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The Intersection Of An Atomic And Planetary Science With A Habitable Star System

Ian Beardsley

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Contents

Abstract…………………………………………………………………3

Important To Point Out……………………………………………….4

Introduction………………………………………………………….…5

1.0. The Theory………………………………………………….……10

2.0 The Proton………………………………………………………..13

3.0 The Solar Formulation………………………………………….17

4.0 Equating The Lunar And Solar Formulations

Yield Our 1 Second Base Unit……………………………………20

5.0 The Solutions For Jupiter and Saturn………………………22

6.0 The Origin of a Second………………………………………..25

7.0 Integrating Analytic and Wave

Solutions of the Solar System……………………………………31

8.0 A Theory For Biological Hydrocarbons…………………….36

9.0 The Radius of a Planet…………………………………….….41

10.0 Intersection Of Optimal Life Conditions

At Earth-like Star Systems……………………………………….43

11.0 Lunar Eclipses………………………………………………..47

12.0 Ideal Way to Model Habitable Star Systems……………53

13.0 Modeling With A Computer……………………………..69

14.0 Perfecting The Ideal Modeling……………………………77!

Appendix 1…………………………………………………………..90

Appendix 2 (Data Used In This Paper)………………………….92!

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Abstract

The rudiments for a theory of planetary and atomic systems bridging macro-scales (planetary

systems) with micro-scales (the atom’s proton) is achieved through a wave equation solution of

our Solar System that is used to describe atoms in quantum mechanics. We find the base unit of

time, the characteristic time, in the solution for our Solar System and the atom comes out to be

the base unit of time developed since ancient times by the ancient Sumerians and Babylonians

of one second. Applying this theory to multiple star systems reveals that the optimal conditions

for life may be intersecting at Earth-like star systems. It is suggested that stars that host planets

that are good hosts for intelligent life to develop begins at around spectral class F stars to

spectral class K stars on the main sequence with the optimal conditions intersecting between

these two around spectral class G stars in the area of our star, the Sun. Moons of planets are

shown to play a key role in the wave solution for the star systems and turn out to define their

ground state if they are habitable. Our moon plays a primary role in allowing for life on the

Earth, stabilizing its orbit allowing for the seasons. It may be the conditions for complex life

maximize during the epoch where a moon perfectly eclipses the star it orbits as seen from the

habitable planet which determines some forms of the equations that give the characteristic time

of the star system at or around 1 second. The characteristic time of 1 second common to the

Solar System and atomic systems appears to be characteristic, as well, of hydrocarbons the

skeletons of life chemistry in a context of sixfold symmetry. An in depth look at the moon

eclipsing the Sun as seen from the Earth over geologic times is given. In section 12.0 and 13.0 we

present the ideal method for modeling habitable star systems with our theory. While this set of

equations diverges with another set and we have to introduce correction factors, we find after

perfecting the method in the next section (14.0) that it works across all of the equations very

well. There is not one kind of planet for any particular orbit that is the only one possible; you can

have many kinds of planets form for a given type of star. We conclude that the intersection of an

atomic and planetary science happens with habitable systems like the one that belongs to our

Sun, a so-called G2V star system and within a range from F to K stars, the G stars being of the

type that includes our Sun, which is in-between the two. This seems to happen in a star system

where there are perfect eclipses of the star by the moon, as occurs in our Solar System.

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Important To Point Out

You can speak of the structure of the solar system even though it changes with time. This is

important to understand when I refer to sizes of the Moon and the planets, and their orbital

distances.

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 (it has a thickness). 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.

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

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Introduction I have a theory that could be considered a partial, crude theory of everything. It

presents a solution to quantum mechanic’s Schrödinger wave equation used for atomic systems,

but for planetary systems, in particular for our star system, the Solar System. Also, the theory

solves the atom’s proton and shows a common characteristic time of about 1 second for both

these systems one on the macroscopic scale, star systems, and the other on the microscopic

scale, atomic systems, through the atom’s proton. Also, in common, it describes hydrocarbons,

the chemical skeleton’s of life, suggesting that life could be part of a universal process in the

Universe. As such, we have a partial theory of everything, but I say partial because it does not, so

far, go into particle physics in depth, for instance in treating the quarks that make up protons, or

other particles such as photons and electrons.

We want to apply the equations in the theory for our Solar System to other star systems, to see

what they don’t explain that they do explain in ours, and from that difference find relationships

between our star system and others which is where I think we will find not just our purpose but

the interconnected purposes between all life in the Universe. I would assume the equations I

have found for our star system where the Sun is a yellow, main sequence, spectral class G star

would apply for most such stars. But here is an example of how the civilization of one kind of

star system could have a relationship with our civilization in terms our star system, the Solar

system…

We consider the HR diagram (below) that plots temperature versus luminosity of stars. We see

the O, B, A stars are the more luminous stars, which is because they are bigger and more

massive and the the F, G stars are medium luminosity, mass, and size (radius). Our Sun is a G

star, particularly G2V, the two because the spectral classes are divided up in to 10 sizes, V for

five meaning main sequence, that it is part of an S shaped curve and is in the phase where it is

burning hydrogen fuel, its original fuel, not the by products. And we see the K and M stars are

the coolest, least massive, least luminous.

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We consider that the radius of our Sun, , is about 1.8 times longer than the orbital radius of

the Moon, . We notice that the mass of a gold atom to the mass of a silver atom is about

1.8 as well which is the same as comparing the molar mass of gold (Au=196.97 g/mol) to the

molar mass of silver (Ag=107.87 g/mol). We have

Gold (Au) and Silver (Ag) have been the primary metals for ceremonial jewelry for Earth

civilizations since ancient times. Our Sun is a spectral Class G2V. One spectral class above that is

spectral class F. A spectral class star has a radius of about 1.7 solar radii, written . We

have the radius of an star

As it would turn out, the mass of a silver atom, Ag, to the mass of a copper atom, Cu (comparing

their molar masses) is about 1.7. If we are to write

4. Then, and we have

Suggesting perhaps the inhabitants of stars are copper and silver type civilizations (where

ceremonial jewelry is concerned) where we, the inhabitants of our star, are a silver and

gold type civilization (where ceremonial jewelry is concerned). This would actually make sense

because F0V stars generally have lower metallicities than stars like ours, G2V. Copper and silver

being lighter than gold are often more easily formed in environments with lower metallicity.

