of 1 14
A Quantum Analog For The Solar System!
Ian Beardsley!
March 7, 2026!
Abstract!
We find if consider the evolved state of the Solar System, that its quantum analog to the Bohr
atom is based on a characteristic time of one-second and the Earth’s Moon as the defining
metric.!
1.0 The Quantum Solution To The Solar System!
The ancient Sumerians (4500 BCE-1900 BCE) used base 60 counting, and divided the Earth
day into 24 hours. The ancient Egyptians (3100 BCE-30 BCE) divided the Earth day into 24
hours as well. Since they both divided the day into 12 hours, and the night into 12 hours and, in
the winter, the night is longer than the day and in the summer, the day is longer than the night,
the hours in a day, or night, can be longer or shorter depending on the time of the year. The
ancient Greeks took the 24 hour day from the ancient Egyptians (Hipparchus, 190 BCE-120
BCE) and and used an hour to be represented by the equinoxes when day equals night,
inventing the equinoctial hour. It was Christiaan Huygens (1629-1695) who took the hour that
had been divided up into 60 minutes, with each minute divided into 60 seconds, from the
ancient Sumerian base 60 counting, and built the first pendulum clock that could measure
down to the second accurately. This was fueled by the need of Newton’s (1642-1727) world
view for gravity and mechanics that needed to measure time down to a unit as small as a
second.!
It is an interesting phenomenon that the Moon near perfectly eclipse the Sun. The eclipse ratio
that allow for this is about 400:!
1. !
where is the radius of the Sun and is the radius of the Moon. is the orbit radius of the
Earth orbit and is the orbital radius of the Moon. The solar radius is about 400 times the
lunar radius; the Earth-Sun distance is about 400 times the Earth-Moon distance.!
The number of seconds in a day are given approximately by:!
2. 86,400 seconds/day=(24 hours)(60 minutes)(60 seconds)!
The number of seconds in a day, 86 400, can be factored as:!
3. 86,400=(6)(6)(6)(400)!
The factor 400 is the eclipse ratio. The factor (216) relates to sixfold symmetry, hexagonal
tiling, and the approximation used by Archimedes as his starting point for calculating .
The appearance of 86 400 in ancient timekeeping thus incorporates the eclipse ratio, whether
by accident or by design.!
R
⊙
R
m
≈ 400 and
r
⊕
r
m
≈ 400
R
⊙
R
m
r
⊕
r
m
6
3
π ≈ 3
of 2 14
Let us suggest that the kinetic energy of the Moon to the kinetic energy of the Earth maps the
24 hour (Earth rotation period) day into 1 second, our basis unit of measuring time:!
4. !
Where is the inclination of the Earth to its orbit.!
!
!
!
Using average orbital velocities. We can get closer to a second using aphelions and perihelions
and perigees and apogees.!
The Moon stabilizes Earth's axial tilt:!
!
!
The Moon stabilizing the Earth’s tilt to its orbit prevents extreme hot and cold on Earth and
allows for the seasons. As such the Moon is key to optimizing conditions for life on the planet.
Perhaps making it possible for intelligent life to evolve.!
We write:!
5. !
We take to be given by:!
6. !
Equation 6 is an approximately 1-second expression for the radius and mass of a proton that
uses a 2/3 fibonacci approximation for , discovered by the author. Thus we see we can see a
possible 1-second invariant that may exist across vast scales from atoms to the Solar System.
We have!
7. !
!
Using Earth’s orbital velocity at perihelion.!
KE
moon
KE
earth
(24#hours)cos(θ ) = 1#second
θ = 23.5
∘
KE
moon
= (7.347673E 22kg)(1022m /s)
2
= 7.6745E 28J
KE
earth
= (5.9722E 24kg)(29,800m /s)
2
= 5.30355E33J
7.6745E 28J
5.30355E33JJ
(86,400s)(cos(23.5
∘
) = 1.1466#seconds ≈ #1#second
θ = 23.5
∘
±
1.3
∘
(with#Moon)
θ = 0
∘
#to#85
∘
(without#Moon,#chaotic)
ℏ
⊙
= (1secon d )KE
earth
ℏ
⊙
1.03351s =
1
3
⋅
h
α
2
c
2
3
⋅
πr
p
Gm
3
p
ϕ
ℏ
⊙
= (1.03351s)(2.7396E33J ) = 2.8314E 33J ⋅ s
KE
Earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E33J
of 3 14
The ground state energy for a hydrogen atom (One electron orbiting a proton) is:!
