The Moon Orbital Triangle (General discussion)(II)

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 Orbital Triangle (General discussion)(II)
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 –8th
December 2020
Abstract
Paper Question:
- Why the moon orbital triangle is an important tool in the moon orbital motion
study?
The Answer:
Because
(1) The moon uses Pythagoras triangle as one of its orbital motion techniques,
based on that, the triangle supports the moon motion equation definition.
(2) There's 2nd
force effects on the moon orbital motion and this force can be
concluded from the moon orbital triangle
(3) There's 2nd
orbit for the moon motion, and this orbit can be discovered by the
moon orbital triangle data analysis
(4) The moon orbital triangle declination angle (1.1 degrees) on the horizontal
level has a massive geometrical effect on the moon motion.
(5) The triangle shows that, the moon daily displacement (88000 km) is defined
based on geometrical rules.
Paper Conclusion
- The moon orbital triangle is a useful tool in the moon orbital motion study

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Contents
Subject Page number
1- Pythagoras Triangle Technique 3
2- Another force (2nd
force) effects on the moon orbital motion 10
3- The moon orbital motion has another orbit (2nd
Orbit) 12
4- The moon orbital triangle declination angle (1.1 degrees) effect 13
5- The moon daily displacement (88000 km) definition. 15
6- The moon orbital triangle data. 16

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1- Pythagoras Triangle Technique
1-1 Preface
1-2 The moon uses Pythagoras triangle in its orbital motion
1-3 The moon motion general description.
1-1 Preface
- The moon uses Pythagoras rule in its orbital motion, this data I have discovered
from very long time because
- The moon 4 basic points are created based on each other by Pythagoras rule,
- Let's prove that in following….
- The 4 basic points are:
o Perigee radius (r=363000 km), this is the most near point the moon can
reach to Earth, means, the moon can't be nearer to Earth than 363000 km
o Apogee radius (r=406000 km), this is the most far point the moon can reach
from Earth, means, the moon can't be far from Earth than 406000 km
o The total solar eclipse radius (r=373000 km). The total solar eclipse
phenomenon is done when the moon be at distance 373000 km from Earth
or shorter
o The moon orbital distance (r=384000 km). this is the registered distance in
the moon data sheet as the moon orbital distance
- These are the 4 basic points of the moon orbital motion, and any radius of these 4
radius can be produced by Pythagoras rule based on another radius and the
dimension 86000 km – let's see that in following….
o (86000 km)2
+(363000km)2
= (373000 km)2
o (86000 km)2
+ (373000 km)2
= (384000 km)2
o (86000 km)2
+ (384000 km)2
= (393500 km)2
o (86000 km)2
+ (393500 km)2
= (406000 km)2
- As the data shows, the 4 basic points are created by Pythagoras rule depends on
each other by using the dimension (86000 km) as basic dimension in all equations

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- Because of this data, I have concluded that the moon uses Pythagoras rule (or
triangle) in its orbital motion – but ….
- Why does the moon use Pythagoras triangle as one of the moon motion
techniques? Why Pythagoras rule is useful for the moon motion? let's answer this
question in the following point….

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1-2 The moon uses Pythagoras triangle in its orbital motion
- AC = 88000 km = The Moon Daily Displacement
- BC = L = 88000 km Cos (θ)
- Angle (C ) I call angle (θ) which is the smallest angle in the triangle…
- The previous triangle tries to explain the useful technique of Pythagoras rule which
the moon uses in its orbital motion, let's try to summarize this intelligent technique
in following….
o The moon daily displacement = 88000 km, the moon moves this distance
per solar day…. But
o During 29.53 days the total distance will be 88000 km x 29.53 days =2.58
mkm
o Through what orbit the moon can move by this distance (2.58 mkm)?! This
orbital circumference (2.58 mkm) its radius is (406000 km)
o So, if the moon moves per a solar day (88000 km) and this distance is
considered as a displacement for the moon, so the moon will have to move
by this displacement through apogee radius only (r= 406000 km) –
o Shortly
o The moon can moves daily a displacement = 88000 km only if the moon
moves through apogee radius (r= 406000 km) in this case the moon can't
move in any orbit nearer to Earth accept apogee radius – means – the moon
will be prisoner in the apogee radius because it moves daily a displacement
= 88000 km.

