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Describing Everything In The Universe From The Big Bang, To The Proton, To The Protoplanetary Disc,

To The Solar System, To Life With One Value, The Unit Of A Second Of Time

By Ian Beardsley

March 12, 2025

of 2 40

Contents

Data Used In This Paper……………………………………3

Abstract…………………………………………………….4

The Theory…………………………………………………4

Jupiter and Saturn………………………………………..11

The Solar Solution………………………………………..14

The Protoplanetary Disc………………………………….17

The Big Bang………………………………………………18

Modeling Star Systems With The Theory……………….21

Discussion………………………………………………….36

1 Second Encodes The Core Element Of Life, Carbon…36

Conclusion…………………………………………………38

of 3 40

Data Used In This Paper

(Proton Mass)

(Planck Constant)

(Proton Radius)

(Gravitational Constant)

(light speed)

(Fine Structure Constant)

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:

m

P

: 1.67262 × 10

−27

kg

h : 6.62607 × 10

−34

J ⋅ s

r

p

: 0.833 × 10

−15

m

G:6.67408 × 10

−11

N

m

2

kg

2

c : 299,792,458m /s

α : 1/137

q

p

= q

e

= 1.6022E − 19coulom bs

k

e

= 8.988E 9

Nm

2

C

2

K E

moon

=

1

2

(7.347673E 22kg)(966m /s)

2

= 3.428E 28J

K E

earth

=

1

2

(5.972E 24kg)(30,290m /s)

2

= 2.7396E 33J

of 4 40

Abstract It is reported here that planetary and atomic scales (the proton) may be described by a

characteristic time of about 1 second, where the second we have today comes from the ancient Sumerians

having divided the Earth rotation period into 24 units (hours) and their base 60 counting system that

ultimately resulted in dividing the hour into 60 minutes and each minute into 60 seconds. We ﬁnd the

ground state of the our solar system is based on the mass of the Moon, and that the algebraic condition for

the Moon nearly perfectly eclipsing the Sun may be a condition for optimal habitability of earth-like

planets in habitable zones in general. It is found from a relationship in Vedic yoga that an equation exists

for the radius of the habitable planet that produces a planetary radius for all spectral-types of stars about

that of the Earth. It is suggested that perhaps this size, owing to its gravity, might be ideal for life

chemistry to take place based on elemental properties. It is found that our characteristic time of one

second, yielding the proton radius closely, and describing our Solar System, may have been inherited

from the Big Bang, and that the the protoplanetary disc that gave rise to the the planets may have

inherited this characteristic time from the Big Bang as well. We ﬁnd the characteristic time of one second

follows from not just the Sumerian quantization of time into (24, 60) for the earth rotation but that this

works with the radius and mass of the Earth, which results in a quantization of angular momentum for the

Solar System.. We also ﬁnd that 1 second encodes for the core element of life, carbon. In short, we can

encode everything in the Universe from the Big Bang, to the proton, to the protoplanetary disc, to the

Solar System, to life with one unit, a second of time.

The Theory In the Bohr atom the radius of the electron’s orbit in the ground state (n=1) is given by

1.

2.

I have extended this to our solar system and found that the ground state of our solar system is based on

the Moon of our Earth. I have written

3.

Yes, interestingly it is equal to 1 second, our base unit of time that ultimately came from the way the

Ancient Sumerians divided up the period of rotation of the Earth (1 Day) into 24 hours, each hour into 60

minutes, each minute into 60 seconds. They used based 60 counting which was handed down to the

Babylonians. is the mass of the Moon, and is the Planck-type constant I have proposed for the

Solar System. I found it is given by

4.

r

1

=

ℏ

2

k e

2

m

e

r

1

≈ 0.529E − 10m

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

λ

moon

=

ℏ

2

⊙

GM

3

m

=

(2.8314E 33)

2

(6.67408E − 11)(7.34763E 22kg)

3

= 3.0281E 8m

λ

moon

c

=

3.0281E 8m

299,792,458m /s

= 1.010secon d s

λ

moon

c

= 1secon d

M

m

ℏ

⊙

ℏ

⊙

= (1secon d )K E

e

of 5 40

Where is the kinetic energy of the Earth in its orbit around the Sun. We see the the third planet from

the Sun, the Earth, determines its Planck constant in terms of the same unit of a second. I also ﬁnd that the

kinetic energy of the Moon to the kinetic energy of the Earth, that quantity times the Earth rotation period

(24 hours, its day) is close to a second, about 1.2 to 1.3 seconds depending whether you use aphelion or

perihelions, and in what combination. That is

5.

