Jupiter Has A Cycle Of 8 Days

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
Jupiter Has A Cycle Of 8 Days
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 – 24th
December 2020
Abstract
Paper hypothesis:
- Jupiter Has A Cycle Of 8 Of Its Days Period
(Jupiter day =9.9 hours, i.e. Jupiter has a cycle each 79.2 hours)
The hypothesis explanation
- Jupiter has a cycle during 79.2 hours = 8 of Jupiter days
- This cycle is created by Jupiter motion but effects also on Saturn and Neptune
motions
Paper Objective:
- The paper tries to prove this fact
Paper Contents:
1- Jupiter Has A Cycle Of 8 Days
2- The Moon Orbital Triangle Discussion

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1- Jupiter Has A Cycle Of 8 Days
1-1 Jupiter Cycle (8 days) Creation 1-2 Jupiter Cycle (8 days) Discussion
1-3 Jupiter Cycle (8 days) Role
1-4 Planet Velocity And Diameter Dependency
1-1 Jupiter Cycle (8 days) Creation
- Jupiter moves during its day period (9.9 hours) distance = 466884 km
- Jupiter Circumference = 449197 km
- The Difference = 17687 km
o During 8 days of Jupiter days (8 x 9.9 hours =79.2 hours), Jupiter moves a
distance = 3735072 km = 8 Jupiter Circumferences + 1 Jupiter Diameter
- Because of (1 Jupiter Diameter) We have concluded that, (8 of Jupiter days)
creates a cycle for Jupiter motion because the produced distance (3735072 km)
creates a defined a defined distance (NO ERROR) i.e.
- Jupiter motion mentioned to pass a distance during (8 of Jupiter days) = 8 Jupiter
Circumferences + 1 Jupiter Diameter
- What do we try to do in this paper?
- We analyze this cycle (8 of Jupiter days) to know how this cycle is created and
Why?
- To make our objective clear as possible, we will analyze 2 lines of data which are
o Jupiter motion during its day period (9.9 h) moves a distance =466884 km
o Jupiter motion during 9.53 h moves a distance = 449197km
o Where ( 449197 km) = Jupiter Circumference
- I want to say that, there are 2 geometrical machines by which these 2 values are
created because the difference between these 2 values (466884 km and 449197
km) creates a cycle for Jupiter done each (8 of Jupiter days), means we have
geometrical mechanism produced this cycle and this cycle is produced for some
geometrical reason …. So we try to see the behind complex interactions based on
which this cycle is created

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1-2 Jupiter Cycle (8 days) Discussion
Part I
Equation no. (a)
π x (10921) s x 13.1 km /s (Jupiter Velocity)= 449197 km (Jupiter Circumference)
Equation no. (b)
7511 seconds x 4.7 km /sec (Pluto Velocity) = 35289 km = 3600 x 9.8
Where
10921 km = The Earth Moon Circumference 13.1 km/s = Jupiter Velocity
7511 km = Pluto Circumference 4.7 km/s = Pluto Velocity
- Equation no. (a) tells us how Jupiter pass a distance = Jupiter circumference, it's a
simple equation, where Jupiter during (10921 seconds) moves a distance = 142984
km = Jupiter diameter and based on that during (π x 10921 seconds) will move a
distance = Jupiter Circumference, i.e. Jupiter uses the Earth Moon Circumference
(10921 km) as a period of time (10921 seconds) and during this period of time
Jupiter moves a distance = its diameter
- We don't try to explain the data now until the data be completed
- Equation no. (b) tells that, Pluto moves during a period (7511 seconds) a distance
= 35289 km where 7511 km = Pluto Circumference
- If we use the distance 35289km as a period of time 35289 seconds = 9.8 hours But
Jupiter day =9.9 hours and the difference =1%
o The previous 2 equations try to investigate how Jupiter Cycle (8 of Jupiter
days) is created, because we have 2 values are defined by 2 different
branches both are used by Jupiter to produce its cycle (8 of Jupiter days)
o To make the discussion as clear s possible , we need to us how the 2 values
(π x 10921 seconds) and (7511 seconds) are used because based on these 2
values Jupiter motion during its day period is different with Jupiter
circumference with 4% and this difference creates a Jupiter diameter during
8 of Jupiter days which causes the cycle

