The Solar Group is A Clock (V)

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 Solar Group is A Clock (V)
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
April 2021
Abstract
Paper hypothesis
- 48 h of the moon motion is equivalent to 1 h of Mercury motion and equivalent to
24 h of Mars Motion and equivalent to 9.23 h of Pluto motion
The hypothesis explanation
- The Solar Group is a great clock
- The Solar group uses energy of a motion for 1 second to cause another motion by
this energy continues for 1 solar day
- Based on that, the solar planets work as gears transport the motion from one gear
to another
- The motion transportation through the solar planets is done by creating different
rates of time between the planets motions –
- Because of that, different rates of time are used among the solar planets
- The paper hypothesis provides one part of the general rate of time – we here uses 4
planets motions and define their motions rates of time relative to each other
- By that, 48 h of the moon motion is equivalent to 1 h of Mercury motion and
equivalent to 24 h of Mars Motion and equivalent to 9.23 h of Pluto motion –
- This definition of the time rates depend on a new suggested rule tells that (Equal
Distances create different rates of time)

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Paper objective
- The paper tries to prove that, these rates of time are truth and be used in these
planets motions

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1- Introduction
- Paper hypothesis tells (48 h of the moon motion is equivalent to 1 h of Mercury
motion and equivalent to 24 h of Mars Motion and equivalent to 9.23 h of Pluto
motion)
- We have discussed before the theoretical description which explains how the solar
group can be a great clock
- Based on that, this paper interests for the suggested rates discussion and prove,
- Because it's a good test for this rule if it really can be a fact
- The 4 planets motions shows different rates of time relative to the moon motion
and what we need to do is that, to test, if these same rates of time are used between
these planets motions among each other –
- Then we need to know what relationship can be suggested between the planet
velocity and its rate of time – or –simply – why different velocities can causes
different rates of time
- And then we need to answer the important question (how the planet can create a
different rate of time by its low velocity? And without needs to use light velocity
of high velocity motion?)
- The paper tries to tell that, this features (which is the rate of time creation by low
velocity) depends on the motion transportation – and both expression depends on
each other needs clear definition - but the short meaning is that – when a motion
be transported from a planet to another planet this transportation of motion is done
by creation of a different rate of time –
- Let's start our discussion trying to see as possible how this process is done and if it
can be a fact

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2- The Earth Moon Rate Of Time
2-1 The Rate Of Time Between The Moon And Mercury Motions
2-2 The Rate Of Time Between The Moon And Mars Motions
2-3 The Rate Of Time Between The Moon And Pluto Motions
2-4 The general Rate Of Time among these 4 planets
2-5 Planet velocity effect on its rate of time
2-1 The Rate Of Time Between The Moon And Mercury Motions
(1)
742 mkm = 88000 km (the moon daily displacement) x 8432 solar days
(2)
742 mkm x 2 = 4.095 mkm (Mercury velocity per solar day) x 365.25 solar days
II – Discussion
Equation No. (1)
742 mkm = 88000 km (the moon daily displacement) x 8432 solar days
- Equation no (1) tells that,
o The moon displacement per solar day during 8432 days be = 742 mkm
o 8432 solar days of the moon motion
o But
o Let's suppose each 2 days of this (8432 days) = 1 hour of Mercury motion
o 8432 hours = 2 x 175.94 days x 23.96 hours
o 175.94 solar days = Mercury Day Period
o 23.9 hours = Earth rotation period
o Where (23.96/23.9) = (361 degrees /360 degrees)
o Based on that, Two Solar Days Of The Moon Motion Be Equivalent To
One Hour Of Mercury Motion