8. !
For the planetary system we would replace (Coulombs’s constant) with (Newton’s
universal constant of gravity). The product of (the charge of an electron squared) and (the
mass of an electron) become a mass cubed. We will choose the mass of the Moon, . We
have the ground state equation is:!
9. !
10. !
Where we have converted meters to seconds by measuring distance in terms of time with the
speed of light ( ). We see the mass of the Moon maps the kinetic energy of the Earth over one
second to 1 second. The Moon is the metric.!
The solution for the orbit of the Earth around Sun with the Schrödinger wave equation can be
inferred from the solution for an electron around a proton in the a hydrogen atom with the
Schrödinger wave equation. The Schrödinger wave equation is, in spherical coordinates!
!
!
Its solution for the atom is as guessed by Niels Bohr before the wave equation existed:!
11. !
12. !
is the energy for an electron orbiting protons and , is the orbital shell for an electron with
protons, the orbital number. I find the solution for the Earth around the Sun utilizes the
Moon around the Earth. This is different than with the atom because planets and moons are
not all the same size and mass like electrons and protons are, and they don’t jump from orbit to
orbit like electrons do. I find that for the Earth around the Sun!
13. !
E = −
ℏ
2
k
e
e
2
m
e
k
e
G
e
2
m
e
M
m
ℏ
2
⊙
GM
3
m
=
(2.8314E33)
2
(6.67408E − 11)(7.34763E 22kg)
3
= 3.0281E8m
ℏ
2
⊙
GM
3
m
1
c
=
3.0281E8m
299,792,458m /s
= 1.010#seconds ≈ 1second
c
−
ℏ
2
2m
[
1
r
2
∂
∂r
(
r
2
∂
∂r
)
+
1
r
2
sinθ
∂
∂θ
(
sinθ
∂
∂θ
)
+
1
r
2
sin
2
θ
∂
2
∂ϕ
2
]
ψ + V(r)ψ = E ψ
E
n
= −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
r
n
=
n
2
ℏ
2
Zk
e
e
2
m
e
E
n
Z
r
n
Z
n
E
n
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
of 4 14
14. !
is the energy of the Earth, and is the planet’s orbit. is the radius of the Sun, is the
radius of the Moon’s orbit, is the mass of the Earth, is the mass of the Moon, is the
orbit number of the Earth which is 3 and is the Planck constant for the solar system.
Instead of having protons, we have the radius of the Sun normalized by the radius of
the Moon. We see that the Moon is indeed the metric, as we said before.!
!
=!
=2.727E33J!
The kinetic energy of the Earth is (using orbital velocity at perihelion):!
!
!
The kinetic energy of the Earth is about equal to the energy of the system, because the orbit of
the Earth is nearly circular. That is!
!
The whole object of developing a theory for the way planetary systems form is that they meet
the following criterion: They predict the Titius-Bode rule for the distribution of the planets; the
distribution gives the planetary orbital periods from Newton’s Universal Law of Gravitation. The
distribution of the planets is chiefly predicted by three factors: The inward forces of gravity
from the parent star, the outward pressure gradient from the stellar production of radiation, and
the outward inertial forces as a cloud collapses into a flat disc around the central star. These
forces separate the flat disc into rings, agglomerations of material, each ring from which a
different planet forms at its central distance from the star. In a theory of planetary formation
from a primordial disc, it should predict the Titius-Bode rule for the distribution of planets
today, which was the distribution of the rings from which the planets formed.!
Also, the Earth has been in the habitable zone since 4 billion years ago when it was at 0.9 AU.
Today it is at 1AU, and that habitable zone can continue to 1.2 AU. So we can speak of the
distance to the Earth over much time. The Earth and Sun formed about 4.6 billion years ago.