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:

If the second were purely arbitrary, we should expect this ratio to not be so close to 1 second. I also ﬁnd

that the proton has a characteristic time of 1 second in two forms that when equated give about the radius

of the proton:

6.

7.

Equating these two yield the radius of a proton, in terms of its mass :

8.

I ﬁnd if I replace 2/3 with , we may have exactly the radius of a proton

9.

This may show an underlying optimization principle where occurs in biological and cosmic structures. I

ﬁnd I can arrive at this radius of a proton another way, With the Planck energy, , given by

frequency of a particle, and from mass-energy equivalence, :

K E

e

K E

m

K E

e

(Ear th Da y) = 1.2 − 1.3secon d s

K E

moon

=

1

2

(7.347673E 22k g)(966m /s)

2

= 3.428E 28J

K E

earth

=

1

2

(5.972E 24kg)(30,290m /s)

2

= 2.7396E 33J

(

1

6 α

2

4πh

G c

)

⋅

r

p

m

p

= 1secon d

(18769)((0.833E − 15m)

6(1.67262E − 27k g)

(6.62607E − 34J ⋅ s)(4π)

(6.674E − 11)(299,792,458m /s)

= 1.00500secon d s

2

3

⋅

π r

p

α

4

G m

3

p

1

3

⋅

h

c

= 1secon d

r

p

m

p

r

p

=

2

3

⋅

h

cm

p

ϕ = 1.618

r

p

= ϕ

h

cm

p

ϕ

E = h f

E = m c

2

of 6 40

We take the rest energy of the mass of a proton :

The frequency of a proton is

Since our theory gave us the factor of 2/3 for the radius of a proton we have:

The radius of a proton is then

This is very close to the value upon which the proton radius converged historically by two independent

methods which was 0.877E-15m. The result from our theory is

The 0.877fm was challenged in 2010 by a third experiment making it 4% smaller and was 0.842E-15m.

We ﬁnd it may be that the radius of a proton is actually

Where is the golden ratio. The most recent value is 0.833E-15m. I guessed since one second

comes from the ancient Sumerians dividing the Earth day (rotation period) into 24 hours, and those into

60 minutes, and those into 60 seconds, that this has to do with the rotational angular momentum of the

Earth, . I found

10.

Where

11.

E = h f

m

p

E = m

p

c

2

f

p

=

m

p

c

2

h

m

p

c

2

h

⋅

r

p

c

=

2

3

≈ ϕ =

m

p

c

h

r

p

m

p

r

p

=

2

3

⋅

h

c

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

ϕ = 0.618

L

earth

L

earth

ℏ

⊙

24 = 60

L

earth

=

4

5

π M

e

f

e

R

2

e

of 7 40

The value is 2.5 which by modeling our Solar System is found to be the exponent in the pressure gradient

for the protoplanetary disc from which our Solar System formed. That is I found

12.

the pressure of the disc as a function of radius. Which suggests that the structure of the protoplanetary

disc could be governed by the same fundamental time of one second in the Earth’s rotation and that the

Earth’s formation process may be encoded in the same number we developed since ancient times to

describe time (24, 60). This is the solution to:

13.

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. It is the pressure gradient of the disc in radial equilibrium balancing the inward gravity and

outward centripetal force. 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:

14.

The surprising result I found was, after applying it to the stars of many spectral types, 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 from the star, which determines their

temperature, but are functions of their size and gravity probably because it is good for life chemistry. Here

are just a few examples using the data for several spectral types:

F8V Star

Mass: 1.18

Radius: 1.221

Luminosity: 1.95

P(R) = P

0

(

R

0

)

−

L

ear th

ℏ

⊙

d P

dr

= − ρ

(

GM

⋆

r

2

−

v

2

ϕ

r

)

v

2

ϕ

/r

GM

⋆

/r

2

ρ

R

planet

= 2

R

2

⋆

r

planet

r

planet

M

⋆

= 1.18(1.9891E 30kg) = 2.347E 30k g

R

⋆

= 1.221(6.9634E 8m) = 8.5023E 8m

r

p

= 1.95L

⊙

AU = 1.3964A U(1.496E11m /AU ) = 2.08905E11m

R

p

=

2R

2

⋆

r

p

= 2

(8.5023E 8m)

2

2.08905E11m

=

6.92076E6m

6.378E6m

= 1.0851Ear th R a dii

of 8 40

F9V Star

Mass: 1.13

Radius: 1.167

Luminosity: 1.66

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. It might be that radius is right for life in terms of gravity and densities for the

elements. I got these these results for the stars from spectral types F5V to K3V.