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First the value 34309 seconds = (π x 10921 seconds)
Equation no. (c)
51118 km = 6.8 km /s x 7511 seconds = 4.7 km/s x 10921 seconds
Where
6.8 km /s Uranus Velocity 51118 km = Uranus Diameter
4.7 km/s Pluto Velocity
- Uranus moves during (7511 seconds) a distance = 51118 km = Uranus diameter
- Pluto moves during (10921 seconds) a distance = 51118 km = Uranus diameter
- Equation no. (c) tells, there's a deep consistency between the moon circumference
(10921 km) and Pluto Circumference (7511 km) to be used as periods of time on a
side and Uranus & Pluto Velocities on the other side, all equations are controlled
by Uranus diameter!
- I want to say that
o To create Jupiter cycle (8 of Jupiter days), there are 2 branches of data are
used, (1st
Branch) uses the period (π x 10921 seconds) and (2nd
branch)
uses the period 7511 seconds, both values (10921 and 7511) are created in
consistency with each other because of Uranus effect but the branches are
extending through the solar system… means … the solar planets are divided
into 2 teams, one team uses (10921 seconds) and the other team uses (7511
seconds) and by that (Jupiter Cycle of its 8 days) express a general cycle in
the solar system motion… we analyze the data to see what planets belong to
what team and by what mechanism and why
o In this point (First) we deal with the period (π x 10921 seconds) and we try
to know what planets move by this period
(1)
- We know that, Jupiter moves during (π x 10921 seconds) a distance = 449197 km
= Jupiter Circumference

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(2)
- The Earth moon moves during (π x 10921 seconds) a distance = 943817 km the
perimeter of the moon orbital triangle (ACE) (Error 1%)
o (Please Review The Moon Orbital Triangle Data)
o Notice (1) , The moon orbital triangle (ACE) whose perimeter = 943817 km
its base (EA) = 449197 km = Jupiter Circumference
o Notice (2), the equation, 13.1 km/s x 10921 s =142984 km, tells that Jupiter
moves during (10921 seconds) a distance = 142984 km = Jupiter diameter,
means, the moon circumference (10921 km) which is used here as a period
of time should be created by a direct effect of Jupiter motion! So, how to
find this consistency between (10921 seconds ) and (7511 seconds)? For
better understanding, we may consider Pluto data (7511 km or seconds) is
created based on Uranus effect but the moon data (10921 km or seconds) is
a point of connection between Pluto and Jupiter, but Pluto is effected by
Uranus and that means, the moon is a connection point between Uranus
(through Pluto) and Jupiter and by that the moon belongs to Jupiter and
Uranus in the same time…
- Let's ask the question clearly as possible, what planet uses (π x 10921 seconds)
in its motion?!
o Jupiter moves during (π x 10921 sec) a distance = Jupiter Circumference
o Pluto move during (π x 10921 sec) a distance = Uranus Circumference
o The moon moves during (π x 10921 sec.) a distance = The Perimeter (ACE)
(Error 1%)
o Venus moves during (10921 sec.) a distance = Saturn Circumference
(Error 1%)
o Earth moves during (10921 sec.) a distance =2 Uranus Circumferences
(Error 1.3%)
- The data tells there are 5 planets use the value (10921 seconds) in their motions.

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- But
- Can the value (π x 10921 km) be produced by planet motion? by what planet?!
o 6.8 km/s x 5040 seconds = (π x 10921 km), the equation tells, Uranus
moves during (5040 seconds) a distance = (π x 10921 km)
o We remember that, Mercury Day needs (5040 sec.) to be 176 solar days
o 5.4 km/s x 6378 seconds = (π x 10921 km), the equation tells, Neptune
(whose velocity 5.4 km/sec) moves during (6378 seconds) a distance = (π x
10921 km)
o We know that, Earth radius = 6378 km
- We notice that, planet diameter and circumference is used easily as period of time,
spite the geometrical mechanism by such using is absent we can't refuse it because
it's found frequently
- The previous 2 equations are distinguish because Uranus equation uses a period of
time (5040 seconds) as a period of time but Neptune equation uses Earth radius as
a period of time, so the near equation to our geometry book is Uranus equation
- That tells, the moon circumference may be defined by Uranus basically
- But
o 29.8 km/s x 365.25 seconds = 10921 km, the equation tells that during
365.25 seconds Earth (whose velocity 29.8 km/sec) moves a distance =
10921 km = the moon circumference
- But
o 35 km/s x 311 seconds = 10921 km, the equation tells that Venus (whose
velocity 35 km/sec) moves during (311 seconds) a distance = 10921 km
And
o 24.1 km/s x 311 seconds = 7511 km, the equation tells that Mars (whose
velocity 24.1 km/sec) moves during (311 seconds) a distance = 7511 km
o I wish the deep interactions are seen in the data, we simply move inside a
network of interactions between different data… the dependency between

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10921 and 7511 are so deep which push us to conclude that these 2 values
(10921 and 7511) are created for geometrical necessities and they together
create an effect greater than our consideration for both planets effects.