As the Sun very slowly loses mass over millions of years as it burns fuel doing fusion, the Earth
slips minimally further out in its orbit over long periods of time. The Earth orbit increases by
r
n
=
2ℏ
2
⊙
GM
3
m
⋅
R
⊙
R
m
⋅
1
n
E
3
r
n
R
⊙
r
m
M
e
M
m
n
ℏ
⊙
Z
R
⊙
/R
m
R
⊙
R
m
=
6.96E8m
1737400m
= 400.5986
E
3
= (1.732)(400.5986)
(6.67408E − 11)
2
(5.972E 24kg)
2
(7.347673E 22kg)
3
2(2.8314E33)
2
KE
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E33J
2.727E33J
2.7396E33J
100 = 99.5 %
E
3
∼ KE
earth
of 5 14
about 0.015 meters per year. The Sun only loses 0.00007% of its mass annually. The Earth is at
1AU=1.496E11m. We have 0.015m/1.496E11m/AU=1.00267E-13AU. So,!
!
The Earth will only move out one ten thousandth of an AU in a billion years. Anatomically
modern humans have only been around for about three hundred thousand years. Civilization
began only about six thousand years ago.!
The Moon slows the Earth rotation and this in turn expands the Moon’s orbit, so it is getting
larger, the Earth loses energy to the Moon. The Earth day gets longer by 0.0067 hours per
million years, and the Moon’s orbit gets 3.78 cm larger per year.!
We suggest the Solar system comes into phase with a possible one second invariant when the
Earth-Sun separation, and Earth-Moon separation, have kinetic energies whose ratio maps the
24 hour day into the 1-second base unit as given by equation 4: !
4. !
That is is when equations 5 and 10 hold:!
5. !
10. !
Something remains to be done. Is there something about the Sun that is common to other
types of stars; stars that are perhaps larger and hotter than the Sun, or perhaps smaller and
cooler, or a different color, like blue or red, instead of yellow? The answer is yes. I actually
found something in ancient Vedic knowledge, in the Hindu traditions. Apparently, in Hindu yoga
the number 108 is an important number. I read that yogis today noticed that the diameter of the
Sun is about 108 times the diameter of the Earth and that the average distance from the Sun to
the Earth is about 108 solar diameters, with 108 being a significant number in yoga. So I wrote
the equivalent:!
15. , !
or for any star and habitable planet: !
16. !
radius of the star. the orbital radius of the habitable planet. We consider the HR
diagram that plots temperature versus luminosity of stars. We see the O, B, A stars are the
more luminous stars, which is because they are bigger and more massive and the the F, G stars
are medium luminosity, mass, and size (radius). Our Sun is a G star, particularly G2V, the two
(1.00267E − 13AU/year)(1E 9years) = 0.0001AU
KE
moon
KE
earth
(24#hours)cos(θ ) = 1#second
ℏ
⊙
= (1secon d )KE
earth
ℏ
2
⊙
GM
3
m
1
c
=
3.0281E8m
299,792,458m /s
= 1.010#seconds ≈ 1second
R
⊕
= 2
R
2
⊙
r
⊕
R
planet
= 2
R
2
⋆
r
habi table
R
⋆
r
habi table
of 6 14
because the spectral classes are divided up in to 10 sizes, V for five meaning main sequence,
that it is part of the S shaped curve and is in the phase where the star is burning hydrogen fuel,
its original fuel, not the by products. And the K and M stars are the coolest, least massive, least
luminous.!
Let us consider the habitable zones of different kinds of stars. In order to get , the
distance of the habitable planet from the star, we use the inverse square law for luminosity of
the star. If the Earth is in the habitable zone, and if the star is one hundred times brighter than
the Sun, then by the inverse square law the distance to the habitable zone of the planet is 10
times that of what the Earth is from the Sun. Thus we have in astronomical units the habitable
zone of a star is given by:!
17. !
the luminosity of the star, the luminosity of the Sun. AU the average Earth-Sun
separation, which is 1. The surprising result I found was, after applying equation 4,
hypothetically predicting the size of a habitable planet, to the stars of all spectral types from F
through K, with their different radii and luminosities (the luminosities determine , the
distances to the habitable zones), that the radius of the planet always came out about the
same, about the radius of the Earth. This may suggest optimally habitable planets are not just a
function of their distance from the star, which is a big factor in determining their temperature,
but are functions of their size and mass meaning the size of the Earth could be good for life
chemistry and atmospheric composition, and gravity. Stars of the same particular luminosities,
temperatures and colors have about the same mass and size (radius). Here are some examples
of such calculations of stars of different sizes, colors, and luminosities using equation 4:!
