Can Earth Motion effect on Pluto Motion?

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
Can Earth Motion effect on Pluto Motion?
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 –19th
June 2021
Abstract
Paper hypothesis
- Pluto And The Moon Motions Are Effected By The Earth Motion.
The hypothesis Explanation
- I use the planets data analysis to discover the solar planets creation and motions.
- The planets data shows the following fact….
- Pluto Day Period be =153.3 hours because Pluto day period depends on The Earth
moon day period (708.7h). That explains why Pluto day period is so long in
comparison with the outer planets days periods. The fact is that, Pluto day period is
created as a function in the Earth Moon Day Period.
- On the other side, the moon orbital apogee radius which should be = 413600 km
became =406000 km because of Pluto motion effect on the moon orbital motion.
- Paper Objective
- Because no physics theory can give an explanation for such dependency found
between 2 small masses planets which are at a huge distance between them, I have
to prove this dependency between the 2 planets by all means. And as a result for
these proves, we have to search after the method by which this dependency can be
created.
Please scan the figure (ORCID)

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1-Introduction
- How to know if any planet motion effects on any other planet motion?
- I think by Planet Data. For example
- Jupiter orbital period 4331 days = Mars orbital period 687 days x 2π
- I consider this data is produced by an effect of Jupiter motion on Mars motion. I
can't explain the geometrical reason by which Mars orbital period be used as a
radius in a circle and Jupiter orbital period be used as this circle circumference,
although, I consider this data expresses Jupiter motion effect on Mars motion.
- In all cases I don't consider this data be created by any pure coincidence. Jupiter
motion effect is seen clearly through. (why not?)
- Because
- I suppose, the solar system is built based on geometrical rules are unknown for us,
by that, we can't neglect the valuable data as a result of our (less) knowledge
- Also …
- The pure coincidence claim doesn't answer (how to know if any planet motion
effect on another planet motion?). the pure coincidence claim simply remove the
planets data and live us in the darkness have no any source of knowledge.
- I consider the planet data is similar to a creature Genes, because of that, the planet
data has the information about this planet creation and motion.
- In fact
- Pluto data shows clearly Earth motion effect on it
- Even if no physics theory can explain this effect,
- The effect is a fact and proved clearly. Whatsoever the claims against this effect,
the data can disprove these claim and shows clearly that Earth motion effect on
Pluto motion.

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2-Methodology
(I)
- I use the solar planets data analysis to discover the planets origin and motions.
- This method is so useful because it creates a comparison between the physics
theories and the planets data to show if the planets data follow the theories really.
- For example, we notice that, Jupiter is the 5th
planet in order to the sun, while the
gravitation equation tells (More Mass Needs Shorter Distance). Jupiter position
doesn't disprove the equation but refers to a contradiction needs explanation.
- That's the useful result of planets data analysis using.
(II)
- I suppose the solar group is a machine of gears and each planet is a gear in it. or it's
one building and each planet is a part of it. this hypothesis leads us to conclude
that, the solar planets motions must create one unified motion. based on that the
solar planets motions can be similar to the train carriages move tougher one
unified motion we call it (Train Carriages Motion)
(III)
- Because the solar planets are parts in the same one building, their data be created
complementary with each and by that their motions be also complementary
creating together one unified motion.
- The best example to explain the (complementary data) is the double production
experiment. Where an electron be created with a positron from gamma ray, that
shows the 2 particles charges are created depending on each other because of the
charge reservation law. But the 2 particles move depend on their charges, means,
the 2 particles motions are still controlled by (Gamma ray features) and they move
a complementary motion.
(IV)
- Based on this hypothesis, the solar planets data should be created complementary
with each other. means, There's One Equation Controls All Solar Planets Data

