Why Does Mercury Day Period Need 5040 Seconds To Be 176 Solar 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
Why Does Mercury Day Period Need 5040 Seconds To Be 176 Solar 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 –23rd
November 2020
Abstract
Paper hypothesis
The Earth moon orbital inclination causes Mercury Day Period to be less than
176 solar days with 84 minutes
Paper Argument
- The paper shows that there are different interactions of motions between the
planets regardless the distances between them..
- Specifically,
- The paper claims that, Uranus has a great effect on the inner plants motions and
the Earth moon orbital inclination effect on Mercury Day Period is one
example of different results cause by Uranus motion interaction with the inner
planets motions
- The Earth moon orbital inclination effect causes Mercury day period to be less
than 176 solar days with 5040 seconds.

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1- Mercury Day Period Analysis
1-1 Mercury Day Period needs 84 minutes to be 176 solar days as a result of the moon
orbital inclination effect on it (the proves discussion)
1-2 Mercury Day Period needs 84 minutes to be 176 solar days as a result of the moon
orbital inclination effect on it (the mechanism explanation)
I- Data
Group No. (1)
Equation No. (I)
5040 seconds = 95.1 x 53
Equation No. (II)
5040 seconds = 116.7 x 43.2
Equation No. (III)
88000 km = 5040 x 17.4 = 5.1 x 17255 km
Equation No. (IV)
5040 degrees = 7 degrees x 720 degrees
Group No. (2)
Equation No. (V)
51118 seconds = 10.14 x 5040 seconds
Mercury moves during 5040 seconds a distance = 2 Saturn diameters
Pluto moves during 51118 seconds a distance = 2 Saturn diameters
1-1 Mercury Day Period needs 5040s, as a result of the moon orbital inclination
(The Proves Discussion)
Equation No. (I)
5040 seconds = 95.1 x 53
- But
o (Earth diameter 12756 km / Pluto diameter 2390 km) =5.33
o Mercury velocity 47.4 km per sec./ Pluto velocity 4.7 km per sec.
o 95.1 degrees = 90 degrees +5.1 degrees (the moon orbital inclination)
- We have seen the value 5.33 in the previous paper which shows the interaction
of motions between (Earth and Mars), this discussion is attached in this paper
to review
- Equation no. (1) tells that, 5040 seconds of Mercury day period is created as a
function in (5.1 deg = The Moon Orbital Inclination)

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Equation No. (II)
5040 seconds = 116.7 x 43.2
- Where
o 116.7 degrees = 90 degrees +26.7 degrees (Saturn orbital inclination)
o But 26.7 degrees = (5.16)2
(and 5.1 deg = the moon orbital inclination)
o The rate 43.2 we know perfectly, it's the period which is needed by Pluto
to move by rotation a distance = 51118 km = Uranus diameter and this
period =1042.5 hours =43.2 days
o The period 43.2 we will review its discussion later in this paper…
Equation No. (III)
88000 km = 5040 x 17.4 = 5.1 x 17255 km
- Where
o 88000 km = the moon daily displacement
o 17.4 degrees= the inner planets orbital inclinations total
o 5.1 degrees = the moon orbital inclination
o 17255 km = 0.5 (π x10921 km), we know that 10921 km = the moon
circumference and we know this distance (π x10921 km) which Pluto
moves during a period =7511 seconds –
o Note Please, the previous discussion is inserted at the end on this paper
to be our discussion reference .. because this paper discussion depends on
that one.
- Equation no. (III) part (1), tells that, the moon daily displacement depends on 2
values, the value 5040 seconds and 17.4 degrees
- Equation no. (III) part (2), tells that, the moon daily displacement depends on
2values, 5.1 deg (the moon orbital inclination) and the value 0.5 (π x10921 km)
but why (0.5) of the value?!
- Because the moon moves another equal displacement (88000 km), that means,
the moon needs (2 values of 5040).