F8V Star!
Mass: 1.18!
Radius: 1.221!
Luminosity: 1.95!
!
!
!
!
F9V Star!
Mass: 1.13!
Radius: 1.167!
Luminosity: 1.66!
!
!
!
r
habi table
r
habi table
=
L
⋆
L
⊙
AU
L
⋆
L
⊙
r
habi table
M
⋆
= 1.18(1.9891E30kg) = 2.347E 30kg
R
⋆
= 1.221(6.9634E8m) = 8.5023E8m
r
p
= 1.95L
⊙
AU = 1.3964AU(1.496E11m /AU ) = 2.08905E11m
R
p
=
2R
2
⋆
r
p
= 2
(8.5023E8m)
2
2.08905E11m
=
6.92076E6m
6.378E6m
= 1.0851Ear th Ra dii
M
⋆
= 1.13(1.9891E30kg) = 2.247683E 30kg
R
⋆
= 1.167(6.9634E8m) = 8.1262878E8m
r
p
= 1.66L
⊙
AU = 1.28841AU(1.496E11m /AU ) = 1.92746E11m
of 7 14
!
G0V Star!
Mass: 1.06!
Radius: 1.100!
Luminosity: 1.35!
!
!
!
!
G1V Star!
Mass: 1.03!
Radius: 1.060!
Luminosity: 1.20!
!
!
!
!
As you can see we consistently get about 1 Earth radius for the radius of every planet in the
habitable zone of each type of star. I have gone through all stars from spectral class A stars to
spectral class M stars and consistency got this result. It may be this radius for a planet is
optimal for life, in particular intelligent life, because given we might, for that, need a material
composition similar to that of Earth, and, in turn, an Earth-like gravity for the right atmosphere,
including atmospheric composition, or planetary mass, the planet might need to be around this
size.!
2.0 The Solar Solution
Our solution of the wave equation for the planets gives the kinetic energy of the Earth from the
mass of the Moon orbiting the Earth, but you could formulate based on the Earth orbiting the
Sun. In our lunar formulation we had:!
1. !
We remember the Moon perfectly eclipses the Sun which is to say!
2. !
R
p
=
2R
2
⋆
r
p
= 2
(8.1262878E8m)
2
1.92746E11m
=
6.852184E6m
6.378E6m
= 1.0743468Ear th Ra dii
M
⋆
= 1.06(1.9891E30kg) = 2.108446E 30kg
R
⋆
= 1.100(6.9634E8m) = 7.65974E8m
r
p
= 1.35L
⊙
AU = 1.161895AU(1.496E11m /AU ) = 1.7382E11m
R
p
=
2R
2
⋆
r
p
= 2
7.65974E8m)
2
1.7382E11m
=
6.751E6m
6.378E6m
= 1.05848Ear th Ra dii
M
⋆
= 1.03(1.9891E30kg) = 2.11E 30kg
R
⋆
= 1.060(6.9634E8m) = 7.381E8m
r
p
= 1.20L
⊙
AU = 1.0954AU(1.496E11m /AU ) = 1.63878589E11m
R
p
=
2R
2
⋆
r
p
= 2
7.3812E 8m)
2
1.63878589E11m
=
6.6491E6m
6.378E6m
= 1.0425Ear th Ra dii
KE
e
= 3
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
R
⊙
R
m
=
r
e
r
m
of 8 14
Thus equation 1 becomes!
3. !
The kinetic energy of the Earth is!
4. !
Putting this in equation 3 gives the mass of the Sun:!
5. !
We recognize that the orbital velocity of the Moon is!
6. !
So equation 5 becomes!
7. !
This gives the mass of the Moon is!
8. !
Putting this in equation 1 yields!
9. !
We now multiply through by and we have!
10. !
The Planck constant for the Sun, , we will call , the subscript for Planck. We have!