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Second The Value 7511 Seconds = (Pluto Circumference 7511 km)
- Pluto motion causes even more puzzled data, so let's move step by step
o 7511 seconds x 4.7 km/s (Pluto velocity) = 35302 km
o But
o 35302 seconds = 3600 x 9.8 hours (Jupiter Day =9.9 hours) (error 1%)
- Let's ask the question clearly as possible, what planets use the value 7511 seconds?
o 6.8 km/sec x 7511 seconds=51118 km, the equation tells, Uranus (whose
velocity 6.8 km/s) moves during 7511 seconds a distance = 51118 km=
Uranus diameter
o 5.4 km/sec x 7511 seconds=40080 km, the equation tells, Neptune (whose
velocity 5.4 km/s) moves during 7511 seconds a distance = 40080 km=
Earth Circumference (error 1%)
Please Note
o The value (π x 10921 km) is produced by motions of 2 planets only which
are (Uranus 6.8 km/s x 5040 seconds = (π x 10921 km)), and (Neptune 5.4
km/s x 6378 seconds = (π x 10921 km))
o So, Uranus and Neptune motions during 2 different periods produced ((π x
10921 km) and used the period (7511 seconds) to produce 2 different
distances
o Even if the geometrical mechanism is absent but we have 2 specific players
to effect on the 2 values we search for (π x 10921 km) and (7511 km)
o That tells the consistency between these 2 values has a geometrical reason
behind

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Let's summarize what we have till now in following:
- Jupiter has a cycle of (8 days of Jupiter days period)
- To create this cycle …..
o Jupiter moves during its day period a distance greater than Jupiter
circumference with 4%
o During 8 of Jupiter days period
o Jupiter moves a distance = 8 Jupiter circumferences + 1 Jupiter diameter
o Because the additional distance (4%) creates a distance = Jupiter diameter in
8 days of Jupiter days, that tells us, Jupiter has a cycle of 8 days
- We analyzed the data to know how this cycle is created, and found that,
o Pluto Circumference (7511 km) is used as a period of time for Pluto velocity
(4.7 km/sec) to produce a distance 35302 km but 35302 seconds = 9.8 hours
which has a difference 1% with Jupiter day period ….And
o The Earth moon circumference 10921 km, if be used as a period of time
10921 seconds, so during (π x 10921 seconds) Jupiter (whose velocity 13.1
km/s) moves a distance = 449197 km = Jupiter Circumference
So
o The cycle 2 data branches are (7511 and 10921) and these 2 branches cause
the cycle to be created
And
o We analyzed the data to see what planets follow what branch and we have
found that, Uranus and Neptune motions effect on both branches (10921 and
7511) by different methods which may cause to create the consistency in
both branches effect on the planets motions
o And we have concluded that, Pluto is created by Uranus effect and the moon
is the connection point between Uranus and Jupiter motions interaction.
That creates the consistency between the moon and Pluto where Neptune
effect is almost seen through Jupiter effect.

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1-3 Jupiter Cycle (8 days) Role
I-Data
(d)
9.9 x 3600 x 13.1 km/sec x 8 = 10.7 x 3600 x 9.7 km /sec x 10 (error 1%)
(e)
10.7 x 3600 x 9.7 km/sec x 8 = 16.1 x 3600 x 5.4 km /sec x 10 (error 4.5 %)
II-Discussion
Equation (d)
9.9 x 3600 x 13.1 km/sec x 8 = 10.7 x 3600 x 9.7 km /sec x 10 (error 1%)
- Equation no. (d) tells us why Jupiter has a cycle in 8 days…. Let's summarize the
idea in following
o Jupiter moves during 8 of its days (8 x 9.9 hours x3600 = 285120 seconds) a
distance = 3735072 km
o This same distance is passed by Saturn during (10) of Saturn days period (10
x 10.7 hours x 3600 = 385200 seconds), during this period Saturn moves a
distance = 3735072 km
o That means,
o Jupiter motion distance in 8 its days = Saturn motion distance in its 10 days
o But
o 3735072 km doesn't = 10 x Saturn Circumference
o Where
o 10 x Saturn Circumference = 3786750 km
o The difference 3786750 km - 3735072 km =51118 km =Uranus Diameter
So
- In 8 of Jupiter days, Jupiter moves a distance = 8 Jupiter circumferences +1 Jupiter
diameter, and this same distance of Jupiter motion = Saturn motion during 10 of its
days and Saturn motion distance = 10 Saturn circumferences +1 Uranus diameter