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o This explanation shows the rate of time (1 hour is equivalent to 48 hours)
and shows how this rate works – in our discussion – it works between the
moon daily displacement (motion) and Mercury Motion
Equation No. (2)
742 mkm x 2 = 4.095 mkm (Mercury velocity per solar day) x 365.25 solar days
- Equation no (b) tells that,
- Mercury moves during 365.25 solar days a distance = 742 mkm x 2
- Let's divide the discussion into basic points as following
o (1st
Point)
o We have to suppose that the periods 365.25 days and (2 x 175.94 days) are
equivalent and by that 1 hour of Mercury motion will be equivalent to 48
hours of the moon motion
o (Why the periods 365.25 days and (2 x 175.94 days) are equivalent? The
question should be discussed later
o (2nd
Point)
o The rule which controls this process tells (Equal Distances Create
Different Rates Of Time) – that means – If Mercury moves 742 mkm and
not 2 x 742 mkm – that will cause the rate of time to be 1 hour of Mercury
motion is equivalent to 24 hours of the moon motion – but because Mercury
moves 2 x 742 mkm that causes the rate to be 1 hour of Mercury motion be
equivalent to 48 hours of the moon motion.
o (3rd
Point)
o (Equal Distances Create Different Rates Of Time) What does this rule
mean? it means the period of each distance is equal to the other period of
time – and that means- the moon (clock) moves 48 hours in during Mercury
(clock) moves 1 hour – both period are equal – the moon second period
=1/48 of Mercury second period

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o That means, the person who lives on the moon will pass (48 hours) of his
life but the person on Mercury will pass only 1 hour of his life – both
periods are equal the definition of second period for each planet is different
from one to answer
o (4th
Point)
o The question is (How different rates of time be created without high velocity
motion?) because
o We have 2 Gears work interacted with Each other which are:
o The moon motion and Mercury motion
o These 2 planets motions are interacted through the motion (energy)
transportation –
o Why these 2 gears are required to be interacted in this motion?
o Because the moon daily displacement = 88000 km where it's necessary to
make this displacement equal (176000 km = 2 x 88000 km) – (in the moon
orbital motion description of the current paper we review why this distance
176000 km is required)
o So, The 2 Gears (Mercury and the moon uses the moon displacement 88000
km daily to produce the required distance 176000 km)
o So, the moon motion distance = 742 mkm and Mercury motion distance = 2
x 742 mkm – similar to that – the moon daily displacement be 88000 km
and be = 176000 km by The Interaction Between Mercury And The Moon
Motion –
o (Because of that, Mercury orbital period =88 days and its day period
=175.95 days, If this data is used through the rate 1000 km = 1 solar day,
these 2 values will be equal the 2 distances of the moon motion)
o (5th
Point)
o Where (23.96/23.9) = (361 degrees /360 degrees)

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o Why we need this rate
o The very small error between (23.96 h and 23.9 h) is rated to the rate
(361/360) and what's this rate
o 360 degrees is the cycle degrees
o 361 degrees is the moon orbit regression degrees – because the moon orbit
regresses during a year (365.25 days) 19 degrees which causes change for
the eclipse calendar with 19 days– so during 19 years – the complete
revolution will be 361 degrees –
o Because of that many data is rated by this rate (361/360) and that means this
small error isn't an error but it's found for a geometrical necessity because
the revolution degrees is not 360 degrees but 361 degrees and that
necessitates geometrical modifications for different data

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2-2 The Rate Of Time Between The Moon And Mars Motions
I- Data
(3)
24.1 (km/s) (Mars Velocity) x 29.53 hours = 2562074.856
(4)
The moon displacement total during 29.53 solar days = 2598693km
The difference between both = 1.4%
II- Discussion
- Equations no. (3 and 4) tell that
- Mars moves during 29.53 hours a distance equal (approximately) the moon
displacements total during 29.53 solar days
- From this data we conclude that
- 1 hour of Mars Motion be equivalent to 24 hours of the moon motion
- We use the same rule (Equal Distances Create Different Rates Of Time)
- But
- There's error 1.4% between the 2 distances, that tells the rate (1 hour = 24 hours) is
correct but in the detailed discussion the accuracy needs to take into consideration
this error 1.4%
- The rule is still solid and correct and it's the rule which control the basic operations
in the solar system motions interaction
Notice
- 1 hour of Mercury Motion be equivalent to 48 hours of the moon motion
- And
- 1 hour of Mars Motion be equivalent to 24 hours of the moon motion
- So
- 2 hours of Mars Motion must be equivalent to 1 hour of Mercury Motion, Is it
true? In point no. (2-4) we should answer this question