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- This idea is clear, because 5040 seconds x 2 = 10080 seconds and Earth motion
during 10080 seconds = light velocity, that’s why 2 values are required for the
Earth and its moon motions.
- Please Note
o 17.4 degrees = 5.1 degrees x 3.4 degrees
o 3.4 degrees = Venus orbital inclination, that tells us the value 17.4 deg =
the inner planets orbital inclinations total depends basically on the moon
and Venus orbital inclination, which shows a collective motion is done
by these 2 planets cooperation.
Equation No. (IV)
- 5040 degrees = 7 degrees x 720 degrees
But
- 720 degrees = 17.4 degrees x 41.4 degrees
o 41 degrees = The Solar Planets Orbital Inclinations Total
o 17.4 degrees = The Inner Planets Orbital Inclinations Total
o 7 degrees = Mercury Orbital Inclination
o 7 degrees = 5.1 deg (the moon orbital inclination) +1.9 deg (Mars
orbital inclination)
- Equation (IV) shows a collective motion done based on Mercury data, i.e.
the collective motion (done by the inner planets) and which is shown by
Venus and the moon cooperation is a part of a greater collective motion
done by all planets which Mercury lead.
A COMMENT
- The data shows that, the value 5040 seconds (mkm or degrees) is related by
different way to the value 5.1
- And that supports the claim, the Mercury day period needs 84 minutes as a
result of the moon orbital inclination effect on it
- Now we will show how that happened! How the moon orbital inclination
effected on Mercury data period to be less than 176 solar days with 5040
seconds?

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1-2 Mercury Day Period needs 84 minutes to be 176 solar days as a result of the
moon orbital inclination effect on it (the mechanism explanation)
Equation No. (V)
51118 seconds = 10.14 x 5040 seconds
Mercury moves during 5040 seconds a distance = 2 Saturn diameters
Pluto moves during 51118 seconds a distance = 2 Saturn diameters
- The produced Saturn diameters are the same, we have an interaction of 2
planets motions
- 10.14 = (Mercury velocity 47.4 km per sec / Pluto velocity 4.7 km per sec)
- This we know perfectly because (14.7/1.44 = 10.14) where 14.7 the rate
between Uranus diameter and the moo diameter, but the rate 1.44 is found
between the moon diameter and Pluto distance, (note please, this same rate is
found between Pluto & Uranus velocities)
- We have clearly interaction in motions between Mercury and Pluto in addition
to Uranus, Venus and the Earth moon..
- I want to say that, this interaction of motions does 2 tasks
o (1st
) produced the moon orbital inclination
o (2nd
) causes Mercury day period to be less than 176 solar days with 5040
seconds
- How that can be happened?
- Because of the light motion which is accompanying with planet motion, the
light motion is a part of planet motion – because of that- the solar planets aren't
separated points of mass from each other – on the contrary – they are knots in
the same robe – they are points found on the same trajectory of energy and
because of that, the interaction of planet motion effects on other planets
motions data.
- Please review the basic argument in following, I inserted the paper completely
in next pages because all the previous data was discussed before in this paper
and can easily support the paper hypothesis.

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Paper Title
Planet Velocity Is Defined As A Function Of Its Diameter (II) (revised)
Abstract
Paper hypothesis
The Earth moon is created by effect of Uranus and Pluto motions
The hypothesis explanation
- Although the Earth moon is created by Planets collisions But
- The Earth moon data is created by an effect of light motion calculations
- That supports the conclusion that (light motion must be a companying with planet
motion and effect on it)
Paper questions
- The paper tries to answer the following question:
o Why Mercury Day needs 5040 seconds to be 176 solar days?
o Why Uranus motion distance during (5040 seconds)= Pluto motion distance
during 7511 seconds.
Paper conclusion
- Light Motion Is Accompanying With Planet Motion.

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1- Uranus & Pluto effect on The Moon
I- Data
Equation No. (1)
6.8 km /s (Uranus Velocity) x 5040 s = 4.7 km /s (Pluto Velocity) x 7511 s
(error 3%)
Where
The Equation result = π x 10921 km (The Earth moon circumference)
7511 km = Pluto Circumference
- Equation No. 1 tells that the same distance is produced by 2 planets motions for 2
defined periods of time. Which is = π x 10921 km (the moon circumference)
- We now will analyze the 2 planets motions to answer 2 questions (1st
) Why the 2
planets move an equal distance? (2nd
) why this (π x10921 km) specifically?
Equation No. (2)
51118 km (Uranus diameter) = 6.8 km/s (Uranus Velocity) x 7511s = 4.7 km /s (Pluto
Velocity) x 10921 s (error less 1%)
- Equation No. 2 (Part I) tells that, Uranus moves during 7511 seconds to pass a
distance = 51118 km = Uranus diameter, but on the other side, this part of equation
tells that Pluto needs to rotate around its axis 6.8 times to pass a distance =51118
(km) (please remember Pluto circumference =7511 km) And (6.8 times of Pluto
rotation =6.8 x 153.3 hours =1042.5 hours =43.3 days)
- Equation No. 2 (Part II) tells that, Pluto moves during 10921 seconds to pass a
distance = 51118 km = Uranus diameter, but on the other side, this part of equation
tells that the moon needs to rotate around its axis 4.7 times to pass a distance
=51118 (km) (please remember The Earth moon Pluto circumference =10921 km)
And (4.7 times of the moon rotation =4.7 x 27.3days =128.3 days or "4.7 x 29.53
days = 138.8 days). (But 128.3/43.3 =3)

