of 1 46
The Solar System’s Quantized Structure Encodes Its Birth Cloud’s Mass And Mirrors The
Atom’s Proton!
By
Ian Beardsley
Copyright © 2025
of 2 46
Contents!
Important To Understand…………………………………………………….3!
Introduction……………………………………………………………………4!
List of Constants, Variables, And Data In This Paper…………………….5!
The Solar System’s Quantized Structure Encodes !
Its Birth Cloud’s Mass And Mirrors The Atom’s Proton…………………..6!
The Unfolding Of The Rudiments For A Theory Of Everything…………..9!
Theory For Inertia…………………………………………………………….17!
The Solar Solution……………………………………………………………24!
Jupiter and Saturn……………………………………………………………28!
Modeling Star Systems With The Theory………………………………….31!
of 3 46
Important To Understand One can speak of the structure of the long term structure of the solar system.
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 (they have widths). 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 microscopically 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 unit of a second becomes important in my theory. We got the second from the rotation period of the
Earth at the time the moon came to perfectly eclipse the Sun. The Moon slows the Earth rotation and this
in turn expands the Moon’s orbit, so it is getting larger as 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.
That is as the Earth’s day gets longer and the lunar orbit grows larger, we got the second at the time that
the Earth day was what it is during the epoch when the Moon perfectly eclipses the Sun, 24 hours.
The near perfect eclipse is a mystery in the sense that it came to happen when anatomically modern
humans arrived on the scene, even before that, perhaps around Homo Erectus and the beginning of the
Stone Age. The Earth day was 18 hours long, long before that, 1.4 billion years ago. Homo Erectus is
around two to three million years ago. In our theory, we suggest conditions for life may be optimized
when the earth day is about 24 hours long, what it is today, because the earth day of 24 hours has a
characteristic time of 1 second in our theory to be presented.
(1.00267E − 13AU/year)(1E 9years) = 0.0001AU
of 4 46
Introduction
My theory predicts the radius of a proton, has a model for inertia that works, and shows the Solar System
to be a direct analog of the wave equation solution for the hydrogen atom. It shows a common thread
between micro scales (atom's proton) and macro scales (the Solar System). The theory is used to model
star systems in general for a wide array of stellar spectral types.
I report here the finding that the atom’s proton and a wave structure of the Solar System lend themselves
to a unifying characteristic time of a duration of 1.0 seconds in common. This is interesting in that the
duration of a second came down to us historically from the ancient Sumerians and Babylonians because
we inherited their base 60 counting. We develop a theory for inertia from this for the proton. We find at
the outset that the connection between the atom’s proton and the Solar System is in the Planck length and
that the Solar System’s quantized structure encodes its birth cloud’s mass. The finding here open the door
to some profound archaeological questions. I further find that the Earth lends itself to base 60 counting in
terms of its mass, size, and rotational period by looking at its rotational angular momentum. This opens
the door to suggesting the possibility that the phenomenon of the Moon perfectly eclipsing the Sun may
be a condition for optimally habitable planets and that habitable star systems can be modeled under a
common idea that might imply that life is part of a Universal natural process.
I am particular about story telling, especially short stories. In the second section I present a short story
about how the rudiments for a theory of everything unfolded. I hope by relating this story you can share in
the joy of discovery that came in this adventure which took place in the mind and on paper. It is a journey
into the microcosmos, and out into the vast expanse of the Universe. What took shape here instills a sense
of mystery in me in that it formulates the magic that is in the Universe in elegant symbolic language,
mathematics.
Some of the intriguing things we will find is that the fundamental particle that makes up atoms, the
proton, has a characteristic time of a second. We will find the ground state of our Solar System is
described by the Earth’s moon and has a characteristic time of one second. We will find that the Planck-
type constant for the Solar System is given by the Earth’s orbital kinetic energy multiplied by 1 second,
where Earth is the planet in the Solar System optimized for life. We will find that the 24 hour day (the
rotation period of the Earth) which is what it is today, has a characteristic time of one second. We will see
that the Earth orbital energy state as a solution to the quantum mechanical wave equation uses the Earth’s
moon to normalize the Sun’s size, giving it a size of 400. We will see that the condition for the Moon
orbiting the Earth such that it perfectly eclipses the Sun may be a condition for optimally habitable
planets in the habitable zones of stars in general. And, we will see the mass, size, and rotational frequency
of the Earth is conducive to the base 60 counting of the ancient Sumerians which is responsible for the
unit of the second we have today as our base unit to measure time. All of this I see as quite magical and
seems to me to suggest that the Earth/Moon/Sun system is this way for a reason, that there may be a
mysterious force behind it.
Let’s begin with a quote from Carl Sagan…
The surface of the Earth is the shore of the cosmic ocean. From it we have learned most of what we know.
Recently, we have waded a little out to sea, enough to dampen our toes, or at most, wet our ankles. The
water seems inviting. The ocean calls. Some part of our being knows this is from where we came. We long
to return. These aspirations are not, I think, irreverent, although they may trouble whatever gods may be.
of 5 46
List of Constants, Variables, And Data In This Paper
(Proton Mass)
(Proton Radius)
(Planck Constant)
(Light Speed)
(Gravitational Constant)
1/137 (Fine Structure Constant)
(Proton Charge)
(Electron Charge)
(Coulomb Constant)
(The Author’s Solar System Planck-Constant)
(Earth Mass)
(Earth Radius)
(Moon Mass)
(Moon Radius)
(Mass of Sun)
(Sun Radius)
(Earth Orbital Radius)
(Moon Orbital Radius)
Earth day=(24)(60)(60)=86,400 seconds. Using the Moon’s orbital velocity at aphelion, and Earth’s
orbital velocity at perihelion we have:
(Kinetic Energy Moon)
(Kinetic Energy Earth)
m
p
: 1.67262E − 27kg
r
p
: 0.833E − 15m
h : 6.62607E − 34J ⋅ s
c : 299,792,458m /s
G : 6.67408E − 11N
m
2
s
2
α :
q
p
: 1.6022E − 19C
q
e
: 1.6022E − 19C
k
e
: 8.988E 9
Nm
2
C
2
ℏ
⊙
: 2.8314E 33J ⋅ s
M
e
: 5.972E 24kg
R
e
: 6.378E6m
M
m
: 7.34767309E 22kg
R
m
: 1.7374E6m
M
⊙
: 1.989E 30kg
R
⊙
: 6.96E 8m
r
e
: 1.496E11m = 1AU
r
m
: 3.844E 8m
K E
m
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
K E
e
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
of 6 46
The Solar System’s Quantized Structure Encodes Its Birth Cloud’s Mass And Mirrors The
Atom’s Proton We have established the possible connection of the atom’s proton to the Solar
System through the characteristic time of a second, a connection between the microcosmos
and the macrocosmos. We want to see what that connection is mathematically. The radius of a
proton to its mass determines the characteristic time of a proton as:!
1. !
2. !
is the fine structure constant, is Planck’s constant, is the universal constant of
gravitation, and is the speed of light. is the radius of a proton and is its mass. These
two equations yield an accurate equation for the radius of the proton!
3. !
The ground state of the solar system we found is based on the Moon orbiting the Earth with a
characteristic time of 1 second:!
4. !
Where is a Planck-type constant for the Solar System and ia raw mass of the Moon.
This comes from the solution of the wave equation for hydrogen describing the ground state as!
5. !
We have!
6. !