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2-3 The Rate Of Time Between The Moon And Pluto Motions
I-Data
(5)
406000 km = 88000 km x 4.61
II-Discussion
Equation no. (5)
406000 km = 88000 km x 4.61 Where
406000 km = Pluto Motion per a solar day
88000 km = the moon displacement per A Solar Day
- Equation no. (5) tells that,
- 1 hour of Pluto motion = 4.61 hours of the moon motion
- Because of that
- The moon day period 708.7 hours = 4.61 x 153.3 h (Pluto day period)
- We still use the same rule (equal distances create different rates of time)
Notice no. (1)
- 48 hours of the moon motion = 1 hour of Mercury motion = 2 hours of Mars
motion, so what about Pluto?
- We have to use the rate 1 hour of Pluto = 4.61 hours of the moon and
- 1 hour of Mars be = 4.61 hours of Pluto
- 1 hour of Mercury be = 9.23 hours of Pluto
- These are the rates which we have to use! Why?
- In the next point (2-4) we should answer
Notice no. (2)
- 24 hours /4.61 = 5.2 hours
- The moon orbital inclination (5.1 degrees) is created by an interaction of Pluto
motion on the moon motion
- This idea is supported by the fact that, Pluto moves during its day period a distance
= the moon displacements' total during 29.53 days

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2-4 The General Rate Of Time Among These 4 Planets
I-Data
(6)
48 hours of the moon motion =1 hour of Mercury Motion = 2 hours of Mars
Motion = 9.23 hours of Pluto Motion
(7)
Mercury (47.4 km/s) moves during its rotation period (1407.6 hours) = 240.2 mkm
Mars (24.1 km/s) moves during Venus day period (2802 hours) = 243.1 mkm
(8)
9.23 x 153.3 hours = 1415 hours
Equation no. (7)
Mercury (47.4 km/s) moves during its rotation period (1407.6 hours) = 240.2 mkm
Mars (24.1 km/s) moves during Venus day period (2802 hours) = 243.1 mkm
- The difference between the distance 240.2 mkm and 243.1 mkm is around 1%
- Equation no. (7) tells that both distances are equal (approximately)
- Let's remember the rule (Equal distances creates different rates of time) means, the
periods of time are equal, so the mercury period (1407.6 hours) is equal Mars
period (2802 hours)
- By that 2 hours of Mars Motion is equivalent to 1 hour of Mercury Motion
- This same rate we have concluded from the moon motion (supposed) interaction
with these 2 planets – that means – we have 2 different data proves this same rate
of time which may shows that it's a fact
Notice
- Mercury velocity (47.4 km/s) = approximately 2 x Mars velocity (24.1 km/s)
- Because of that Mercury and Mars motion rate to the moon motion distance equal
(2) and to their motions distance equal (2) also – means –the secret is the velocities
rate, because of that, the claim that it create a different rate of time (1h = 2 h) is not
strong claim!

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- The first objection against this idea is that – the period 2802 hour = Venus day
period and that means this period is defined geometrically through the solar system
and not be limited to Mars and Mercury motion
- 1 hour of Pluto motion = 4.61 hours of the moon motion
- And
- We suppose this same rate should be used between Pluto and Mars motions (later
we discuss Why)
- Means
- 1 hour of Mars Motion = 4.61 hours of Pluto motion
- But
- (Mars Velocity 24.1 km/s) / (Pluto velocity 4.7 km /s) = 5.13
- But
- (24 hours/ 4.61) = 5.2
- That tells the interaction is not some numbers found by pure coincidences – it's
clearly directed data – The interaction is complex and found deeply between Mars
and Pluto and the moon which causes the rate 5.1 to be in different units (as a rate
between velocities, or rate of periods of time and also as the moon orbital
inclination 5.1 deg – it's strong geometrical machine which we don't know its rules
and we need to discover them)
- The rates of time between the planets motions is a geometrical tool based on which
the solar system is designed and created – it's a treasure we should analyze till
discover how this wonderful machine is created.
- Also
- (Pluto orbital distance 5906 mkm/ Mars orbital distance 227.9 mkm)1/2
= 5.1