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- Equation no. (2) shows some deep dependency in data between Uranus, Pluto and
the Earth moon…
But
- To make this situation more clear we need to see what's happening with the other
planets concerning this same distance (π x 10921 km (the moon circumference))
o (i) We know Pluto uses 7511 seconds to pass a distance = π x 10921 km
o (ii) And Uranus uses 5040 seconds to pass a distance = π x 10921 km
o (iii) But Neptune uses 6378 seconds to pass a distance = π x 10921 km
(6378 km = Earth Radius)
o (iv) Also Saturn uses 3535 seconds to pass a distance = π x 10921 km
(3535 km =Mars ancient Radius!)
- Other planets give us more puzzles
o (v) Mars uses 1433 seconds to pass a distance = π x 10921 km
(1433 mkm = Mars orbital circumference!)
o (vi) Earth uses 1151 seconds to pass a distance = π x 10921 km
(1151 mkm = the planets motions distances total during their days periods)
- Neptune and Saturn give us similar data as we have received from Uranus and
Pluto, other planets give more puzzles as we have seen in numbers (v and vi)
Equation No. (3)
2 x 38025 (Venus Circumference) = 6.8 km/sec x 10921 seconds (error 2.4%)
- Equation No. (3) tells that, Uranus during 10921 seconds moves a distance = 2
Venus Circumference
- Please remember why the equation uses (2)! Because (Uranus diameter / the moon
diameter) = 14.7 but the light motion based on which Uranus diameter is created
produced the rate 29.4 = 14.7 x 2
- Means, Each time Uranus will use the moon diameter (or circumference), the rate
(2) can be used accordingly – now let's remember how Uranus diameter is created
based on this rate 29.4 in following

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o 97.8 seconds x 0.3 mkm/sec (light know velocity) = 29.4
o 2872.5 mkm = Uranus orbital distance = 97.8 mkm x 29.4
o i.e. Uranus orbital distance by ((97.8)2
x 0.3 mkm/sec)
o (Uranus diameter)2
= Mars Uranus Distance (error 1.2%)
Equation No. (4)
304 mkm = 51118 km (Uranus diameter) x 6052 km (Venus Radius)
- 304 mkm = 3475 km (the moon diameter) x 88000 km (the moon displacement)
- Let's review what we are doing
- The equal distance of Uranus motion during 5040 seconds and Pluto motion during
7511 seconds produced (π x 10921 km),
- i.e. The moon circumference was the motions goal
- based this moon circumference (10921 km), equation no. (3) produced Venus
Circumferences by Uranus motion
- by Uranus & Venus (equation No. 4) produces the value 304 million km which is
defined the moon daily displacement as a function of its diameter…
Equation No. (5)
2 x 378675 (Saturn Circumference) = 4.7 km/sec x 160592 seconds
- where 160592 km = (Uranus Circumference)
- We need to see the 2 equations (3 and 5) beside each other ….
Equation No. (3) (revision)
2 x 38025 (Venus Circumference) = 6.8 km/sec x 10921 seconds (error 2.4%)
- Saturn diameter = 9.96 x Venus diameter …..But
- The right part of Equation (5) has (4.7 km/sec) and 160592
- The right part of Equation (3) has (6.8 km/sec) and 10921
o (6.8/4.7) = 1.446
o (160592/10921) = 14.7
o So (14.7/1.446) = 10.2 (error 0.3%)
o 10.2 /2 = 5.1 degrees (The Moon Orbital Inclination)

