The moon uses pythagoras triangle technique in its orbital motion (part no.3)

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
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Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
The Moon Uses Pythagoras Triangle Technique In Its Orbital Motion (Part No. 3)
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 – 18th
December 2020
Abstract
Paper Argument:
- There are 3 new discovered tools can be used in the moon orbital motion study and
analysis, where these 3 tools can help to figure the moon motion trajectory and
define the moon position in its orbit, these 3 tools are
1. The concept (the moon uses Pythagoras triangle in its orbital motion)
2. The equation (θ1= θ0 + 1.7 degrees) which can define the moon position in
its orbit, where this equation depends on the previous concept.
3. The Moon Orbital Triangle which is created as a result of the moon using
of Pythagoras triangle technique as one of the moon motion techniques
Paper objective:
- The paper summarizes the advantages of these 3 tools and how they can help to
define the moon position in its orbit as near s possible.
- Also
- The paper provides the basic difficulties to figure the moon motion trajectory and
the solutions which are provided by the 3 new discovered tools.

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Contents
The Subject Page No.
1- The Moon Uses Pythagoras Triangle As One Of The Moon
Motion Technique (The Concept)
3
2- The Equation Discussion (θ1= θ0 + 1.7 degrees) 10
3- The Moon Motion Trajectory Difficulties 18
4- The Moon Orbital Triangle Discussion 19
5- Appendix No. 1 29

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1- The Moon Uses Pythagoras Triangle As One Of The Moon Motion Technique
(The Concept)
1-1 Preface
1-2 Why Does The Moon Use Pythagoras Triangle In Its Motion?
1-3 How Does The Moon Use Pythagoras Triangle In Its Motion?
1-4 The Moon Orbital Motion
1-1 Preface
- The moon motion 4 basic points were the method by which I have discovered that
the moon uses Pythagoras triangle in its orbital motion.. these 4 points are
o Perigee radius (r=0.363 mkm), which is the most near point the moon can
reach to Earth.
o Apogee radius (r=0.406 mkm), which is the most far point the moon can
reach from Earth.
o T. S. Eclipse (r= 0.373 mkm), the moon creates total solar eclipse when the
moon be at this distance from Earth or Shorter.
o Orbital distance (r=0.384 mkm), this is the registered distance in the moon
data sheet as the moon orbital distance
- These 4 points are defined based on each other by Pythagoras rule:
o (363000 km)2
+ (86000 km)2
= (373000 km)2
o (373000 km)2
+ (86000 km)2
= (384000 km)2
o (384000 km)2
+ (86000 km)2
= (393000 km)2
o (393000 km)2
+ (86000 km)2
= (406000 km)2
(Error 1%)
- Based on this data, the concept is created, that the moon uses Pythagoras triangle
as one of the moon motion techniques
- But
- Why does the moon use Pythagoras triangle as one of its motion techniques?

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1-2 Why Does The Moon Use Pythagoras Triangle In Its Motion?
- Let's try to summarize the answer in following:
o The moon uses Pythagoras triangle basically to decrease its displacement
through its orbit
o The moon daily displacement = 88000 km and the moon has to move this
distance every day without any decreasing (later we will know why!)
o But
o If the moon moves by this displacement as its orbital displacement the moon
will revolve around Earth through its apogee orbit only (r=0.406 mkm)
o For that reason
o The moon creates an angle between its motion direction and its orbit
horizontal level to create a displacement through its orbit less than (88000
km)
o As a result of this technique, the moon can revolve around Earth through
more near orbits than apogee orbit (r=0.406 mkm)
o Let's explain this intelligent technique with some data and details to show
the useful result of using Pythagoras triangle by the moon motion….

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1-3 How Does The Moon Use Pythagoras Triangle In Its Motion?
- 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 orbit horizontal
level, by that the real displacement through the moon orbit will be (L= 88000 km
cos (θ)), and by that, spite the moon moves 88000 km but the orbital horizontal
displacement 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, the real displacement (L) = 77237 km, So, if the moon real
displacement daily be (77237 km) during 29.53 days, the moon will pass a
distance = 2.28 million km and this will be the moon orbital circumference, where
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 where 2.598 mkm = 2π x (0.413 mkm)
- The Moon Orbital Apogee Radius =0.406 mkm
- So the moon will move along month revolving around 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.