Which is close to the orbital radius of the Moon around the Earth. We find our Planck-type
constant for the Solar System is given by the kinetic energy of the Earth in its orbit and a
characteristic time of 1-second. We have!
7. !
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
ϕ ⋅
πr
p
α
4
Gm
3
p
1
3
⋅
h
c
= 1secon d
α
h
G
c
r
p
m
p
r
p
= ϕ ⋅
h
cm
p
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1secon d
ℏ
⊙
M
m
r
1
=
ℏ
2
ke
2
m
e
ℏ
2
⊙
GM
3
m
=
(2.8314E33)
2
(6.67408E − 11)(7.34763E22kg)
3
= 3.0281E8m
ℏ
⊙
= (1secon d )KE
e
of 7 46
Combining equation1, 4, and 7 we have using as well the kinetic energy of the Earth
, and the orbital velocity of the Earth :!
8. !
9. !
10. !
!
2.6566-E16kg/m=2.889E-16kg/m=92% accuracy!
We used at perihelion. Where is the mass of the Sun, is the mass of the Earth, is
the Earth’s orbit (1AU), and is the mass of a proton. is the cross-sectional area of a
proton and is the Planck length given by!
11. !
Where is the reduced Planck constant. This equation (equation 10) is in kilograms
per meter . We have!
12. !
This is on the order of the mass of a star-forming nebula from which a smaller one breaks away
and the protoplanetary disc and the star form to form the Solar System. We see it’s inverse
multiplied by the square of the Earth’s orbital radius is on the left. We see that the Planck
length action on a mass in terms of the Earth, Moon, and Sun gives the mass of a proton to its
radius. It would seem the Solar System’s quantized structure encodes its birth cloud’s mass
and mirrors the atom’s proton. The Planck length may be the smallest unit of space below
which space is no longer described by classical physics, that space does not have an infinite
number of points, but rather has pixels per square meter determining its definition or degree of
resolution of space.!
Star systems form from giant molecular clouds (GMC’s). A cloud that is a portion of the GMC
collapses under its rotation and flattens out into a rotating disc from which the planets and
central star form. Good examples of GMC’s from which star systems form that have masses on
the order of what is given in equation 12:!
KE
e
= (1/2)M
e
v
2
e
v
e
= GM
⊙
/r
e
4
2π
⋅
M
3
m
M
2
⊙
M
2
e
r
2
e
=
1
3α
2
⋅
πr
2
p
m
p
. ℓ
p
M
3
m
M
2
⊙
M
2
e
r
2
e
= π
2
12α
2
⋅
r
p
m
p
. ℓ
p
M
2
⊙
M
2
e
M
3
m
⋅
ℓ
p
r
2
e
=
6 2
π
m
p
r
p
α
2
(5.972E 24kg)
2
(1.989E30kg)
2
(7.34767E 22kg)
3
(1.6162E − 35m)
(1.471E11m)
2
=
(8.485)
π
1.67262E − 27kg
0.833E − 15m
1
(18769)
r
e
M
⊙
M
e
r
e
m
p
πr
2
p
ℓ
p
ℓ
p
=
ℏG
c
3
ℏ = h /2π
M
2
e
M
2
⊙
M
3
m
= 3.5568E41kg
of 8 46
!
are,!
The Orion Molecular Cloud Complex!
Total Mass: ~ (~ )!
Includes: Orion A (M42, the Orion Nebula ~ ) and Orion B (~ )!
NGC 604 —- Triangulum Galaxy (M33)!
Mass: ~ (~ )!
The characteristic time of 1 second should be in the protoplanetary disc from which the planets
formed. I would guess it would be in the time between collisions of particles in the
protoplanetary disc. Ultimately, the planets form from these collisions. The time between
collisions in the protoplanetary disc are given by!
1. Particle number density n ( the number of particles per unit volume).!
2. Relative velocity between particles .!
3. Particle cross-section (related to particle size).!
For micron to millimeter sized grains in a dense inner region of the protoplanetary disc (like
about 1 AU from the star, which is the Earth orbit) the range of these values are:!
1. particles per meter cubed (from disc models)!
2. Particles sizes are meters.!
3. Relative velocities of particles are as driven by Brownian motion,
turbulence, and gas drag.!
We can imagine a scenario where this yields 1 second for the time between collisions of
particles in the formation of planets by using typical values!
!
It could be that the Earth/Moon/Sun system is a fossil of the protoplanetary disc’s initial
conditions."
M
2
e
M
2
⊙
M
3
m
= 3.5568E41kg
200,000M ⊙
4E41kg
100,000M ⊙
80,000M ⊙
150,000 − 200,000M ⊙
3 − 4E41kg
v
rel
σ = π r
2
n ≈ 10
10
− 10
15
r ≈ 10
−6
− 10
−3
v
rel
≈ 1 − 10m /s
t
c
≈
1
nσ v
rel
=
1
(0.32 × 10
12
m
−3
)(π(10
−6
m)
2
(1m /s)
= 1secon d
of 9 46
The Unfolding Of The Rudiments For A Theory Of Everything
It all started when I was trying to work out a theory for inertia, that property of matter to resist
change in motion: when you push on it, it pushes back. The more of it there is, the more it
pushes back. I had decided to start with the constants, like the gravitational constant, because
I figured they measured the properties of space and time. I eventually wrote an expression that
to my surprise was equal to 1 second:!
!
the radius of a proton, its mass, the constant of gravitation, the speed of light,
Planck’s constant, and the fine structure constant. I found that interesting and figured if the
proton was characterized by the second, and the second came from the ancient Sumerians
dividing-up the rotation period of the Earth into 24 hours, each hour into 60 minutes, and each
minute into 60 seconds from their base 12 and base 60 mathematics, that it had to have
something to do with the celestial motions and periods they observed in the sky, that if the
second was natural, then it would be in the motions of the Earth, Moon, Sun, and stars. My first
guess, which panned out, was that the kinetic energy of the Moon to the kinetic energy of the
Earth times the 24 hour day, should be one second, or close to it. At first I found it was close to
it, but then I made an adjustment for the Earth’s tilt to its orbit of and it came out
exact for all practical purposes. I got!
!
the kinetic energy of the Moon, the kinetic energy of the Earth, and EarthDay is 24
hours (86,400 seconds, the rotation period of the Earth). I then thought this was quantum
mechanical and that I should make a Planck-type constant for the Solar System, . I found it
was in this very equation because it is in joule-seconds which could be the kinetic energy of
the earth times one second in the above equation, so I had:!
!
I then thought I don’t need to solve the Schrodinger Wave equation of quantum mechanics for
the Solar System, but just look at the equations of Niels Bohr for the Bohr model of the atom,
which he wrote down before the Schrodinger equation existed from suggesting the proton had
discreet orbitals for the electrons and was quantized by , the Planck constant. He didn’t know
why or how it quantized like this by integer multiples of , but he found it worked. He was
inspired to do this by the emission spectra of hydrogen for different energies, he suggested
after the electron jumped from one orbital to another by adding energy, that when it fell back in
it would emit light of a particular frequency. So I looked at his equation for the energies of
orbitals and their orbital distances :!
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
r
p
m
p
G
c
h
α
θ = 23.5
∘
KE
m
KE
e
(Ear th Day)cos(θ ) = 1.0second s
KE
m
KE
e
ℏ
⊙
ℏ
⊙
= (1secon d )KE
e
ℏ
ℏ
E
n
r
n
E
n
= −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
of 10 46
It later turned out for the hydrogen atom the case of Z=1 (1 proton) that the Schrodinger equation had
these as the solutions. That equation in spherical coordinates is:
This equation can be solved for the hydrogen atom giving the results of the the Bohr equations, but that is
for Z=1, for other elements, like Z=2,3,4…The equation becomes too difficult to solve. But Bohr’s
equation works for these other values of Z.
