Uranus is perpendicular on earth moon orbit (viii) (revised)

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
Uranus Is Perpendicular On Earth Moon Orbit (VIII) (revised)
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 –21st
November 2020
Abstract
Paper hypothesis
The moon and Pluto motions are transported from one to another through the
solar planets motions.
Paper Investigation
- We analyze the moon orbital triangle to see this motion transportation
- Let's remember our basic argument:
- Uranus axial tilt (97.8 deg) creates an angle (91.1 deg) with the moon axial tilt (6.7
deg), but Uranus axial tilt is perpendicular on the moon orbit, because Uranus
creates a decline between the moon orbital triangle and the horizontal level =1.1
degrees, by that the angle became 90 degrees (perpendicularity)
- The angle 1.1 degrees Uranus created by stuffing consisted of 2 parts, the 1st
part is
Jupiter and Saturn interaction which created an angle =0.6 deg and seen in the
orbital triangle by the green rectangle under the triangle, the 2nd
part, An angle (0.5
degrees) is consumed by the moon diameter (3475 km) and means, Uranus axial
tilt angle above the moon surface should be = 90.6 deg
- Because Uranus orbital inclination =0.8 degrees and vertically =90.8 deg, so we
tried to make the angle (90.6 deg) = to (90.8 deg)! how?
- We cut inside the moon diameter because it consumes (0.5 deg) and reach to
smaller moon diameter (1390 km) which consumes only (0.2 deg) so the angle
above this smaller diameter surface is 90.9 deg.
- The rest diameter = 1042.5 km x 2, which we need to analyze deeply!
Paper conclusion
- Planet diameter is created based on this planet motion
- There's a transportation of motion between the Earth moon and Pluto.
- The planets do this motion transportation, that causes this motion features to be
seen on all planets motions
- The argument proves that, the planets can't be considered as separated masses
points, on the contrary they are knots in the same rope, where all solar planets are
created from the same one trajectory of energy and move by it.

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1- The Moon And Pluto Motions Transportations
1-1 The Moon diameter division 1-2 Planet diameter &velocity relationship analysis
1-3The period 1042.5 h analysis 1-4 Why the moon diameter is divided?
1-1 The Moon diameter division
Figure No. (1)
(The moon orbital triangle is inserted in the paper end to be our discussion reference)
Let's summarize this figure idea
- (BTM) is Uranus axial tilt line (97.8 degrees), and (0.5 degrees) of it is consumed
for the moon diameter, so the angle on (T) = 90.6 degrees
- We cut into the moon diameter to above the small circle (smaller moon) its
diameter =1390 km (r= 695 km) – it consumes only (0.2 degrees), and the angle
above it = 90.9 degrees, from which 90 deg (perpendicularity), 0.8 deg (Uranus
orbital inclination) and 0.1 deg (the change in Mars orbital inclination form 1.8
deg to 1.9 deg)
- We study here the motion transportation between the moon and Pluto, and we need
from this figure is the value 1042.5 km, where the rest diameter = 2085 km and
its radius = 1042.5 km.
- Why this division for the moon diameter is correct? Because
o 38041 km (Venus circumference) = 27.3 x 1390 km (where 27.3 days is the
moon orbital period) that tells us this division is correct because the value
1390 km is used as any other registered data.
o 12014 km (Venus diameter) = 695 km (r) x 17.4 (where 17.4 deg = the
inner planets orbital inclinations total), the tested data is used as the
registered one and because of that we know that, this data is created really
by the geometrical structure, and not invented numbers.
o The value 1042.5 hours is used geometrically for different planets motions,
and the planets diameters and circumferences are used as periods of time
frequently in the solar system – so this value 1042.5 km- is a crucial proof
that the moon diameter division is a correct one – and this paper dedicated a
complete point of discussion to analyze this value 1042.5 km.

