next day (19/6/2020) should be on 395628.4 km
The registered value on 19/6/2020 (395303 km)
The difference = 395303 km – 395628.4 km = (-) 325.4 km

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
(12) 19 – June -2020
Jun 19 395303 km Jun 20 391409 km (The Next Day)
On 19/6/2020 the moon Earth distance was 395303 km
So we need to know where the moon will be on the next day (20/6/2020)?
Let's try to do that….
- The radius =395303 km so C (orbital circumference) = 395303 x 2π=2.483 mkm
- The moon moves it during 29.53 days ( L0= 2.483 mkm/29.53) = 84109.8 km
- L0= 84109.8 km but L0 = 88000 km Cos θ0
- θ0= 17.1 degrees
- θ1= θ0+ 1.6 = 17.1 +1.6 = 18.7 degrees
- L1 = 88000 km (Cos 18.7 degrees) = 83354.5 km
- During 29.53 days = L1 x29.53 = 83354.5 km x 29.53 = 2.461 mkm
- 2.461 mkm = 2π x 391753.3 km
- i.e. The moon on next day (20/6/2020) should be on 391753.3 km
The registered value on 20/6/2020 (391409 km)
The difference = 391409 km – 391753.3 km = 345 km

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
(13) 20 – June -2020
Jun 20 391409 km Jun 21 387432 km (The Next Day)
On 20/6/2020 the moon Earth distance was 391409 km
So we need to know where the moon will be on the next day (21/6/2020)?
Let's try to do that….
- The radius =391409 km so C (orbital circumference) = 391409 x 2π=2.459 mkm
- The moon moves it during 29.53 days ( L0= 2.459 mkm/29.53) = 83281.3 km
- L0= 83281.3 km but L0 = 88000 km Cos θ0
- θ0= 18.848 degrees
- θ1= θ0+ 1.6 = 18.848 +1.6 = 20.448 degrees
- L1 = 88000 km (Cos 20.448 degrees) = 82455 km
- During 29.53 days = L1 x29.53 = 82455 km x 29.53 = 2.4349 mkm
- 2.4349 mkm = 2π x 387526.1 km
- i.e. The moon on next day (21/6/2020) should be on 387526.1 km
The registered value on 21/6/2020 (387432 km)
The difference = 387432 km – 387526.1 km = (-) 94 km

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
(14) 21 – June -2020
Jun 21 387432 km Jun 22 383607 km (The Next Day)
On 21/6/2020 the moon Earth distance was 387432 km
So we need to know where the moon will be on the next day (22/6/2020)?
Let's try to do that….
- The radius =387432 km so C (orbital circumference) = 387432 x 2π=2.434 mkm
- The moon moves it during 29.53 days ( L0= 2.434 mkm/29.53) = 82435 km
- L0= 82435 km but L0 = 88000 km Cos θ0
- θ0= 20.485 degrees
- θ1= θ0+ 1.6 = 20.485 +1.6 = 22.085 degrees
- L1 = 88000 km (Cos 22.085 degrees) = 81543.1 km
- During 29.53 days = L1 x29.53 = 81543.1 km x 29.53 = 2.4079 mkm
- 2.4079 mkm = 2π x 383240.4 km
- i.e. The moon on next day (22/6/2020) should be on 383240.4 km
The registered value on 22/6/2020 (383607 km)
The difference = 383607 km – 383240.4 km =367 km

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
(15) 22 – June -2020
Jun 22 383607 km Jun 23 380110 km (The Next Day)
On 22/6/2020 the moon Earth distance was 383607 km
So we need to know where the moon will be on the next day (23/6/2020)?
Let's try to do that….
- The radius =383607 km so C (orbital circumference) = 383607 x 2π=2.41 mkm
- The moon moves it during 29.53 days ( L0= 2.41 mkm/29.53) = 81621.2 km
- L0= 81621.2 km but L0 = 88000 km Cos θ0
- θ0= 21.9495 degrees
- θ1= θ0+ 1.6 = 21.9495 +1.6 =23.5495 degrees
- L1 = 88000 km (Cos 23.5495 degrees) = 80671 km
- During 29.53 days = L1 x29.53 = 80671 km x 29.53 = 2.382 mkm
- 2.382 mkm = 2π x 379141 km
- i.e. The moon on next day (23/6/2020) should be on 379141 km
The registered value on 23/6/2020 (380110 km)
The difference = 380110 km – 379141 km =970 km