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:

15.

Also, the theory utilizes the fact that the Moon as seen from the Earth perfectly eclipses the Sun as a

possible condition for optimal habitability of the planet, which is

M

⋆

= 1.13(1.9891E 30kg) = 2.247683E 30k g

R

⋆

= 1.167(6.9634E 8m) = 8.1262878E 8m

r

p

= 1.66L

⊙

AU = 1.28841AU(1.496E11m /AU ) = 1.92746E11m

R

p

=

2R

2

⋆

r

p

= 2

(8.1262878E 8m)

2

1.92746E11m

=

6.852184E6m

6.378E6m

= 1.0743468Ear th Ra dii

M

⋆

= 1.06(1.9891E 30kg) = 2.108446E 30kg

R

⋆

= 1.100(6.9634E 8m) = 7.65974E 8m

r

p

= 1.35L

⊙

AU = 1.161895AU(1.496E11m /AU ) = 1.7382 E11m

R

p

=

2R

2

⋆

r

p

= 2

7.65974E 8m)

2

1.7382E11m

=

6.751E6m

6.378E6m

= 1.05848Ear th Ra dii

M

⋆

= 1.03(1.9891E 30kg) = 2.11E 30k g

R

⋆

= 1.060(6.9634E 8m) = 7.381E 8m

r

p

= 1.20L

⊙

AU = 1.0954A U(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 R a dii

r

planet

r

planet

=

L

⋆

L

⊙

AU

of 9 40

16.

Orbital radius of the planet to that of the moon is radius of the star to that of the moon. It is known that

the Moon has a lot to do with the conditions for life on Earth being good because its orbit holds the Earth

at its inclination to Sun its orbit preventing temperature extremes and allowing for the seasons. The

Schrödinger Wave Equation must be solved to determine the energies and orbitals of the electron in the

hydrogen atom. In spherical coordinates it is

17.

It has the solutions

18.

19.

I ﬁnd the solutions are for the Earth orbiting the Sun are:

20.

21.

is the solar radius, that of the Moon. For Earth , third planet. For a star brighter than the Sun,

more massive, larger, it may be that in many cases the habitable zone, which is further out, still is the n=3

orbit for planets because the distribution of the planets might be stretched out putting the n=3 planet in the

habitable zone. The same may be true of stars that are dimmer, less massive and smaller, because the

planet distribution might be tighter. Since the n=3 habitable zone in such a scenario is closer in, n=3

might be in the habitable zone. While we don’t have complete data for brighter stars, like F stars, we do

for dimmer stars because it is easier to obtain and a good example of this is the M2V star TOI 700. We

have found four exoplanets around this star and TOI 700 e is an Earth-like planet in its habitable zone

which is also the third planet. With the equations so far we can solve planetary systems with a method that

is independent of orbit number, n. To solve the wave equation in the case of a protoplanetary disc you

would want the wave equation to be in cylindrical coordinates, to solve it looks like this, the time

independent Schrödinger equation is

22.

r

planet

r

moon

=

R

star

R

moon

−

ℏ

2

2 m

[

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

E

n

= −

Z

2

(k

e

2

)

2

m

e

2ℏ

2

n

2

r

n

=

n

2

ℏ

2

Z k

e

2

m

e

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

R

⊙

R

m

n = 3

−

ℏ

2

2M

e

∇

2

Ψ(ρ , ϕ, z) + V(ρ, z)Ψ(ρ, ϕ, z) = E Ψ(ρ, ϕ, z)

of 10 40

The Laplacian in cylindrical coordinates is

23.

The gravitational potential is

24.

Where is the mass of the central protostar. We can refer to the variables as such:

We separate the variables according to:

25.

Substituting 23 and 25 into 1 we have

26.