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1-2 Planet Diameter And Velocity Relationship Analysis
- Let's discuss the following data:
- 142984 km = 13.1 km/sec (Jupiter velocity) x10921 seconds, this equation tells
that, Jupiter needs to move during 10921 seconds to pass a distance =142984 km
(= Jupiter diameter). But on the other side, the equation tells, the Earth moon needs
to rotate 13.1 times around its axis to pass by rotation a distance = 142984 km (the
Earth moon circumference 10921 km) (The moon needs 27.3 days to rotate around
its axis and by that the 13.1 rotations needs 13.1 x 27.3 = 358 days = 1 year)
- 51118 km = 6.8 km/sec (Uranus velocity)x 7511 seconds, this equation tells that,
Uranus needs to move during 7511 seconds to pass a distance =51118 km
(=Uranus diameter). But on the other side, the equation tells, Pluto needs to rotate
6.8 times around its axis to pass by rotation a distance = 51118 km (Pluto
circumference 7511 km) (Pluto 153.3 hours to rotate around its axis and by that the
6.8 rotations needs 6.8 x 153.3 = 1042.5 hours)
- 51118 km = 4.7 km/sec (Pluto velocity) x 10921 seconds, this equation tells that,
Pluto needs to move during 10921 seconds to pass a distance =51118 km
(=Uranus diameter). But on the other side, the equation tells, the Earth moon needs
to rotate 4.7 times around its axis to pass by rotation a distance = 51118 km (The
moon needs 27.3 days to rotate around its axis and by that the 4.7 rotations needs
4.7 x 27.3 = 128.3 days).
- 142984 km = 4.7 km/sec (Pluto velocity) x 30589 seconds, this equation tells
that, Pluto needs to move during 30589 seconds to pass a distance =142984 km
(=Jupiter diameter). But on the other side 30589 days = Uranus orbital period.
- 378674 km = 35 km/sec (Venus velocity) x 10921 seconds, this equation tells,
Venus needs to move during 10921 seconds to pass a distance =378674 km
(=Saturn Circumference). But on the other side, the equation tells, the Earth moon
needs to rotate 35 times around its axis to pass by rotation a distance = 378674 km
(the moon needs 27.3 day for each rotation, so for 35 it needs 35 x27.3 =955.5
days). (NOTE Please, Saturn diameter = Venus circumference x π, this
relationship will change our vision for the equation, because in this case Saturn
circumference can be considered as relating to Venus Circumference by the rate
(π2
) and that makes Venus is similar to the previous 2 planets, which use their
velocities to pass their own diameters or circumferences distances – but the close
relationship between Venus and Saturn causes the exchange of their data.
- 40080 km = 5.4 km/sec (Neptune velocity) x 7511 seconds (error 1%), this
equation tells that, Neptune needs to move during 7511 seconds to pass a distance
=40080 km (=Earth Circumference). But on the other side, the equation tells, Pluto
needs to rotate 5.4 times around its axis to pass by rotation a distance = 40080 km
(Pluto needs 153.3 hour for each rotation time, so for 5.4 it needs 5.4 x153.3
=827.8 hours =34.5 days).

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- 10921 km = 4.7 km/sec (Pluto velocity) x 2390 seconds (error 2.8%), this
equation tells, Pluto needs to move during 2390 seconds to pass a distance =10921
km (= the Moon Circumference). (where 2390 km = Pluto diameter).
- 2 x 160592 km = 29.8 km/sec (Earth velocity) x 10921 seconds (error 1.3%),
this equation tells, Earth needs to move during 10921 seconds to pass a distance =
2 x160592 km (2 Uranus Circumferences) (where10921 km= the moon Diameter).
on the other side, the moon needs to rotate around its axis 29.8 times to move a
distance = 2 Uranus Circumferences (where 29.8 times x 27.3 = 813.5 days)
- 2 x 38041 km = 6.8 km/sec (Uranus velocity) x 10921 seconds (error 2.4%), this
equation tells, Uranus needs to move during 10921 seconds to pass a distance = 2
x38041 km (2 Venus Circumferences) (where10921 km= the moon Diameter). on
the other side, the moon needs to rotate around its axis 6.8 times to move a
distance = 2 Venus Circumferences (where 6.8 times x 27.3 = 185.6 days)
- 2 x 120536 km = 4.7 km/sec (Pluto velocity) x 51118 seconds, this equation tells,
Pluto needs to move during 51118 seconds to pass a distance = 2 x 120536 km
(2 Saturn diameters) (where 51118 km = Uranus Diameter).
- (Please Note, Mercury day period needs 5040 seconds to be 176 solar day, but
(51118 seconds/5040 seconds) = (Mercury velocity / Pluto velocity). To add more
interesting information, Mercury moves during 5040 seconds a distance = 2 Saturn
diameters (error 1%). But Mars moves during (5040 seconds) a distance = Saturn
diameter!
- Also this equation can be seen as following
- 2 x 378674 km (Saturn circumference) = 4.7 km/s x 160592 seconds, (where
160592 km = Uranus circumference) – and the equation in this form is interesting
because Saturn circumference = Earth moon distance at total solar eclipse radius,
and the 2 values of it means the orbital diameter.
- And Uranus diameter is defined by Perigee radius as we have discussed before
(2.58 mkm Earth motion distance per solar day = 7.1 x 0.363 km Perigee radius,
and 0.363 mkm = 51118 km Uranus diameter).