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- Let's discuss an example for better explanation.
- The Diameter =12430
- There's no a planet in the solar group its diameter =12430km, this number I have
invented it. let's test it to know if it has any effect on the solar system motion.
o The (supposed) diameter 12430 km in the one middle value between Venus
diameter (12104 km) and earth diameter (12756 km).
o (Earth diameter / its moon diameter) = (Jupiter diameter) / (12430 x π)
o Mars (24.1 km/s) moves during (12430 seconds) a distance = 300000 km =
light (0.3 mkm/s) motion distance for 1 second.
o Saturn (13.1 km/s) moves during (12430 seconds) a distance = 120536 km
= Saturn Diameter
- We compare between 2 visions.
- The classical vision tells that, the solar planets are rigid separated bodies revolve
around the sun independent in their creation and motion data from one another.
- We provide a new vision…
- The solar system is created based on one geometrical design, the planets are parts
in one building and their data (must) be complementary one another, Because they
are created based on geometrical rules. As in a triangle we know 2 angles (60 deg
and 80 deg) what's the third one?
- The diameter 12430 km proves this vision. It behaves perfectly as similar as any
planet (real) diameter. let's remember
o Jupiter (13.1 km/s) moves during 10921 sec a distance = 142984 km (=
Jupiter diameter) where (10921 km = the Earth moon circumference)
o Uranus (6.8 km/s) moves during 7510 sec a distance = 51118 km (=Uranus
diameter) where (7510 km = Pluto circumference)
- Based on that, the (supposed) diameter 12430 km is designed based on the solar
system geometrical design but be not created.
- Which supports the paper vision about the solar system creation and motion.

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3- Pluto Data Analysis
3-1 Preface
3-2 Pluto Day Period Definition
3-3 Pluto Motion Distance During Its Day Period
3-1 Preface
- The paper hypothesis tells us about 2 different processes are done depend on one
another which are, (1st
) Pluto Data Creation (2nd
) the moon orbital apogee radius
definition. Let's analyze these 2 process in short in following…
(1st
) Pluto Data Creation
- Pluto Day Period = 153.3 days
- (Jupiter day =9.9 h, Saturn day =10.7 h, Uranus day =17.2 h, Neptune day =16.1h)
- Pluto day period 153.3 h =53.9 h (the outer planets days total) x 2.84
- Pluto Day period is belonged to the inner planets and not to the outer Planets.
Because some of inner planets have long days.
- Pluto day period is rated (frequently) and in tenth of data with the moon day period
as we will see in the data discussion.
(2nd
) The Moon Orbital Apogee Radius Definition
- The moon daily displacement =88000 km
- The displacements total during the moon day period (29.53 days) = 2598693 km,
this distance should be equal the moon apogee orbital circumference, by that, the
moon apogee orbital radius should be = 413600 km
- But
- It's not true, the moon apogee radius =406000 km
- This decreasing in the moon apogee orbital radius is done by Pluto motion effect
on the moon motion.
- Because of that, Pluto moves during the solar day a distance =406000 km.

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3-2 Pluto Day Period Definition
I- Data
(1)
(406000 km /88000 km) = (708.7 h/ 153.3 h) = 4.61
(2)
Tan (12.2) x 708.7 hours = 153.3 hours
(3)
Tan (13.17) x 655.7 hours = 153.3 hours
(4)
413600 km cos (10.96 deg) = 406000 km
(5)
(10.96 hours /1.7 deg) = (153.3 h /24 h) (error 1%)
II- Discussion
The previous 5 data shows that, Pluto day period is created as a function in the moon
day period. It's hard to create pure coincidences in the 2 planets (all) data. Logically
this data is created based on geometrical reason and mechanism. Let's discuss this
data one by one in following…
Equation no. (1)
(406000 km /88000 km) = (708.7 h/ 153.3 h) = 4.61
- Where
- 406000 km = Pluto motion distance per a solar day
- 88000 km = The moon displacement per a solar day
- 708.7 hours = The Moon Day Period
- 153.3 hours = Pluto Day Period
- Equation no. (1) shows that, the 2 planets days periods are rated with their motions
distances per a solar day, showing that, both planets motions per a solar day effect
in their days periods definition process.