Solving it all the way is beyond the scope of this paper. We have hindsight, so we were able to write out

the solutions by form and dimensional analysis. Let us now compute the value of and show that the

units work…

∇

2

=

1

ρ

∂

∂ρ

(

ρ

∂

∂ρ

)

+

1

ρ

2

∂

2

∂ ϕ

2

+

∂

2

∂z

2

V(ρ, z) = −

GM

s

ρ

2

+ z

2

M

s

Ψ(ρ , ϕ, z) = R(ρ) ⋅ Φ(ϕ) ⋅ Z(z)

−

ℏ

2

2M

e

[

1

ρ

d

dρ

(

ρ

d R

dρ

)

Φ ⋅ Z +

1

ρ

2

R ⋅

d

2

Φ

d ϕ

2

⋅ Z + R ⋅ Φ ⋅

d

2

Z

d z

2

]

+ V (ρ, z)RΦZ = E RΦZ

ℏ

⊙

ℏ

⊙

= (hC )K E

e

of 11 40

Where

Where is the radius of a proton, is the mass of a proton, is the speed of light, and is the ﬁne

structure constant. We found this gives the characteristic time of one second in terms of a proton. We

guess the planetary scale is connected to the proton scale because the planets formed from the

protoplanetary disc and are made of different combinations of protons. We derive the value of our solar

Planck constant:

=

27.

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.

hC = 1secon d

C =

1

3

⋅

1

α

2

c

2

3

⋅

π r

p

G m

3

p

r

p

m

p

c

α

C =

1

3

⋅

1

α

2

c

1

3

⋅

2 π r

p

G m

3

p

1

3

⋅

18769

299792458

1

3

⋅

2 π (0.833E − 15)

(6.67408E − 11)(1.67262E − 27)

3

1.55976565E 33

s

m

kg

3

⋅

s

2

kg

m

3

=

s

m

s

2

kg

2

m

2

=

s

m

⋅

s

kg ⋅ m

=

1

kg

⋅

s

2

m

2

1

C

= k g

m

2

s

2

=

1

2

mv

2

= e n erg y

hC = (6.62607E − 34)(1.55976565E 33) = 1.03351secon d s ≈ 1.0secon d s

hC =

(

kg

m

s

2

⋅ m ⋅ s

)

(

1

kg

⋅

s

2

m

2

)

(

kg

m

2

s

)(

1

kg

⋅

s

2

m

2

)

= secon d s

K E

earth

=

1

2

(5.972E 24kg)(30,290m /s)

2

= 2.7396E 33J

ℏ

⊙

= (hC )K E

earth

= (1.03351s)(2.7396E 33J ) = 2.8314E 33J ⋅ s

of 12 40

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

28.

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 22k g)

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

= 1.006pr oton s ≈ 1.0pr oton s = h ydrogen(H )

Z = Z

H

E

5

=

Z

H

5

G

2

M

2

j

M

3

m

2ℏ

2

⊙

Z

H

n = 5

K E

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(2.8314E 33)

2

of 13 40

=

The equation for Saturn is then

29.

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

=

=2.727E36J

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

30.

Where

In this case the element associated with the Earth is calcium which is Z=20 protons.

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 22k g)

3

2(2.8314E 33)

2

K E

e

=

1

2

(5.972E 24kg)(30,290m /s)

2

= 2.7396E 33J

2.727E 36J

2.7396E 33J

100 = 99.5 %

E

3

= 3

Z

2

Ca

G

2

M

2

e

M

3

m

2ℏ

2

⊙

R

⊙

R

m

→ Z

2

of 14 40

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:

31.

We remember the Moon perfectly eclipses the Sun which is to say

32.

Thus equation 31 becomes

33.

The kinetic energy of the Earth is

34.

Putting this in equation 33 gives the mass of the Sun:

35.

We recognize that the orbital velocity of the Moon is

36.

So equation 35 becomes

37.

This gives the mass of the Moon is

38.

Putting this in equation 31 yields

39.

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 15 40

We now multiply through by and we have

40.

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:

41.

We can write 41 as

42.

Where we say

Let us see how accurate our equation 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

L

p

= r

e

v

m

M

e

= r

e

v

m

M

e

= (1.496E11m)(1022m /s)(5.972E 24k g) = 9.13E 38k g ⋅

m

2

s

L

2

p

= r

2

e

v

2

m

M

2

e

= 7.4483E 77J ⋅ m

2

⋅ k g = 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 16 40

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:

This gives:

43.