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o This is the basic point for which the moon uses Pythagoras triangle
technique, because the intelligent moon can move in more nearer orbits by
using Pythagoras triangle technique…
o The moon moves 88000 km and this value doesn't change, the moon moves
it in all cases
o But
o This distance isn't considered as a displacement through the moon orbit,
means… the moon moves 88000 km but the real displacement through the
orbit is less than this value! How
o As in the triangle ….. the moon moves along the right triangle hypotenuse
(AC), which equal the displacement =88000 km, but the moon creates an
angle (θ) between its motion direction and the horizontal level (BC) by that,
the real displacement through the moon orbit is (L) = 88000 Cos (θ)
o That creates another displacement through the orbit different from the moon
motion target per solar day (88000km) and because the real displacement
(L) is shorter than the moon daily target (88000 km), the moon could move
through orbits more near to the Earth
o But,
o The angle (θ) has a massive effect on the moon motion
o Why….?!
o We know that, the full displacement 88000 km per solar day needs to be
passed through a wide orbit (its radius r= 406000 km), and
o If the angle (θ) =29 degrees, the real displacement (L) = 88000 Cos (θ) will
be = 77000 km and this displacement can be passed narrow orbit as perigee
orbit (its radius r= 363000 km)
So
o If the angle (θ) be smaller than (29 degrees), so the real displacement will be
greater (longer) and that needs more wide orbits to be passed through, and

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the wider orbits need the moon to move through higher orbits above perigee
orbit....
o That means
o The change in the angle (θ) causes to change the real displacement and the
height of moon motion above perigee radius… and by that the angle (θ)
control almost the moon orbital motion basic features –because it defines the
real displacement long (L = 88000 Cos (θ)) and based on that change the
moon motion height above perigee radius… i.e.
o The intelligent moon uses Pythagoras triangle in its orbital motion to create
ways to move in more near orbit to Earth without change its motion daily
target 88000 km but in using this intelligent technique the moon put himself
under the angle (θ) control where the angle defines the real displacement
and the moon motion height above perigee.
o Based on that the moon orbital motion equation can be defined based on this
angle (θ)…. The following equation shows that
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
o This equation expresses the moon motion changes, let's discuss it …
o The angle (θ) defines the real displacement and the moon motion height
above perigee, by that, the angle (θ) is the most important component in
Pythagoras triangle which is used by the moon orbital motion, so the
equation defines the change in this angle day after day and by that the
change in the moon motion can be defined by the definition of the angle (θ)
changes… the equation supposes that, the angle is increased by 0.98562
degrees per solar day based on a hypothesis tells us that the moon motion
daily depends on this angle 0.98562 degrees - but why? let's answer in
following…

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1-3 The Moon Motion General Description.
- The moon moves per a solar day a typical motion to Earth motion, to avoid the
separation from the Earth through their motions…i.e.
- The moon moves during a solar day a distance = 2.58 million km with an angle
=0.98562 degrees on the horizontal level revolving around the sun along the Earth
orbital circumference- typically to Earth motion –
- If there's no Lorentz Length Contraction Phenomenon in the solar system motion,
the moon motion would to be seen as a parallel line to Earth motion trajectory.
- But
- Because of Lorentz Length Contraction Phenomenon effected on the moon motion
(2.58 million km) by the rate 1.0725 and causes this distance to be contracted to
be = (2.41 million km)
- The difficulties are started here, because the different distance (0.17 mkm) will
cause inevitably the separation between the moon and Earth motions, and because
of that the moon has to move an additional distance = this different distance = 0.17
mkm to cover this difference on the same day period to avoid the separation from
Earth through their motions…
- So, The moon has found that, there's a target (0.17 mkm) of motion in its waiting.
- Because of that, the moon moves its daily displacement (88000 km) depending on
Earth gravity forces …. But
- This displacement (88000 km) isn't enough, and the moon still needs another
displacement (88000 km) to cover the different distance
- Now this description has 3 important information let's write them in following
o (1st
) The angle 0.98562 degrees defines the daily change in the moon orbital
motion because this angle is the moon motion angle and not because it's the
Earth motion angle, the point here is that, the moon moves a motion
typically to Earth motion and because that, this angle is the moon motion