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- The cycle is seen clearly, it's a quantum of energy and because of that it creates a
cycle in Jupiter motion and a cycle in Saturn motion and these cycles are defined
by the planets diameters (Jupiter and Uranus)
- Let's remember here equation no (a and c)
o (a) 13.1 km x 10921 seconds = 142984 km = Jupiter diameter
o (c) 6.8 km x 7511 seconds = 51118 km = Uranus diameter
o These 2 equations are specific 2 equations and no other equations are similar
because the planet uses its velocity to move a distance = its diameter
i.e.
o Jupiter during (10921 s) moves a distance = Jupiter Diameter
And
o Uranus during (7511 s) moves a distance = Uranus Diameter
o The complex interaction should be seen in more light with the equation no.
(c)
o (c) 6.8 km x 7511 seconds = 51118 km = 4.7 km/s x 10921 s
o By this equation, a harmony is created between Jupiter and Uranus motions
based on the deep interaction between Pluto and the moon data
Here
o Uranus diameter (51118 km) is a real player in this interaction and has a
great effect on the solar system motion…
Let's move to the next equation

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Equation (e)
10.7 x 3600 x 9.7 km/sec x 8 = 16.1 x 3600 x 5.4 km /sec x 10 (error 4.5 %)
- This equation uses the same concept which is used in the previous one (no. d)
- I suppose that, Saturn motion distance during 8 of its days = Neptune Motion
distance during 10 days, so I built this equation based on this same concept but it
has an error 4.5%!
- Because
- Neptune moves during 10 of its days period a distance = 3129840 km
- Saturn motion distance during 8 of its days = 2989152 km
- The difference around 142984 km (Jupiter Diameter) (error 1.6%)
- Neptune moves a distance during (10 of its days) greater than Saturn motion
distance during (8 of its days)…!
o This information causes a serious problem, because I suppose that, the
motion energy is transported from Jupiter to Saturn by a rate 80% for that
reason (8 of Jupiter days) of motion produces a distance = Saturn motion
distance during (10 of its days)
o By this same concept I suppose the motion energy is transported from
Saturn to Neptune! Here 2 questions are created
(1st
) why Saturn doesn't transport its motion energy to Uranus at first?
(2n
) how Neptune motion distance can be greater than Saturn if
Saturn is the source of energy of Neptune motion?!
- The questions answers need more analysis
- Any way the error in Neptune equation (no. e) doesn't disprove the idea but Uranus
data completely disproves it, Uranus motion during 10 of its days completely
different from Saturn motion distance during 8 days and that tells the energy is
transported from Saturn to Neptune and by some trick Neptune could move a
distance = 142984 km = Jupiter diameter as additional distance to its defined by
Saturn motion energy.

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Jupiter Cycle (8 days) General Discussion
- How to explain the previous data?
- The planets aren't separated rigid bodies revolve around the sun independently
from each other
- The data shows that this description is incorrect, instead:
o The solar system is consisted of one trajectory of energy
o This trajectory has 2 ends (or poles) and these poles are Jupiter and Uranus
o The solar system is similar to a piece of textile it's made by 2 threads used to
weave this piece of textile one thread is perpendicular on the other, the first
one is Jupiter and the second is Uranus
o The weaving between the 2 threads created the other planets as knots in this
same thread
o The weaving process is done based on geometrical rules
o So planets velocities are created depends on each other by geometrical rules
o The basic difficulty is the planet diameter or circumference using as a period
of time, such using is found widely in the solar system motions, let's explain
why…!
o 142984 km =13.1 km/s x 10921 seconds, in this equation, Jupiter is created
before the Earth moon, so Jupiter diameter is known and also Jupiter
velocity, what's not know is the moon circumference (10921 km), so Jupiter
creates a period of time (10921 second) based on its own data to create a
planet circumference by using 1 km for each 1 second! why? because the
born planet data will be in harmony with Jupiter motion data, it's a giving
birth process, the parent gives birth for a child who's born with data similar
to its parents, that will provide the general harmony of motion – but this
explanation doesn't solve the question by what geometrical mechanism the
planet velocity can use planet diameter or circumference as a period of time
because light velocity can do that but planet with love velocity can't?