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- The second objection is that
- Because 1 hour of Mars motion = 4.61 hours of Pluto motion and
- 1 hour of Mercury motion = 2 hours of Mars motion
- That means
- 1 hour of Mercury motion should be equal 9.23 hours of Pluto motion
- Now let's see the equation no. (8)
Equation no. (8)
9.23 x 153.3 hours = 1415 hours
- Where
- 153.3 hours = Pluto Rotation Period
- 1407.6 hours = Mercury Rotation Period (the error 0.5% with 1415)
- So, both 2 planets rotation periods depends on this same rate 9.23
- We have a reason to consider that, this data is created based on a geometrical
machine and that tells the different rates of time is almost a fact tool used in the
solar system geometry.
Notice
- But the rate 4.61 doesn't define Mars motion with Mercury motion
- Where
- 4.61 x 26.4 hours (Mars rotation period) = 113.4 hours
- And
- Mercury rotation period =1407.6 hours?
- So the rate 4.61 is a relationship between mars and Pluto motions but doesn't work
between Mars and Mercury motion
- It's not understandable why in some data the rates work and in another data the
rates don't work? how to understand that?
- To understand we need to study the relationship between planet velocity and its
rate of time – let's do that in following

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2-5 Planet Velocity Effect On Its Rate Of Time
- Let's summarize the idea in following
o The solar system uses 1 second of motion to causes a motion for 1 solar day
o If this rate of time depends on the planets velocity –as we have discussed
before – the difference between the planet velocities should be so great – but
not – the greatest velocity is mercury (47.4 km/s) and the smallest is Pluto
(4.7 km/s) the rate between them is (47.4/4.7) = 10.08
o So the maximum rate the solar planets can do is that,
o 1 hour can be equivalent to 10.08 hours
o But the solar group machine has greater rates of time than this one
o How that is done?
o By Velocities Accumulation
o Let's use one example
o The 3 planets (Mercury + Venus + Earth) velocities total = 112.2 km/s
o Pluto velocity = 4.7 km/s
o The rate between them (112.2/4.7) = 24
o That's the way by which the solar system creates rates of time greater than
the planets velocities
o The basic difficulty is that, we don't know how the 3 plants velocities total
be used in comparison with Pluto - because
o We see each planet independent from the other planets we can't understand
the teamwork system because of this vision
o The 3 planets move on one side and Pluto moves on the other side by that
they create the rate 1 hour (of the 3 planets motions total) = 24 hours of
Pluto motion –
o This rate is used by Mars and the moon motions proportionality – that
means both the moon and Mars motions depends on the 4 planets motions –
but we still need to discover the mechanism behind

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o For that reason
o Pluto moves during its day period a distance = earth motion distance
during its day period = the moon displacements total during its days
period.
o Because of that also
o Earth Velocity / Pluto velocity = (Pluto day period/ Earth day period)

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3- The Moon Orbital Motion Description
3-1 Why Does The Moon Use Pythagorean Triangle In Its Motion?
3-2 How Does The Moon Use Pythagorean Triangle In Its Motion?
3-3 The Moon Orbital Motion Analysis
3-4 The Moon Orbital Motion Equation

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3-1 Why Does The Moon Use Pythagorean Triangle In Its Motion?
- Let's summarize this question answer in following:
o The moon uses Pythagorean triangle basically to decrease its displacement
daily through its orbit
o The moon daily displacement = 88000 km and the moon has to move this
distance every day without any decreasing (later we will know why!)
o But
o If the moon moves by this displacement as its orbital displacement the moon
would revolve around Earth through its apogee orbit only (r=0.406 mkm)
o For that reason
o The moon creates an angle between its motion direction and its orbit
horizontal level to create a displacement through its orbit less than (88000
km)
o As a result of this technique, the moon can revolve around Earth through
more near orbits than apogee orbit (r=0.406 mkm)
o Simply, because the moon uses this technique the moon can revolve around
Earth through perigee orbit (r=0.363 mkm)
o Let's explain this intelligent technique with some details to show the useful
result of using Pythagorean triangle by the moon orbital motion….