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angle and because of that this angle 0.98562 deg is used in the moon orbital
motion equation definition, as we have discussed before
o (2nd
) Because the moon needs to move 2 displacement each equal (88000
km) where one displacement only isn't enough… that tells us, there's one
more orbit for the moon motion because where the moon moves this (2nd
)
displacement – the moon must move it through some orbit, but we see only
one displacement (88000 km), so if this displacement is found, it must be
passed through another orbit of the moon motion !
o But …Why we don't see this another displacement (88000 km)? why we see
only one displacement if the moon moves 2 of them?!
o In the moon original motion (2.58 mkm), which is contracted to be (2.41
mkm), the moon moves a typical motion to Earth motion, in distance and in
angle where we don't see this motion because the moon moves it’s a long
the Earth orbital circumference revolving around the sun, and we can't see
the motion on this trajectory of motion – and way- this motion (2.58 mkm)
is a fact regardless our vision for it, because Earth moves daily a distance =
2.58 mkm, and if the moon doesn't move this same distance per solar day
the moon would be separated from Earth motion, by that, our vision for
motion can't be the only proof for it, because this motion is not seen but it's
fact more than any other seen motion….
o (3rd
) There's must be 2nd
force effect on the moon orbital motion, that
because the moon daily displacements (2 x 88000 km) is a distance required
to cover one distance and if this motion is done by one force that
necessitates this displacement to be 170000 km in one motion, but it's not by
that, on the contrary, the moon does 2 displacements per solar day each
=88000 km and because of that, each displacement expresses about one
force and these 2 forces effects are equal to each other and because of that
the displacements are equal…

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2- Another force (2nd
force) effects on the moon orbital motion
- The moon orbital triangle is discussed in full details in point no. (6) of this paper
(6- The Moon Orbital Triangle Data) but I have brought the triangle figure here
to facilitate the explanation..
- The apogee point in this triangle at (POINT D) where ED =406000 km
- As we see
- The great triangle 3 points are (E, C and A), point E is Earth and the distance
between E and C = 373000 km = the distance between Earth and moon at total
solar eclipse radius
- But
- What's this point (A)? because this point is found after apogee point (r=406000
km) with a distance =43000 km
- The triangle will be destroyed without this point (A)! but?
- Earth gravity effect reaches to apogee radius only, and the moon can't move
betony apogee radius (r=406000 km),so what is the force which effect on this
point (A)??
- The moon orbital triangle is an easy method to discover this (2nd
) force effects on
the moon orbital motion….

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- Let's try to discover this (2nd
force)…
- We have 2 reasons at least to suppose that, there's 2nd
force effects on the moon
orbital motion these 3 reasons are
o (1st
reason) The Point (A) in the moon orbital triangle tells us that there's
another force effects on the moon orbital motion
o (2nd
reason)The 2 displacements (2 x 88000 km) according to the previous
analysis tells us that there's one more force effect on the moon orbital
motion
o (3rd
reason) Because Earth Mass = the 4 inner planets masses total (error
1%), this equality suggests that, there's one more force equal Earth gravity
can effect on the moon orbital motion- the distances between these inner
planets and the moon may prevent the gravity effect, but I see this is a
difficulty to be solved but the chance still potential because 2 masses are
equal so they can produce similar to gravity forces and can cause motion for
2 displacements each =88000 km.
2nd
force effect on the moon orbital motion
- 149.6 mkm (Earth orbital distance) = 14984 km (Jupiter diameter) x 1047
- 1047 = The Sun Mass / Jupiter Mass
- The equation tells that, Earth orbital distance is defined based on some interaction
found between the sun mass gravity and Jupiter mass gravity, I want to say that,
Jupiter effects on the sun gravity to define Earth orbital distance..
- By this same rule,
- The sun and Jupiter masses gravity has interactive effect on the Earth moon
motion and by this interactive effect with Earth gravity, another force is created
which effects on the moon orbital motion,
- So this second force is created by an interaction between the sun, Earth and Jupiter
masses gravity
- But, Jupiter gravity is supported by the inner planets gravity in this process

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3- The moon orbital motion has another orbit (2nd
Orbit)
- This conclusion is a simple and clear
- Because
- The moon moves 2 daily displacement each =88000 km under effect of 2 forces,
so one displacement we see through the moon orbit around Earth but the other
displacement we don't
- There must be another orbit for the moon motion an the moon moves its second
displacement (88000 km) through this second orbit
- Where is this second orbit?
- We have 2 features to find this second orbit
o (1st
Feature), the distance between The Point (A) (in the moon orbital
triangle) and the end of the lunar umbra length (1.392 mkm) – let's explain
that….
o The lunar umbra length =1.392 mkm = the sun diameter … but
o From E (Earth) to The point (A) = 449197 km = Jupiter Circumference
o The difference 1.392 mkm – 449197 km = 942803 km
o The triangle ACE perimeter = 942803 km
o That means, The point (A) separates between 2 equal values or 2 equal
triangles, one which we know (the moon orbital triangle) and the second is
seen as the triangle perimeter in the distance between these 2 points (The
point (A) and the lunar umbra length end point) – the data tells us that, the
2nd
orbit is found and neighbor for the 1st
orbit and found between the first
orbit and the lunar umbra length end point…
o (2nd
Feature), There's some angle 0.8 degrees between these 2 orbit, as we
have discussed that before with Venus axial tilt discussion.