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- i.e. the interaction between Uranus, Pluto and Venus Motions created the moon
orbital inclination (5.1 deg) …But
- Where we have seen the moon orbital inclination as (10.2degrees =2 x 5.1)?
Equation No. (6)
10.2 degrees = 3 x 3.4 degrees (Venus orbital inclination)
1.446 = 6.8 km per second / 4.7 km per second
= (the moon diameter 3475 km / Pluto diameter 2390 km)
= The moon orbit regression monthly (1.44 degrees)
Equation No. (7)
14.7 x1.446 = 21.3 where
5.6 degrees x 2 x 1.9 degrees (Mars orbital inclination) = 21.3 degrees
Where
- 5.6 degrees = 5.1 degrees (the moon orbital inclination) + 0.5 degree (the moon
angular diameter).
- Equation no. (7) tells that, Mars orbital inclination is created as 2nd
product for this
same effect of Uranus and Pluto motions effect on the moon motion.
Equation No. (8)
304 days = 7 x 43.3 days
- Please remember 43.3 days is needed by Pluto rotation to move a distance =
Uranus diameter, (equation no. 2) (the period 1042.5 hours =43.3 days)
- We know that, 304 million km is produced by Uranus and Venus Interaction
- Equation no. (8) shows how Mercury orbital inclination (7 degrees) is created, that
means this value (7 degrees) depends on the value 304 and Uranus effect –
- Please remember the distance of distance 304 mkm where
o 304 mkm x 0.8 = 243 mkm
o If 304 mkm is a distance between the sun and a point (S) in Space
o So the distance from this point (S) to Mercury =243 mkm (1.3%)
o i.e. Uranus orbital inclination (0.8 degrees) works as a rate to divide this
distance 304 mkm into 2 parts based on the rate 0.8 and by this distribution
Uranus defined Mercury orbital distance (304 -243) – and that shows
Uranus effect on Mercury creation data which supports the conclusion that
Mercury orbital inclination (7 deg) is created by this equation (No. 8)

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More Discussion
- Based on the previous analysis, the moon orbital inclination (5.1 degrees) is
created by Uranus and Pluto motions effect on the moon motion.
- But
- Why Pluto & Uranus orbital inclinations aren't produced by this same effect,
specially, why they aren't produced from the number (5.1 x 2 =10.2 degrees)?
- Indeed they are created – let's show that in following:
o 10.2 degrees +7 degrees (Mercury orbital inclination) = 17.2 degrees
(Pluto orbital inclination)
o 122.5 degrees (Pluto Axial Tilt) = 17.2 degrees x 7.1
o 97.8 degrees (Uranus Axial Tilt)/ 122.5 degrees (Pluto Axial Tilt) = 0.8
degrees (Uranus orbital inclination).
- What's the rate 7.1 which is used in the second equation? It's the contraction rate
and we know it perfectly – let's remember it
o 2.58 mkm = Earth motion distance per solar day = 7.1 x 0.363 mkm
o 0.363 mkm =Perigee radius, which is the most near point can the moon
reach to Earth.
o 0.363 mkm =7.1 x 51118 km (Uranus diameter)
o The previous tells us that, the distance 2.58 mkm is contracted by (Lorentz
length Contraction effect with rate 7.1) to produce 0.363 mkm (perigee
radius), then this new distance (0.363 mkm) is contracted again by the rat
7.1 to produce 51118 km (Uranus diameter)
o But The contraction moves from the longer distance to the shorter distance,
so how 17.2 mkm (=17.2 deg) be equal 122.5 mkm (=122.5 deg) by length
contraction phenomenon? It's one more question needs to be answer with
extension for the theory because such using is repeated in the solar system
data – or may be I follow wrong way in explanation.

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What conclusion can we reach here?
Conclusions
- The planet orbital inclination is created as a rate between the diameters rate and
velocities rate as we have seen in the moon orbital inclination (5.1 deg x 2)
Please Note
Equation No. (6) (revision)
10.2 degrees = 3 x 3.4 degrees (Venus orbital inclination)
1.446 = 6.8 km per second / 4.7 km per second
= (the moon diameter 3475 km / Pluto diameter 2390 km)
= The moon orbit regression monthly (1.44 degrees)
- Equation No (6) shows how the moon orbital inclination is created depending on
Venus orbital inclination.
- The rate (3) we have seen in Equation no. (2) (where 128.3/43.3 =3)
- Why?
- Because Equation no. (2) defines the time required from Pluto and the moon to
rotate a distance = 51118 km= Uranus diameter…because of that the rate (3) is
found by Pluto and Uranus motions interaction – and this rate is multiplied with
Venus orbital inclination to produce the moon orbital inclination (2x 5.1deg = 10.2
deg = 3.4 deg x 3 = 10.2 degrees).