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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
(r=0.406 mkm)
o That means, the angle (θ) changes the real displacement (L) and also
changes the distance between the moon to perigee or to apogee, shortly, the
angle (θ) defines the moon position (as a ship) between 2 river banks….
- The angle (θ) defines the moon orbital motion basic features and we have to
discuss is deeply with the moon orbital motion equation (θ1= θ0 + 1.7 degrees)
- Any way here we need to refer to one important notice in following
Notice
o We know that (363000)2
+ (86000)2
= (373000)2
o In Pythagoras triangle with dimensions (363000 km, 373000km, 86000 km),
what's the angle (θ)? The angle (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, we will need
this notice later in our discussion

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1-4 The Moon Orbital Motion
- The moon moves per a solar day a motion typical to the Earth motion to avoid the
separation from Earth through their motions, based on this rule, the moon moves
per a solar day 2.58 million km with an angle declines on the horizontal level
0.98562 degrees as typical as Earth motion
- If there's no Lorentz Length Contraction Phenomenon effect on the moon motion,
the moon motion trajectory would to be a parallel line to Earth Motion Trajectory,
but because Lorentz Length Contraction effects on the moon motion daily distance
(2.58 mkm) with a rate 1.0725 and causes this distance to be contracted (2.41
mkm)
- The moon difficulties are started here, because the difference between both
distances (0.17 mkm) will cause the moon to be separated from Earth motion
inevitably
- We should notice that, these motions are done far from our observation, means, we
see nothing of this motion distance, because the moon moves on the Earth orbital
circumference revolving around the sun, but, even if we can't observe this motion
distance the motion is still fact and proved by its power because the Earth moves
per a solar day 2.58 mkm and if the moon doesn't move this same distance every
solar day that necessities the moon to be separated from the earth through their
motions course – based on that- the facts prove this motion regardless our
observation ability for it.
- Now the moon has n additional distance to be passed (0.17 mkm) and the moon
has to pass this distance on the same solar day to avoid the separation from the
Earth during their motions
- Because of that, the moon moves its daily displacement (88000 km) depends on
Earth gravity force, but the different distance (0.17 mkm) to be covered still needs
the moon to move one more displacement (= 88000 km)

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- The previous explanation tells that, the moon has to move 2 displacements each =
88000 km, while we see one displacement only because it's done through the moon
orbital around Earth but the other displacement should be done also because this
total distance (0.17 mkm) is required to cover the different distance and create the
total (2.58 mkm) which saves the moon and Earth motion accompanying.
- Now we have 2 basic information about the moon orbital motion
o (1st
information) the moon uses Pythagoras triangle in its orbital motion
o (2nd
information) the moon has to move 2 displacements each =88000 km
and their total distance =0.17 mkm which is a required distance necessary to
cover the difference between the moon and Earth motions distances.
- This explanation helps us to understand why the moon uses Pythagoras triangle in
its motion, because the moon can't decrease the actual motion distance (88000 km)
because the moon needs this distance to cover the different distance between its
contracted motion distance (2.41 mkm) and Earth motion distance (2.58 mkm), so
the moon needs to move this distance perfectly, but if it's used as a full
displacement the moon would be prisoner in the apogee orbit (r=0.406 mkm)
always as we have discussed before, because of that, the moon creates Pythagoras
triangle technique by which the moon moves actually 88000 km but the real
displacement through the moon orbit became less (L = 88000 Cos θ) and by that
the moon can achieve 2 objectives, first to pass the required distance and second to
move in near orbits to Earth, that shows the intelligent moon motion techniques…
- (Notice, Lorentz Length Contraction Effect Discussion is in Appendix No. 1 of
this current Paper)

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The Moon Orbital Motion Needs One More Orbit
- The previous explanation tells that, the moon moves 2 displacements each =88000
km, we see one of these 2 displacements but where's the other displacement?!
- We know that, the moon original motion (2.58 mkm) which is contracted to be
(2.41 mkm) isn't seen by us because the moon moves this distance revolving with
Earth around the sun along the Earth Orbital Circumference
- We may accept also that, the 2nd
displacement the moon does on this same
trajectory and isn't seen by us
- So,
- There must be one more orbit for the moon to move through this 2nd
displacement
regards our observation ability
- That means, there's 2nd
orbit for the moon motion
- But
- How can we discover this second orbit if we can't observe it?
- Because of the moon orbital geometrical structure, we can discover this second
orbit by analyze the moon orbit geometrical data, based on that with the analysis of
the moon orbital triangle geometrical structure we should discuss the second orbit
of the moon motion.

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2- The Equation Discussion (θ1= θ0 + 1.7 degrees)
2-1 The Equation Concept
2-2 The Equation Test and Accuracy
2-1 The Equation Concept
The moon orbital motion equation
(θ1= θ0 + 1.7 degrees)
- The moon orbital motion equation is created depending on the concept which we
have discussed in point No. (1)
- The moon uses Pythagoras triangle in its orbital motion, and by this intelligent
technique the moon be under control by the angle (θ) change
- As we have discussed, the angle (θ) defines almost all the moon motion features,
this massive effect of the angle (θ) is found as a result of the moon using for
Pythagoras triangle concept…
- The moon uses Pythagoras triangle because the moon daily displacement (88000
km) is so long and if the moon moves it as a displacement through its orbit the
total distance during 29.53 days will cause the moon to revolve around Earth
through its apogee orbit (radius =0.406 mkm)
- For that reason the moon uses Pythagoras triangle, and by this intelligent technique
the moon moves 88000 km but the moon real displacement (L) through its orbit
will be less than it because (L =88000 Cos θ), by that the moon moves 88000 km
and can revolve around Earth in more near orbit than apogee orbit (radius =0.406
mkm)