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4- The moon orbital triangle declination angle (1.1 degrees) effect
- As we know, The base AE declines on the horizontal level with an angle 1.1
degrees, and this angle has many significant effects on the moon orbital motion, so
let's remember them in following
o Earth angle =13.3 degrees and if we add the angle 1.1 deg which is found
under E, so the total will be 14.4 degrees (where this value is effective
greatly in the moon orbital motion)
o Also
o We know the angle 90.9 degrees which is found inside the moon diameter
(Please review the point no. 6-3 for the full details)… so
o 90.9 degrees x 1.1 degrees = 100 degrees the result (100 deg) has a
massive effect on many geometrical interaction occurred during the moon
orbital motion – first let's know how this angle is created (100 deg)?
o 100 degrees = 90.9 degrees +9.09 degrees
o So what's happened is that, 0.1 degree defines the value (9.09 deg) from the
origin 90.9 deg and then both are added to each other…
o Why 100 deg is useful? because Venus axial tilt was 1.774 degrees (where
Venus diameter =1.774 x Mars diameter) but after the creation of the angle
100 deg, Venus axial tilt 1.774 deg became 177.4 degrees and based on that
the moon orbital inclination (5.1 deg) is created because (360 deg –(177.4
deg x 2) = 5.2 deg (5.1 deg + 0.1 deg)

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o Where Mars orbital inclination was 1.8 degrees and became 1.9 degrees, this
information can easily be known also because (Earth diameter/7070 km)
=1.8 but (Earth diameter /6792 km Mars diameter) =1.9, and that means,
Mars ancient diameter (7070 km) creates an orbital inclination =1.8 deg, but
after Mars Migration, and Mars diameter is changed to be (6792 km)
because of the collisions with planets, so the rate became 1.9 and because of
that Mars orbital inclination =1.9 degrees
o Please remember we know that Mars ancient diameter =7070 km because it
meets the equation (D= Rx 1092
) where Mars ancient orbital distance = 84
mkm, so mars ancient diameter =7070 km.
o Please note
o The angle 1.1 degrees has many other massive effects on the moon orbital
motion and other planet motions geometrical interactions.

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5- The moon daily displacement (88000 km) definition.
- In The Triangle (BCZ)
- Please review this triangle in the moon orbital triangle,
- The triangle hypotenuse (CZ) = 88000 km
- The angle (ZCB) = 12.195 degrees
- This data provides some interesting information, because the triangle hypotenuse
(CZ) defines a distance =88000 km where the moon daily displacement = 88000
km …..and
- The angle =12.195 degrees where (13.18 degrees – 0.98562 deg =12.195 deg)
- We know that, 13.18 degrees = the moon motion degrees per a solar day
- And 0.98562 degrees is the daily motion degrees for Earth and the moon motions
- Why this data is interesting?
- Because the moon daily displacement is defined by Pythagoras rule depends on the
angle (12.195 degrees), that shows the moon daily displacements (2 x 88000 km)
is found based on this angle (12.195 deg) where the basic motion distance (2.58
mkm) which is contracted to be (2.41 mkm) depends on the angle (0.98562 deg)
that proves my claim, that the moon moves a typical motion to Earth motion and
the displacements (88000 km x 2) the moon moves by necessary because of the
contraction.

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6- The Moon Orbital Triangle Data
6-1 Preface
The Moon Orbital Motion (Revision)
- Why does the moon use Pythagoras Rule in its motion? and why do we need the
moon orbital triangle?
o The moon daily displacement =88000 km but the moon doesn’t use it as its
real displacement but instead the moon uses Pythagoras triangle to define
another real displacement
o Based on that
o The moon uses the right triangle dimension (L= 88000 km Cos θ) where
this (L) is the moon real displacement through its orbit daily
o The angle (θ) is the smallest angle in the right triangle, and it effects on the
moon real displacement and its motion height above perigee radius (r=0.363
mkm)
Why?
o The daily displacement = 88000 km and during 29.53 days, the total
distance the moon passes = 2.59 mkm which can be done only through
apogee orbit whose radius (r=0.406 mkm) –
That means –
o If the moon uses only 88000 km as its real displacement daily, the moon
would move only through the apogee orbit (r= 0.406 mkm)
o Because the moon tries to move in more near orbits to the Earth, the moon
uses Pythagoras triangle technique to produce smaller displacements though
its orbit while the moon move this same distance (88000 km) as a real
motion -
o So, this real displacement technique (L= 88000 km Cos θ) gives the moon
the ability to move through lower orbits to the Earth, and the change of the
angle θ changes also its real displacement