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Important Notice
- We need to review this equation again (14.7 = 4.7 x π) where 14.7 is the rate
between Uranus and the moon diameter and (4.7 km/sec) is Pluto velocity….
- Shortly
- The Diameters Rate Is A Function Of A Planet Velocity …how can this be
understood?
- From Lorentz Transformations Discussion we have concluded that, particle
properties are created as a function of this particle motion, because of that, if this
particle travels with high velocity motion, the particle length will be contracted, its
mass will be increased and its rate of time will be changes
- W agreed that, the matter dimensions must be created as a function of its motion,
- The previous equation (14.7 = 4.7 x π) is a very good touch for this principle –
when the picture be so clear, I hope to write that, (Because Earth velocity =x, so
its diameter = y and its axial tilt =z) this is the direct application for this concept,
but because the solar system isn't consisted of separated planets from one another,
because the solar system is consisted of one rope and the planets are knots on this
same rope, or the solar system is one trajectory of energy, and each planet is a
point on it – because of that- no planet has independent data – accordingly – Pluto
motion effects on the moon diameter – but Pluto diameter will be effected by
another planet motion – it's a network of thousands of points from which we see
only 9 knots and all other connections are hidden and need analysis to be
discovered – But – are these connections will be hidden from us forever? Of
course not – they must be concluded from the seen motions – the hidden part of the
solar system is a complementary to the seen part of it and because of that the
correct geometrical mechanism will discover both of them.

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Equation No. (9)
5.1 degrees – 1.44 degrees = 3.66 degrees
Where
o 5.1 degrees = the moon orbital inclination
o 1.44 degrees = the moon orbit regression monthly
o 3.66 = (Earth Diameter / The Moon Diameter)
- This equation may help our investigation somehow
- We have 2 motions and the rate between them is a rate between 2 diameters, it's
perfectly typical the previous equation meaning….
- Equation No. (9) tells that, Earth and moon diameters are created based on
some motions rate…
- And, tells also that, the moon orbit regression depends on the Earth and moon
diameters on one side and on the moon orbital inclination on the other side! Why?
what deep meaning is found behind?
Equation No. (10)
(12104 km /2390 km) = (17.2 degrees /3.4 degrees) = 5.1
o 12104 km = Venus Diameter
o 2390 km = Pluto Diameter
o 17.2 degrees = Pluto orbital inclination
o 3.4 degrees = Venus orbital inclination
o 5.1 degrees = The Moon orbital inclination
- Equation No (10) tries to help us greatly, in our previous investigation we had 2
basic conclusions which are
o (1st
Conclusion) The moon orbital inclination 5.1 degrees is create by
interaction of motions between Uranus, Pluto and Venus
o (2nd
Conclusion) the planets diameters rate depends on planets motions.
Which support the claim that (the matter dimensions are created as function
in this matter motion).

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- Equation no. (10) shows that greatly, where Venus & Pluto diameters are rated to
each other based on their orbital inclinations (motions), and this proportionality
between them produce the rate (5.1) which = 5.1 degrees (the moon orbital
inclination).
- First question, Why their rate produce specifically (5.1 degrees)?
- As we have seen, the moon circumference is used in most equations in this paper,
the most planets use the moon circumference (10921 km) as a period of time
(10921 seconds), including Uranus, Pluto and Venus
- So based on the conclusion, the planets diameters are created based on their
motions, indeed, Venus and Pluto diameters are created based on their motions!
- More Data can support this discussion
- 2390 km (Pluto diameter) = π2
x 243 km …. And
- 120536 km (Saturn diameter) = π2
x 12104 km (Venus diameter)
o 243 days =Venus Rotation Period – Venus Cycle Period is used by Pluto
as distance (or diameter)
o We have seen that, planet diameter and circumference is used frequently as
periods of time – so the using of a period of time as a distance or diameter
value is a logical using –please note- we look at a planet data through a
mirror and we see what the geometrical structure gives us to see – so I can't
discover where's the light beam by which the periods of time are used as
distances and vice versa, because the light beam is hidden behind the planet
motion, and because of that we see light motion features are found in planet
motion but we can't discover the light motion which caused them.
o But, we have planets data are depending on each other and this dependency
can be explained by light motion features. But we don't catch the light
motion behind – Spite of that the dependency is a real and can't b denied.
- But Why Not Earth Diameter? where Earth is the moon parent an should be in
Venus place in the previous equation!