I then looked at Bohr’s equations and changed the Coulomb constant to the gravitational constant ,
which changed charge (electron charge) into mass , put in my Planck-type constant for the Solar
System, , did the dimensional analysis and wrote:
They worked correctly to better than 99% for the Earth/Moon/Sun system, which was the n=3 orbit. I
knew I couldn’t use , which would ridiculously be the number of protons in the Sun, so since this was
gravity not electric fields, I used the radius of the Sun, . But, since the Moon of the Earth was
mysteriously appearing in my equations, I used it to normalize the radius of the Sun, which gave it a
radius of 400 (400 moon radii). This made it work. I then realized since we were dealing with the Earth,
the planet most optimally suited for life in the Solar System, and thought since this was because the Moon
optimizes the conditions for life on Earth by holding the Earth at its inclination to its orbit around the Sun
allowing for the season’s and preventing temperature extremes, that the Earth perfectly eclipsing the Sun
as seen from the Earth had meaning, and I wrote the conditions for a perfect eclipse, which are the radius
of the Sun to the radius of the Moon equals the orbital radius of the Earth to the orbital radius of the
Moon. I then thought this might be a condition for habitable planets to be optimally successful and wrote
it:
Later I used this in my solutions for habitable star systems by spectral type of the stars. Finally, I wanted
to formulate the ground state for the solar system and used the equation from the Bohr atom, which is
!
r
n
=
n
2
ℏ
2
Z k
e
e
2
m
e
−
ℏ
2
2 m
[
1
r
2
∂
∂r
(
r
2
∂
∂r
)
+
1
r
2
si n θ
∂
∂θ
(
si n θ
∂
∂θ
)
+
1
r
2
si n
2
θ
∂
2
∂ ϕ
2
]
ψ + V(r)ψ = E ψ
k
e
G
e
m
ℏ
⊙
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
r
n
=
2ℏ
2
⊙
GM
3
m
⋅
R
⊙
R
m
⋅
1
n
Z
R
⊙
r
planet
r
moon
=
R
star
R
moon
r
1
=
ℏ
2
k e
2
m
e
r
1
≈ 0.529E − 10m
of 11 46
Thus writing it with the mass of the Moon, .
Which is about the Moon’s orbital distance (3.844E8m). Noticing this in meters, is about the speed of
light in meters per second, if I multiplied by one over , the speed of light, it would be 1 second:
It was at this point that I was sure I was onto something. Later, playing with the equations, I ended up
with another equation for a second using the proton. It was
Where is the golden ratio (0.618). Equating the left of this with left of my first equation got the ball
rolling
I found it yielded the radius of the proton as
It was at this point that I realized I had the rudiments for a kind of theory of everything that bridged the
microcosms to the macrocosmos. In the next section we will put forth a theory for the mass of a proton.
With what we have so far we can almost apply it to stars of other spectral types (masses, sizes, colors, and
temperatures) for formulating how other stars might structure the mechanics of their habitable planets.
In order to apply this to other star systems, we have to be able to predict the radius of the habitable planet,
presumably in the n=3 orbit. I found the answer to be in the Vedic literature of India. They 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:
radius of the star.The surprising result I found was, after applying it 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 the distance
M
m
ℏ
2
⊙
GM
3
m
=
(2.8314E 33)
2
(6.67408E − 11)(7.34763E 22kg)
3
= 3.0281E 8m
c
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1secon d
ϕ ⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
ϕ
(
1
6 α
2
4πh
G c
)
⋅
r
p
m
p
= 1secon d
r
p
= ϕ ⋅
h
cm
p
R
planet
= 2
R
2
⋆
r
planet
R
⋆
r
planet
of 12 46
from the star, which determines their temperature, but are functions of their size and mass probably
because they are good for life chemistry atmospheric composition, and gravity when they are the size and
mass of the Earth.
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, and the luminosity of the Sun. We see our theory has applications to
archaeology because the second came to us historically from the ancient Sumerians because they divided
the Earth day (rotation period) into 24 hours, and, because each hour and minute got further divisions by
60 because their base 60 counting system was inherited by the ancient Babylonians who were the ultimate
source of dividing the hour into minutes and the minutes into seconds. I have found this system is given
by the rotational angular momentum of the Earth, , in terms the solar system Planck-type constant,
because, as I already pointed out:
This base 60 counting combined with dividing the day into 24 units is mathematically optimal because
the rotational angular momentum incorporates not just the day (rotation period of the Earth) but the mass
and size of the Earth. We can say we are touching on archaeoastronomy. This is because 60/24=2.5 and
the Scottish engineer, Alexander Thom, found ancient megalithic (stone) observatories throughout Europe
may have been based on a unit of length he called the megalithic yard and that the separations between
stones, that align with celestial positions and cycles, are recurrently separated by 2.5 megalithic yards.
Like in Stonehenge.
You will notice the 4/5 in the above equation can be written as 2/5, meaning the equation becomes:
the rotational frequency of the Earth, Since , where is the angular velocity of the Earth, we
have:
We notice , the ancient Sumerian factors defining the second from the rotation period of the
Earth, We see the rotation, mass and radius (size) of the Earth are optimally formulated with ancient
r
planet
r
planet
=
L
⋆
L
⊙
AU
L
⋆
L
⊙
L
eart h
L
eart h
ℏ
⊙
24 = 60
L
eart h
=
4
5
π M
e
f
e
R
2
e
L
eart h
=
2
5
M
e
2 π f
e
R
2
e
f
e
2 π f
e
= ω
e
ω
e
L
eart h
=
2
5
M
e
ω
e
R
2
e
2/5 = 24/60
of 13 46
Sumerian base 12 and base 60 mathematics (they divided the day into 12 hours, and the night into 12
hours).
I have found that the pressure gradient of the protopanetary disc, as a function of radius, that gave birth to
our solar system, is given by:
the radius of the disc where pressure is . This is the solution to:
The protoplanetary disc that evolves into the planets has two forces that balance its pressure, the
centripetal force of the gas disc due to its rotation around the protostar and the inward gravitational
force on the disc from the protostar , and these are related by the density of the gas that makes
up the disc.