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3-2 How Does The Moon Use Pythagorean Triangle In Its Motion?
- The moon moves daily (88000 km) on the right triangle hypotenuse (AC), but the
moon creates an angle (θ) between its motion direction and its orbit horizontal
level, by that the real displacement through the moon orbit will be (L= 88000 km
cos (θ)), and by that, spite the moon moves 88000 km, but the real orbital
horizontal displacement be less than (88000 km) and this is the objective for which
the moon uses Pythagorean triangle –
As an example,
- If (θ) =28.63 degrees, the real displacement (L== 88000 km cos (θ)) = 77237 km,
So, if the moon real displacement daily be (77237 km), during 29.53 days the
moon will pass a distance = 2.28 million km and this will be the moon orbital
circumference, where 2.28 mkm = 2π x (0.363 mkm)
- The Moon Orbital Perigee Radius =0.363 mkm
- That means, the moon by a real displacement =77237 km can move around Earth
through the perigee orbit (radius =0.363 mkm), this is the useful result the moon
performs by using Pythagorean triangle,
- Now let's suppose the moon doesn't use Pythagorean triangle, what would happen?
- The moon daily displacement = 88000 km, during 29.53 days the moon moves a
distance = 2.598 mkm where 2.598 mkm = 2π x (0.413 mkm)
- The Moon Orbital Apogee Radius =0.406 mkm
- So the moon will move along month revolving around Earth through its apogee
orbit (or even far from apogee orbit) because the total distance can't be passed
through any more near orbit around Earth…
- The data shows how Pythagorean triangle is so useful for the moon orbital motion.

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The Angle θ
- The angle (θ) should get our attention for its specific effect…let's summarize the
idea in following
o The angle (θ) changes the real displacement (L = 88000 cos (θ)), through the
moon orbit..
o We know that, when the real displacement (L) be shorter the moon can
move through near orbits to Earth and by that the moon can be near or at
Perigee radius (0.363 mkm)
o When the real displacement (L) be greater the moon has to move through
orbits far from Earth and by that the moon can be near or at apogee orbit
(r=0.406 mkm)
o That means, the angle (θ) changes the real displacement (L) and also
changes the distance between the moon to perigee or to apogee, shortly, the
angle (θ) defines the moon position (as a ship) between 2 river banks….
- The angle (θ) defines the moon orbital motion basic features and we have to
discuss is deeply with the moon orbital motion equation (θ1= θ0 + 1.7 degrees),
but before we need to analyze the moon orbital motion
Notice
o We know that (363000)2
+ (86000)2
= (373000)2
o In Pythagoras triangle with dimensions (363000 km, 373000km, 86000 km),
what's the angle (θ)? The angle (θ) = 13.33 degrees
o Also (396800)2
+ (86000)2
= (406000)2
the angle (θ) = 12.229 degrees
o I have used (363000 km and 406000 km) because they are the perigee and
apogee radiuses between which the moon moves.
o The difference between angles = 1.1 degrees
i.e.,
The angle (1.1 deg.) controls the moon motion from perigee to apogee, we will need
this notice later in our discussion