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Equation No. (11)
(12756 km /2390 km) = 5.33 = (227.9/43.3)
Where 12756 km = Earth Diameter
o 2390 km = Pluto Diameter
o 227.9 mkm = Mars orbital distance
- Earth doesn’t produce the rate 5.1 in its diameter rate to Pluto because Earth is
loaded by Mars Data…
- The rate (5.33) which is found in place of (5.1), is found because of Mars motion
effect on Earth motion, the equation has some difficulty let's try to explain it in
following:
- 227.9 mkm = Mars orbital distance, but here it's used as 227.9 days and we know
that the value 43.3 days is created by Pluto rotation (6.8 times) to pass Uranus
diameter (51118 km), shortly… the rate 5.33 is produced because of interaction
between Mars and Pluto in the moon orbit which necessitates Earth to be a player
with them. To make this equation more clear let's see the following one
Equation No. (12)
90560 days (Pluto Orbital Period) = 43.3 days x 2090
2090 mkm =Jupiter Uranus Distance
- The distance 2090 mkm is used here as a rate of as a period of time, why? because
Pluto orbital period is an original one in the solar system, we see it as a period of
time but not all planets do the same… and because of that equation no (11) has this
difficulty where the period of time (43.3 days) is interacted with a distance (227.9
mkm) in one equation.
- Please Note
- (2090 seconds x 0.3 mkm /s light known velocity) =627 mkm, where 627 mkm =
Earth Jupiter distance – that shows the distance 2090 mkm is used clearly as a
period of time.

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Equation No. (13)
37 = 5.1 x 7.25
- 5.1 degrees = The Moon Orbital Inclination
- 7.25 degrees = The Sun Obliquity to ecliptic
- 37 x π =116.2 and 116.2 x π=365.25 days
- Because (Earth orbital period =365.25 days)
- That shows a basic interaction between The Moon Orbit And Sun Data.
- Please Note
o The sun circle Earth during 365.25 days and gives always the same face,
that shows the value 365.25 days is created by the previous interaction
between Earth (thought its moon orbit) and the sun motions
o The rate of time (1 to 365) which is created in the moon orbit gives another
proof for the interaction between the moon orbit and the sun data….
o That again tells us the Earth isn't a common moon in the solar system but
very specific one.

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Let's review The Earth Moon Orbital Triangle because we use it
Figure No. (1) (my figure)
Please Note
(1) SZ = 7665 km ZF = 2414 km
- CZS = 77.8 degrees CZF =102.195 degrees
(2) DY = 3475 km BCY = 28.39 degrees
(3) XB = 16203 km XCB = 10.67
- XCE = 66 degrees CX = 87513 km

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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
AB = 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
)
- i.e. the triangle (ODB) angles are 26.564 degrees and 63.435 degrees
NOTE
- for these 2 angles (26.564 deg and 63.435 deg) we have searched, because these 2
angles will correct many data in the orbital triangle.
(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 = 373000 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)
- BS= (the moon Circumference) =10921 km
- BZ = 18586 km BF =21000 km
- BD = DA = 43000 km (BY =46475 km)
- BA = BC = 86000 km
- CS = = 86690 km
- CZ= (the moon daily displacement) =88000 km
- CF= 88526.8 km CD =96150.9 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 = 118.92 deg) (
- (Uranus axial tilt = 97.8 degrees = FSC 0.6 degrees)
- Angle under (E) = 13.33 degrees 1.1 degrees = 14.43 degrees

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20
The Moon Orbital Motion
- Please remember why 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 its
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 height in motion above perigee radius!
o Why?
o Because the displacement 88000 km during 29.53 days is a great distance
can be provided only by the apogee orbit whose radius (r=0.406 mkm) so if
the moon uses only 88000 km as a real displacement daily, the moon would
move only through apogee radius
o But
o Because the moon uses real displacement technique (L= 88000 km Cos θ)
so the moon has the ability to move through lower orbits with the Earth, and
based on that, when the angle θ be smaller the real displacement be greater
and needs more wide orbit to be performed which force the moon to move
in high orbits 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 Then by more analysis, we have discovered that, a 2nd
force effects on the
moon orbital motion and this force effects on the point (A) in the moon
orbital triangle – where this point is an essential part of the triangle while it's
far from apogee radius with 43000 km
o The 2nd
force is a result of interaction gravity forces between the sun, Earth
and Jupiter on 2 points (Earth and its moon), and because of this interaction
Jupiter causes some gravity force (10% of Earth gravity force) to be effected
on the point (A) and causes the moon motion to apogee radius.
o Then Uranus axial tilt perpendicularity effects analysis gives us the
suggestion that another orbit must be found for the moon motion and this
orbit is found under the first one as described in the following figure.

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A Model For The Moon Motion 2 Orbits

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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

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