KE
e
= 3
r
e
r
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
KE
e
=
1
2
⋅
GM
⊙
M
e
r
e
M
⊙
= 3r
2
e
⋅
GM
e
r
m
⋅
M
3
m
ℏ
2
⊙
v
2
m
=
GM
e
r
m
M
⊙
= 3r
2
e
⋅ v
2
m
⋅
M
3
m
ℏ
2
⊙
M
3
m
=
M
⊙
ℏ
2
⊙
3r
2
e
v
2
m
KE
e
=
R
⊙
R
m
⋅
G
2
M
2
e
M
⊙
2r
2
e
v
2
m
M
2
e
/M
2
e
KE
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2r
2
e
v
2
m
M
2
e
ℏ
⊙
L
p
p
of 9 14
!
!
We write for the solution of the Earth/Sun system:!
11. !
We can write 11 as!
12. !
Where we say!
!
!
Let us see how accurate our equation is:!
!
= !
= !
!
!
We have that the kinetic energy of the Earth is!
!
Our equation has an accuracy of!
L
p
= r
e
v
m
M
e
= r
e
v
m
M
e
= (1.496E11m)(1022m /s)(5.972E 24kg) = 9.13E 38kg ⋅
m
2
s
L
2
p
= r
2
e
v
2
m
M
2
e
= 7.4483E 77J ⋅ m
2
⋅ kg = 8.3367E 77kg
2
⋅
m
4
s
2
KE
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
KE
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
ℏ
⊙
= 9.13E38J ⋅ s
h
⊙
= 2π ℏ
⊙
= 5.7365E39J ⋅ s
KE
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
R
⊙
R
m
(6.67408E − 11)
2
(5.972E 24kg)
4
(1.9891E30kg)
2(8.3367E 77kg
2
⋅
m
4
s
2
)
R
⊙
R
m
(6.759E30J )
R
⊙
R
m
=
6.957E8m
1737400m
= 400.426
KE
e
= 2.70655E33J
KE
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E33J
of 10 14
!
Which is very good.!
Let us equate the lunar and solar formulations:!
!
!
!
This gives:!
13. !
We remember that!
!
!
And since, !
14. !
!
Equation 14 becomes!
2.70655E33J
2.7396E33J
= 98.79 %
KE
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
KE
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
3
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
L
p
=
M
2
e
M
⊙
M
3
m
3
⋅ ℏ
⊙
ℏ
⊙
= (hC )K E
e
hC = 1secon d
KE
e
=
1
2
M
e
v
2
e
2v
m
=
v
2
e
r
e
(1secon d )
M
2
e
M
⊙
M
3
m
3
M
2
e
M
⊙
M
3
m
3
=
(5.972E 24kg)
2
(1.9891E30kg)
(7.34763E 22kg)
3
(1.732)
= 321,331.459 ≈ 321,331
of 11 14
15. !
The condition of a perfect eclipse gives us another expression for the base unit of a second.
is another version of the Planck Constant, which is intrinsic to the the solar formulation as
opposed to the lunar formulation. We want to see what the ground state looks like and what its
characteristic time is, if it is 1 second like it is for the lunar formulation. Looking at the equation
for energy:!
!
We see the ground state should be:!
16. !
And, it is equal to 1 second. You will notice where in the derivation for the energy we lost
, we have to put it in the ground state equation. The computation is:!
!
3.0 Jupiter and Saturn !
We want to find what the wave equation solutions are for Jupiter and Saturn because they
significantly carry the majority of the mass of the solar system and thus should embody most
clearly the dynamics of the wave solution to the Solar System. We also show here how well the
solution for the Earth works, which is 99.5% accuracy.!
I find that as we cross the asteroid belt leaving behind the terrestrial planets, which are solid,
and go to the gas giants and ice giants, the atomic number is no longer squared and the
square root of the the orbital number moves from the numerator to the denominator. I believe
this is because the solar system here should be modeled in two parts, just as it is in theories of
solar system formation because there is a force other than just gravity of the Sun at work,
which is the radiation pressure of the Sun, which is what separates it into two parts, the
terrestrial planets on this side of the asteroid belt and the gas giants on the other side of the
asteroid belt. The effect the radiation pressure has is to blow the lighter elements out beyond
the asteroid belt when the solar system forms, which are gases such as hydrogen and helium,
while the heavier elements are too heavy to be blown out from the inside of the asteroid belt,
allowing for the formation of the terrestrial planets Venus, Earth, and Mars. The result is that our
equation has the atomic number of the heavier metals such as calcium for the Earth, while the
equation for the gas giants has the atomic numbers of the gasses. We write for these planets!