They would be more accessible for a civilization around an F0V star, making them favored for

craftsmanship. We also have since (equation 2)

Which is to say that the orbital radius of the F0V system moon is the radius of our Sun, .

Something that is further interesting here is that the habitable zone of an F0V star is about as far

from the star as the center of the asteroid belt is from our star, the Sun. This is because F0V are

more massive and more luminous, as well as larger. Such a star has a mass of 1.61 solar masses

(about the golden ratio of the Earth’s Sun), a radius of 1.728 solar radii (which is about Ag/Cu),

and a luminosity of about 7.24 solar luminosities. By the inverse square law for luminosity such

a star has its habitable zone at

⊙

OurMoon

⊙

OurMoon

F 0V

1.7R

⊙

F 0V

F0Vstar

= 1.7R

⊙

F0Vstar

⊙

F0Vstar

Our Moon

F05star

Our Moon

F 0V

G2V

F0Vstar

= 1.7R

⊙

1.7R

⊙

= 1.7r

F0Vmoon

⊙

= r

F0Vmoon

⊙

habitable

= 7.24L

⊙

= 2.7AU

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The habitable zone our star, the Sun, is where Earth is, 1AU. The asteroid belt in our solar

system is at 2.2 to 3.2 AU. Its center is then at 2.7AU. This may represent a shift in concept for a

habitable planet around the F0V star. The asteroid belt in our solar system is where a planet

can’t form. The same distance from an F0V star is where not just a planet can form, but where a

habitable planet can form.

Stars like F0V stars would not be as good for hosting life as G2V stars like our Sun because they

are bigger and thus hotter meaning they burn up quick leaving less time for intelligence to

evolve and become technological. G2V stars like our Sun come just after them, however it is

possible that even better than G stars like our Sun are those that come after them in spectral

type, the so-called K stars known as orange dwarfs which have even better stability and longer

lifespans. Interestingly, we can’t apply our theory of ceremonial jewelry to them because these

elements, group 11 elements, started before the G star with the F star with copper and silver,

which were followed by silver and gold for G stars like our Sun and to go to a K star we have to

go to the next element heavier than gold in group 11 and there is only one more which is

Roentgenium (Rg). It is extremely radioactive and can only be created in the laboratory. If we

move along to the next spectral class, the M stars, they are brown dwarfs and while they can

have life and are stable for even longer than K stars (are less massive and cooler) their habitable

zones are so close in to the star that when they orbit the star they become tidally locked meaning

their orbital period equals their rotation period, which is saying they always have the same side

facing the star, and the same side turned away. This means they are very hot on one side and

very cold on the other meaning life is optimal only at the edge between one side and the other, in

the twilight between always day and always night. This would not be good for doing astronomy

and learning about the Universe and your place in it. M red dwarfs are the most common stars

in the Galaxy, if not Universe, and because they are not very bright it is easier to detect planets

around them. Most of the planets we have found so far are in such star systems. However, we

hope with the advent of the James Webb Space Telescope we will be able to detect Earth-like

planets around Sun-like stars like ours. Thus we would conclude that planets with intelligent life

are mostly in the F/G/K region of the HR diagrams. We might suggest that copper-silver F star

life is less sophisticated than silver-gold G star life like ourselves, but that we are less

sophisticated than K star life, which might be the most sophisticated kind. We can look at the

stars that come before F stars, the A stars, such as Vega. These stars can have gas giants in their

habitable zones with moons that have habitable conditions. We want the radius of the moon in

the F0V star system for the habitable planet. We can get this value because we are guessing, as is

true in our Solar System that the moon as seen from the habitable planet perfectly eclipses the

star as our moon does with our Sun as seen from the Earth. We also know that the orbital radius

of the Moon in this system is equation 6, .

The condition for a perfect eclipse is

Which becomes

Which is

⊙

= r

F0Vmoon

planet

moon

star

moon

F0Vplanet

F0Vmoon

F0Vstar

F0Vmoon

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10.

Which is exactly the value we wanted to be able to determine because as we will see when I

present my theory that the moon of the habitable planet plays a key role in the wave solution to

the star system. We see that the Moon of the Earth is a natural yardstick. For instance it gives

the radius of our Sun in lunar radii as 400. We see this computation from the radius of the moon

of the F0V star is correct because

11.

We also see

12.

We have that (F0V moon 1.2 times bigger than ours).

We have as well

13.

Making the inhabitants of these F stars our cousins, where we are of the inhabitants of the G

stars. We now want to know how the calendar of F star copper-silver civilizations work. To do

that we need to know the orbital period of their planet, and the orbital period of their moon. We

use Kepler’s law.

The mass of the star is interestingly about the golden ratio of our Sun’s mass, which goes to

show another connection between the copper-silver civilizations and silver-gold civilizations. We

have

(1.61)(1.989E30kg)=3.2E30kg

F0Vmoon

= R

F0Vstar

F0Vmoon

F0Vplanet

= 2.7AU(1.496E11m /AU ) = 4E11m

F0Vstar

= (1.7)(6.96265E 8m) = 1.18365E 9m

⊙

= r

F0Vmoon

= 6.96265E 8m

F0Vmoon

= (1.18365E 9m)

(6.96265E8m)

4E11m

= 2.060E6m

F0Vstar

F0Vmoon

1.18365E 9

6.96265E8m

= 1.69999 ≈ 1.7 =

⊙

Our Moon

6.96265E8m

3.84E8m

= 1.8132 =

F0Vmoon

Our Moon

2.060

1.74

= 1.184 ≈ 1.2

F0Vmoon

Our Moon

4π

F0Vstar

F0Vplanet

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Its planet has year of about 3 and half Earth years. Now we find the orbital period of its moon:

I find if taking the mass of the planet to be a little less than the mass of the Earth — the mass of

the earth is 5.972E24kg — we get 16 moons per the planet’s year. I think this is what we want. If

we use the Earth mass we get 18.82 moons. I find that value is 5E24 kg.

