The Moon Uses Pythagoras Triangle Technique In Its Orbital Motion (Part No. 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
1
The Moon Uses Pythagoras Triangle Technique In Its Orbital Motion (Part No. 2)
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 – 17th
December 2020
Abstract
Paper Claim:
- The moon uses the equation (θ1= θ0 + 1.7 degrees) in its orbital motion, because
of that, this equation should be considered a useful tool to define the moon
position through its orbital motion.
The Claim Explanation:
- The equation depends the conclusion (The moon uses Pythagoras triangle in its
orbital motion to decrease its daily displacement (88000 km) by creating an angle
(θ) between its motion direction and the orbit horizontal level, which creates a real
displacement through the orbit (L =88000 Cos θ), that enables the moon to move
through more near orbits around the Earth, because the moon daily displacement
(88000 km) if be used as a real displacement daily will cause the moon to move
through apogee orbit only (r=0.406 mkm).
Paper Question:
- Why the Equation uses the angle (1.7 degrees)? The moon axial tilt declines on
the Earth ecliptic line with 1.6 degrees which can be more consistent to be used in
the equation, BUT the equation uses 1.7 degree! It's a questionable using which
necessitates an answer.
- The paper tries to answer this question

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1- Introduction
- This paper is dedicated to answer the paper question…
Notice No. (1)
- The moon using Pythagoras triangle concept is discussed deeply in the previous
paper – please review it
The Moon Uses Pythagoras Triangle Technique In Its Orbital Motion (Revised)
https://www.academia.edu/s/3eb5edfeef
or
https://www.academia.edu/44707217/The_Moon_Uses_Pythagoras_Triangle_Techni
que_In_Its_Orbital_Motion_Revised_
or
https://www2.slideshare.net/Gergesfrancis/the-moon-uses-pythagoras-triangle-
technique-in-its-orbital-motion-revised
Notice No. (2)
- The moon orbital motion equation is discussed and tested in the previous paper
The Moon Motion Basic Equation (modified) (Part II)
https://www.academia.edu/s/30cd42a65d
or
https://www.academia.edu/44725481/The_Moon_Motion_Basic_Equation_modified_Part_II_
or
https://www2.slideshare.net/Gergesfrancis/the-moon-motion-basic-equation-modified-part-ii
(The Tests Contain The Moon Position Definition During 42 Days, and the equation
produced very good results)
- This paper will discuss only the question, Why the Equation uses the angle (1.7
degrees)?

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2- Why the Equation uses the angle (1.7 degrees)?
2-1 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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2-2- Why the Equation uses the angle (1.7 degrees)?
I-Data
(1)
45 degrees = 26.6 degrees x 1.7
(2)
92.8 degrees = 27.3 x 3.4 degrees (but 92.8 degrees =91.1 deg +1.7 degrees)
(3)
45 degrees = 1.6 x 28.3 degrees (but 28.3 degrees = 26.6 deg +1.7 degrees)
(4)
7.3 degrees = 5.6 degrees + 1.7 degrees (but 29.2 degrees =7.3 degrees x 4)
(5)
26.6 degrees = 18.4 degrees x 1.44 degrees
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!

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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)!
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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Equation No. (2)
92.8 degrees = 27.3 x 3.4 degrees (but 92.8 degrees =91.1 deg +1.7 degrees)
- 3.4 degrees = Venus orbital inclination
- 91.1 degrees = the line BC extends inside the moon diameter s seen in the moon
orbital triangle and the angle is 91.1 degrees because there's an angle 1.1 degrees
between the moon orbital triangle bases (EA) and the horizontal level (90+1.1
=91.1 degrees)
- Equation no. (2) tells that, the moon cycle (27.3) depends on the geometrical
structure (92.8 degrees) and on Venus orbital inclination.
Equation No. (3)
45 degrees = 1.6 x 28.3 degrees (but 28.3 degrees = 26.6 deg +1.7 degrees)
- 28.3 degrees = Neptune Axial Tilt
- 1.6 degrees = the moon axial tilt declines on the Earth ecliptic with 1.6 deg
- Equation no. (3) tells that, there's some complex geometrical interaction behind
which uses Neptune Axial Tilt the moon motion definition from perigee to apogee
(angle 26.6 degrees)…
Please remember
- 1.44 deg (the moon orbit regresses monthly) = 0.8 deg (Uranus orbital inclination)
x 1.8 deg (Neptune orbital inclination). That shows one more feature of this
complex geometrical interaction between Neptune Data and the moon motion.
Equation No. (4)
7.3 degrees = 5.6 degrees + 1.7 degrees (but 29.2 degrees =7.3 degrees x 4)
- We know that Earth moves during 29.53 days (29.2 degrees) and the moon moves
during the same period 389.2 degrees (360 deg +29.2 deg)
- 5.6 degrees = 5.1 deg +0.5 deg, and that means, the moon orbital inclination (5.1)
is measured above the moon diameter (0.5),
- Equation no. (4) shows the value 1.7 deg. is used frequently on the moon motion