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- This technique caused the moon motion be under control of the angle (θ), because,
when the angle (θ) be decreased the real displacement (L) will be longer and that
necessitates the moon to move through orbits far from perigee orbit (r=0.363
mkm) to find more wide orbits to move through (more far from Earth)
- When the angle (θ) be greater the real displacement (L =88000 km Cos (θ)) will be
shorter and the moon can be more near to perigee orbit (more near to Earth)
- By that, the angle (θ) almost controls the moon motion (all) features
- For that reason the moon orbital motion equation uses the angle (θ) with a constant
- Let's see The Moon Orbital Motion Equation
(θ1= θ0 + 1.7 degrees)
- θ1= The Pythagoras triangle angle for today
- θ0= The Pythagoras triangle angle for yesterday
- 1.7 degrees = Constant
- How does this equation work?

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How to use this equation?
- Perigee Radius =0.363 mkm, so Its Orbital Circumference =2.28 mkm
- Suppose the moon will revolve around Earth through perigee orbit only during
29.53 days, so
- (2.28 mkm /29.53 days) = 77237 km
- This is (the real displacement = L = 88000 km Cos θ = 77237 km),
- What's the angle θ value? the angle θ = 28.63 degrees
- Suppose the moon stand yesterday on this point with the angle (θ) =28.63 degrees,
where the moon will move today?
- From Perigee (the most near point to Earth) the moon will move in Ascending
motion because it moves from perigee (0.363 mkm) to apogee (0.406 mkm)
- In Ascending motion we use (-1.7 degrees) because the angle (θ) is decreased
where the real displacement (L) is increased – so let's do that in following
o (θ1= θ0 + 1.7 degrees)
o (θ1= 28.63 degrees - 1.7 degrees) = 26.93 degrees
o L = 88000 Cos (26.93 degrees) = 78454 km
o During 29.53 days so (78454 km x 29.53 days = 2.316 mkm)
o 2.316 mkm = 2π x 368722 km
That means
o The moon be today on Perigee radius (r=0.363 mkm) and will start its
motion displacement 88000 km through its orbit. For day motion the
equation uses 1.7 degrees, that means, the moon on perigee uses
Pythagoras triangle with angle (28.63 degrees) and during one solar day the
moon uses - 1.7 degrees and by that the angle will be (28.63 degrees -1.7
degrees = 26.93 degrees)…. The angle 1.7 degrees expresses The Moon
Daily Motion
o By using Pythagoras triangle its angle (θ) = 26.93 degrees, the displacement
(88000 km) will create a real displacement through the moon orbit = 78454

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km and the moon will finish its motion today at a distance 368722 km
means the moon is far from perigee radius with (368722 km-363000 km
=5722 km )
o So, the moon tomorrow will be at the point 368722 km and will have the
Pythagoras triangle its angle 26.93 degrees.
The Descending Motion
o When the moon moves from apogee to perigee, so the angle (1.7 degrees)
will be positive (+1.7 degrees) because the angle (θ) is increased because the
real displacement (L = 88000 Cos (θ)) be shorter.
o So if the moon in apogee radius (r=0.406 mkm), what's the Pythagoras angle
(θ)?
o The apogee orbital circumference = 0.406 mkm x2π =2.55 mkm = 29.53
days x 86400 km, the angle (θ) = 11 degrees
o The moon moves from apogee to perigee (descending motion)
o (θ1= θ0 + 1.7 degrees) = (θ1= 11 degrees + 1.7 degrees) = 12.7 degrees
o L = 88000 Cos (12.7 degrees) = 85847 km
o During 29.53 days so (85847 km x 29.53 days = 2.535 mkm)
o 2.535 mkm = 2π x 403467 km
o So after one day the moon will be on 403467 km far from apogee (406000
km) with 2540 km
Now let's see this equation test and efficiency in following

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2-2 The Equation Test and Accuracy
- I have tested the Equation with real data for 2 months June 2020 and October 2020
- The results are very good and I provide the results here for better vision
concerning the equation sufficiency
1st
Test June 2020
Day Registered Data The Results (1.7) Difference
6-6-2020 369418 km
7-6-2020 373729 km 374772.5 - 1044
8-6-2020 378917 km 378821.5 96
9-6-2020 384534 km 383667.7 867
10-6-2020 390096 km 388890 1206
11-6-2020 395156 km 394000 1156
12-6-2020 399345 km 398604.2 741
13-6-2020 402395 km 402361.3 34
14-6-2020 404153 km 405052.8 -900
15-6-2020 404574 km ---- ---
16-6-2020 403718 km 401848.5 1870
17-6-2020 401733 km 400876.1 857
18-6-2020 398840 km 398640.7 200
19-6-2020 395303 km 395417.4 115
20-6-2020 391409 km 391521.2 -113
21-6-2020 387432 km 387273.4 159
22-6-2020 383607 km 382968.4 639
23-6-2020 380110 km 378852 1258
24-6-2020 377044 km 375107 1937
25-6-2020 374451 km 371836.5 2615
26-6-2020 372338 km 369077 3262
27-6-2020 370703 km 366855.6 3847
[

