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:
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:
The number of seconds in a day, 86 400, can be factored as:
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.
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:
1.1
R
⊙
R
m
≈ 400 and
r
⊕
r
m
≈ 400
R
⊙
R
m
r
⊕
r
m
1.2 86,400 seconds/day = (24 hours)(60 minutes)(60 seconds)
1.3 86,400 = (6)(6)(6)(400)
6
3
π ≈ 3
π
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 form a Planck-type constant for the Solar System:
We take to be given by:
Equation 6 is an approximately 1-second expression for the radius and mass of a proton that uses
a 2/3 fibonacci approximation for $\phi$, 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
Using Earth's orbital velocity at perihelion.
The ground state energy for a hydrogen atom (One electron orbiting a proton) is:
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:
1.4
K E
moon
K E
earth
(24 hours)cos(θ ) = 1 second
θ = 23.5
∘
K E
earth
= (5.9722E 24 kg)(29,800 m/s)
2
= 5.30355E 33 J
7.6745E28 J
5.30355E33 J
(86,400 s)cos(23.5
∘
) = 1.1466 seconds ≈ 1 second
θ = 23.5
∘
±
1.3
∘
(with Moon)
θ = 0
∘
to 85
∘
(without Moon, chaotic)
1.5 ℏ
⊙
= (1 second) K E
earth
ℏ
⊙
1.6 1.03351 s =
1
3
⋅
h
α
2
c
2
3
⋅
π r
p
Gm
3
p
1.7 ℏ
⊙
= (1.03351 s)(2.7396E 33 J) = 2.8314E 33 J ⋅ s
K E
Earth
=
1
2
(5.972E 24 kg)(30,290 m/s)
2
= 2.7396E 33 J
1.8 r
1
= −
ℏ
2
k
e
e
2
m
e
k
e
G
e
2
m
e
M
m
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:
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
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.
The kinetic energy of the Earth is (using orbital velocity at perihelion):
1.9
ℏ
2
⊙
GM
3
m
=
(2.8314E 33)
2
(6.67408E − 11)(7.34763E 22 kg)
3
= 3.0281E8 m
1.10
ℏ
2
⊙
GM
3
m
1
c
=
3.0281E8 m
299,792,458 m/s
= 1.010 seconds ≈ 1 second
c
−
ℏ
2
2m
[
1
r
2
∂
∂r
(
r
2
∂
∂r
)
+
1
r
2
sin θ
∂
∂θ
(
sin θ
∂
∂θ
)
+
1
r
2
sin
2
θ
∂
2
∂ϕ
2
]
ψ + V(r)ψ = E ψ
1.11 E
n
= −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
1.12 r
n
=
n
2
ℏ
2
Zk
e
e
2
m
e
E
n
Z
r
n
Z
Z
n
1.13 E
n
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
1.14 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
=
6.96E8 m
1737400 m
= 400.5986
E
3
= (1.732)(400.5986)
(6.67408E − 11)
2
(5.972E24 kg)
2
(7.347673E22 kg)
3
2(2.8314E33)
2
=
= 2.727E 33 J
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
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:
That is is when equations 5 and 10 hold:
K E
earth
=
1
2
(5.972E 24 kg)(30,290 m/s)
2