We remember that

And since,

44.

K E

earth

=

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 )K E

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 22k g)

3

(1.732)

= 321,331.459 ≈ 321,331

of 17 40

Equation 44 becomes

45.

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.

The Protoplanetary Disc If the characteristic time for both the solar system

and the proton

are 1 second, then 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 by using typical values:

Our equation

1secon d = 2r

p

v

m

v

2

p

M

3

m

3

M

2

p

M

⋆

L

p

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

(

1

6 α

2

4πh

G c

)

⋅

r

p

m

p

= 1secon d

2

3

⋅

π r

p

α

4

G m

3

p

1

3

⋅

h

c

= 1secon d

v

rel

σ = π r

2

n ≈ 10

10

− 10

15

r ≈ 10

−6

− 10

−3

v

rel

≈ 1 − 10m /s

t

c

≈

1

(0.32 × 10

12

m

−3

)(π (10

−6

m)

2

(1m /s)

= 1secon d

of 18 40

suggests that the proton’s fundamental structure encodes a natural unit of time, the presence of G, h, and c

may emerge from a balance between gravity, quantum mechanics, and relativistic effects. Since the

equation for the characteristic time of the solar system

indicates that the solar system is quantized by planetary formation processes in a way that maintains a

fundamental unit of periodicity of 1 second. So we are connecting the formation of entire planetary

systems to the fundamental structure of matter itself (the protons). A micron-sized dust grain at about

10E-12 grams, has about 10E11 protons, a millimeter sized dust grain, about 10E-6 grams, has about

10E17 protons. But where did these dust grains come from? They formed in the protoplanetary gas cloud

from elements, mostly hydrogen and helium, and elements like C, O, Si, and Fe, that formed from

hydrogen and helium in stars by nucleosynthesis. The heavier elements are made from hydrogen and

helium in stars then later expelled into space by supernovae. But where did the hydrogen (and some of the

helium) come from? They were made in the Big Bang, so we should be able to track the characteristic

time of the solar system, and proton back to the Big Bang and the formation of the Universe.

The Big Bang In other words, if the planetary system inherits a characteristic time of 1 second from the

dust grains in the protoplanetary disc, and the dust grains inherit a characteristic time of 1 second from the

elements, and elements inherit the characteristic time of 1 second from the proton, then we would guess

that the proton inherits the characteristic time of 1 second from its origins in the Big Bang that gave birth

to the Universe. And, indeed it does:

Big Bang

(time=1second)

Around one second after the Big Bang, the universe had cooled enough that neutrinos decoupled, and

protons and neutrons were forming in equilibrium, this is the moment when baryons (protons/neutrons)

become stable, linking the 1-second time unit to matter itself.

(time=1-3 minutes)

At this time the ﬁrst atomic nuclei (H, He, Li) form.

Thus the 1-second time unit marks when the proton’s number became ﬁxed in the Universe, given by our

equations

(

1

6 α

2

4πh

G c

)

⋅

r

p

m

p

= 1secon d

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

(

1

6 α

2

4πh

G c

)

⋅

r

p

m

p

= 1secon d

of 19 40

which give the radius of a proton when set equal to one another

Or,

Where we say

Where

is a little over a second, and

is a little under a second. These equations contain the fundamental constants related to gravity (G),

quantum mechanics (h), relativity (c), and electromagnetism ( ), constants that cover the fundamental

forces that shaped the early Universe. G, h, c, and control the rate of interaction in the early universe,

including weak interactions, and gravity, which govern proton stability and neutrino decoupling. Thus,

our equations describe the fundamental physics at t=1 second, conﬁrming that this time is deeply

embedded in the structure of the Universe. This means the 1-second characteristic time was imprinted at

the birth of the Universe then inherited by the proton through fundamental constants, planetary systems

(through particles interactions in the protoplanetary discs), galaxies and cosmic evolution because protons

make up most of the universe’s baryonic matter. My equations link proton properties to 1-second, and

protons were ﬁxed in the Universe at 1 second, meaning we could be seeing a universal clock that has

inﬂuenced everything since the Big Bang.