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
(16) 23 – June -2020
Jun 23 380110 km Jun 24 377044 km (The Next Day)
On 23/6/2020 the moon Earth distance was 380110 km
So we need to know where the moon will be on the next day (24/6/2020)?
Let's try to do that….
- The radius =380110 km so C (orbital circumference) = 380110 x 2π=2.388 mkm
- The moon moves it during 29.53 days ( L0= 2.388 mkm/29.53) = 80877 km
- L0= 80877 km but L0 = 88000 km Cos θ0
- θ0= 23.2112 degrees
- θ1= θ0+ 1.6 = 23.2112 +1.6 = 24.8112 degrees
- L1 = 88000 km (Cos 24.8112 degrees) = 79877.2 km
- During 29.53 days = L1 x29.53 = 79877.2 km x 29.53 = 2.358 mkm
- 2.358 mkm = 2π x 375410.5 km
- i.e. The moon on next day (24/6/2020) should be on 375410.5 km
The registered value on 24/6/2020 (377044 km)
The difference = 377044 km – 375410.5 km =1634 km

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
(17) 24 – June -2020
Jun 24 377044 km Jun 25 374451 km (The Next Day)
On 24/6/2020 the moon Earth distance was 377044 km
So we need to know where the moon will be on the next day (25/6/2020)?
Let's try to do that….
- The radius =377044 km so C (orbital circumference) = 377044 x 2π=2.369 mkm
- The moon moves it during 29.53 days ( L0= 2.369 mkm/29.53) = 80224.8 km
- L0= 80224.8 km but L0 = 88000 km Cos θ0
- θ0= 24.266 degrees
- θ1= θ0+ 1.6 = 24.266 +1.6 = 25.866 degrees
- L1 = 88000 km (Cos 25.866 degrees) = 79183.9 km
- During 29.53 days = L1 x29.53 = 79183.9 km x 29.53 = 2.338 mkm
- 2.338 mkm = 2π x 372152 km
- i.e. The moon on next day (25/6/2020) should be on 372152 km
The registered value on 25/6/2020 (374451 km)
The difference = 374451 km – 372152 km =2300 km

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
(18) 25 – June -2020
Jun 25 374451 km Jun 26 372338 km (The Next Day)
On 25/6/2020 the moon Earth distance was 374451 km
So we need to know where the moon will be on the next day (26/6/2020)?
Let's try to do that….
- The radius =374451 km so C (orbital circumference) = 374451 x 2π=2.352 mkm
- The moon moves it during 29.53 days ( L0= 2.352 mkm/29.53) = 79673 km
- L0= 79673 km but L0 = 88000 km Cos θ0
- θ0= 25.126 degrees
- θ1= θ0+ 1.6 = 25.126 +1.6 = 26.726 degrees
- L1 = 88000 km (Cos 26.726 degrees) = 78599 km
- During 29.53 days = L1 x29.53 =78599 km x 29.53 = 2.321 mkm
- 2.321 mkm = 2π x 369402 km
- i.e. The moon on next day (26/6/2020) should be on 369402 km
The registered value on 26/6/2020 (372338 km)
The difference = 372338 km – 369402 =2936 km