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The 1st
Test Results Analysis:
- The Total Results Are 20 Values
(1st
Category)
o 15 values, defines the moon position in range 1300 km (Error 3%)
(2nd
Category)
o 2 values, defines the moon position in range 1300-2000 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 (1300 km) =
error (3%), that means, (15 values of 20) defines the moon position with
error (3%) only (Small Error Range)
o 2nd
Category of results defines the moon position in error range from (1300
km to 2000 km) = error (4.5%), that means (2 values of 20) defines the
moon position with error (4.5%) (Average Error Range)
o 3rd
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

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2nd
Test October 2020
Day Registered Data Results (1.7) Difference
5-10-2020 405,690 km --- ---
6-10-2020 404,171 km 403125.3 km 1046 km
7-10-2020 401,649 km 401390 km 259 km
8-10-2020 398,073 km 398545.6 Km - 473 km
9-10-2020 393,464 km 394568.8 km -1105 km
10-10-2020 387,944 km 389510 km -1567 km
11-10-2020 381,763 km 383520 km -1758 km
12-10-2020 375,302 km 376875.3km -1574 km
13-10-2020 369,063 km 369981km -919 km
14-10-2020 363,617 km 363363.4km 254 km
15-10-2020 359,530 km 357612 km 1918 km
16-10-2020 357,269 km 353307 km 3962 km
17-10-2020 357,105 km ---- --
18-10-2020 359,048 km --- --
19-10-2020 362,851 km 364979.7 km - 2129 km
20-10-2020 368,058 km 368579.3 km -522 km
21-10-2020 374,101 km 373492.4 km 609 km
22-10-2020 380,412 km 379168.3 Km 1244 Km
23-10-2020 386,497 km 385059.3Km 1438 km
24-10-2020 391,989 km 390694.3 km 1295 km
25-10-2020 396,659 km 395729.5 km 930 km
26-10-2020 400,395 km 399958.7 km 437 km
27-10-2020 403,181 km 403299 km 112 km
28-10-2020 405,059 km 405738.5 km -680 km
29-10-2020 406,104 km 407359.4 km -1256 km
[

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The Test Results Analysis:
- The Total Results Are 22 Values
(1st
Category)
o 15 values, defines the moon position in range 1300 km (Error 3%)
(2nd
Category)
o 5 values, defines the moon position in range 1300-2000 km (Error 4.6 %)
(3rd
Category)
o 2 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 (1300 km) =
error (3%), that means, (15 values of 22) defines the moon position with
error (3%) only (Small Error Range)
o 2nd
km to 2000 km) = error (4.5%), that means (5 values of 22) defines the
moon position with error (4.5%) (Average Error Range)
o 3rd
km to 3500 km) = error (8%), that means (2 values of 22) 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

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3- The Moon Motion Trajectory Difficulties
- From the previous analysis we can conclude the basic difficulties in the moon
motion trajectory, basically there are 2 difficulties let's refer to them in following
o (1st
Difficulty) the moon unstable motion, and this a difficulty can be
solved and the moon orbital motion equation tries to solve it as possible.
The moon doesn't move equal distances (from perigee to apogee) every day,
on the contrary, some days the moon pass (2000 km) from perigee to apogee
through its day motion and in some other days the moon can jump 6000 km
in one day only
o The moon orbital motion equation tries to solve this difficulty by using the
constant 1.7 degrees, and this constant enables the equation to pass long
distance daily, for example (4000 km per solar day from perigee to apogee)
and by that if the moon moves 3000 km only the difference will be (-1000
km) and that's not so much difference, and if the moon move 5000 km the
difference will be the same (1000 km)
o (2nd
Difficulty) the moon stayed in perigee and apogee points for long
time! That means, when the moon reaches to perigee radius (r=0.363 mkm)
or apogee radius (r=0.406 mkm) the moon stayed there and doesn't move by
the Pythagoras triangle for 2 or 3 days, the moon moves on the same
horizontal level for example (363000 km) the moon can stay on this point
for 2 sequent days and move each day 88000 km but doesn't leave this point
(363000 km) as the equation tells us
o This difficulty is occurred with perigee and apogee points only any because
of that there are 2 or 3 days aren't accounted in the previous (2) tests because
the moon stayed in these points all these days without change its orbital
radius

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4- The Moon Orbital Triangle Discussion
4-1 The Moon orbital triangle Data (Revision)
Figure No. (1) (my figure)
Let's Review The Moon Orbital Triangle Data
(1st
Point)
- The figure I brought from internet to use in the Explanation -
- We have supposed that the inner circle is Perigee orbit and
the outer circle is apogee orbit – and we have calculated the
tangent DB = 181843 km
- AB = 363686 km (= perigee radius approximately)
- Perigee radius r =0.363 mkm Apogee radius r =0.406 mkm
- Based on that, the triangle (ODB) is a specific Pythagoras
triangle (1, 2 and 51/2
)
- The triangle (ODB) angles are 26.564 deg. and 63.435 deg.