How is it ﬁgured that the protons and neutrons stopped converting into one another and their numbers in

the Universe became set? The idea is that neutrino decoupling (neutrinos stop interacting with one

2

3

⋅

π r

p

α

4

G m

3

p

1

3

⋅

h

c

= 1secon d

r

p

=

2

3

⋅

h

cm

p

r

p

= ϕ

h

cm

p

ϕ ⋅

π r

p

α

4

G m

3

p

1

3

⋅

h

c

= 1secon d

2

3

⋅

π r

p

α

4

G m

3

p

1

3

⋅

h

c

= 1secon d

ϕ ⋅

π r

p

α

4

G m

3

p

1

3

⋅

h

c

= 1secon d

α

of 20 40

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 happens when the temperature of the universe is which occurs at 1 second after the Big

Bang. We know this because temperature evolves with time as

Meaning that the universe had cooled to 1MeV after 1 second. Before decoupling, the universe was so

dense that protons and neutrons were constantly interconverting but because weak interaction stops

maintaining neutron-proton equilibrium at 1 second, the proton to neutron ratio freezes in. Neutrons are

unstable and decay with a half life of about 10 minutes, however when a few minutes after the bang they

start forming, with protons, helium-4 nuclei, they become stable. The protons don’t decay rapidly so you

end up with, after freezing, 6 times more protons than neutrons in the Universe, this explains why 25% of

mass of the universe is helium. It is about 75% hydrogen.

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

G

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

T

decoupling

= (G

2

F

M

Pl

)

−1/3

T = 1MeV

T ∝ t

−1/2

H

2

=

8π G

3

ρ

ρ ∝ T

4

H ∝

T

2

M

Pl

M

Pl

≈ 2.4E18G eV

of 21 40

Since

we have

We said protons and neutrons are set in the universe when it has cooled in its expansion to 1MeV. We

have

This was done in Planck units where time can be expressed in inverse energy. Since in Planck units

we have

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

t ∝

1

H

t ∝

M

Pl

T

2

t ∝

2.4E18GeV

(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

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

L

earth

=

4

5

π M

e

f

e

R

2

e

P(R) = P

0

(

R

0

)

−L

ear th

ℏ

⊙

ℏ

⊙

= t

c

K E

e

t

c

= 1secon d

L

earth

ℏ

⊙

= p

ℏ

⊙

t

c

ℏ

⊙

= t

c

K E

e

of 22 40

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). The radius of the planet’s moon we suggested is given

by a perfect eclipse:

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:

v

e

=

GM

e

r

e

K E

earth

=

1

2

M

e

v

2

e

t

c

t

c

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

M

m

L

earth

R

e

=

2R

2

⊙

r

e

r

m

= R

⊙

Ag

Au

=

R

⋆

(1.8)

R

⋆

R

m

=

r

p

r

m

L

earth

=

4

5

π M

e

f

e

R

2

e

L

e

=

4

5

π (5.972E 24k g)

1

(86400secon d s)

(6.378E6m)

2

7.07866672E 33 J ⋅ s

v

p

=

GM

⋆

r

p

=

(6.674E − 11)(1.989E 30k g)

(1.496E11m)

= 29,788.24m /s

of 23 40

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:

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

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

=

(7.07866672E 33 J ⋅ 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

R

m

= R

⋆

r

m

r

p

= (6.957E 8m)

3.865E 8m

1.496E11m

= 1.79738E 6m

V

m

=

4

3

π R

3

m

=

4

3

π (1.79738E 6m)

3

= 2.432E19m

3

ρ

m

=

7.213E 22k g

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

⊙

of 24 40

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.

//

// 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);

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 22k g)

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 22k g)(1,015.5m /s)

2

= 3.719E 28J

K E

m

K E

e

(Pl a n etDa y) =

(3.719E 28J )

(2.65E 33J )

(86,400sec) = 1.2secon d s

of 25 40

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;

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;

of 26 40

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("Angular Momentum of Planet: %f E33 \n", L_p/

1E33);

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("hbarstar: %f E33 Js \n", hbarstar/1E33);

printf("characteristic time: %f seconds\n", t_c);

printf("\n");

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");

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;}}

of 27 40

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

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. The average pressure exponents by spectral

class are given in the following table…

r

m

= R

⊙

Ag

Au

=

R

⋆

(1.8)

R

e

=

2R

2

⊙

r

e

of 28 40

Now we run the program using this for examples of three spectral types…

of 29 40

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 30 40

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 31 40

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 32 40

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,…

1secon d =

K E

m

K E

e

(Ear th D a y)

ℏ

⊙

= (1secon d )K E

e

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

of 33 40

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 34 40

The results are

of 35 40

of 36 40

Discussion 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, 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.