24. 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
24
(19) 26 – June -2020
Jun 26 372338 km Jun 27 370703 km (The Next Day)
On 26/6/2020 the moon Earth distance was 372338 km
So we need to know where the moon will be on the next day (27/6/2020)?
Let's try to do that….
- The radius =372338 km so C (orbital circumference) = 372338 x 2π=2.339 mkm
- The moon moves it during 29.53 days ( L0= 2.339 mkm/29.53) = 79223.4 km
- L0= 79223.4 km but L0 = 88000 km Cos θ0
- θ0= 25.8 degrees
- θ1= θ0+ 1.6 = 25.8 +1.6 = 27.4 degrees
- L1 = 88000 km (Cos 27.4 degrees) = 78128 km
- During 29.53 days = L1 x29.53 = 78128 km x 29.53 = 2.307 mkm
- 2.307 mkm = 2π x 367188.3 km
- i.e. The moon on next day (27/6/2020) should be on 367188.3 km
The registered value on 27/6/2020 (370703 km)
The difference = 370703 km –367188.3 =3515 km
Note Please
- This day is the last day in the moon descending motion, after it the moon stay in
perigee 3 days!! (28-6-2020 369571 km), (29-6-2020 369005 km), (30-6-2020
369106 km) –
- That tells, when the moon changes its motion from ascending to descending and
vice versa, the moon consumes long time in this process which causes great error
for the calculations…
- i.e. there's specific reason at perigee and apogee point causes the moon to stay
longer time there than expected…
- based on that, we have to analyze the equation test results for 20 days only

25. 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
25
2-2 The Test Results Discussion And Analysis
Let's take a look on the different distances between the equation results and the
registered data in following
Day Registered Data Equation Results Difference
6-6-2020 369418 km --------------- --
7-6-2020 373729 km 374466.6 (-738 km)
8-6-2020 378917 km 378531.3 386 km
9-6-2020 384534 km 383397.4 1137 km
10-6-2020 390096 km 388644 1452 km
11-6-2020 395156 km 393780.5 1375 km
12-6-2020 399345 km 398411 934 km
13-6-2020 402395 km 402193.6 202 km
14-6-2020 404153 km 404906.3 754 km
15-6-2020 404574 km ----- -------
16-6-2020 403718 km 402018.6 1700 km
17-6-2020 401733 km 401053 680 km
18-6-2020 398840 km 398832.4 8 km
19-6-2020 395303 km 395628.4 (-326 km)
20-6-2020 391409 km 391753.3 345 km
21-6-2020 387432 km 387526.1 94 km
22-6-2020 383607 km 383240.4 367 km
23-6-2020 380110 km 379141 970 km
24-6-2020 377044 km 375410.5 1634 km
25-6-2020 374451 km 372152 2300 km
26-6-2020 372338 km 369402 2936 km
27-6-2020 370703 km 367188.3 3515 km
[

26. 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
26
The Test Results Analysis:
- The Total Results Are 20 Values
(1st
Category)
o 13 values, defines the moon position in range 1150 km (Error 2.7%)
(2nd
Category)
o 4 values, defines the moon position in range 1150-2000 km (Error 4.6 %)
(3rd
Category)
o 3 values, defines the moon position in range 2000-3500 km (Error 8 %)
- The Results Explanation
- The distance from perigee to apogee =43000 km…
o 1st
Category of results defines the moon position in error range (1150 km) =
error (2.7%), that means, (13 values of 20) defines the moon position with
error (2.7%) only (Small Error Range)
o 2nd
Category of results defines the moon position in error range from (1150
km to 2000 km) = error (4.5%), that means (4 values of 20) defines the
moon position with error (4.5%) (Average Error Range)
o 3rd
Category of results defines the moon position in error range from (2000
km to 3500 km) = error (8%), that means (3 values of 20) defines the moon
position with error (8%) (Great Error Range)
- The Equation Accuracy
o The previous explanation shows that, the equation has a good range of
accuracy and its error is in the acceptable error range
The Conclusion
The Equation Is correct and trustee
And
It's a useful tool to define the moon position daily

27. 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
27
References
The Moon Motion Trajectory Analysis (II)
https://www.academia.edu/44368860/The_Moon_Motion_Trajectory_Analysis_II_
The Moon Motion Basic Equation (III) (Test No.2)
https://www.academia.edu/s/e493aec33a
or
https://www.slideshare.net/Gergesfrancis/the-moon-motion-basic-equation-iii-test-no2
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_
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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