I want 16 moons because my guess is that there are two arrays upon which Nature is founded: 12

(3 by 4) and 16 (4 by 4). This occurs not just in physics but in music. The dynamic 12 of

flamenco which is a rhythm of two three’s plus three two’s, and the rumba is four four’s, as well

as the crux of North Indian classical music which is tin tal played in four, four times, or we also

have the 12 bar blues. We have the standard model of particles physics, pictured below, the

particles are 6 quarks (lavender) and six leptons (green) make six two times is twelve, and we

have the four exchange particles responsible for the forces (in red). Adding these four to the 12

make 16. So the F-star copper-silver civilizations would be 16 moons per year, and the G-star

silver gold civilizations would be 12 moons per year, which is what we are.

The twelve moons divided by the 16 moons is 0.75=3/4 and 3

times four is 12. Three to four would be the relationship between

copper-silver civilizations and silver-gold civilizations. There may

be a reason for it, and it may be rooted in our intertwined

destinies that we will learn about when we achieve interstellar

space travel and visit these worlds.

4π

F0Vstar

4π

(6.674E − 11)(3.2E 30kg)

= 1.8485E − 19

F0Vplanet

= (1.8485E − 19)(4E11m)

= 1.087676E 8secon d s

F0Vplanet

= 3.44664years

4π

F0Vplanet

F0Vmoon

4π

(6.674E − 11)(5E 24kg)

(6.96E8m)

= 6315615secon d s

F0Vplanet

F0Vmoon

1E8secon d s

6315615second s

≈ 16m oons

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1.0 The Theory In my theory, that will be presented in this section and the next, the solutions

to the Schrödinger wave equation for the atom applied to the Earth/Moon/Sun system, where

the Schrödinger wave equation in spherical coordinates is:

1.1

and has solutions for the atom as guessed at by Bohr before the wave equation was discovered

1.2

1.3

are

1.4

1.5

Where G is the universal constant of gravitation, is the kinetic energy of the Earth in its

orbit around the Sun, is the mass of the Earth, is the mass of the Moon, is the orbital

radius of the planet, the orbital number (is three for Earth), is the radius of the Sun, is

the radius of the Moon, and is a Planck constant associated with the Solar System and is

given by

1.6

1.7

Where

1.8

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

fine structure constant. I found this gives the characteristic time of one second in terms of a

proton (Equation 1.7). We guess the planetary scale is connected to the proton scale because the

planets formed from the protoplanetary disc are made of different combinations of protons. We

derive the value of our solar Planck constant

−

ℏ

[

∂

∂r

(

∂

∂r

)

sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

sin

∂

∂ϕ

]

ψ + V(r)ψ = E ψ

= −

)

2ℏ

ℏ

Z k

K E

= n

⊙

⋅

2ℏ

⊙

2ℏ

⊙

⋅

⊙

⋅

K E

⊙

ℏ

⊙

ℏ

⊙

= (hC )K E

hC = 1secon d

C =

⋅

π r

G m

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1.9

Where we used the maximum orbital velocity of the Earth (perihelion). We notice the kinetic

energy of the Earth is given by not just the mass of the Earth, but by the mass of the Moon

looking at equation 1.4:

is the orbital number of the Earth, which is 3. You will notice the radius of the sun, , and the

radius of the Moon, , factors into the solution, which is because we are dealing with not just

an atom, but a planetary orbital system, which forms not just from the inverse square law of the

Newton’s gravity due to the Sun like occurs in the atom with electric fields, but has the

rotational centripetal force of the protoplanetary cloud from which the planets formed, and the

outward radiation pressure from the Sun. We guessed that the mass of the the Moon figured in

instead of the Sun, and that the characteristic time for determining the Planck-type constant for

the solar system in terms of the kinetic energy of the Earth was 1 second because I had found

that it is approximately true that:

C =

⋅

2π r

G m

⋅

18769

299792458

⋅

2π (0.833E − 15)

(6.67408E − 11)(1.67262E − 27)

1.55976565E 33

⋅

kg ⋅ m

⋅

= kg

= energ y

hC = (6.62607E − 34)(1.55976565E33) = 1.03351secon d s ≈ 1.0secon d s

hC =

(

⋅ m ⋅ s

)

(

⋅

)

(

)(

⋅

)

= secon d s

K E

earth

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

= 2.7396E 33J

ℏ

⊙

= (hC )K E

earth

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

K E

= n

⊙

⋅

2ℏ

⊙

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1.10

Where is the kinetic energy of the Moon and EarthDay is the the rotation period of the

Earth (about 24 hours). It in fact may be that the fact that the Moon as seen from the Earth

nearly perfectly eclipses the Sun is a condition for optimal habitability of the Earth. That is for

habitable planets:

1.11

That is the orbital radius of the planet (Earth) to the orbital radius of its moon (The Moon) is

about equal to the the radius of the star (The Sun) to the radius of its moon (The Moon). We say

the system is quantized by the Moon and the base unit of one second. It is a fact that the Moon

orbiting the Earth optimizes the conditions for life on Earth because it holds the Earth at its

inclination to its orbit around the Sun allowing for the seasons, preventing extreme hot and

extreme cold.