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1-3The Period 1042.5 h analysis
- The period 1042.5 hours is created by Pluto rotation to move a distance = Uranus
diameter = 51118 km, because (51118 km = 6.8 x 7511 km) so Pluto needs to
rotate around its axis 6.8 times to pass the distance 5118 km and each rotation of
Pluto needs 153.3 hours, so the 6.8 rotations needs 1042.5 hours.
- But what about the 1042.5 seconds?
- During 1042.5 seconds Mercury moves a distance =49528 km = Neptune Diameter
- During 1042.5 s Mars moves a distance =25123 km = Uranus Radius (1.7%)
- During 1042.5 s Pluto moves a distance =4879 km = Mercury Diameter
- During 1042.5 s Uranus moves a distance =7070km = ancient Mars Diameter
- I try to show that, the motion is transported from planet to another in different
motion data – the value 1042.5 seconds (or hours or even km) it's some clear value
used in the transportation process and because of that, we notice it easily., the
other data express the motion transportation with more hidden description and
because of that we consider this data is independent from each other – but it's not
true.
Notice no. (1)
- One of the basic difficulties in the motion transportation explanation is that, the
planet data uses light motion features, as we have seen frequently… the distance
values are used as periods of time which is a light motion feature and can't be
explained by any planet motion – also – the different rates of time – while we deal
with 1042.5 hours we found here also 1042.5 seconds are working – such changes
in the rates of time can be done only by high velocity motion (light motion) – but
the feature is seen in planet motion and not in light motion and that causes a basic
difficulty for explanation – no one can say Earth moves by light velocity (0.3
mkm/sec) but the relativistic effects are observed in each data around – Earth
velocity is 29.8 km/sec but the light motion feature is found clearly –
- The point is that, light motion is accompanying with planet motion, but hidden
behind it.
Notice no. (2)
- Let's discuss the value 1042.5 km which we have found in the moo diameter in
following…

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The value 1042.5 km
Figure No. (1) (revision) ….Why the value 1042.5 km is specific in this figure?
Equation No. (a)
- The moon (rotation) to move (51118 km) needs 128.3 days BUT Pluto (rotation)
to move (51118 km) needs 1042.5 hours (=43.32 days) i.e. (128.3/43.3) = 3
Equation No. (b)
- 0.406 mkm (Pluto motion per solar day) /51118 km (Uranus diameter) = (346.6
days nodal year/ 43.32 days
- Equation No. (b) tells that, the nodal year is created by the effect of this value
(1042.5 hours) or it's produced by an effect of Uranus & Pluto motions interaction
on the moon orbit.
- That means, the moon orbit regression is created by Uranus & Pluto motions
interactions effect on the moon orbit…
- 97.8 degrees (Uranus axial tilt) = 5.1 degrees (the moon orbital inclination) x 19
- That means practically, Uranus effect creates the moon orbital inclination to be 5.1
degrees in comparison with 19 degrees (the moon orbit regresses yearly 19 deg)
- Also …17.4 degrees = 5.1 degrees x 3.4 degrees (Venus orbital inclination)
- But
- 17.2 degrees (Pluto orbital inclination) = 17.4 deg x 0.99
- i.e. Uranus and Pluto motions interaction effect on the moon orbit created the
moon orbital inclination (5.1 degrees) and this effect is done through Venus
contribution with them (please remember 12104 km Venus diameter =17.4 x 695
km (r))
- That again tells us, the moon orbital inclination is the most powerful data in the
moon motion
Equation No. (c)
- 1042.5 km = 5.1 x 205 km
- 205 km /sec = The Solar Planets Velocities Total
- The equation tells that, the value 1042.5 km transports the solar planets motions
energies total to the moon motion.
- That tells the moon motion expresses A Total Motion Of The Solar Planets