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Equation No. (5)
26.6 degrees = 18.4 degrees x 1.44 degrees
- 26.6 degrees is the angle (BCA) and 18.4 degrees is the angle CDA in the same
triangle (BCA)
- 1.44 degrees = the moon orbit regression per month
- Equation no. (5) tries to show that, there's a deep interaction between the moon
orbital motion and the moon orbital regression (19 degrees) per sidereal year.
BUT
- The previous explanation tries to tell one important information that, the moon
orbital cycle (27.3 days) is created based on the point (A), and spite this creation
depends on Venus orbital inclination but still the main force behind is Jupiter data,
this information is important for 2 reasons which are:
o (1st
reason) to prove that the point (A) is a real point effects on the moon
orbital motion, an that will increase our confidence in the claim that, another
force (2nd
force) effects on the moon orbital motion
o (2nd
reason) to show that the origin of this second force effects on the moon
orbital motion, where the point (A) represents Jupiter data and tells that
Jupiter gravity interaction with the sun gravity causes another force (2nd
force) to be created (on this point (A)) and effects on the moon orbital
motion – this is the main objective behind the paper question answer… we
need here to see how Jupiter gravity effects on the moon motion and causes
the moon orbital period to be created, this conclusion we can reach by the
reference that the distance between the Earth and the point (A) = Jupiter
Circumference, spite the geometrical interaction obscurity
o In following we should discuss these 2 reasons… first to prove that Jupiter
causes the moon orbital period (27.3 days) to be created and second to
discover by what force Jupiter does that…

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2-3 Jupiter Causes The Moon Orbital Cycle Top Be Created
Equation No. (6)
4331 days x 2 = 27.3 days x 318
- 4331 days = Jupiter Orbital Period
- 27.3 days = The moon orbital period
- 318 = Jupiter Mass / Earth Mass
- I claim that, planets masses rate shows an interaction of gravity effect, and by that
the rate 318 shows Jupiter mass effect on earth mass and that shows the force by
which Jupiter effects to create the moon orbital period (27.3 days)
Please remember
- 142984 km (Jupiter diameter) x 1047 = 149.6 mkm (Earth orbital distance),
where (1047 = the sun mass / Jupiter mass), this equation tells that Earth orbital
distance is defined by interaction between the sun and Jupiter masses, that tells us
spite the distance Jupiter gravity practices a real effect on the Earth orbital distance
definition,
- The masses rate effect we should discuss in the next point,
- Equation no. 6 tells that the moon orbital period is defined based on Jupiter gravity
effect.
Equation No. (7)
(177.4 deg/3.1 deg) = (180 deg/π)
- 177.4 deg = Venus Axial Tilt
- 3.1 degrees = Jupiter Axial Tilt
- Equation no. (7) shows a deep interaction between Jupiter and Venus which enable
to perform the same task together (the moon orbital period creation)

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The Planets Masses Rate
I-Data
(a)
(the sun mass / Jupiter mass) x Jupiter diameter = Earth orbital distance
(b)
(the sun mass / Jupiter mass) x Jupiter Circumference = Mars orbital diameter (1%)
(c)
(the sun mass / Saturn mass) x Venus diameter = Earth Venus Distance (1%)
(d)
(the sun mass / Uranus mass) x earth diameter = Earth orbital diameter (2.4%)
(e)
(Uranus mass / Earth mass) x Uranus Circumference = the moon orbital
circumference
II-Discussion
- The previous data shows that, planets masses rate create clear effects on the data
creation, so the distance can't completely remove the planet mass gravity effect
whatsoever the distance is
- This notice I have discovered by analyze the solar planets data, the planets show
that planets masses rates are used frequently in different data creation, and there's
no logical understanding behind the pure coincidence claim, because of that, this
data forces us to find one more expression for masses rate gravity effect spite of
the huge distance that because the planets data is created in consistency with such
rates as the previous data tells us
- Based on this claim, Jupiter mass interaction with the sun mass creates a direct
effect on the Earth orbital distance definition and creates another force effects on
the moon orbital motion which is seen effected on the point (A) in the moon
orbital triangle and by this force Jupiter gravity interaction effects on the moon
orbital motion to create its cycle =27.3 days

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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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