We find our unit of one second pans out nearly perfect in deriving the ground state for the Solar

System by writing

1.12

Where c is the speed of light. We consider equation 1.4

1.4

We want to eliminate from the equation because we don’t know if it is the same for star

systems other than ours, the Solar System. 1.12 and 1.4 yield

1.13

Using the fact the the orbital velocity of the Earth is

And,

We have

1second =

K E

(Ear th Da y)

K E

planet

moon

≈

star

moon

ℏ

⊙

1second

⋅

K E

= n

⊙

⋅

2ℏ

⊙

ℏ

⊙

K E

= 3

⊙

⋅

(1sec)c

⊙

K E

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1.14

We use equation 1.11

It is

1.15

Which can be written

1.16

So equation 1.14 becomes

1.17

Which can be written

1.18

We now want to write this in a general form not just Earth mass but planet mass, and not just

solar mass but stellar mass. We have

1.19

2.0 The Proton I said my theory predicts the Earth/Moon/Sun system and the atom’s proton

to have a characteristic time of a second in common. Let’s show this more explicitly than I

presented so far. I found 1 second is the base unit of the orbital dynamics of the solar system. I

find I can create three spacetime operators, one that acts on the radius of the proton to its mass,

and the other that acts on the radius of the Sun to its mass, and one that acts on angular

momentum to speed. is Planck’s constant, is the universal constant of gravitation, is the

speed of light, is the fine structure constant, is the radius of a proton, is the mass of a

proton, is the mass of the Moon, is its radius, and the EarthDay is the rotation period of

the Earth:

⋅

⊙

= (1secon d )c

planet

moon

≈

star

moon

≈

⊙

= R

⊙

⋅

= (1secon d )c

(1sec)c

⋅

⊙

⋅ r

(1sec)c

⋅

⋆

⋅ r

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2.1

2.2

2.3

They seem to suggest that the Earth orbit might by quantized in terms of the Moon and a base

unit of one second, which we will see can be considered a characteristic time of the Universe.

Equation 2.2 is actually 1.3 seconds but rounds to one second. However it is derived from

another equation that gives 1.2 seconds. Namely, equation 1.10:

It says one second is the Earth Day (completion of one rotation of the Earth) adjusted by the

kinetic energy of the Moon, to the kinetic energy of the Earth.

We see the spacetime operators solve the atom by giving us the radius of a proton. We set

equation 2.1 equal to equation 2.3

2.3

2.1

These two yield

2.4

I find this is close to the experimental value of the radius of a proton. I find I can arrive at this

radius of a proton another way, energy is given by Planck’s constant and frequency

We have

(

6α

4πh

)

⋅

= 1secon d

(

(Ear th Da y)

)

⋅

⊙

= 1secon d

⋅

π r

G m

⋅

= 1secon d

1second =

K E

(Ear th Da y)

⋅

π r

G m

⋅

= 1secon d

(

6α

4πh

)

⋅

= 1secon d

⋅

E = h f

f = 1/s, h = J ⋅ s, h f = (J ⋅ s)(1/s) = J

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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

2.5

Where is the golden ratio constant (0.618…). This is more along the lines of more recent

measurements. Both equations 2.4 and 2.5 for the radius of the proton can be right depending

on the dynamics of what is going on; the radius of a proton is not precisely defined, it is more of

a fuzzy cloud of subatomic particles. Thus we have solved the atom with our spacetime

operators by producing the radius of a proton. I began working on this theory when the proton

radius was 0.833 fm so it is what I have been using in this paper. We continue to be honing in on

its experimental value every year with more experiments. The vast gap between the historical

0.877fm and the 2010 0.842fm is known in physics as the proton puzzle.

Indeed over the full range of values for the radius of a proton, the characteristic time is very

close to one second. The large value they got a long time ago gives:

E = J = Joules = energ y

E = m

⋅

≈ ϕ =

⋅

6.62607E − 34

(299,792,458)(1.67262E − 27)

= 0.88094E − 15m

= ϕ ⋅

= 0.816632E − 15m

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The current value, which they believe they have really closed in on as very accurate in two

different experiments in 2019 using different methods, which is , gives a

time of 1.00500 seconds.

Let us consider the radius of a proton is

If we can say that the radius of a proton is always the same value, which we can’t because it

depends on what is going on, we might say it is usually this like the classical radius of an

electron is used in most problems. However, here we are using it as it applies to the formation of

the planets from the protoplanetary disc made-up of these protons. If and when the proton

radius is this, then the spacetime operators are

2.1

2.2

2.3

In which case

=0.9950796 seconds ~ 1 second

(

6α

4πh

)

⋅

= 1secon d

(18769)((0.88094E − 15m)

6(1.67262E − 27kg)

(6.62607E − 34J ⋅ s)(4π)

(6.674E − 11)(299,792,458m /s)

= 1.06284secon d s

= 0.833E − 15m

= ϕ ⋅

= 0.816632E − 15m

(

6α

4πh

)

⋅

= 1secon d

(

(Ear th Da y)

)

⋅

⊙

= 1secon d

ϕ ⋅

π r

G m

⋅

= 1secon d

ϕ ⋅

π r

G m

⋅

= 1secon d

(0.618)

(352,275,361)π (0.833E − 15m)

(6.674E − 11)(1.67262E − 27)

(6.62607E − 34)

(299792458)

of 17 93

The condition for a perfect eclipse:

Which says the radius of the Sun to the radius of the Moon equals the Earth orbital radius to the

lunar orbital radius are equal does indicate that the Earth/Moon/Sun system converges on the

basis unit of one second of time for the solution of the Schrödinger wave equation during the

time we are in where such an eclipse happens. That is the basis unit of time in the solution is

given by

2.6.

It may be this eclipse condition is the condition for the optimization of the habitability of the

Earth. We can show that the epoch that this condition holds gives the unit of a second as the

solution by using it to form a solar solution to the wave equation and equating it with our lunar

solution. Let’s do that now…

3.0 The Solar Formulation 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:

3.1

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

3.2

Thus equation 3.1 becomes

3.3

The kinetic energy of the Earth is

3.4

Putting this in equation 3.3 gives the mass of the Sun:

3.5

We recognize that the orbital velocity of the Moon is

⊙

moon

earth

moon

ℏ

⊙

⋅

= 1secon d

K E

= 3

⊙

⋅

2ℏ

⊙

K E

= 3

⋅

2ℏ

⊙

K E

⋅

⊙

= 3r

⋅

ℏ

⊙

of 18 93

3.6

So equation 3.5 becomes

3.7

This gives the mass of the Moon is

3.8

Putting this in equation 3.1 yields

3.9

We now multiply through by and we have

3.10

Thus the Planck constant for the Sun, , in this the case the star is the Sun, is angular

momentum quantized, the angular momentum we will call , the subscript for Planck. We

have

We write for the solution of the Earth/Sun system:

3.11

We can write 3.11 as

3.12

We say. . That is

⊙

= 3r

⋅ v

⋅

ℏ

⊙

ℏ

⊙

K E

⊙

⋅

⊙

K E

⊙

⋅

⊙

ℏ

⊙

= r

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

= r

= 7.4483E 77J ⋅ m

⋅ kg = 8.3367E 77kg

⋅

K E

⊙

⋅

⊙

K E

⊙

⋅

⊙

2ℏ

⊙

ℏ

⊙

= ℏ

⊙

/2π

ℏ

⊙

= 9.13E 38J ⋅ s

of 19 93

Let us see how accurate our equation is:

We have that the kinetic energy of the Earth is

Our equation has an accuracy of

Which is very good.

We call the same in our solar solution. But we now want our solution for the solar

formulation. is the mass of the Earth and is the mass of the Sun. We have our solution

might be!

3.13 !

Where is !

We have

3.14

⊙

= 2π ℏ

⊙

= 5.7365E 39J ⋅ s

K E

⊙

⋅

⊙

2ℏ

⊙

(6.67408E − 11)

(5.972E 24kg)

(1.9891E 30kg)

2(8.3367E 77kg

⋅

)

⊙

(6.759E 30J )

⊙

6.957E8m

1737400m

= 400.426

K E

= 2.70655E 33J

K E

earth

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

= 2.7396E 33J

2.70655E 33J

2.7396E 33J

= 98.79 %

ℏ

⊙

ℏ

⊙

⋅

⊙

ℏ

⊙

= 9.13E 38J ⋅ s

(9.13E 38)

(6.67408E − 11)(5.972E 24kg)

⋅

400.5986

= 1.4638E11m

of 20 93

This has an accuracy of

Thus the solutions to the wave equation!

for the solar formulations are!

3.15

3.16

4.0 Equating The Lunar And Solar Formulations Yield Our 1 Second Base Unit

Let us equate equation 1.4 with equation 3.12:

1.1

3.12

This gives:

4.1

We remember that

1.4638E11m

1.496E11m

100 = 97.85 %

−

ℏ

[

∂

∂r

(

∂

∂r

)

sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

sin

∂

∂ϕ

]

ψ + V(r)ψ = E ψ

K E

⊙

⋅

⊙

2ℏ

⊙

ℏ

⊙

⋅

⊙

ℏ

⊙

= 9.13E 38J ⋅ s

⊙

= 2π ℏ

⊙

= 5.7365E 39J ⋅ s

K E

= n

⊙

⋅

2ℏ

⊙

K E

⊙

⋅

⊙

2ℏ

⊙

⋅

2ℏ

⊙

⋅

⊙

⋅ ℏ

⊙

ℏ

⊙

= (hC )K E

of 21 93

This gives

4.2

We have

4.3

This equates the orbital velocity of the Moon with the centripetal acceleration of the Earth in

terms of one second by way of the mass of the Earth, the mass of the Sun, the mass of the Moon,

and the orbital number of the Earth. Let us compute

4.4

Let us see how well equation 19 works. at aphelion is 966 m/s and .

. We have

That is an accuracy of

Equation 4.3 can be written:

4.5

hC = 1secon d

C =

⋅

2π r

G m

6α

h 4π r

= 1.004996352secon d s

6α

⋅

h 4π

⋅

⊙

(1second )

⊙

(5.972E 24kg)

(1.9891E 30kg)

(7.34763E 22kg)

(1.732)

= 321,331.459 ≈ 321,331

= 29,800m /s

= 1AU = 1.496E11m

2(966m /s) =

(29,800m /s)

1.496E11m

(1sec)(321,331.459)

1,907m /s = 1,932m /s

1907

1932

= 98.7 %

1second = 2r

⊙

of 22 93

In terms of other star systems this is

4.5

5.0 The Solutions For Jupiter and Saturn

Now 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.

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 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 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

1second = 2r

⋆

E =

2ℏ

⊙

K E

(1.89813E 27kg)(13720m /s)

= 1.7865E 35J

E =

(6.67408E − 11)

(1.89813E 27kg)

(7.347673E 22kg)

2(2.8314E 33)

E =

(3.971E 35J ) = Z

(1.776E 35J )

1.7865E 35J

1.776E 35J

= 1.006pr otons ≈ 1.0protons = hydrogen(H )

Z = Z

of 23 93

5.1

Where is the atomic number of hydrogen which is 1 proton, and for the orbital

number of Jupiter, . Now we move on to Saturn…

The equation for Saturn is then

5.2

It makes sense 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

2ℏ

⊙

n = 5

K E

(5.683E 26kg)(10140m /s)

= 2.92E 34J

E =

(6.67408E − 11)

(5.683E 26kg)

(7.347673E 22)

2(2.8314E 33)

2.45

(3.5588E 34J ) = Z(1.45259E 34J )

Z(1.45259E 34J ) = (2.92E 34J )

Z = 2pr oton s = Helium(He)

2ℏ

⊙

= n

⊙

2ℏ

⊙

6.96E8m

1737400m

= 400.5986

= (1.732)(400.5986)

(6.67408E − 11)

(5.972E 24kg)

(7.347673E 22kg)

2(2.8314E 33)

K E

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

= 2.7396E 33J

of 24 93

Which is very good, about 100% for all practical purposes. The elemental expression of the

solution for the Earth would be

5.3.

Where

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

2.727E 36J

2.7396E 33J

100 = 99.5 %

= 3

2ℏ

⊙

→ Z

of 25 93

6.0 The Origin of a Second It is worth looking at ancient systems of looking at time because

that is where the origins of our current systems began. Indeed our system came from the ancient

Sumerians and Babylonians. They divided the day into 12 units where the day is given by the

rising and setting of the Sun, which in turn is given by the period of the rotation of the Earth.