= 2.7396E 33 J
2.727E33 J
2.7396E33 J
100 = 99.5 %
E
3
∼ K E
earth
(1.00267E − 13 AU/year)(1E 9 years) = 0.0001 AU
1.4
K E
moon
K E
earth
(24 hours)cos(θ ) = 1 second
1.5 ℏ
⊙
= (1 second)K E
earth
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:
or for any star and habitable planet:
the 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
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:
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
1.10
ℏ
2
⊙
GM
3
m
1
c
= 1.010 seconds ≈ 1 second
1.15 R
⊕
= 2
R
2
⊙
r
⊕
,
1.16 R
planet
= 2
R
2
⋆
r
habitable
R
⋆
r
habitable
r
habitable
1.17 r
habitable
=
L
⋆
L
⊙
AU
L
⋆
L
⊙
r
habitable
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
G0V Star
Mass: 1.06
Radius: 1.100
Luminosity: 1.35
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
M
⋆
= 1.18(1.9891E 30 kg) = 2.347E 30 kg
R
⋆
= 1.221(6.9634E8 m) = 8.5023E8 m
r
p
= 1.95L
⊙
AU = 1.3964 AU (1.496E11 m/AU) = 2.08905E11 m
R
p
=
2R
2
⋆
r
p
= 2
(8.5023E8 m)
2
2.08905E11 m
=
6.92076E6 m
6.378E6 m
= 1.0851 EarthRadii
M
⋆
= 1.13(1.9891E 30 kg) = 2.247683E 30 kg
R
⋆
= 1.167(6.9634E8 m) = 8.1262878E8 m
r
p
= 1.66 AU = 1.28841 AU (1.496E11 m/AU) = 1.92746E11 m
R
p
=
2R
2
⋆
r
p
= 2
(8.1262878E8 m)
2
1.92746E11 m
=
6.852184E6 m
6.378E6 m
= 1.0743468 EarthRadii
M
⋆
= 1.06(1.9891E 30 kg) = 2.108446E 30 kg
R
⋆
= 1.100(6.9634E8 m) = 7.65974E8 m
r
p
= 1.35 AU = 1.161895 AU (1.496E11 m/AU) = 1.7382E11 m
R
p
=
2R
2
⋆
r
p
= 2
(7.65974E8 m)
2
1.7382E11 m
=
6.751E6 m
6.378E6 m
= 1.05848 EarthRadii
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:
We remember the Moon perfectly eclipses the Sun which is to say
Thus equation 2.1 becomes
The kinetic energy of the Earth is
Putting this in equation 2.3 gives the mass of the Sun:
We recognize that the orbital velocity of the Moon is
So equation 2.5 becomes
This gives the mass of the Moon is
Putting this in equation 2.1 yields
2.1 KE
e
= 3
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
2.2
R
⊙
R
m
=
r
e
r
m
2.3 KE
e
= 3
r
e
r
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
2.4 KE
e
=
1
2
⋅
GM
⊙
M
e
r
e
2.5 M
⊙
= 3 r
2
e
⋅
GM
e
r
m
⋅
M
3
m
ℏ
2
⊙
2.6 v
2
m
=
GM
e
r
m
2.7 M
⊙
= 3 r
2
e
⋅ v
2
m
⋅
M
3
m
ℏ
2
⊙
2.8 M
3
m
=
M
⊙
ℏ
2
⊙
3 r
2
e
v
2
m
2.9 KE
e
=
R
⊙
R
m
⋅
G
2
M
2
e
M
⊙
2 r
2
e
v
2
m
We now multiply through by and we have
The Planck constant for the Sun, , we will call , the subscript for Planck. We have
We write for the solution of the Earth/Sun system:
We can write 2.11 as
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
Which is very good.