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(2nd
Point) The Moon Orbital Triangle Data Correction
- EB = Perigee radius = 363000 km
- ED = Apogee radius = 406000 km
- EA= (Jupiter Circumference) =449197 km
- AC = (Saturn diameter) =121620 km (error 1%)
- ES = total solar eclipse radius = 373900 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)
- CX= =87521 km
- CS = = 86690 km
- CZ= (the moon daily displacement) =88000 km
- CF= 88526.8 km CD =96150.9 km CY= 97766 km
- BA = BC = 86000 km
- BS= (the moon Circumference) =10921 km
- BZ = 18586 km BF =21000 km
- BD = DA = 43000 km
- BY = = 46475 km
- SZ = 7665 km ZF= 2414 km
- DY = 3475 km BX= 16203 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 = 118.3 deg)
- BCY = 28.39 degrees ECZ= 88.9 degrees
- XCE = 66 degrees
- CZS = 77.8 degrees
- CZF =102.195 degrees
- XCB = 10.67deg
- (Uranus Axial Tilt = 97.8 degrees = FSC 97.2 degrees + 0.6 degrees) (i.e. the
angle under FSC)
- Angle under (E) = 13.33 degrees 1.1 degrees = 14.43 degrees
- Ecliptic Line creates 0.5 degrees with the moon orbital triangle base (EA)

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4-2 The Moon Orbital Triangle Details Discussion
- How to draw The Moon Orbital Triangle….?
- The first horizontal black thick line which is under all triangle details and has zero
angle with the horizontal level this black line is the Moon Axial Tilt (6.7 degrees)
- The triangle base (red) thick line declines on the horizontal level (the black line)
with an angle =1.1 degrees
- Point E represents the Earth
- Point B represents Perigee radius (r=0.363 mkm)
- Point D represents Apogee radius (r=0.406 mkm)
- Point A represents a point in space far from Apogee radius with 43000 km at the
same horizontal level, means no angle between these points (E,B,D,A)
- The Ecliptic Line which is seen in the triangle has an angle = 0.5 degrees between
it and the moon orbital triangle base (The red line), why?!
- Because 1.6 degrees is found between the Earth Ecliptic & The Moon Axial Tilt
- The moon orbital motion is ranged between the point (B) (Perigee radius r=0.363
mkm) and the point D (Apogee radius =0.406 mkm).
- We will discuss the triangle details in full analysis one after one – but – at first
- Our basic discussion triangle is the triangle BCD because it contains the moon
orbital motion from perigee (Point B) to apogee (Point D)
- Please Note, the triangle (BCD) is a similar to the general triangle we have
discussed separately in page no.6 (the triangle DOB) where the dimensions are
rated (406000km , 363000 mkm and 181843 km) and (96151 km, 86000km and
43000km), because of that the angles are equal, which makes both triangles are
similar, both are typical to Pythagoras triangle (1,2, (5)1/2
)
- Our consideration now should be directed to the line BC =86000 km, this is the
value which we have found in the moon motion 4 points definition and we have
asked why all points use this dimension (86000 km) which is not found in the
moon orbital motion data, let's consider it in following

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The Dimension 86000 km
- The moon orbital triangle is a vertical triangle, the line BC is perpendicular on
the base EA (=449197 km)
- By that
- While the moon motion is done from perigee (B) to apogee (D) on (x-y plain) the
line BC is found on (z-axis) perpendicular on the base EA.
- Based on that,
- The line CE =373000 km = The Total Solar Eclipse Radius …… BUT
- The line CE Is NOT the Total Solar Eclipse Radius Because
- The line CE is found vertical level (z=axis) while the moon moves on (x-y plain)
- Shortly
- The moon orbital triangle is a Pythagoras triangle found on the vertical level
(z=axis) and this triangle defines the moon orbital motion points using Pythagoras
rule….
- The dimension 86000 km is found on the vertical level (z-axis)…
- What does that tell us?
- The distance EC =373000 km has an angle =13.33 degrees with the horizontal
base (EA) because the point (C) is on the vertical axis (BC) (z-axis) but when this
angle 13.33 deg be not found, the distance EC =373000 km on the horizontal level
will = the total solar eclipse radius..
Means
- The moon orbital triangle (Pythagoras triangle) defines the moon orbital motion
points vertically but the moon uses the (vertical) definition by its horizontal motion
an by that, the points definition which is done by the vertical triangle is used by the
moon horizontal motion