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o The smaller real displacement can be used for more near orbits to Earth (as
perigee radius r=0.363mkm) but the longer real displacements need more
wide orbit to be passed through, which forces the moon to move in higher
orbits above perigee.
o That means, the angle θ changes the real displacements and the height of the
moon motion above perigee radius (r= 0.363 mkm).
o Then based on that I have suggested the moon motion equation which is
Gerges Equation For The Moon Orbital Motion
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
o θ is the smallest angle in Pythagoras triangle used for the moon real
displacement (L= 88000 km Cos θ),
o 0.985 degrees = Earth daily motion degrees.
o With more analysis, a 2nd
force effects on the moon orbital motion is
discovered, this 2nd
force effects on the point (A) in the moon orbital triangle
– where this point is an essential part of the triangle but it's far with 43000
km beyond apogee radius (r= 0.406 mkm).
o This 2nd
force effects on the moon orbital motion is produced by gravity
forces interaction between the sun and Jupiter on Earth and its moon – So
this 2nd
force pulls the moon to apogee radius but Earth gravity tries to keep
the moon at perigee radius. (This 2nd
force effects on the point (A) in the
moon orbital triangle)
Notice
o 149.6 mkm Earth orbital distance = 142984 km (Jupiter diameter) x 1047
(The sun mass/ Jupiter mass), because of this data, we conclude that, Earth
orbital distance is defined by gravity forces interaction between the sun and
Jupiter masses. Which supports the claim that 2nd
force can be produced by
the sun and Jupiter gravities interaction effect on the Earth moon.

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6-2 The Moon Orbital Triangle Details
Let's review The Earth Moon Orbital Triangle because we use it
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

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6-3 The Moon Diameter Division
Figure No. 2 (my figure) (The Moon Diameter Division)
- Uranus Axial Tilt (97.8 degrees) has an angle (91.1 degrees) with The Moon Axial
Tilt (6.7 degrees), so The blue vertical line is Uranus axial tilt and the angle in the
circles center = 91.1 degrees
- This value 91.1 degrees is decreased by 0.5 degrees which is consumed by the
moon diameter (the circles express the moon diameter with some division), also
0.6 degrees is consumed by the green box (I suppose Saturn & Jupiter motions
interaction effect is done here to create this 0.6 degrees)
- So the angle between the blue vertical line and the horizontal red line 90 degrees
- The red horizontal line is the moon orbital triangle base (AE = 449197 km)
- The blue vertical line angle above the moon directly =90.6 degrees
- But
- In this figure I cut layers from the moon diameter to create smaller moon diameters
to consume smaller angles than (0.5 degrees), and based on that we measure the
angle of The blue vertical line, let's explain this moon diameter division:
- The Blue Circle its diameter R1= 1390 km and r1 =695 km
- The Red Circle its diameter R2= 2085 km and r2 =1042.5 km
- The Black Circle its diameter R3= 2185 km =π x 695 km and r3 =1092.5 km
- The Brown Circle its diameter R4= 3208 km and r4 =1604 km =5040/π
- The Orange Circle is the moon itself its diameter R5= 3475 km = 5 x 695 km