Let us equate the lunar and solar formulations:
M
2
e
/M
2
e
2.10 KE
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2 r
2
e
v
2
m
M
2
e
ℏ
⊙
L
p
p
L
p
= r
e
v
m
M
e
= (1.496E11 m)(1022 m/s)(5.972E 24 kg) = 9.13E 38 kg ⋅
m
2
s
L
2
p
= r
2
e
v
2
m
M
2
e
= 7.4483E 77 J ⋅ m
2
⋅ kg = 8.3367E 77 kg
2
⋅
m
4
s
2
2.11 KE
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
2.12 K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
ℏ
⊙
= 9.13E 38 J ⋅ s
h
⊙
= 2π ℏ
⊙
= 5.7365E 39 J ⋅ s
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
=
R
⊙
R
m
(6.67408E − 11)
2
(5.972E24 kg)
4
(1.9891E30 kg)
2(8.3367E 77 kg
2
⋅
m
4
s
2
)
=
R
⊙
R
m
(6.759E 30 J)
R
⊙
R
m
=
6.957E8 m
1737400 m
= 400.426
K E
e
= 2.70655E 33 J
K E
earth
=
1
2
(5.972E 24 kg)(30,290 m/s)
2
= 2.7396E 33 J
2.70655E33 J
2.7396E33 J
= 98.79 %
This gives:
We remember that
And since,
Equation 2.14 becomes
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 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:
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:
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
K E
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
2.13 L
p
=
M
2
e
M
⊙
M
3
m
3
⋅ ℏ
⊙
ℏ
⊙
= (hC ) K E
e
hC = 1 second
K E
e
=
1
2
M
e
v
2
e
2.14 2v
m
=
v
2
e
r
e
(1 second)
M
2
e
M
⊙
M
3
m
3
M
2
e
M
⊙
M
3
m
3
=
(5.972E24 kg)
2
(1.9891E30 kg)
(7.34763E22 kg)
3
(1.732)
= 321,331.459 ≈ 321,331
2.15 1 second = 2r
e
v
m
v
2
e
M
3
m
3
M
2
e
M
⊙
L
p
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
2.16
L
2
p
GM
2
e
M
⊙
⋅
3
c
= 1 second
n = 3
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 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
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, $n=5$. Now we move on to Saturn...
(9.13E38 J ⋅ s)
2
(6.674E − 11)(5.972E24 kg)
2
(1.989E30 kg)
⋅
3
c
= 1.0172 seconds
E =
Z
n
G
2
M
2
m
3
2ℏ
2
⊙
K E
j
=
1
2
(1.89813E 27 kg)(13720 m/s)
2
= 1.7865E 35 J
E =
Z
H
5
(6.67408E − 11)
2
(1.89813E27 kg)
2
(7.347673E22 kg)
3
2(2.8314E33)
2
E =
Z
H
5
(3.971E 35 J) = Z
H
(1.776E 35 J)
Z
H
=
1.7865E35 J
1.776E35 J
= 1.006 protons ≈ 1.0 protons = hydrogen (H)
Z = Z
H
E
5
=
Z
H
5
G
2
M
2
j
M
3
m
2ℏ
2
⊙
Z
H
n = 5
The equation for Saturn is then
It is nice 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
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 protons.
K E
S
=
1
2
(5.683E 26 kg)(10140 m/s)
2
= 2.92E 34 J
E =
Z
6
(6.67408E − 11)
2
(5.683E26 kg)
2
(7.347673E22)
3
2(2.8314E33)
2
=
Z
2.45
(3.5588E 34 J) = Z(1.45259E 34 J)
Z(1.45259E 34 J) = (2.92E 34 J)
Z = 2 protons = Helium (He)
E
6
=
Z
He
6
G
2
M
2
s
M
3
m
2ℏ
2
⊙
E
n
= n
R
⊙
R
m
G
2
M
2
e
M
3
m
2ℏ
2
⊙
R
⊙
R
m
=
6.96E8 m
1737400 m
= 400.5986
E
3
= (1.732)(400.5986)
(6.67408E − 11)
2
(5.972E24 kg)
2
(7.347673E22 kg)
3
2(2.8314E33)
2
=
= 2.727E 33 J
K E
e
=
1
2
(5.972E 24 kg)(30,290 m/s)
2
= 2.7396E 33 J
2.727E33 J
2.7396E33 J
100 = 99.5 %
E
3
= 3
Z
2
Ca
G
2
M
2
e
M
3
m
2ℏ
2
⊙
R
⊙
R
m
→ Z
2
Z = 20
References
Beardsley, I. (2025) Theory For The Solar System And The Atom's Proton; Linking Microscales
To Macroscales, DOI: 10.13140/RG.2.2.19296.34561
Beardsley, I. (2026) How Physics and Archaeology Point to a Natural Constant of 1-Second,
https://doi.org/10.5281/zenodo.18829259
Beardsley, I. (2026) The Sublime and Mysterious Place of Humans in the Cosmos; A Work in
Exoarchaeology, https://doi.org/10.5281/zenodo.18715148