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o The previous data pushes us far from this concept inside the mechanism
details… We know that, Uranus effect on Pluto greatly, so Uranus and Pluto
can be considered as one team…. Then Uranus and Pluto team will meet
Jupiter on the Earth moon circumference… that makes the Earth moon is so
specific point of interaction between Jupiter and Uranus motions…
Notice
- Pluto Axial Tilt 122.5 degrees x 0.8 degrees (Uranus orbital inclination) =97.8 deg
o (97.8 degrees = Uranus Axial Tilt)
- We consider that, Uranus axial tilt (97.8 deg) is the most great force effected on
the inner planets and specifically on the moon orbit
- And this equation tells us that, Pluto and Uranus behave together as one team and
specifically the value (97.8 degrees) (Uranus axial tilt) can be used by Uranus data
or by Pluto data as the equation tells us

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1-4 Planet Velocity And Diameter Dependency
I- Data
(1)
VelocityPlutokm/sec4.7
VelocityJupiterkm/sec13.1
DiameterUranuskm51118
DiameterJupiterkm142984
=
(2)
VelocitySaturnkm/sec9.7
ceCircumfernVenuskm38026
=
(3)
VelocityUranuskm/sec6.8
DiameterMarskm3475
DiameterMarskm6792
= (1.5%)
(4)
VelocityNeptunekm/sec5.4
DiameterNeptunekm49528
DiameterSaturnkm120536
=
(5)
VelocityMoonThekm/sec27.7
VelocityVenuskm/sec35
VelocityNeptunekm/sec5.4
ceCircumfernEarthkm40080
==
(6) (Error 1.3%)
VelocityMarskm/sec24.1
ceCircumfernEarthkm40080
=
(7)
VelocityEarthkm/sec29.8
DiameterSaturnkm120536
=
(8)
VelocityMoonThekm/sec27.78
RadiusPlutokm1195
DiameterMercurykm4879
=
(9)
VelocityPlutokm/sec4.7
DiameterPlutokm2390
DiameterMoonThekm3475
==
(10)
VelocitySaturnkm/sec9.7
DiameterMercurykm4879
DiameterVenuskm12104
=
(11)
VelocityEarthkm/sec29.8
VelocityMercurykm/sec47.4
DiameterMarskm6792
ceCircumfernMoonThekm10921
= (1%)

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II- Discussion
General Notices
- Planets diameters are created as function in the planets velocities (or vice versa,
Planets Velocities are created depending on their diameters)
- The outer planets velocities are created based on Jupiter Velocity as functions in
the planets diameters, that because of the motion transportation from Jupiter to
Saturn to Neptune….
- The inner planets velocities (Venus, The Moon and Mars) are created as functions
in Earth and Uranus diameters!
- Earth velocity is created with Venus Velocity as functions in Saturn and Jupiter
diameters
- Mercury velocity is created based on Earth velocity as function in the moon and
Mars diameters.
- The previous data tells that
- Planet velocity can't be created independent from other planets data and that
supports the one geometrical building description for the solar system, where the
solar system is one building and each planet is a part of this same building

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2- The Moon Orbital Triangle Discussion
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.
(2nd
Point) The Moon Orbital Triangle Data Correction

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- 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 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 in the small figure separately (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
and 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 moon orbital triangle (Pythagoras
triangle) is a vertical triangle on the base (EA), because of that the point (C) is
found on (Z-axis) by geometrical necessities and form one of the triangle basic
points. BUT
- What's the point (A)? how this point can be created and can effect on the moon
orbital triangle and motion?! Where this point is far from apogee radius with
43000 km and the moon can't move beyond the apogee radius, and can never reach
to this point (A). So how this point is found and why it has an effect on the moon
orbital triangle and motion?! So this point (A) raises an essential question in the
moon orbital triangle geometrical structure analysis.
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) and we have to find the reason by which this point (A) is created here
and causes such massive effect on the moon orbital triangle and motion.
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), If 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
- 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 effect 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..
Shortly.. 2 points of gravities total are created, 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 effects on the moon orbital motion and that causes the
distance (EA) to be = 449197 km = Jupiter Circumference
1047 = (The Sun Mass /Jupiter Mass)

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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 (2 equal distances)
- 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 of 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 here 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 Orbit 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 tracks as 2 prisoners 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 the 2
cars are connected with chain of steel the working car pulls the other one
and both move equal distances through the 2 tracks… that perform 2
distances by one car motion.
o What do 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
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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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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