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1-4 Why the moon diameter is divided?
The figure shows cross section in the moon body, the moon diameters are divided into
4 diameters each one is greater its previous – their value as following
- The smallest diameter (the green ball) = 1390 km (R1)
- The next one (the blue circle) = 1492km (R2)
- The next one (the red circle) = 1604 km (R3)
- The greatest one (the black circle) = 3475 km (R4)
Why and how this division is done?
Where 449197 km = (Jupiter circumference)
A. 449197 km x 3475 km = 2 x 778.6 mkm (Jupiter Orbital Distance)
B. 449197 km x 1604 km = 720.7 mkm (Jupiter Mercury Distance)
C. 449197 km x 1492 km = 670 mkm (Jupiter Venus Distance)
D. 449197 km x 1390 km = 627 mkm (Jupiter Earth Distance)
Now let's ask (Can This Division Be A Real One?!)
- The data proves that, this division is a real one let's provide some proves to shows
that in following
o The radius 1604 km x π = 5040 km this value is belonged to Mercury
perfectly because Mercury day needs 5040 seconds to be 176 solar days and
we know that Mercury during its day period moves a distance =720.7 mkm
= Mercury Jupiter distance, that shows the equation no. (B) is created based
on geometrical mechanism clearly and strongly.
o The radius 1492 km = 47.4 kmx10 x π, where 47.4 km /sec is Mercury
velocity, and the value 10 is the rate between Mercury & Pluto velocities.
The equation shows that, Venus Jupiter distance is defined by a direct effect
of Mercury motion. and this is a logical data, Because Venus data is created
almost by Mercury motion – for example – Mercury moves during its
rotation period (58.66 solar days) a distance = 243 mkm but Venus rotation
period = 243 solar days. That shows also the equation no. (C) is created
based on geometrical mechanism.
o W know the last (smallest) value 1390 km is created based on geometrical
mechanism to produce the angle 90.9 degrees of Uranus axial tilt, means the
data is created – But let's ask
But by what mechanism these parts of the moon are related to these distances?

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Light Motion Effect
Equation No. (I)
0.3 mkm/sec x 6939.75 seconds = 2088 mkm (Jupiter Uranus Distance)
Equation No. (II)
0.3 mkm/sec x 2088 seconds = 627 mkm (Jupiter Earth Distance)
Equation No. (III)
1.16 mkm/sec x 627 seconds = 720.7 mkm (Jupiter Mercury Distance)
Equation No. (IV)
1.16 mkm/sec x 670 seconds = 778.6 mkm (Jupiter Orbital Distance)
Equation No. (V)
1.16 mkm/sec x 580 seconds = 670 mkm (Jupiter Venus Distance)
Where
0.3 mkm/sec (light known velocity)
1.16 mkm/sec (light supposed velocity)
Discussion
- How the previous description can be real or effective in the planets motions?
- Let's summarize the idea in following:
o The moon different radiuses or diameters values are used as periods of time
where 1 km = 1 second
o That means, the previous division of the moon diameters is a division for
periods of time
o Now where the periods of time are defined, the light will use it
o The light uses these periods of time for its motions and by light motion these
periods of time create defined distances from each other – and because of
that – these different radiuses are created in proportionality with different
distances
o I want to say that
o The solar system is a geometrical machine has one engine which is the light
motion – it creates the matter and space in proportionality and harmony with
the light motion – the matter does effect on this geometrical mechanism
almost through the time periods definition – why?
o Because the time is the matter cycle measurement, for that reason, the
matter defines its required time to do its cycle and based on that the light
does its motion.
A Conclusion
o The light motion is accompanying with planet motion and part of it.

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More Analysis For The Moon Diameter
- There's one more question is left behind – let's ask it Now
- What are the diameters of these radiuses (1604 km And value 1492 km)? and
these values are radiuses because we know the last (1390 km) is a diameter and its
radius = 695 km? let's answer this question in separated points
(1st
Point The Radius 1604 km)
- Let's start with the first radius (1604 km).. why we know this radius is a right one?
Why is it not invented data? Because
o 1604 km x π =5040 km
o Mercury Day Period (4222.6 hours) needs a period 5040 seconds to be 4224
hours or 176 solar days –
o As it's suggested, 1 km is used by light motion as 1 second – because of that
this value is real, specially because (1604 km x 449197 km =720.7 mkm)
Where mercury moves during its day period a distance = 720.7 mkm =
Mercury Jupiter distance
o So this consistency of many data in on direction pushes to us conclude that,
the value 1604 km causes the value 5040 sec in Mercury Day Period
o But
o Why we need 5040 sec x 2 =10080 seconds? Why we need the diameter of
this radius value?
o Because
(1) 720.7 million km = Mercury Jupiter distance and the orbital
diameter between the 2 planets = 720.7 mkm x 2
o More Important Reason
(2) Light known velocity (0.3mkm/sec) = Earth velocity x 10080 sec
that means, Earth motion during 10080 seconds is equivalent to light
motion during 1 second – now this rate is saved by Mercury data,
Mercury works as the gear which causes earth velocity to be equal
light velocity- and the rate between them is saved in Mercury Data for
the geometrical structure which operates the machine.
That means, the value 5040 seconds, is found basically to save a
velocity of light beam to be used by a planet and this planet is Earth,
so each 30 km per second Earth moves receives a support from
Mercury Geometrical machine by (100)2
times, to create a total
velocity = light known velocity (0.3mkm/sec).
o But we still need to know … how this parts of the moon are divided and
distributed geometrically? We need more deep vision based on this new
radius 1604km?