Thus the day was given by the period from the rising of the Sun to its setting, which was divided

into 12 units which we call, today, hours. Thus the day from sunrise to sunrise, or sunset to

sunset, is 24 hours. Why they chose 12 units could come from the fact that 12 is the smallest

abundant number, which means it has a lot of divisors: 1, 2, 3, 4, 6. Abundant means their sum

is greater than 12 itself: 1+2+3+4+6=16. We know for certain they chose 12 because there are

three sections on each finger, so with 4 such fingers your can touch each such section with your

thumb to count to twelve. The Babylonians got base 60 from the Sumerians and further divided

each hour into 60 minutes, and each minute into 60 seconds. Why base 60 was chosen is that it

has a lot of divisors as well, including the first 6 integers. It is the smallest number that does

this.

We see dividing the day into 24 hours and dividing that further with base 60 leads to the

duration of a second we have today. Only on planet Earth would we have the primitive, ancient

origins of our mathematics in the end line up with modern physics in that, as it would turn out,

it gave us our basis unit of a second to be a natural constant. Though we could guess that on

other planets the ancient civilizations when first inventing mathematics and astronomy, would

use base 60 because it is so convenient for doing math being evenly divisible by the first 6

integers. However, we did that and combined it with divisions of 24 units. Not necessarily would

any planet do that. However, we had other reasons to do that; the Moon orbits the Earth close to

12 times in the time it takes the Earth to go once around the the Sun. However, there is an

ancient system where the people didn’t divide the day into 24 units, but rather into 60 units,

meaning that their hour was 24 minutes long. That is

and . This

was the Vedic time-keeping system in ancient India. Leave it to the Hindu Indians to have

extraterrestrial intelligence in their ancient beginnings. So let’s go into this. We will see that

base 60 combined with 24 describes the angular momentum of the Earth. That means it doesn’t

just include the rotation period of the Earth, but the size of the Earth (its radius) and the mass of

the Earth.

Indeed if the dynamics of the factors the ancients gave us to create the duration of a second are

connected to the dynamics of star systems then the second should define the rotational angular

momentum of the Earth since we divide the rotation period of the Earth into these factors to get

the unit of second, and should be connected to our Planck constant for the solar system which is

in units of angular momentum as well, as is Planck’s constant for the atom.

The angular momentum of the Earth with respect to the Sun is 2.66E40 kg m2/s. The rotational

angular momentum is 7.05E33 kg m2/s. In orbit angular momentum is given by

For a uniform rotational sphere it is given by

We found our solar system Planck constant was

(24h ours)(60min /h our) = 1440min /d a y

1440minutes /60 = 24minutes /h our

L = 2π M f r

L =

π M f r

ℏ

⊙

= 2.8314E 33J ⋅ s

of 26 93

This gives

6.1.

We are now equipped to show that the ancient Sumerians were right in dividing the rotation

period of the Earth (the day) into 24 units (the hour) because

That is

6.2.

Which is to say the angular momentum of the Earth to its Planck constant gives the base 60

counting in terms of the 24 hour day, the 60 of 60 minutes in an hour, and 60 seconds in a

minute, that determine our base unit of duration we call a second that happens to be, as I have

shown, the base unit of the wave solution to the atom and the Earth/Moon/Sun system.

Our equation in this paper for the Earth energy as a solution of the wave equation 1.4

1.4

does not depend on the Moon’s distance from the Earth, only its mass. 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. Equation 6.2

6.2.

only specifies divide the day into 24 units, and hours into 60 minutes and minutes into 60

seconds, regardless of what the Earth rotational velocity is. But it was more or less the same as it

is now when the Sumerians started civilization. But it may be that it holds for when the Earth

day is such that when the Moon perfectly eclipses the Sun, which we said might be a condition

for the optimization of life preventing extreme hot and cold. That is when the following holds

Which holds for today and held for the ancient Sumerians and holds for when the Earth rotation

gives the duration of a second we have today.

We want the basis set of equations for the solar system. We have

earth

ℏ

⊙

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

2.5(24) = 60

earth

ℏ

⊙

24 = 60

K E

= n

⊙

⋅

2ℏ

⊙

earth

ℏ

⊙

24 = 60

planet

moon

≈

star

moon

ℏ

⊙

= 3.0281E 8m

of 27 93

We have from equation

and,

From these it becomes clear that

6.3.

6.4.

Thus combining equation 6.2 with the following…

6.2.

We write

From 6.3 we have

Which gives us

6.5.

This is very accurate to give us a second. But notice 6.262 is approximately . We see that

moon

ℏ

⊙

⋅

= 1.0secon d s

ℏ

⊙

= (hC )K E

hC = 1secon d

ℏ

⊙

K E

ℏ

⊙

earth

ℏ

⊙

24 = 60

ℏ

⊙

= (1secon d )K E

ℏ

⊙

= (1secon d )

ℏ

⊙

ℏ

⊙

= (1secon d )GM

1second =

(

)

earth

(

)

(7.05E 33)

(6.67408E − 11)((7.347673E 22kg)

(299792458m /s)

(

)

6.262sec = 1.002secon d s = 1.00secon d s

2π

of 28 93

6.6.

is the circumference of a unit circle, it could be that the circumference of the Earth orbit, can

be taken as 1 (unity) and essentially we have the mystery of base 60 and the 24 hour day of the

ancient Sumerians is solved, it connects the circumference of a unit circle to 1 (unity).

6.7.

6.8.

The Hindu Vedic System has a day is 60 ghatika of 24 minutes each, each ghatika is divided into

60 palas of 24 seconds each, and each pala is divided into 60 vispalas, each vispala of 0.4

seconds each. So where our system has a base unit of 1 second, theirs has a base unit of 0.4

seconds, so that could be an advantage to their system, a smaller unit of time is more refined.