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3-3 The Moon Orbital Motion
- The moon moves per a solar day a motion typical to the Earth motion to avoid the
separation from Earth through their motions, based on this rule, the moon moves
per a solar day 2.573 million km with an angle declines on the horizontal level
0.98562 degrees as typical to Earth motion
- If there's no Lorentz Length Contraction Phenomenon effect on the moon motion,
the moon motion trajectory would to be a parallel line to Earth Motion Trajectory,
But Lorentz Length Contraction effects on the moon motion daily distance (2.573
mkm) with a rate 1.0725 and causes this distance to be contracted (2.399 mkm)
- The moon difficulties are started here, because the difference between both
distances (0.17 mkm) will cause the moon to be separated from Earth motion
inevitably
- We should notice that, these motions are done far from our observation, means, we
see nothing of this motion distance, because the moon moves on the Earth orbital
circumference revolving around the sun, but, even if we can't observe this motion
distance the motion is still fact and proved by its power, because the Earth moves
per a solar day 2.573 mkm and if the moon doesn't move this same distance every
solar day that necessities the moon to be separated from the Earth through their
motions course – based on that- the facts prove this motion regardless our
observation ability for it.
- Now the moon has an additional distance to be passed (0.17 mkm) and the moon
has to pass this distance on the same solar day to avoid the separation from the
Earth during their motions.
- Because of that, the moon moves its daily displacement (88000 km) depends on
Earth gravity force (by which we see the moon in the Earth sky), but the different
distance (0.17 mkm) to be covered still needs the moon to move one more
displacement (= 88000 km)

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- The previous explanation tells that, the moon has to move 2 displacements each =
88000 km, while we see one displacement only because it's done through the
moon orbital motion around Earth but the other displacement should be done also
because this total distance (0.17 mkm) is required to cover the different distance
and create the total (2.573 mkm) which saves the moon and Earth motions
accompanying.
- Now we have 2 basic information about the moon orbital motion
o (1st
information) the moon uses Pythagorean triangle in its orbital motion
o (2nd
information) the moon has to move 2 displacements each =88000 km
and their total distance =0.17 mkm which is a required distance necessary to
cover the difference between the moon and Earth motions distances.
- This explanation helps us to understand why the moon uses Pythagorean triangle
in its motion, because the moon can't decrease its daily displacement (88000 km)
because the moon needs this distance to cover the different distance between its
contracted motion distance (2.399 mkm) and Earth motion distance (2.573 mkm),
So the moon needs to move this displacement perfectly, but if it's used as a
displacement through the moon orbit, the moon would be always a prisoner in the
apogee orbit (r=0.406 mkm) as we have discussed before, because of that, the
moon creates Pythagorean triangle technique by which the moon moves actually
88000 km daily but the real displacement through the moon orbit became less (L =
88000 Cos θ) and by that the moon can achieve 2 objectives, First to pass the
required distance (88000 km) and Second to move in near orbits to Earth, that
shows the intelligent moon motion technique…
- (Notice, Lorentz Length Contraction Effect Discussion is in Appendix No. 1)

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The Moon Orbital Motion Needs One More Orbit
- The previous explanation tells that, the moon moves 2 displacements each =88000
km, we see one of these 2 displacements but where's the other displacement?!
- We know that, the moon original motion (2.573 mkm) which is contracted to be
(2.399 mkm) isn't seen by us because the moon moves this distance revolving with
Earth around the sun along the Earth Orbital Circumference
- We may accept that, the 2nd
displacement the moon does on this same trajectory
and isn't seen by us.
- So,
- There must be one more orbit for the moon to move through this 2nd
displacement.
means,
- There's 2nd
Orbit For The Moon Motion
- But
- How can we discover this second orbit if we can't observe the 2nd
displacement
motion?
- We can discover this 2nd
orbit by the moon orbit data analysis. So we should
depend on the moon orbital triangle data analysis to define this 2nd
orbit position.
- For that we have to discuss the moon 2nd
orbit in our deep analysis of The Moon
Orbital Triangle Geometrical Structure.