1 Second Encodes The Core Element Of Life, Carbon I have found that the basis unit of one second is

not just a Natural constant for physical systems like the atom and the planets around the Sun but for the

basis of biological life, carbon, that 1 second produces carbon, six protons and in turn the hydrocarbons,

the chemical skeletons of life. I found

1).

2).

(1.00267E − 13AU/year)(1E 9years) = 0.0001AU

1secon d =

1

6 α

2

⋅

r

p

m

p

4πh

G c

1secon d =

K E

m

K E

e

(Ear th Da y)

of 37 40

Where is the ﬁne structure constant, is the radius of a proton, is the mass of a proton,

is Planck’s constant, is the constant of gravitation, is the speed of light, is the kinetic energy of

the Moon, is the kinetic energy of the Earth, and is the rotation period of the Earth is

one day.

The ﬁrst can be written:

3).

4).

From which instead of saying the left sides of these equations are seconds, we say they are proton-

seconds by not letting the units of cancel with the bodies of these equations on the left, which are in

units of mass, but rather divide into them, giving us a number of protons. I say this is the biological

because these are the hydrocarbons the backbones of biological chemistry. We see they display sixfold

symmetry. I can generate integer numbers of protons from the time values from these equations for all of

the elements with a computer program. Some results are:

A very interesting thing here is looking at the values generated by the program, the smallest integer value

1 second produces 6 protons (carbon) and the largest integer value 6 seconds produces one proton

(hydrogen). Beyond six seconds you have fractional protons, and the rest of the elements heavier than

carbon are formed by fractional seconds. These are the hydrocarbons the backbones of biological

chemistry. And carbon is the core element of life. We see the duration of the base unit of measuring time,

1 second, given to us by the ancients (the base 60, sexagesimal, system of counting of the Sumerians who

invented math and writing and started civilization), is perfect for the mathematical formulation of life

chemistry. Here is the code for the program, it ﬁnds integer solutions for time values, incremented by the

program at the discretion of the user:

α = 1/137

r

p

m

p

h

G

c

K E

m

K E

e

Ear th Da y

1

α

2

m

p

h 4π r

2

p

G c

= 6pr oton ⋅ secon d s = carbon(C )

1

α

2

m

p

h 4π r

2

p

G c

= 1pr oton ⋅ 6secon d s = hydr ogen(H )

m

p

of 38 40

#include <stdio.h>

#include <math.h>

int main(int argc, const char * argv[]) {

int n;

ﬂoat value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11, c=299792458,protons[100],r=0.833E-15;

do

{

printf("By what value would you like to increment?: ");

scanf("%f", &increment);

printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");

scanf("%i", &n);

}

while (n>=101);

{

for (int i=0; i<n;i++)

{

protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));

int intpart=(int)protons[i];

ﬂoat decpart=protons[i]-intpart;

t=t+increment;

if (decpart<0.25)

{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);

}}}}

We have that

Since this 6 seconds is also proton-seconds we have

5). is carbon (C)

6). is hydrogen (H)

Conclusion We see that the proton and the planetary system can be described by a common characteristic

time of about 1 second to conveniently predict the proton’s radius and Earth orbit, possibly bridging

macroscopic and microscopic scales. We see this happens if the ground state of the solar system is given

by the mass of the Earth’s moon. It may be this is because the Earth-Moon system is the pivotal point in

our solar system where angular momentum is quantized for the whole system. We see that the condition

for the Moon perfectly eclipsing the Sun may have something to do with the optimal habitability of our

Earth, as the relationship is inherent in the mathematical structure of the Solar System. In short, we can

encode everything in the Universe from the Big Bang, to the proton, to the protoplanetary disc, to the

Solar System, to life with one unit, a second of time.

1

α

2

⋅

r

p

m

p

4π h

G c

= 18769

0.833E − 15

1.67262E − 27

4π (6.62607E − 34)

(6.67408E − 11)(299,792458)

= 6.029978s ≈ 6s

1

6pr ot o n s

⋅

1

α

2

⋅

r

p

m

p

4π h

G c

= 1secon d

1

1pr ot o n

⋅

1

α

2

⋅

r

p

m

p

4π h

G c

= 6secon d s

of 39 40

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The Author