I have computed my Planck-type constant, , as such:
Where
this characteristic time of one second is not just in the Solar System, and atom’s proton, but in the basis of
life chemistry, carbon, and the hydrocarbons, the skeletons of life chemistry. I found
is carbon (C)
is hydrogen (H)
Which is to say that six protons, which is carbon, the basis of life as we know it, has a characteristic time
of one second because in the first equation above, we have a mass divided by the mass of a proton, times
seconds, giving six protons times a second (6 proton-seconds) which means 6 protons (carbon, the basis
P(R ) = P
0
(
R
R
0
)
−
L
earth
ℏ
⊙
R
0
P
0
d P
dr
= − ρ
(
GM
⋆
r
2
−
v
2
ϕ
r
)
v
2
ϕ
/r
GM
⋆
/r
2
ρ
ℏ
⊙
ℏ
⊙
= (hC )KE
e
hC = 1secon d
C =
1
3
⋅
1
α
2
c
2
3
⋅
π r
p
G m
3
p
ℏ
⊙
= (hC )KE
eart h
= (1.03351s)(2.7396E 33J ) = 2.8314E 33J ⋅ s
1
6proton s
⋅
1
α
2
⋅
r
p
m
p
4πh
G c
= 1secon d
1
1proton
⋅
1
α
2
⋅
r
p
m
p
4πh
G c
= 6secon d s
of 14 46
of life) has a characteristic time of one second. This means that 1 proton, hydrogen, has a characteristic
time of six seconds. Hydrogen is the most fundamental element in the periodic table of the elements
which was theoretically created in the so-called big bang that gave birth to the universe, and is the
element from which all of the other heavier elements were made by stars. This six-fold symmetry that is
in hydrocarbons, the skeletons of biological chemistry, is fundamental to defining the periodic table of the
elements because it has been found that the six protons of carbon and their respective charges, interact
with its six electrons, their respective charges, to balance to make carbon the most stable element
mathematically in which to describe the rest of the atoms in the periodic table. This is no doubt related to
the regular hexagon, a six-sided polygon which tessellates (tiles a surface without leaving gaps) because it
has its radii equal in length to its sides. This hexagonal tessellating property is used by bees to make their
honeycombs. So we see our theory now goes beyond the atom and the solar system. That it goes to
biological chemistry. But, it does not stop there. It seems to go into cosmology, the study of the origin and
fate of the universe. We see this because my equations link proton properties to 1-second, and protons
were fixed in the universe at 1 second after it, meaning we could be seeing a universal clock that has
influenced everything since the Big Bang.
The idea is that neutrino decoupling (neutrinos stop interacting with one another) happens when the
reaction rate of weak interactions falls below the Hubble parameter, the expansion rate of the universe
. The reaction rate per particles is given by
is the Fermi constant is about , and is the temperature of the Universe. The
expansion rate of the universe is given by
Where is the Plank mass is about 1.22E19GeV. and have units of inverse time ( ). Neutrino
decoupling happens when
This is when the number of protons in the universe was set in place which, as it would turn out, is close to
one second in rough estimate.
The expansion rate of the Universe is governed by the Friedmann equation
Where is the energy density of the Universe. It is
The Hubble expansion rate is
Γ
H
Γ ≈ G
2
F
T
5
G
F
1.166E − 5G eV
−2
T
H ≈
T
2
M
Pl
M
Pl
Γ
H
s
−1
G
2
F
T
5
=
T
2
M
Pl
H
2
=
8π G
3
ρ
ρ
ρ ∝ T
4
of 15 46
Since
we have
We said protons and neutrons are set in the universe when it has cooled in its expansion to about 1MeV.
We have
This was done in Planck units where time can be expressed in inverse energy. Since in Planck units
we have
This theory seems, then, to have applications at the core of cosmology, astrobiology (the study of life in
the universe in general), solar system mechanics, particle physics, theories of common structure between
micro-scales and macro-scales, and biology.
To compute the moon’s orbital radius I just use
Where Ag is the molar mass of silver and Au is the molar mass of silver, a connection to the 1.8 that
appears in our Solar System. We use this because we know it works for our Solar System.
This theory has a magical, mystical feel to it in that it involves the condition for the Moon perfectly
eclipsing the Sun as seen from the Earth, the ratio the Solar radius to the lunar orbital radius being in the
ratio of a gold atom mass to a silver atom mass, where gold and silver have been the metals of ceremonial
jewelry on Earth since ancient times, the Sun gold in color, the Moon silver in color, and in that the basis
unit of time in the theory is the second, which came to us from the ancient Sumerians who first settled
H ∝
T
2
M
Pl
M
Pl
≈ 2.4E18G eV
t ∝
1
H
t ∝
M
Pl
T
2
t ∝
2.4E18G eV
(1E − 3G eV )
2
= 2.4E 24G eV
−1
1G eV
−1
= 5.39E − 25s
t ≈ (2.4E 24)(5.39E − 25)
t ≈ 1.3secon d s
r
m
= R
⊙
Ag
Au
=
R
⋆
(1.8)
of 16 46
down from following the herds, and crafting spearpoints from stone to hunt, to invent agriculture,
architecture, mathematics, and writing.
Perhaps one day, humans will travel to other star systems and find other worlds whose mathematical
relationships are as beautiful and exciting as they are here in our star system, the Solar System. Perhaps
we will learn of other civilizations on other planets, and learn their ancient histories and how they became
who they are today, what kind of music and art they have, and did they build pyramids or tipis in that
these things may be universal in their geometries.
of 17 46
Theory For Inertia I had two equations that gave the radius of a proton with characteristic times of one
second each. I had to break down the equations in terms of their operational parameters as described by a
geometric model. This is what I came up with, a proton is a 4d hypersphere who's cross-section is a
sphere. Of course occupying the dimension of time (4th dimension in drawing) is the vertical component
of the drawing. I have to draw this 3d cross-section as a circle (we cannot mentally visualize four
dimensions). The proton is moving through time at the speed of light (vertical component in the drawing)
it is a bubble in space. The normal force holding it in 3d space is proportional to the inertia
created by the pliability of space measured by G. So when we push on it (Force applied in drawing) there
is a counter force explaining Newton's action/reaction.
I think you could look at this another way: the cross sectional area of the proton moving against space is
in the opposite direction of the force applied and h is the granularity of space, G still its pliability. That is
to say, the flux of a normal force to a hemisphere is over the area of the cross-section of the sphere.
It is the goal of this opening section to provide a theory for inertia, that quality of a mass to resist change
in motion. We want the the theory to include not just the quantum mechanics constant for energy over
time Planck’s constant, but to include the universal constant of gravitation G, the constant the speed of
light from relativity, and the fine structure constant for theories of electric fields so as to bring together
the things that have been pitted against one another: quantum mechanics, relativity, classical physics,
electric fields, and gravitational fields. Towards these ends we will suggest a proton is a 3D cross-section
of a 4D hypersphere held in place countering its motion through time by a normal force that produces its
inertia (measured in mass in kilograms) much the same way we model a block on an inclined plain
countered by friction from the normal force to its motion. The following is the illustration of such a
proton as a cross-sectional bubble in space:
To get the ball rolling, I had found a wave solution to the Earth/Moon/Sun system where the Earth
orbiting the Sun is like an electron orbiting a proton with a quantum mechanical solution. I found this
solution had a characteristic time of one second. But, I found as well, I could describe the proton as
having a characteristic time of one second, and that this yielded the radius of a proton very close to that
obtained by modern experiments. So, it is now before me to come up with a theory for the proton in terms
of these characteristic times before I present my theory for a wave solution of the Solar System.
F
n
= h /(ct
2
1
)
h
c
α
of 18 46
The expressions for the characteristic times of 1-second for the proton that I found, were:
1.
2.
Where is the golden ratio, is the radius of a proton, and is the mass of a proton. We find
these produce close to the most recent measurements of the radius of a proton, if you equate the left sides
of each, to one another:
3.
4.
To derive this equation for the radius of a proton from first principles I had set out to do it with the Planck
energy, , given by frequency of a particle, and from mass-energy equivalence, :
We take the rest energy of the mass of a proton :
The frequency of a proton is
We see at this point we have to set the expression equal to . So we need to come up with a theory for
inertia that explains it:
The radius of a proton is then
(
1
6α
2
4πh
G c
)
⋅
r
p
m
p
= 1secon d
ϕ ⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
ϕ = 0.618
r
p
m
p
r
p
= ϕ
h
cm
p
r
p
= 0.816632E − 15m
E = h f
E = m c
2
E = h f
m
p
E = m
p
c
2
f
p
=
m
p
c
2
h
ϕ
m
p
c
2
h
⋅
r
p
c
= ϕ =
m
p
c
h
r
p
m
p
r
p
= ϕ ⋅
h
c
of 19 46
In order to prove our theory for the radius of a proton as incorporating , we will apply our model
outlined involving a normal force, to a 3d cross-section of a 4d hypersphere countering its direction
through time, t. We begin by writing equation 1 as:
5.