Further their day is divided into 60 units, ours into only 24, so their hour, the ghatika is only 24

minutes long, and ours is 60 minutes long, The proponents of this system in India say since

when working and doing chores we do a few chores in an hour, they do about one per ghatika is

24 minutes, which makes the measure of time more manageable. That could be an advantage I

think. They say our hour is so long because the lines had to be far apart on the Egyptian Sun Dial

so the shadow cast by the Sun didn’t cross-over onto another line. But today, with modern

technology we can make clock lines marking hours closer together and measure them with a

pointer hand pointing to them without any problems, and the result is we would have a smaller

more refined hour (60 of them in a day as opposed to 24). They suggest this method of

measuring time would work better in science and engineering as well, that we have to get away

from the way sun dials had to be made in Egypt in ancient times.

But they further point out that their system describes Nature. They say theirs are 108,000

vispalas in a day, and 108,000 vispalas in a night giving 216,000 vispalas in a 24 hour day. The

diameter of the Sun is 108 that of the Earth, and the average distance from the Sun to the Earth

is 108 solar diameters, and the average distance from the Moon to Earth is 108 lunar diameters.

108(10)(10)(10)=108,000. Ourselves and them use base 10 counting, and that is probably

because we have ten fingers to count on.

The incredible thing here is that the wonderful equations:

unify the Mesopotamian system with the Hindu system because

1 =

(

)

2π

cos(45

∘

) = cos(π /4) =

earth

ℏ

⊙

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

earth

ℏ

⊙

24 = 60

1vispal a = 0.4secon d s

(0.4s)(2.5) = 1secon d

of 29 93

This suggests that the Hindu system is just as Natural as the Western system.

We have said the ancient Sumerians gave us the duration of a second we have today by dividing

up the rotation period of the Earth into 24 units we call hours, which the Babylonians divided

into 60 minutes and 60 seconds that they got from the Sumerian base 60. We have found in my

theory that the 1 second is a natural constant because it is the base unit of our solution to both

the solar system and the atom. We further found the Hindu’s used similar numbers in an

inverted way to the way of the West that came from the Sumerians, and that their system relates

not just to the West’s but to our theory in a dynamic way concerning the angular momentum of

the Earth and our Planck constant for the Solar system.

The Sumerians of Mesopotamia and the Hindus of the Indus Valley Civilization were of the first

civilizations, but so was Egypt. If the second was in Mesopotamia and in India, it was only

necessary to find it in Egypt. Since the Earth spin has been the natural clock for which the

humans first measured time since ancient times, the rising and setting of the Sun due to it, I

looked at how much distance through which the earth rotates at its equator in one second. It is,

since this distance is

where theta is in radians, and the the radius of the Earth is 6.378E6m. The earth rotates through

360 degrees per 24 hours, or per 86,400 seconds = 0.004167 degrees per second = 7.27E-5

radians. We have

If the second was to exist in ancient times in Egypt we should see it in the Great Pyramids. The

three largest are in a line and are separated by

Khufu to Khafre (0.25 miles)

Khafre to Menkaure (0.3 miles)

We see the second and third pyramids built, Khafre and Menkaure, which were built in the 25th

century BC and were of the the 4th dynasty of the old kingdom, are 0.3 miles apart, the distance

through which the Earth rotates in one second, the one second we found to be a natural constant

at the basis of the the atom and solar system.

The ancient Sumerians say they got there mathematics in their writings from the Gods who

came to Earth from the sky they called the Anunaki. Some have suggested the Anunaki were

ancient aliens. The Ancient Egyptians built enormous pyramids from heavy stones weighing

tons, somehow lifted 50 feet to be stacked in near perfect mathematical proportions. This has

been a deep mystery in Archaeology. Again some suggest ancient aliens here. I have shown all

three of these civilizations have a system of measurement (the Egyptians had a 24 hour day as

well) that is intrinsic to the laws of nature describing the atom and the solar system in modern

times. It can be suggested that the second was given to them by ancient aliens.

A really good find is the following…

s = r θ

s = (6.378E6m)(7.27E − 5ra d ) = 464m = 0.464k m = 0.2883mi ≈ 0.3mi

of 30 93

My theory suggests the Sumerians could have ultimately got the unit of a second for measuring

time from ancient Aliens, they called the Anunaki who the say came from the sky. The same

second I have found is characteristic of our solar system and the proton. Scholars have puzzled

over the the depiction in Sumerian art of the strange watch-like bracelets around the wrists of

the Anunaki gods, who they say came from the sky and gave them mathematics. They have

twelve divisions like our clocks today have because the 12 and 6 o’clock positions have two

pointers running together as one.

of 31 93

7.0 Integrating Analytic and Wave Solutions of the Solar System We would like to see

how our wave solution for the solar system figures into the classical analytic theory of the

formation of our solar system.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. The pressure gradient of

the disc in radial equilibrium balancing the inward gravity and outward centripetal force is

7.1.

We can solve this for pressure in the protoplanetary disc as a function of r, distance from the

star, as follows: Assume the gas is isothermal, meaning the temperature T is constant so we can

relate pressure and density with

Where is the speed of sound in the gas which depends on its temperature. We take the gas to

be in nearly Keplerian rotation. That is the rotation is given by Newtonian gravity:

And we take into account that the rotational velocity is slowed down by gas pressure using the

the parameter which is less than one:

We can say for a protoplanetary disc like that from which our solar system originated that its

density varies with radius as a power law:

is the reference density at and s is the power law exponent. We can write

We have from 7.1:

7.2.

Since , we have that which gives from 2:

⋆

d P

= − ρ

(

⋆

−

)

P = c

⋆

= v

(1 − η)

ρ(r) = ρ

(

)

−s

= v

(1 − η)

≈

⋆

(1 − 2η)

d P

= − ρ

(

⋆

⋅ 2η

)

P = c

d P/dr = c

dρ /dr

of 32 93

We integrate both sides:

And, we have

7.3

We take

as small because is small and r is large so we can make the approximation . We

have

7.4.