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3-4 The Moon Orbital Motion Equation
3-4-1 The Equation Concept
3-4-2 The Equation Test and Accuracy
3-4-1 The Equation Concept
The Moon Orbital Motion Equation
(θ1= θ0 + 1.7 degrees)
- The moon orbital motion equation is created depending on the concept we have
discussed, which is (the moon uses Pythagorean triangle in its orbital motion)
- The moon uses Pythagorean triangle and by this intelligent technique the moon be
under control of the angle (θ) change
- The angle (θ) defines almost all the moon motion features.…
- The moon uses this technique, aiming to create a real displacement shorter than its
actual displacement (88000 km) based on the equation (L =88000 cos (θ)) and by
that while the moon moves a displacement =88000 km but the real displacement
(L) through its orbit be shorter than 88000 km and by that the moon can revolve
around Earth through more near orbits than its apogee orbit (r=0.406 mkm).
- The moon orbital motion equation depends on this concept and, the equation
uses (the constant) 1.7 degrees as the moon daily motion degrees, and the equation
uses the previous day angle (θ0) to produce the today angle (θ1)
- We have 3 questions in this equation study which are:
o How does this equation work?
o Is this equation trustee and correct?
o Why does the equation use the angle 1.7 degrees for the moon daily motion?
Let's try to answer….

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

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means the moon is far from perigee radius with (368722 km-363000 km
=5722 km )
o So, the moon after 1 day motion will be at the point 368722 km and will
have the Pythagorean triangle its angle 26.93 degrees.
The Descending Motion
o When the moon moves from apogee (0.406 mkm) to perigee (0.363 mkm),
so the angle (1.7 degrees) will be positive (+1.7 degrees) because the angle
(θ) is increased and the real displacement (L = 88000 Cos (θ)) be shorter.
So
o If the moon in apogee radius (r=0.406 mkm), what's the angle (θ)?
o The apogee orbital circumference = 0.406 mkm x2π =2.55 mkm = 29.53
days x 86400 km, the angle (θ) = 10.96 degrees (=11 deg approx.)
o The moon moves from apogee to perigee (descending motion)
o (θ1= θ0 + 1.7 degrees) means (θ1= 11 degrees + 1.7 degrees) = 12.7 deg.
o L = 88000 Cos (12.7 degrees) = 85847 km
o During 29.53 days so (85847 km x 29.53 days = 2.535 mkm)
o 2.535 mkm = 2π x 403467 km
So
o After one day the moon will be on 403467 km far from apogee (406000 km)
with 2540 km
Now let's see this equation test and efficiency in following

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3-4-2 The Equation Test and Accuracy
- I have tested the Equation with real data for 2 months June 2020 and October 2020
- The results are very good and I provide the results here for better vision
concerning the equation efficiency
1st
Test June 2020
Day Registered Data The Results (1.7) Difference
6-6-2020 369418 km
7-6-2020 373729 km 374772.5 - 1044
8-6-2020 378917 km 378821.5 96
9-6-2020 384534 km 383667.7 867
10-6-2020 390096 km 388890 1206
11-6-2020 395156 km 394000 1156
12-6-2020 399345 km 398604.2 741
13-6-2020 402395 km 402361.3 34
14-6-2020 404153 km 405052.8 -900
15-6-2020 404574 km ---- ---
16-6-2020 403718 km 401848.5 1870
17-6-2020 401733 km 400876.1 857
18-6-2020 398840 km 398640.7 200
19-6-2020 395303 km 395417.4 115
20-6-2020 391409 km 391521.2 -113
21-6-2020 387432 km 387273.4 159
22-6-2020 383607 km 382968.4 639
23-6-2020 380110 km 378852 1258
24-6-2020 377044 km 375107 1937
25-6-2020 374451 km 371836.5 2615
26-6-2020 372338 km 369077 3262
27-6-2020 370703 km 366855.6 3847
[