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The Point (A)
- The moon orbital triangle geometrical structure depends on 3 points (E, C and A),
E is Earth (by its gravity the moon revolves around it) and C is a vertical point
found by geometrical necessity because the Pythagoras triangle is vertical on the
base (EA), I want to say, if the Pythagoras triangle technique using by the moon
for its orbital motion necessities this triangle to be vertical on the bases (EA) that
necessitates to create the point (C) on a vertical axis (z-axis) as it's now, means,
this point (C) is found by geometrical necessities. But
- What's the point (A)? how this point can be created and can effect on the moon
orbital triangle?! Because this point is far from apogee radius with 43000 km and
the moon can't move beyond the apogee radius, means, this point (A) is found in
space and should have no effect on the moon orbital motion! so to find this point
(A) in the moon orbital triangle geometrical structure that creates a question needs
to be solved!
- But geometrically the point (A) is one pillar of the moon orbital triangle pillars,
means, the geometrical structure forces us to accept the massive importance of the
point (A) where no clear reason we have to explain why this point has such
massive importance?!
The Ecliptic Line
- The ecliptic line is seen in the figure creates an angle = 0.5 degrees with the
triangle base (red line), because the moon axial tilt declines on the Earth ecliptic
with (1.6 degrees).
- Please Note, this angle (0.5 degrees) is its right triangle hypotenuse =396800 km,
so its dimension will be =3475 km (the moon diameter), but if its right triangle
hypotenuse =1.392 mkm (lunar umbra length), so its dimension will be =12104 km
(Venus diameter).

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2nd
Force Effects On The Moon Orbital Motion
- we have discussed the Point (A) before
- This point (A) is one pillar in the moon orbital triangle but there's no explanation
how this point can be created… it's found basically for geometrical necessities
- I suggest that, there's another force effects on the moon orbital motion and this
point (A) is a proof for this second force existence
- There's one more reason to suppose this (2nd
) force …. Because, the moon motion
needs 2 displacements (2 x 88000 km) to cover the distance (0.17 mkm), but the
moon moves one displacement seen by us through its orbit, where's the other
displacement be passed?
- That creates 2 reasons support the same conclusion that, there's one more force
effect on the moon orbital motion…
- This 2nd
force I claim, is found by interaction between the sun and Jupiter masses
gravities effect on the moon motion, because the sun gravity on the moon is
greater than Earth gravity effect on the moon, and Jupiter effect here creates an
interaction with the sun gravity to create specific effect on the moon motion.. this
interactive effect is effected also by Earth gravity… shortly.. that creates 2 points
of gravities total, Earth and its moon (these are the 2 points) and these 2 points are
effected by (the sun & Jupiter) gravities interactive effect, by that, another force is
created and be effective on the point (A) found on 43000 km from apogee radius
(0.406 mkm).
Please remember
- 149.6 mkm (Earth orbital distance) = 1047 x 142984 km (Jupiter diameter),
this equation shows that (the sun /Jupiter) masses rate effects on the Earth orbital
distance definition and that tells Jupiter effects on Earth orbital distance, and by
this same effect Jupiter effect on the moon orbital motion and that causes the
distance (EA) to be = 449197 km = Jupiter Circumference

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2nd
Orbit Is Found For The Moon Motion
- The moon needs to move one more displacement =88000 km, to cover the
distance 0.17 mkm.
- The moon moves its first displacement (88000 km) through its orbit around Earth,
but where the moon moves the second displacement?
- There must be 2nd
orbit for the moon motion, but where this orbit?
- We know that, the lunar eclipse umbra length = 1.392 mkm = the sun diameter
- And
- The distance EA = 449197 km = Jupiter diameter
- The distance from the point (A) to the end of the lunar umbra length = 942803 km
- But
- The Triangle EAC Perimeter = 942803 km
- That tells us, the point (A) separates between 2 equal values
- The first value is the triangle EAC perimeter and the second value is the distance
from the point (A) to the end of lunar umbra length…
- That tells us, the moon 2nd
orbit is a neighbor one to the first orbit, simply the point
(A) separates between the moon 2 orbits…
- But how to understand that?
- Can the moon be out of apogee radius (0.406 mkm)?! Because Earth gravity
prevents the moon to move out apogee radius… but
- The 2nd
orbit position is defined clearly by the previous data analysis, and we have
to accept that this definition is a correct one and there's one more orbit found
beyond the point (A), even if the moon can't move through this orbit
- So we have some dilemma her because the data tells a second orbit must be found
for the moon motion but there's no way to use this second orbit by the moon
because the moon is a prisoner behind the apogee radius (r=0.406 mkm)