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Equation no. (2)
Tan (12.2) x 708.7 hours = 153.3 hours
- Where
- The angle 12.2 degrees = 13.177 deg – 0.98562 deg
- 13.177 degrees = The Moon Motion Degrees Per A Solar Day
- 0.98562 degrees = The Earth Motion Degrees Per A Solar Day
- Equation no. (2) shows that the 2 days periods are created based on each other
geometrically. The fact can't be denied. Pluto day period is created as a function in
the moon day period based on the angle (12.2 deg) which is the difference between
the moon motion degrees per a solar day and Earth motion degrees per a solar day.
- The data supports the idea clearly.
Equation no. (3)
Tan (13.17) x 655.7 hours = 153.3 hours
- Where
- 13.177 degrees = The Moon Motion Degrees Per A Solar Day
- Equation no. (3) shows that, Pluto rotation period is created depending on the
moon rotation period based on the angle (13.177 deg).
- The idea is proved clearly and strongly, no pure coincidence can use all planet data
to produce another planet (all) data, it's illogical idea, the fact is seen in the data
clearly as the day light. Even if no physics theory can help to explain how this data
is created, the fact tells, the data is created depends on each other geometrically
- We try here to discover how this process is done…
Equation no. (4)
413600 km cos (10.96 deg) = 406000 km
- Where
- 413600 km = the moon apogee (supposed) Radius

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- As we have discussed before, the moon displacements total 2598693 km should be
the moon apogee orbital circumference, by that, the moon apogee orbital radius
should be =413600 km
- But
- It's not true, the moon apogee orbital radius =406000 km
- Why?
- I claim by an effect of Pluto motion on the moon motion. and I use the distance
406000 km as a proof because Pluto moves during a solar day 406000 km.
- Now, how the distance 413600 km became 406000 km
- By using the angle (10.96 degrees)
- This angle we have studied before deeply because it’s the main angle in the moon
orbital geometrical structure. Let's remember that:
o The moon orbital perigee radius =363000 km
o The moon orbital apogee radius = 406000 km
o If we draw these 2 radiuses in 2 circles as in the figure
o The tangent DB = 181843 km, also
o The perigee orbital circumference = 2.28 mkm
o (181843 km /2.28 mkm) = 0.08
o 137 degrees x 0.08 = (10.96 degrees)
o As we remember that 137 degrees = 95.1 deg x 1.44 deg where
o 95.1 deg =90 deg + 5.1 deg (the moon orbital inclination)
o 1.44 deg = the moon orbit regression degrees per month.
o We have supposed that the angle 137 degrees is created based on the moon
orbital inclination (degrees) and the moon orbit regression angle monthly.
And this angle (137 deg) is the basic angle in the moon orbital geometrical
structure (hypothesis).

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o The calculations of the tangent DB = 181843 km aimed to explain one basic
idea which is, the moon orbit is built geometrically to create the rate (0.08)
by which the moon orbital inclination and regression will create the angle
(10.96 deg).
o Let's explain that more clearly
o The moon orbit inclination and regression controls the moon orbital motion.
this is a hypothesis, Now, the moon orbital motion creates the moon orbit
based on a geometrical design, the moon orbital motion uses the moon orbit
geometrical design to create the angle (10.96 deg) from the basic one (137
deg). by that the original angle is (137 deg) which is created by the moon
inclination and regression unification, and the born angle is (10.96 deg)
which is the motor of the machine.
o What we need to refer is that, the moon orbit geometrical design is used in
this process. That's why the distance 406000 km is defined based on this
process. By that, the main player is the angle (10.96 deg).
Equation no. (5)
(10.96 hours /1.7 deg) = (153.3 h /24 h) (error 1%)
- Where
- 1.7 deg is the angle used in the moon orbital motion equation (θ1= θ0 + 1.7 deg)
- 153.3 hours = Pluto Day Period
- 24 hours = Earth Day Period
- Equation no. (5) shows the facts, let's discuss it in following…
- The angle (1.7 degrees) defines the moon daily motion which we will discuss in
the next point (Point no. 4 - the moon orbital motion analysis)
- The angle (10.96 deg) is rated with 1.7 deg by the same rate between Earth and
Pluto days periods.
- The equation shows that, the moon apogee orbital radius be 406000 km because of
Pluto motion effect on the moon motion.