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- The moon consumes 0.5 degrees for its diameter (3475 km) …So
- R1 = 1390 km, the consumed angle will be 0.2 degrees i.e. the angle above The
Blue Circle = 90.9 degrees.
- R2= 2085 km (60% of the moon diameter) and because of that, the consumed angle
will be only 0.3 degrees, means, the angle above The Red Circle =90.8 degrees
- Note Please, 90.8 degrees = 90 degrees + 0.8 degrees (Uranus orbital inclination),
by that the rest degree of Uranus axial tilt = its orbital inclination vertically.
Notice No. 1
- Pluto Axial Tilt 122.5 degrees = 7.1 x 17.2 degrees (Pluto orbital inclination),
- 7.1 a rate created by Lorentz Length Contraction Phenomenon…
- That tells us, the values (122.5 deg and 17.2 deg) are equivalent values, how? It's a
contraction, the energy of (122.5 deg) in contracted by 7.1 and the rest 17.2 deg is
created from this same energy (122.5 deg). they are 2 equivalent values but one of
them passed through a different frame and faced a contraction phenomenon and
because of that its value (122.5 deg) became (17.2 deg)
- This notice is provided here to show that in the solar system there's some way to
create A Planet Axial Tilt = Its Orbital Inclination…. What's the important
result for this equality? We have seen this equality in Uranus & Pluto data, where
we know that (122.5 deg x 0.8 deg = 97.8 deg)– there's some geometrical machine
found behind this equality…
Notice No. 2
- Are these divided diameters real diameters or invented numbers? Let's test them :
(1)
- 149.6 mkm (Earth Orbital Distance) = 3475 km (the moon diameter) x 43000 km
(the distance from perigee to apogee)
- 149.6 mkm (Earth Orbital Distance) = 2085 km (R2) x 71492 km (Jupiter Radius)

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22
- 149.6 mkm (Earth Orbital Distance) = 1047 x 142984 km (Jupiter diameter) (1047
= The Sun Mass / Jupiter Mass)
(2)
- R4= 3208 km x 466884 km = 149.6 mkm x π2
- 466884 km = Jupiter motion distance during its day period.
(3)
- R3= 2185 km x π x 21346.6 km (Mars Circumference) = 149.6 mkm (2%)
(4)
- R2= 2085 km x 71492 km (Jupiter Radius) =149.6 mkm
(5)
- 4900 mkm = 4222.6 x 1.16 mkm = 10921 km x 449197 km
- 302 mkm = 4222.6 x 71492 km (Jupiter Radius)
- 305 mkm = 88000 km x 3475 km = 25.2 x (3475 km)2
- 177.4 degrees -90 degrees = 87.4 degrees (+0.6 degree = 88 degrees)
- 3475 s x 0.3 mkm/sec =1042.5 mkm but
- 2085 mkm = 3475s x 0.3 mkm/s x2
Discussion
- The points (1-4) show, Earth orbital distance 149.6 mkm, is defined as a function
in the moon diameter, This same feature is used for the smaller diameters, which
proves these diameters are real values
- Point (No.5) shows light motions behind Planet Data – Let's remember 1st
hypothesis (Planet Motion depends On Light Motion), the show some of these
light motions which caused to create the provided data.

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References
The Moon Motion Trajectory Analysis (II)
https://www.academia.edu/44368860/The_Moon_Motion_Trajectory_Analysis_II_
or
https://www.slideshare.net/Gergesfrancis/the-moon-motion-trajectory-analysis-ii
Light Motion Features Are Discovered in Planet Motion
https://www.slideshare.net/Gergesfrancis/light-motion-features-are-discovered-in-planet-motion
or
https://www.academia.edu/44286772/Light_Motion_Features_Are_Discovered_in_Planet_Motion
Can Different Rates Of Time Be Found In The Solar System Motion?(II)
https://www.academia.edu/44334645/Can_Different_Rates_Of_Time_Be_Found_In_The_Solar_System_Motion_II_
Does Particle Data Depend on Its Motion? (Lorentz Transformations Analysis)
https://vixra.org/abs/1912.0134
Dr. Budochkina, Svetlana Aleksandrovna
Associate professor - Candidate of physico-mathematical sciences (2005)
http://www.mathnet.ru/eng/person22119
List of publications on Google Scholar
List of publications on ZentralBlatt
https://mathscinet.ams.org/mathscinet/MRAuthorID/757317
http://elibrary.ru/author_items.asp?spin=6087-3245
http://orcid.org/0000-0003-3447-0425
http://www.researcherid.com/rid/G-7453-2014
http://www.scopus.com/authid/detail.url?authorId=6507007003
https://www.researchgate.net/profile/Svetlana_Budochkina
Full list of
publications:
http://web-local.rudn.ru/web-
local/prep/rj/index.php?id=2944&p=15209
Mr.Gerges Francis Tawdrous +201022532292
Physics Department- Physics & Mathematics Faculty
Gerges Francis Tawdrous +201022532292
Curriculum Vitae http://vixra.org/abs/1902.0044
E-mail mrwaheid@gmail.com
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

The Moon Orbital Triangle (General discussion)(II)

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