Where , the constant of gravitation measures the pliability of space, and the granularity of space, and
c the speed of propagation. measures the inertia endowed in a proton. We write equation 2 as:
6.
We now say that and that the normal force is
7.
This gives us:
8.
=
Since , we have
9.
This gives
10.
is the cross-sectional area of the proton countering the normal force, , against its motion through
time, this is measured by the constant of gravitation. It is to say that
11.
r
p
= ϕ ⋅
h
cm
p
ϕ
F
n
m
p
=
1
6α
2
4πh
G c
⋅
r
p
1secon d
G
h
m
p
1 =
ϕ
9
⋅
π r
p
α
4
G m
3
p
⋅
h
c(1secon d )
2
⋅
h
c
t
1
= 1secon d
F
n
=
h
ct
2
1
1 =
ϕ
9
⋅
π r
p
α
4
G m
3
p
⋅
h
c
⋅ F
n
π
9α
4
⋅
F
n
G
⋅
r
p
m
2
p
(
ϕ
h
cm
p
)
r
p
= ϕ
h
cm
p
1 =
π
9α
2
⋅
F
n
G
⋅
r
2
p
m
2
p
m
p
=
1
3α
2
⋅
π r
2
p
F
n
G
π r
2
p
F
n
G
m
p
∝
Ar eaCrossSect ion Pr oton ⋅ F
n
G
of 20 46
And, the coupling constant is
12.
Let us see if this is accurate:
We used the experimental value of a proton . And we have demonstrated that our
model of a proton as a 3D cross-section of a 4D hypersphere countering the normal force against its
motion through time gives its inertia that can counter a force at right angles to its motion through time and
the normal force.
It is thought that the proton does not have an exact radius, but that it is a fuzzy cloud of subatomic
particles. As such depending on what is going on can determine its state, or effective radius. It could be
that the proton radius is as large as
Which it was nearly measured to be before 2010 in two separate experiments. Or as small as
Which is closer to current measurements, which have decreased by 4% since 2010, and could get smaller.
In which case the characteristic time, , could be as large as
Using 2/3 as a fibonacci approximation to . Or, it could be as small as
C =
1
3α
2
F
n
=
h
ct
2
1
=
6.62607E − 34J ⋅ s
(299,792,458m /s)(1s
2
)
= 2.21022E − 42N
m
p
=
18769
3
⋅
π (2.21022E − 42N )
6.674E − 11N
m
2
kg
2
(0.833E − 15m) = 1.68E − 27kg
r
p
= 0.833E − 15m
r
p
=
2
3
⋅
h
cm
p
r
p
=
2
3
⋅
6.62607E − 34
(299,792,458)(1.67262E − 27)
= 0.88094E − 15m
r
p
= ϕ ⋅
h
cm
p
= 0.816632E − 15m
t
1
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1.03351secon d s
ϕ
ϕ ⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= (0.618)
(352275361)π (0.833E − 15m)
(6.674E − 11)(1.67262E − 27kg)
3
⋅
1
3
⋅
6.62607E − 34
299792458
of 21 46
=0.995 seconds
Or perhaps more often it is in the area of:
But, what this tells us is that the unit of a second might be a natural constant. And, since the second comes
from dividing the Earth rotation period into 24 hours, and each hour into 60 minutes, and each minute
into 60 seconds, which ultimately comes to us from the ancient Sumerians who first settled down from
hunting, wandering, and gathering and flaking stones into spearpoints to invent agriculture, writing, and
mathematics, that this might be related to the mechanics of our Solar System. We find if we take the
second as natural we have a wave mechanics solution to our Solar System with a characteristic time of
one second that is connected to the characteristic time of the proton, thus connecting macro scales (the
solar system) to micro scales (the atom).
Why Is Used In The Equation For The Radius Of A Proton
We ask why the golden ratio is used to derive the radius of a proton. We start with our equation
1:!
!
This can be written!
13. !
Where . We notice is the force between two protons separated by the
radius of a proton. Of course two such protons cannot overlap by current theories. But it would
seem this gives rise to the proton’s inertia. We will call it . We also notice is the
normal force that gives rise to the proton’s inertia, . We have!
14. !
Now we look at equation 2. It is!
1
6α
2
m
p
h 4π r
2
p
G c
= 1.004996352secon ds
ϕ
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
Gm
2
p
r
2
p
=
h
c
⋅
1
t
2
1
⋅
4π
36α
4
t
1
= 1secon d
Gm
2
p
r
2
p
F
pp
h
c
⋅
1
t
2
1
F
n
F
pp
= F
n
⋅
4π
36α
4
of 22 46
!
It can be written!
15. !
We see that is the inverse of the potential energy between the two protons
separated by the radius of a proton, we will call such a potential energy . We write 15 as!
16. !
Where !
!
Is the normal potential.!
17. !
Where is the golden ratio. Now we notice from equations
14 and 16 that!
18. !
Or!
19. !
And this should explain it. The gravitational force and its potential is in the normal force and
normal potential in time . The golden ratio is to divide a line such that the whole is to the
greater part as the greater part is to the lesser. What that means is the normal and the action
ϕ ⋅
πr
p
α
4
Gm
3
p
1
3
⋅
h
c
= 1secon d
(
1
9
⋅
ϕπ
α
4
)
(
r
p
Gm
2
p
)(
h
2
c
2
⋅
1
m
p
⋅
1
t
2
1
)
= 1
(
r
p
Gm
2
p
)
U
pp
(
1
U
pp
)
(
U
n
)
(
1
9
⋅
ϕπ
α
4
)
= 1
U
n
=
(
h
2
c
2
⋅
1
m
p
⋅
1
t
2
1
)
4π
36α
4
1
9
ϕπ
α
4
= Φ
Φ = 1/ϕ = ( 5 + 1)/2 = 1.618...
F
pp
F
n
= Φ
U
n
U
pp
(
F
pp
)(
U
pp
)
=
(
F
n
) (
U
n
)
Φ
ct
of 23 46
are in the ratio of which being the most irrational number, there is no repetition over cycles,
there is minimal interference between the two for equation 19. Verifying using :!
"
Φ
2.7E − 34N
2.21E − 42N
⋅
2.92E − 57J
2.24E − 49J
= 1.6 = Φ
of 24 46
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.
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.
K E
e
= 3
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
R
⊙
R
m
=
r
e
r
m
K E
e
= 3
r
e
r
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
K E
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
K E
e
=
R
⊙
R
m
⋅
G
2
M
2
e
M
⊙
2r
2
e
v
2
m
of 25 46
We now multiply through by and we have
10.
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:
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
M
2
e
/M
2
e
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2r
2
e
v
2
m
M
2
e
ℏ
⊙
L
p
p
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
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
ℏ
⊙
= 9.13E 38J ⋅ s
h
⊙
= 2π ℏ
⊙
= 5.7365E 39J ⋅ s
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
R
⊙
R
m
(6.67408E − 11)
2
(5.972E 24kg)
4
(1.9891E 30kg)
2(8.3367E 77kg
2
⋅
m
4
s
2
)
R
⊙
R
m
(6.759E 30J )
R
⊙
R
m
=
6.957E 8m
1737400m
= 400.426
K E
e
= 2.70655E 33J
of 26 46
Our equation has an accuracy of
Which is very good.