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- How to solve this dilemma?
- The power which is provided by the moon orbital triangle is the geometrical rules,
the triangle shows that the moon orbital motion is done based on geometrical rules
perfectly an by that, the geometrical structure shows many data about this motion
- So, what advantage can be provided by the moon orbital geometrical structure
- Let's imagine a simple description in following
o Imagine a car connected by a chain of steel with another car both are the
same type and manufacturer and production date, simply 2 typical cars
o One moves by its motor and the other doesn't
o Both cars are on 2 different traces as 2 prisoner cars, no one can be out of its
track
o The only available motion is to forward… now the working car moves by its
motor and can pass the track but the other can't move …. But because they 2
cars are connected with chain of steel the working car pulls the other and
both move equal distances through the 2 tracks… that perform 2 distances
by one car motion
o What we need to perform this experiment?
o We need a suitable geometrical structure only…
o The moon doesn't move beyond apogee radius (r=0.406 mkm) but the other
displacement (88000 km) is passed through the other orbit (behind the point
A) how? Because the moon orbital geometrical structure provides this
chance for the moon orbital motion and this is the basic positive result of the
complex geometrical structure of the moon orbital triangle
o Shortly
o The second force which effects on the moon orbital motion, effects to create
a geometrical structure interactive with the moon orbital motion and create a
parallel displacement in the (2nd
orbit) as a result for the moon displacement
in its orbit around Earth and by that the moon moves both displacements and
creates the total distance (0.17 mkm) to cover the difference.

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Why the Equation uses the angle (1.7 degrees)?
I-Data
45 degrees = 26.6 degrees x 1.7
II-Discussion
Equation No. (1)
45 degrees = 26.6 degrees x 1.7 degrees
- Why does the Equation use 1.7 degrees (θ1= θ0 + 1.7 degrees)? Equation No. 1 is
the reason and the basic equation in this data, let's analyze it in following:
o 45 degrees is the angle (A) in the moon orbital triangle, this angle is a very
basic one, because it's used for 2 triangles, the greatest triangle (ECA) and
the triangle (BCA), the angle (A) is an angle in both triangles..
o If the triangle (BCA) is created as a part of the greatest one (ECA) that
means, the angle (A) is a factor in this creation, means, the greatest triangle
(ECA) is found before and depending on the angle (A) the triangle (BCA) is
created…. we interest for the triangle (BCA) because it contains the moon
orbital motion from perigee (Point B) to apogee (Point D).
o 26.6 degrees is the angle (BCA) which defines the moon motion from
perigee (point B) to apogee (point D), that means, the angle (26.6 degrees)
defines the moon orbital motion because it defines the distance from perigee
to apogee, in more clear words, this angle define The Moon Orbital Period,
and based on this angle the moon creates its cycle.
o 1.7 deg is the moon orbital motion equation value (θ1= θ0 + 1.7 deg.), this
equation tells that, the angle of today (θ1) = the angle of yesterday (θ0) + 1.7 deg!
o So, What's 1.7 deg? It's The Daily Motion Angle, means, this value (1.7 deg)
expresses the moon daily motion…
o What does Equation No.(1) tell us?
o Equation no. (1) tells that, the moon orbital period is defined based on the moon
daily motion and the angle (A) =45 degrees, means, the point (A) causes to
create the moon orbital period (27.3 days)!

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Let's remember the question
o Why does the equation (θ1= θ0 + 1.7 deg.) use this value 1.7 degrees? This
question need to be reformed…. As following… if the moon orbital period
=27.3 days and its distance from perigee to apogee depends on the angle
(BCA) =26.6 degrees, So, the moon daily motion should depend on the
angle 1.7 degrees
o The orbital period (27.3 days) and the angle 26.6 degrees force the moon to
use 1.7 degrees in its orbital motion equation!
o Let's use this conclusion as a hypothesis and see if really the moon be forced
to use this value 1.7 degrees! But, in all cases, if the moon be forced to use
it, why this specific value 1.7 degrees?! Because (the moon orbital
inclination 5.1 degrees – Venus orbital inclination 3.4 degrees =1.7 degrees)
o We consider that, Earth should effect on the moon and because of that the
value 1.6 degrees is more consistent to be used because the moon axial tilt
declines on the Earth ecliptic with 1.6 degrees but in fact Venus effect in
this point is more deep than Earth on the moon motion, and because of that
the moon uses the difference between both orbital inclinations…. Also it's
noticeable that, the moon orbital inclination =(2/3) Venus orbital inclination