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The 1st
Test Results Analysis:
- The Total Results Are 20 Values
(1st
Category)
o 15 values, defines the moon position in range 1300 km (Error 3%)
(2nd
Category)
o 2 values, defines the moon position in range 1300-2000 km (Error 4.6 %)
(3rd
Category)
o 3 values, defines the moon position in range 2000-3500 km (Error 8 %)
- The Results Explanation
- The distance from perigee to apogee =43000 km…
o 1st
Category of results defines the moon position in error range (1300 km) =
error (3%), that means, (15 values of 20) defines the moon position with
error (3%) only (Small Error Range)
o 2nd
Category of results defines the moon position in error range from (1300
km to 2000 km) = error (4.5%), that means (2 values of 20) defines the
moon position with error (4.5%) (Average Error Range)
o 3rd
km to 3500 km) = error (8%), that means (3 values of 20) defines the moon
position with error (8%) (Great Error Range)
- The Equation Accuracy
o The previous explanation shows that, the equation has a good range of
accuracy and its error is in the acceptable error range
The Conclusion
The Equation Is correct and trustee
And
It's a useful tool to define the moon position daily

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

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The Test Results Analysis:
- The Total Results Are 22 Values
(1st
Category)
o 15 values, defines the moon position in range 1300 km (Error 3%)
(2nd
Category)
o 5 values, defines the moon position in range 1300-2000 km (Error 4.6 %)
(3rd
Category)
o 2 values, defines the moon position in range 2000-3500 km (Error 8 %)
- The Results Explanation
- The distance from perigee to apogee =43000 km…
o 1st
Category of results defines the moon position in error range (1300 km) =
error (3%), that means, (15 values of 22) defines the moon position with
error (3%) only (Small Error Range)
o 2nd
km to 2000 km) = error (4.5%), that means (5 values of 22) defines the
moon position with error (4.5%) (Average Error Range)
o 3rd
km to 3500 km) = error (8%), that means (2 values of 22) defines the moon
position with error (8%) (Great Error Range)
- The Equation Accuracy
o The previous explanation shows that, the equation has a good range of
accuracy and its error is in the acceptable error range
The Conclusion
The Equation Is correct and trustee
And
It's a useful tool to define the moon position daily

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3-4-3 The Value 1.7 degrees
- The 3rd
question was, why the equation uses 1.7 degrees?
Because
1.7 degrees = 0.98562 degrees + 0.712 degrees
Where
- 0.98562 degrees = Earth motion daily degrees, and it equals the moon daily
motion degrees because the moon has to move an equal distance to Earth motion
daily distance to save their motions accompanying
- This question and the angle 0.712 degrees is discussed deeply before.

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The Moon Motion Difficulties
- There are 2 basic difficulties are observed in the moon orbital motions, let's refer
to them in following:
o (1st
Difficulty) The moon moves per day different distances from perigee to
apogee…..
o We know the moon moves from perigee to apogee (go and back) during
Anomalistic month (27.55 solar days)
o (43000 km x 2) / 27.55 days = 3122 km
o The moon doesn't use this rate (3122 km) in its motion, instead the moon
can move (6000 km) on one day only and on another day may move only
2500 km (or even less)!
o The moon orbital equation tries to solve this difficulty by using the rate 1.7
degrees in the equation (θ1 = θ0 + 1.7 degrees), the value 1.7 degrees is a
great number and enables the moon to move around (5000 km) per solar day
and by that if the moon moves per solar day 4000 km the different distance
will be 1000 km and if the moon moves 6000 km the different will be
– 1000 km, it’s the same difference, and by that, the error be minimized as
possible enabling the equation to be more efficient..
o (2nd
Difficulty) The moon stays in perigee and apogee points long time….
o That means, while the moon be on perigee or apogee, the moon doesn't use
the equation and doesn't change its distance to perigee or apogee for long
days…we may notice that in the equation tests, when the moon reach to
perigee or apogee the equation stops its work and stays 2 or 3 days to return
to its work… because the moon consumes long time to leave the points
(perigee and apogee)…

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References
The Moon Orbital Motion Geometry (II)
https://www.academia.edu/45181646/The_Moon_Orbital_Motion_Geometry_II_
or
https://www.slideshare.net/Gergesfrancis/the-moon-orbital-motion-geometry-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
The Moon Motion Trajectory Analysis (II)
https://www.academia.edu/44368860/The_Moon_Motion_Trajectory_Analysis_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
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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