Let us equate the lunar and solar formulations:
This gives:
13.
We remember that
And since,
14.
Equation 14 becomes
K E
eart h
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
2.70655E 33J
2.7396E 33J
= 98.79 %
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
L
p
=
M
2
e
M
⊙
M
3
m
3
⋅ ℏ
⊙
ℏ
⊙
= (hC )KE
e
hC = 1secon d
K E
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
(1.9891E 30kg)
(7.34763E 22kg)
3
(1.732)
= 321,331.459 ≈ 321,331
of 27 46
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:
1secon d = 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
L
2
p
GM
2
e
M
⊙
⋅
3
c
= 1secon d
n = 3
(9.13E 38J ⋅ s)
2
(6.674E − 11)(5.972E 24kg)
2
(1.989E 30kg)
⋅
3
c
= 1.0172secon d s
of 28 46
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
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…
E =
Z
n
G
2
M
2
m
3
2ℏ
2
⊙
K E
j
=
1
2
(1.89813E 27kg)(13720m /s)
2
= 1.7865E 35J
E =
Z
H
5
(6.67408E − 11)
2
(1.89813E 27kg)
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
E =
Z
H
5
(3.971E 35J ) = Z
H
(1.776E 35J )
Z
H
=
1.7865E 35J
1.776E 35J
= 1.006proton s ≈ 1.0pr oton s = h ydr ogen(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
K E
S
=
1
2
(5.683E 26kg)(10140m /s)
2
= 2.92E 34J
of 29 46
=
The equation for Saturn is then
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
E =
Z
6
(6.67408E − 11)
2
(5.683E 26kg)
2
(7.347673E 22)
3
2(2.8314E 33)
2
Z
2.45
(3.5588E 34J ) = Z(1.45259E 34J )
Z(1.45259E 34J ) = (2.92E 34J )
Z = 2pr oton s = Heliu m(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
=
6.96E 8m
1737400m
= 400.5986
E
3
= (1.732)(400.5986)
(6.67408E − 11)
2
(5.972E 24kg)
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
K E
e
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
2.727E 33J
2.7396E 33J
100 = 99.5 %
E
3
= 3
Z
2
Ca
G
2
M
2
e
M
3
m
2ℏ
2
⊙
of 30 46
In this case the element associated with the Earth is calcium which is Z=20 protons.
R
⊙
R
m
→ Z
2
of 31 46
Modeling Star Systems With The Theory The best way to solve star systems with our theory would be
to use
1.0 , 3.0
2.0. , 4.0.
Where here . Equation 2.0 becomes
5.0.
Where for Earth p=2.5, the exponent in the pressure gradient for its protoplanetary disc. From this we get
. We now get the characteristic time, , from
by using
6.0. and 7.0.
And we have . We can now put in equation 1.0
Which here is 1 second for the Earth, to get the mass of the moon, . But to use equation 5, we need
from equation 3.0. This requires the mass of the Earth, the frequency of the earth, which we get
from the planet’s day (Its rotation rate) and the radius of the planet. We have all of these values for
habitable planets in a K2V star and an M2V star, but they are tidally locked. We have the frequencies
because the planets are tidally locked, so their planets rotation periods are equal to their orbital periods. If
we can’t measure the planet’s radius in another star system, we might obtain it from:
8.0.
Which works for the Earth. We can get the orbital radius of our Moon from
9.0.
It is given by the ratio of silver (Ag) to gold (Au) by molar mass is equal to . The radius of the
planet’s moon we suggested is given by a perfect eclipse:
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1secon d
L
eart h
=
4
5
π M
e
f
e
R
2
e
P(R ) = P
0
(
R
R
0
)
−L
earth
ℏ
⊙
ℏ
⊙
= t
c
K E
e
t
c
= 1secon d
L
eart h
ℏ
⊙
= p
ℏ
⊙
t
c
ℏ
⊙
= t
c
K E
e
v
e
=
GM
e
r
e
K E
eart h
=
1
2
M
e
v
2
e
t
c
t
c
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1secon d
M
m
L
eart h
R
e
=
2R
2
⊙
r
e
r
m
= R
⊙
Ag
Au
=
R
⋆
(1.8)
r
m
/R
⊙
of 32 46
10.0.
We have already applied our theory to the Earth/Moon/Sun System and it worked out nicely. Now we
want to apply this ideal approach we just outlined, to it, so we can test it. We start with the angular
momentum of the Earth. It is given by
Or,…
=
The orbital velocities and kinetic energies of the Earth are given by:
We can now determine :
This is correct for our solar system’s Planck constant. We have the characteristic time is
Which is correct as well. Now we compute the mass of our moon…
This is also very accurate (actual value: 7.347673kg. Now we compute the orbital radius of the Moon…
This is accurate too (actual value: 3.84E8m). From this we have the radius of the Moon:
R
⋆
R
m
=
r
p
r
m
L
eart h
=
4
5
π M
e
f
e
R
2
e
L
e
=
4
5
π (5.972E 24kg)
1
(86400secon d s)
(6.378E6m)
2
7.07866672E 33J ⋅ s
v
p
=
GM
⋆
r
p
=
(6.674E − 11)(1.989E 30kg)
(1.496E11m)
= 29,788.24m /s
K E
p
=
1
2
M
p
v
2
p
=
1
2
(5.972E 24kg)(29,788.24m /s)
2
= 2.65E 33J
ℏ
⋆
ℏ
⋆
=
L
p
p
=
(7.07866672E 33J ⋅ s)
2.5
= 2.831467E 33J ⋅ s
t
c
=
ℏ
⋆
K E
p
=
(2.831467E 33J ⋅ s)
2.65E 33J
= 1.068secon d s ≈ 1secon d
M
3
m
=
(2.831467E 33J ⋅ s)
2
(6.674E − 11)(299,729,458m /s)(1.068secon d s)
M
m
= 7.213E 22kg
r
m
= R
⋆
Ag
Au
= R
⋆
/(1.8) =
6.957E 8m
1.8
= 3.865EE 8m
of 33 46
This is pretty accurate, too. The actual value is 1.7374E6m
Now to get the density of the Moon…
This is good, our Moon is about 3.344g/cm3. Now we want to check
So this gives the correct characteristic time for the Earth/Moon/Sun system. Let’s compute the planet day
characteristic time…
We see the system for modeling star systems works.
We can make a program to model star systems in general given the spectral class of the Star. So HR
diagrams plot mass versus luminosity to give spectral types of stars. So F Stars would be more luminous
blue stars, G stars would be yellow medium luminosity stars, and K stars would be less luminous orange
stars, and so on. There are ten divisions of each, and a “V” meaning “five” indicates the star is on the
main sequence. So our Sun is a G2V star. A medium luminosity, yellow star. Here is a my program in C
that does that for the method we just outlined.