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Appendix No.1
Is There Lorentz Length Contraction Effect In The Solar System?
i.e.
(Are There Relativistic Effects In The Solar System?)
Lorentz Length Contraction Effect is the near possible answer to explain the planets
data, in following I provide one example of such planets data to prove that, this
conclusion is the most near one to explain it.
I- Data (A)
Why These Distances Are Equal?
(1)
Saturn Orbital Distance = Saturn Uranus Distance
= Mars Orbital Circumference
= Pluto Neptune Distance
= Pluto eccentricity Distance
= Neptune Orbital Distance/π
= Uranus Orbital Distance /2
= Mercury Jupiter Distance x 2
(2)
Mercury Neptune Distance = Saturn Pluto Distance
Jupiter Pluto Distance = Uranus Neptune Circumference
Earth Neptune Distance = Mercury Saturn Circumference (0.5%)
(3)
Jupiter Mercury Distance = 2 Mercury Orbital Circumference
Jupiter Venus Distance = Venus Orbital Circumference (1.5%)
Jupiter Earth Distance = Earth Orbital Circumference (1.2%)
(Earth and Jupiter at 2 different sides from the sun)
(4)
Jupiter Mercury Distance = Mars Orbital Distance x π (0.6%)
Jupiter Uranus Distance = Venus Jupiter Circumference (0.8%)
Pluto Orbital Distance = Earth Orbital Circumference x 2π
II- Discussion (A)
The previous distances form around 50% of all distances found in the solar system
(All orbital and internal distances)… Why These Distances Are Equal One Other?
We may notice that – the distances equality can be produced more easily by light
motion than the rigid body motion - for example – when we push a ball toward a wall
the ball after collision with the wall will return a distance (NOT) equal the original
one - because the collision causes to decrease the ball motion momentum – but the
light can be reflected at equal distances easily – means – equal distances can be
produced by light motion more easy than the Rigid Body Motion.

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I-Data (B)
Why These Distances Are NOT Equal?
1. 0725.1
mkm2.41nceCircumfereOrbitalMoon
mkm2.58MotionDailyEarth
=
2. 1.0725
km)(378500radiusEclipseSolarTotal
km)(406000radiusorbitalApogee
=
3. 0725.1
distanceMercuryJupitermkm720.3
DistanceOrbitalJuppitermkm6.778
= (Error 0.7%)
4. 1.0725
DistanceVenusJupitermkm670
distanceMercuryJupitermkm720.3
=
5. 1.0725
DistanceEarthJupitermkm629
DistanceVenusJupitermkm670
= (0.6%)
6. 1.0725
mkm)(1325.3DistanceVenusSarurn
mkm)(1433.5DistanceOrbitalSaturn
= (0.8%)
7. 1.0725
mkm)(1205.6DistanceMarsSarurn
mkm)(1284DistanceEarthSaturn
= (0.7%)
8. 1.0725
mkm)(2644DistanceMarsUranus
mkm)(2872.5DistanceOrbitalUranus
= (0.7%)
9. 1.0725
mkm)(4495.1DistanceOrbitalNeptune
mkm)(4894nceCircumfereOrbitalJupiter
= (1.5 %)
(10)
I-Discussion (B)
The same rate (1.0725) is used for all equations (around 18 distances = 40% of all
solar system distances) – why?
Suppose the equal distances are produced by light reflection and that cause these
distances to be equal – as I have supposed in the previous point (A).
Now suppose– part of these equal distances – is passed through another frame relative
to us – so this part of distances will suffer from Lorentz Length Contraction Effect
which is seen in the rate 1.0725
(Another frame can be found in the solar system because we deal with light motion) –
This explanation can answer why some distances are equal and others are rated with
the same rate (1.0725) – it's simply a feature of light motion.
0725.1
T.AxailEarth23.4
T.AxailMars25.2
T.AxailMars25.2
T.AxailSatrun26.7
TiltAxailSatrun26.7
TiltAxailNeptune28.3
===

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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
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
Can Different Rates Of Time Be Found In The Solar System Motion?(II)
https://www.academia.edu/44334645/Can_Different_Rates_Of_Time_Be_Found_In_The_Solar_System_Motion_II_
Does Particle Data Depend on Its Motion? (Lorentz Transformations Analysis)
https://vixra.org/abs/1912.0134
Dr. Budochkina, Svetlana Aleksandrovna
Associate professor - Candidate of physico-mathematical sciences (2005)
http://www.mathnet.ru/eng/person22119
List of publications on Google Scholar
List of publications on ZentralBlatt
https://mathscinet.ams.org/mathscinet/MRAuthorID/757317
http://elibrary.ru/author_items.asp?spin=6087-3245
http://orcid.org/0000-0003-3447-0425
http://www.researcherid.com/rid/G-7453-2014
http://www.scopus.com/authid/detail.url?authorId=6507007003
https://www.researchgate.net/profile/Svetlana_Budochkina
Full list of
publications:
http://web-local.rudn.ru/web-
local/prep/rj/index.php?id=2944&p=15209
Mr.Gerges Francis Tawdrous +201022532292
Physics Department- Physics & Mathematics Faculty
Gerges Francis Tawdrous +201022532292
Curriculum Vitae http://vixra.org/abs/1902.0044
E-mail mrwaheid@gmail.com
Linkedln https://eg.linkedin.com/in/gerges-francis-86a351a1
Facebook https://www.facebook.com
Researcherid https://publons.com/researcher/3510834/gerges-tawadrous/
ORCID https://orcid.org/0000-0002-1041-7147
Quora https://www.quora.com/profile/Gerges-F-Tawdrous
Google https://scholar.google.com/citations?user=2Y4ZdTUAAAAJ&hl=en
Academia https://rudn.academia.edu/GergesTawadrous
List of publications http://vixra.org/author/gerges_francis_tawdrous

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