!
1secon d = 2r
e
v
m
v
2
e
M
3
m
3
M
2
e
M
⊙
L
p
KE
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
L
2
p
GM
2
e
M
⊙
⋅
3
c
= 1secon d
n = 3
(9.13E38J ⋅ s)
2
(6.674E − 11)(5.972E 24kg)
2
(1.989E30kg)
⋅
3
c
= 1.0172secon ds
E =
Z
n
G
2
M
2
m
3
2ℏ
2
⊙
of 12 14
So, for Jupiter we have (And again using the maximum orbital velocity which is at perihelion):!
!
!
!
!
Jupiter is mostly composed of hydrogen gas, and secondly helium gas, so it is appropriate that
. !
Our equation for Jupiter is!
!
Where is the atomic number of hydrogen which is 1 proton, and for the orbital
number of Jupiter, . Now we move on to Saturn…!
!
!
= !
!
!
The equation for Saturn is then!
!
KE
j
=
1
2
(1.89813E 27kg)(13720m /s)
2
= 1.7865E35J
E =
Z
H
5
(6.67408E − 11)
2
(1.89813E 27kg)
2
(7.347673E 22kg)
3
2(2.8314E33)
2
E =
Z
H
5
(3.971E35J ) = Z
H
(1.776E35J )
Z
H
=
1.7865E35J
1.776E35J
= 1.006protons ≈ 1.0protons = hydrogen(H )
Z = Z
H
E
5
=
Z
H
5
G
2
M
2
j
M
3
m
2ℏ
2
⊙
Z
H
n = 5
n = 5
KE
S
=
1
2
(5.683E 26kg)(10140m /s)
2
= 2.92E 34J
E =
Z
6
(6.67408E − 11)
2
(5.683E 26kg)
2
(7.347673E 22)
3
2(2.8314E33)
2
Z
2.45
(3.5588E34J ) = Z(1.45259E34J )
Z(1.45259E 34J ) = (2.92E 34J )
Z = 2proton s = Helium(H e)
E
6
=
Z
He
6
G
2
M
2
s
M
3
m
2ℏ
2
⊙
of 13 14
It is nice that that Saturn would use Helium in the equation because Saturn is the next planet
after Jupiter and Jupiter uses hydrogen, and helium is the next element after hydrogen. As well,
just like Jupiter, Saturn is primarily composed of hydrogen and helium gas.!
The accuracy for Earth orbit is!
!
!
=!
=2.727E33J!
The kinetic energy of the Earth is!
!
!
Which is very good, about 100% accuracy for all practical purposes. The elemental expression
of the solution for the Earth would be!
!
Where!
!
In this case the element associated with the Earth is calcium which is Z=20 protons.!
References!
1. Beardsley, I. (2025) Theory For The Solar System And The Atom’s Proton; Linking
Microscales %
To Macroscales!
2. Beardsley, I. (2026) How Physics and Archaeology Point to a Natural Constant of 1-Second!
3. Beardsley, I. (2026) The Sublime and Mysterious Place of Humans in the Cosmos; A Work
in Exoarchaeology
E
n
= n
R
⊙
R
m
G
2
M
2
e
M
3
m
2ℏ
2
⊙
R
⊙
R
m
=
6.96E8m
1737400m
= 400.5986
E
3
= (1.732)(400.5986)
(6.67408E − 11)
2
(5.972E 24kg)
2
(7.347673E 22kg)
3
2(2.8314E33)
2
KE
e
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E33J
2.727E33J
2.7396E33J
100 = 99.5 %
E
3
= 3
Z
2
Ca
G
2
M
2
e
M
3
m
2ℏ
2
⊙
R
⊙
R
m
→ Z
2
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