R
m
= R
⋆
r
m
r
p
= (6.957E 8m)
3.865E 8m
1.496E11m
= 1.79738E6m
V
m
=
4
3
π R
3
m
=
4
3
π (1.79738E6m)
3
= 2.432E19m
3
ρ
m
=
7.213E 22kg
2.432E19m
3
= 2.96587g /c m
3
≈ 3g /c m
3
1secon d = 2r
e
v
m
v
2
e
M
3
m
3
M
2
e
M
⊙
v
m
=
GM
p
r
m
=
(6.674E − 11)(5.972E 24kg)
(3.865EE 8m)
= 1,015.5m /s
2(1.496E11m)
1,015.5m /s
(29,788.24m /s)
2
(7.213E 22kg)
3
(1.732)
(5.972E 24kg)
2
(1.989E 30kg)
= 1.03648sec ≈ 1secon d
1secon d ≈
K E
m
K E
e
(Ear th Da y)
K E
m
=
1
2
(7.213E 22kg)(1,015.5m /s)
2
= 3.719E 28J
K E
m
K E
e
(Pla netDa y) =
(3.719E 28J )
(2.65E 33J )
(86,400sec) = 1.2secon d s
of 34 46
//
// main.c
// modelsystem
//
// Created by Ian Beardsley on 2/9/25.
//
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
float R_p, M_p, R_s, M_s, t_c, M_m, rho_m, rho_p, PlanetDay,
V_p,StarRadius, PlanetRadius, PlanetMass, StarLuminosity, PlanetOrbit,
StarMass, r_p, T_p, p, L_p, KE_p, v_p, T_m,Tmoon, C_m;
float G=6.674E-11, hbarstar, PDCT,Tsquared,T,PlanetYear;
float r_m, R_m, V_m, MoonDensity, part1, part2, part3,v_m, KE_m;
int i;
printf ("What is the radius of the star in solar radii? ");
scanf ("%f", &StarRadius);
printf ("What is the mass of the star in solar masses? ");
scanf ("%f", &StarMass);
printf ("What is the luminosity of the star in solar luminosities? ");
scanf ("%f", &StarLuminosity);
PlanetOrbit=sqrt(StarLuminosity);
r_p=PlanetOrbit*1.496E11;
M_s=1.9891E30*StarMass;
Tsquared=((4*3.14159*3.14159)/(G*M_s))*r_p*r_p*r_p;
T=sqrt(Tsquared);
PlanetYear=T/31557600;
printf("Do you want us to compute the planet radius, 1=yes, 0=no? ");
scanf("%i", &i);
R_s=6.9364E8*StarRadius;
if (i==1)
{
R_s=6.9364E8*StarRadius;
R_p=2*(R_s*R_s)/r_p;
PlanetRadius=R_p/6.378E6;
}
else
{
printf("What is the planet radius in Earth radii?: ");
scanf("%f", &PlanetRadius);
R_p=PlanetRadius*6.378E6;
}
printf("What is the mass of the planet in Earth masses? ");
scanf("%f", &PlanetMass);
M_p=PlanetMass*5.972E24;
printf ("What is the planet day in Earth days? ");
scanf ("%f", &PlanetDay);
T_p=PlanetDay*86400;
of 35 46
printf("That is %f seconds \n", T_p);
{
printf("What is p the pressure gradient exponent of the
protoplanetary disc? ");
scanf("%f", &p);
M_s=1.9891E30*StarMass;
r_m=R_s/1.8;
v_p=sqrt(G*M_s/r_p);
L_p=0.8*3.14159*M_p*(1/T_p)*R_p*R_p;
KE_p=0.5*M_p*v_p*v_p;
hbarstar=L_p/p;
t_c=hbarstar/KE_p;
part1=cbrt(hbarstar/(t_c));
part2=cbrt(1/G);
part3=cbrt(hbarstar/299792458);
M_m=part1*part2*part3;
R_s=StarRadius*6.9634E8;
R_m=R_s*r_m/r_p;
V_m=1.33333*3.14159*R_m*R_m*R_m;
rho_m=(M_m/V_m);
MoonDensity=rho_m*0.001;
V_p=1.33333*3.14159*R_p*R_p*R_p;
rho_p=(M_p/V_p)*0.001;
printf("\n");
printf("\n");
printf("Angular Momentum of Planet: %f E33 \n", L_p/
1E33);
printf("\n");
printf("\n");
printf("PlanetYear: %f years \n", PlanetYear);
printf("PlanetYear: %f seconds \n", T);
printf("planet orbital velocity: %f m/s \n", v_p);
printf("planet mass: %f E24 kg \n", M_p/1E24);
printf("planet mass: %f Earth masses \n", M_p/5.972E24);
printf("planet radius %f meters \n", R_p);
printf("planet radius: %f Earth Radii \n", PlanetRadius);
printf("planet orbital radius: %f E11 m \n", r_p/1E11);
printf ("planet orbital radius: %f Earth distances \n",
r_p/1.496E11);
printf("planet KE: %f E33 J \n",KE_p/1E33);
printf("planet density: %f g/cm3 \n", rho_p);
printf("\n");
printf("\n");
printf("hbarstar: %f E33 Js \n", hbarstar/1E33);
printf("characteristic time: %f seconds\n", t_c);
printf("\n");
printf("\n");
of 36 46
printf("Orbital Radius of Moon: %f E8 m \n", r_m/1E8);
printf("Orbital Radius of Moon: %f Moon Distances \n",
r_m/3.84E8);
printf("Radius of Moon: %f E6 m \n", R_m/1E6);
printf("Radius of Moon: %f Moon Radii \n", R_m/1.7374E6);
printf("Mass of Moon: %f E22 kg \n", M_m/1E22);
printf("Mass of Moon %f Moon Masses \n", M_m/
7.347673E22);
printf("density of moon: %f g/cm3 \n", MoonDensity);
printf("\n");
printf("\n");
v_m=sqrt(G*M_p/r_m);
KE_m=0.5*M_m*v_m*v_m;
PDCT=(KE_m/KE_p)*(T_p);
printf("Orbital Velocity of Moon: %f m/s \n", v_m);
printf("PlanetDay Characteristic Time: %f seconds \n",
PDCT);
C_m=2*3.14159*r_m;
T_m=C_m/v_m;
Tmoon=T_m*(1.0/24)*(1.0/60)*(1.0/60);
printf("Lunar Orbital Period: %f seconds \n", T_m);
printf("Lunar Orbital Period: %f days \n", Tmoon);
return 0;}}
Now we show running the program (3 examples) for a wide spread of spectral types including F, G, and
K-type stars. We will need to input in the program not just the mass of the star, its luminosity, and size,
but the pressure gradient exponent for the disc from which the star’s planets formed.