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The Moon Diameter Description:
- We have a new radius (1604km) and its diameter = 3208 km, so the rest part is
(the moon diameter 3475 km – 3208 km = 267 km)
- That means,
- This new radius diameter (1604km x 2) consumes almost the moon diameter and
the rest is only 267 km
- We accept that, the matter dimension is used as a period of time – so we have 267
km which can be used as 267 seconds
Equation No. (VI)
1.16 mkm/sec (light supposed velocity) x 267 seconds = 304 million km.
- This distance (304 mkm) starts from the sun to beyond Mars with 76.1 km
- This distance is so specific distance and very distinguished, because
o (a) 304 mkm = 88000 km (the moon daily displacement) x 3475 km
o (b) 304 mkm 0.8 (Uranus orbital inclination) = 243.2 million km
- Equation No. (a) tells that, the distance 304 million km is a distance produced by
the moon own data, the moon daily displacement (88000 km) x its diameter
(3475km), so the distance 304 mkm must be very familiar for the moon data
- Equation No. (b) tells some interesting data, because…let's the distance 304 mkm
from the sun creates a point (S) in Space – so from Mercury to this point (S) the
distance = 243 mkm (error 1.2%), that means, the rate (0.8) divided the distance
304 mkm into 2 distances, these 2 distances start from the point (S) and the longer
distance reaches to the sun (304 mkm) but the shorter distance reaches to Mercury
(a distance 243 mkm where mercury orbital distance =57.9mkm) –
- The previous explanation must be so important because
o Mercury moves during its rotation period (58.66 days) a distance =243 mkm
o Mars moves during Venus day period (116.75 days) a distance =243 mkm
o But
o 243 solar days = Venus Rotation Period
- That tells us clearly, the distance 304 mkm is a part of the geometrical machine by
which the moon does its orbital motion and because of that the values of the
motion diameter division be in consistency with this distance.
- Note Please, Mars orbital distance (227.9 mkm) = 3 x 76.1 mkm
- Where the rate 3 we have found before in the discussion of the distance 1042.5
mkm as a period 1042.5 hours (or seconds). Means this interaction is used her by
Mars to define its position relative to the other inner planets influences ranges.

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(2nd
Point The Diameter 1390 km)
- We remember the value 1290 days or degrees, but we can't find it here?
- Let's see how to find this distance 1290 km in the moon diameter division in
following (then we will remember the using of this value 1290 days)
We have 2 values which are
(1)
A Diameter 1390 Km The Moon Rest = 2085 Km
(2)
A Diameter 1290 Km The Moon Rest = 2185 Km
Notice (1)
- 1390 km = 2 x 695 km
- But
- 2185 km = π x 695 km
- We have seen a similar relationship before
- 2872.5 mkm Uranus orbital distance = 2x 1433.5 mkm (Saturn orbital distance)
- 4495.1 mkm Neptune orbital distance = π x 1433.5 mkm (Saturn orbital distance)
- Let's see the next equations
o 3208 km (the red diameter) = 2185 km x 1.461
o 2π x 346.6 km = 2185 km (error 0.4%)
Equation No. (VII)
1.16 mkm/sec (light supposed velocity) x 1390 seconds = 1612 million km.
- Where
- 1622.7 mkm (= Uranus Neptune Distance)
- Also
- The value 1612 mkm = 1 mkm x 1604 km (approximately)
- Where (1 mkm = the moon orbital diameter)
- Equation No. (VII) shows that the relationship between the moon parts is created
based on the light motion, because the radiuses are used as periods of time and
then based on these periods the light motion creates the distances and based on the
distances the time periods are created…

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- Let's remember the 1290 days using:
- I have suggested that, the moon moves a vertical motion… in this motion the
moon moves vertically without any horizontal displacement… based on that
- The moon will move from perigee (0.363 mkm) to apogee (0.406 mkm) vertically
and without horizontal displacement through the moon orbit – so in motion from
perigee to apogee (43000 km) and return back, the moon consumes a distance =
86000 km of its daily displacement =88000 km, so the rest is 2000 km
- Now we suppose that the moon will move this 2000 km through its orbit
horizontally
- The moon orbital circumference at apogee radius (r=0.406mkm), equal=2.58 mkm,
so if the moon moves horizontally through its orbit only 2000 km per solar day, so
the moon needs 1290 days to cover its orbital circumference (2.58 mkm), from this
description the value 1290 days is created.

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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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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.

I do this research
2nd
16
A Model For The Moon Motion 2 Orbits

I do this research
2nd
17
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

Uranus is perpendicular on earth moon orbit (viii) (revised)

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