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3-3 Pluto Motion Distance During Its Day Period
I-Data
(A)
- Earth (29.8 km/s) moves during a solar day (24 h) a distance = 2574720 km
- Pluto (4.7 km/s) moves during its day (153.3) a distance = 2593836 km
- The moon displacements total during (29.53 days) = 2598693 km
- Pluto and the moon motion distances be equal but they different from Earth motion
distance with (1%).
- What geometrical result can be produced by these equal distances?
- Why do the 3 planets move equal distances in their days periods?
(B)
- (Earth velocity / Pluto velocity) = (Pluto Day period/ earth day period) (0.8%)
- Pluto orbital distance 5906 mkm = Earth orbital circumference 940 mkm x 2π
- Pluto orbital period 90560 days = Earth Cycle (1461 days) x 2π3
II-Discussion
- Earth and Pluto motions data be created in full harmony and proportionality with
one another. when we put the moon data with Earth data in comparison with Pluto
data, we have to conclude that this data is created based on a clear effect of Earth
motion on Pluto motion.
- The discussion shows that, Pluto motion effect on the moon motion is a fact can't
be denied, and by this fact the moon apogee orbital radius be 406000 km in place
of 413600 km.
- We have to search after this data to know by what method this effect be created.

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

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4-1 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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4-2 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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4-3 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…

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Notice
- The moon motion distance daily (2.574 mkm) be decreased with the rate (1.0725)
to be 2.399 mkm. I suppose this rate be found by Lorentz length contraction
phenomenon. But whether this hypothesis is real or imaginary idea it has no effect
on the result. Because the rate (1.0725) effects on around 40% of all solar planets
distances and based on that (40%) of the distances be contracted with this rate
(1.0725). This data is discussed deeply with Jupiter motion analysis (point no. 7)
and in The Appendix No. 1 (at end of this paper).
- The idea is that, this rate (1.0725) effects on the distances and causes to contract
them. And the moon motion distance is one of these distances are effected by this
rate and then the moon has to suffer to repair this rate (1.0725) effect on the moon
daily motion distance.

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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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4-4 The Moon Orbital Motion Equation
4-4-1 The Equation Concept
4-4-2 The Equation Test and Accuracy
4-4-1 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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4-4-2 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

I do this research
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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
[

I do this research
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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

I do this research
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26
4-4-3 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
- Also
- Because of Pluto effect on the moon motion please remember equation no. (5)
in point no. (3-2)
Please read my paper
(The Moon Orbital Motion Geometry (II))
https://www.researchgate.net/publication/351858672_The_Moon_Orbital_Motion_Geometry_II
https://www.academia.edu/45181646/The_Moon_Orbital_Motion_Geometry_II_
https://www.slideshare.net/Gergesfrancis/the-moon-orbital-motion-geometry-ii

I do this research
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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 and Biography
The Solar System Motion Analysis
https://www.researchgate.net/publication/352151822_The_Solar_System_Motion_Analysis
Did Lorentz Ask (How The matter is Created)?
https://www.academia.edu/48987206/Did_Lorentz_Ask_How_The_matter_is_Created_
The Moon Orbital Motion Geometry (II)
https://www.academia.edu/45181646/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
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
Twitter https://twitter.com/GergesTawdrous1
Facebook https://www.facebook.com/gergis.tawadrous/
Linkedln https://eg.linkedin.com/in/gerges-francis-86a351a1
Quora https://www.quora.com/profile/Gerges-F-Tawdrous
Researchgate https://www.researchgate.net/profile/Gerges-Tawdrous
Researcherid https://publons.com/researcher/3510834/gerges-tawadrous/
ORCID https://orcid.org/0000-0002-1041-7147
Google https://scholar.google.com/citations?user=2Y4ZdTUAAAAJ&hl=en
box https://app.box.com/s/47fwd0gshir636xt0i3wpso8lvvl8vnv
Academia https://rudn.academia.edu/GergesTawadrous
List of publications http://vixra.org/author/gerges_francis_tawdrous
Also https://www.slideshare.net/Gergesfrancis?utm_campaign=profiletracking&utm_medium=sssite&utm_source=ssslideview
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Can Earth Motion effect on Pluto Motion?

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