To compute the moon’s orbital radius I just use
Where Ag is the molar mass of silver and Au is the molar mass of silver, a connection to the 1.8 that
appears in our Solar System. We use this because we know it works for our Solar System. I compute the
radius of the planet using
But, give the option of putting in your own radius. I have run the program for F5V stars, through GV
stars, to K3V stars and I use this equation to compute the radii of the planets because, again, we know it
works for our star system, and further we found given the way the radius of a star varies with with
luminosity in the HR diagram, this equation always gives a planet around the size of the Earth. I feel this
size is ideal for planets with sophisticated life because of the laws of chemistry determining a functional
density for the planet having water and the right gravity. As such I always use the planet day as one Earth
day, which again I feel is optimal for life in terms of climate. So these values all constant, we only vary
star mass, size, and luminosity as they work on the HR diagram. I also vary the pressure gradient
exponent now using the average theoretical values it has for each spectral class. The trend is that it
steadily decreases on average with mass and luminosity of the star though it can go up and down
depending on the peculiarities of the system. One of the reasons is that while for a G2V star it can range
on average from p= 1.7-2.1, for our Sun, a G2V star, it is actually high, it is 2.5. However, here we will
r
m
= R
⊙
Ag
Au
=
R
⋆
(1.8)
R
e
=
2R
2
⊙
r
e
of 37 46
model stars with everything constant, as we said, but the pressure gradient will gradually decrease with
spectral class, and when we do a G2V star, we won’t use the Sun’s data, but the average value for G2V
stars. We will do lot’s of models, allowing no gaps in the data for a plot, so we can get a well defined
curve. We will use the upper value for p in instances here. We will also use
Instead of
Thus taking the tilt of the planet to its orbital as and not what it is in particular for the Earth
, which may actually be optimal for the most habitable types of planets around stars like our
Sun. The average pressure exponents by spectral class are given in the following table…
K E
moon
K E
planet
(Pla netDa y)cos(0
∘
) ≈ 1secon d
K E
m
K E
e
(Pla netDa y)cos(23.5
∘
) = 1secon d
θ = 0
∘
θ = 23.5
∘
of 38 46
Now we run the program using this for examples of three spectral types…
of 39 46
F5V Star
What is the radius of the star in solar radii? 1.473
What is the mass of the star in solar masses? 1.33
What is the luminosity of the star in solar luminosities? 3.63
Do you want us to compute the planet radius, 1=yes, 0=no? 1
What is the mass of the planet in Earth masses? 1
What is the planet day in Earth days? 1
That is 86400.000000 seconds
What is p the pressure gradient exponent of the protoplanetary disc? 2.4
Angular Momentum of Planet: 9.321447 E33
PlanetYear: 2.280109 years
PlanetYear: 71954776.000000 seconds
planet orbital velocity: 24888.847656 m/s
planet mass: 5.972000 E24 kg
planet mass: 1.000000 Earth masses
planet radius 7325190.000000 meters
planet radius: 1.148509 Earth Radii
planet orbital radius: 2.850263 E11 m
planet orbital radius: 1.905256 Earth distances
planet KE: 1.849692 E33 J
planet density: 3.627237 g/cm3
hbarstar: 3.883936 E33 Js
characteristic time: 2.099775 seconds
Orbital Radius of Moon: 5.676287 E8 m
Orbital Radius of Moon: 1.478200 Moon Distances
Radius of Moon: 2.042695 E6 m
Radius of Moon: 1.175720 Moon Radii
Mass of Moon: 7.107576 E22 kg
Mass of Moon 0.967323 Moon Masses
density of moon: 1.990782 g/cm3
Orbital Velocity of Moon: 837.955261 m/s
PlanetDay Characteristic Time: 1.165595 seconds
Lunar Orbital Period: 4256210.000000 seconds
Lunar Orbital Period: 49.261688 days
Program ended with exit code: 0
of 40 46
G3V Star
What is the radius of the star in solar radii? 1.002
What is the mass of the star in solar masses? 0.99
What is the luminosity of the star in solar luminosities? 0.98
Do you want us to compute the planet radius, 1=yes, 0=no? 1
What is the mass of the planet in Earth masses? 1
What is the planet day in Earth days? 1
That is 86400.000000 seconds
What is p the pressure gradient exponent of the protoplanetary disc? 2.1
Angular Momentum of Planet: 7.393050 E33
PlanetYear: 0.989814 years
PlanetYear: 31236148.000000 seconds
planet orbital velocity: 29789.738281 m/s
planet mass: 5.972000 E24 kg
planet mass: 1.000000 Earth masses
planet radius 6523625.500000 meters
planet radius: 1.022833 Earth Radii
planet orbital radius: 1.480965 E11 m
planet orbital radius: 0.989950 Earth distances
planet KE: 2.649862 E33 J
planet density: 5.135297 g/cm3
hbarstar: 3.520500 E33 Js
characteristic time: 1.328560 seconds
Orbital Radius of Moon: 3.861263 E8 m
Orbital Radius of Moon: 1.005537 Moon Distances
Radius of Moon: 1.819172 E6 m
Radius of Moon: 1.047066 Moon Radii
Mass of Moon: 7.754257 E22 kg
Mass of Moon 1.055335 Moon Masses
density of moon: 3.074905 g/cm3
Orbital Velocity of Moon: 1015.987427 m/s
PlanetDay Characteristic Time: 1.304901 seconds
Lunar Orbital Period: 2387924.250000 seconds
Lunar Orbital Period: 27.638012 days
Program ended with exit code: 0
of 41 46
K3V Star
What is the radius of the star in solar radii? 0.755
What is the mass of the star in solar masses? 0.78
What is the luminosity of the star in solar luminosities? 0.28
Do you want us to compute the planet radius, 1=yes, 0=no? 1
What is the mass of the planet in Earth masses? 1
What is the planet day in Earth days? 1
That is 86400.000000 seconds
What is p the pressure gradient exponent of the protoplanetary disc? 1.5
Angular Momentum of Planet: 8.340819 E33
PlanetYear: 0.435785 years
PlanetYear: 13752343.000000 seconds
planet orbital velocity: 36167.078125 m/s
planet mass: 5.972000 E24 kg
planet mass: 1.000000 Earth masses
planet radius 6929175.500000 meters
planet radius: 1.086418 Earth Radii
planet orbital radius: 0.791609 E11 m
planet orbital radius: 0.529150 Earth distances
planet KE: 3.905860 E33 J
planet density: 4.285367 g/cm3
hbarstar: 5.560546 E33 Js
characteristic time: 1.423642 seconds
Orbital Radius of Moon: 2.909435 E8 m
Orbital Radius of Moon: 0.757665 Moon Distances
Radius of Moon: 1.932263 E6 m
Radius of Moon: 1.112158 Moon Radii
Mass of Moon: 10.277222 E22 kg
Mass of Moon 1.398704 Moon Masses
density of moon: 3.400867 g/cm3
Orbital Velocity of Moon: 1170.438843 m/s
PlanetDay Characteristic Time: 1.557185 seconds
Lunar Orbital Period: 1561850.125000 seconds
Lunar Orbital Period: 18.076969 days
Program ended with exit code: 0
of 42 46
Here is a plot of these results from F5V Stars to K3V stars:
We see the characteristic time decreases as a curve to intersect close
to a characteristic time of one second with a planet day
characteristic time of one second as a straight line at a G3V star, a
G5V star, and a G8V, which is a region near where our Sun is and may
have something to do with it being so optimal for life. We note here
that this uses the average value for a G2V star, and our Sun comes
closer to a second because its pressure exponent is higher than on
average, it is very high, or steep (p=2.5) which means the pressure of
its disc drops rapidly with distance.
Planet Day Characteristic Time:
Characteristic time: ,
Here we fit the curves for characteristic time and planet day characteristic time. We name the spectral
types with number for input according to the following scheme.
F5V is 1.5, F6V is 1.6, F7V is 1.7,…G0V is 2.0, G1V is 2.1,…
1sec on d ∼
K E
m
K E
e
(Ear t h D a y)
ℏ
⊙
= (1seco n d )K E
e
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1seco n d
of 43 46
For the characteristic time we fit the curve with a power law decay
For the planet day characteristic time we fit the curve with a straight line
Where we have chosen in ,
We intend to fit the curves in the following plot:
y = 2.8x
−
3
2
x
+ 1.1
y = 0.168x + 0.913595
y = m x + b
m =
G3V − F 9V
2.3 − 1.9
= 0.168
b = 1.165595 − 0.168(1.5)
of 44 46
The results are
of 45 46
of 46 46
The Author