What we can get out of this is since the deviation parameter, , is given by

7.5. and

7.6.

dρ

= −

2ηGM

⋆

∫

dρ

= −

2ηGM

⋆

∫

(

)

2ηGM

⋆

(

−

)

ρ(r) = ρ

ex p

2ηGM

⋆

(

−

)

= c

ex p

2ηGM

⋆

(

−

)

2ηGM

⋆

(

−

)

≈ 1 + x

≈ P

1 +

2ηGM

⋆

(

−

)

= c

η = −

(

)

dlnP

dln R

μm

of 33 93

Where, is the Boltzmann constant, is the molecular weight of

hydrogen, and is the mass of hydrogen is basically the mass of a proton is 1.67E-27kg. Since

for a protoplanetary cloud at Earth orbit T is around 280 degrees Kelvin we have

Typically in discs the pressure decreases with radius as a power law

Where , so

7.7.

So, essentially, by the chain rule

to clarify things. The reason 7.7 is significant is that equation

Where

The spacetime operators are…

= 1.38E − 23J/K

μ ≈ 2.3

= 1k m /s

P(R) ∝ R

−q

q ≈ 2.5

dlnP

dln R

≈ − 2.5

η = −

(

1k m /s

30k m /s

)

(−2.5) = 1.5E − 3

dlnP

dln R

dlnP

d R

⋅

d R

dln R

d P

d R

⋅ R =

d P

d R

earth

ℏ

⊙

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

earth

π M

ℏ

⊙

= 2.8314E 33J ⋅ s

moon

ℏ

⊙

⋅

= 1.0secon d s

of 34 93

So we have

7.8.

We have

7.9.

Integrate both sides

If is the reference pressure at the reference radius then the pressure as a function of radius

for the protoplanetary disc from which our solar system formed is:

7.10. or,

Where

from

(

6α

4πh

)

⋅

= 1secon d

(

(Ear th Da y)

)

⋅

⊙

= 1secon d

⋅

π r

G m

⋅

= 1secon d

d P

d R

= −

earth

ℏ

⊙

d P

= −

earth

ℏ

⊙

d R

∫

d P

= −

earth

ℏ

⊙

∫

d R

lnP = −

earth

ℏ

⊙

ln R + C

P(R) = P

(

)

−

ear th

ℏ

⊙

P(R) = P

(

)

ex p

[

−

earth

ℏ

⊙

]

earth

π M

ℏ

⊙

= GM

c(1secon d )

moon

ℏ

⊙

⋅

= 1.0secon d s

of 35 93

(

6α

4πh

)

⋅

= 1secon d

of 36 93

8.0 A Theory For Biological Hydrocarbons 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, that it is in the sixfold Nature of the chemical skeletons from

which life is built, the hydrocarbons. I found

1).

2).

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:

1second =

6α

⋅

4πh

1second =

K E

(Ear th Da y)

α = 1/137

K E

Ear th Da y

h 4π r

= 6pr oton ⋅ secon d s = carbon(C )

h 4π r

= 1pr oton ⋅ 6second s = hydrogen(H )

of 37 93

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:

#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;

{

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)

⋅

4πh

G c

= 18769

0.833E − 15

1.67262E − 27

4π (6.62607E − 34)

(6.67408E − 11)(299,792458)

= 6.029978s ≈ 6s

6pr oton s

⋅

4πh

G c

= 1secon d

1pr oton

⋅

4πh

G c

= 6secon d s

of 38 93

Our Theory For Hydrocarbons With all that has been said we are equipped to proceed. We

want to consider the radius of a hydrogen atom and the radius of a carbon atom. The radius of

a carbon atom given in your periodic table of the elements is often 70 to 76 picometers. The

covalent radius of hydrogen is given as 31 picometers. The atomic radius of hydrogen is 53

picometers and the atomic radius of carbon is 67 picometers. We want to consider the atomic

radii of both, because the covalent radius, determined by x-ray diﬀraction for diatomic

hydrogen, is the size of two hydrogen atoms joined H2 divided by two, where it is measured

that way, joined, in the laboratory. Carbon is C2 divided by 2. We are interested in the single

carbon and hydrogen atoms, because we want to know what our theory for their six-fold

symmetry with one another in their representations in proton-seconds says about the way they

combine as the skeletons of life chemistry. We start with the Planck constant, which is like

ﬂux, a mass (perhaps a number of particles) per second over an area. That is it is kilograms per

second over square area:!

We have equations 1 and 2:!

1). is carbon (C)

2). is hydrogen (H)

We can write these

3).

4). !

We have from 3 and 4…!

5). !

We ﬁnd the ratio between the surface areas of the hydrogen and carbon atoms:!

h = 6.62607E − 34J ⋅ s = 6.62607E − 37

⋅ m

6pr oton s

⋅

4πh

G c

= 1secon d

1pr oton

⋅

4πh

G c

= 6secon d s

(

6.62607E − 37

⋅ m

)

6secon d s

(

6.62607E − 37

⋅ m

)

6secon d s

1.67262E − 27kg

= 2.37689E − 6m

(

6.62607E − 37

⋅ m

)

1secon d

(

6.62607E − 37

⋅ m

)

1secon d

6(1.67262E − 27kg)

= 6.602486E − 8m

⋅

(6seconds)

⋅

(1second)

2.37689E − 6m

6.602486E − 8m

= 35.9999 ≈ 40

of 39 93

6). !

7). !

It is the golden mean. These atomic radii are the radii between the nucleus of the atoms and

their valence shell, which is what we want because the valence shell is the outermost electrons

responsible for the way the hydrogen and the carbon combine to make hydrocarbons. We will

write this!

8). !

I am guessing the reason we have the golden mean here is that it is the number used for

closest packing. But what we really want to do is look at the concept of action, for hydrogen

given by six seconds and carbon given by 1 second. We take equations 1 and 2:!

1). is carbon (C)

2). is hydrogen (H)

And, we write

9).

Where is the radius of the atom, and t its time values given here by equations 1 and 2. We have for

hydrogen

10).

This is actually very close to the radius of a hydrogen which can vary around this depending on how you

are looking at it, which we said is given by 5.3E-11m. For carbon we have:

11).

And this is actually very close to the radius of a carbon atom which is 6.7E-11m. The thing is if we consider

the bond length of the simplest hydrocarbon CH4, methane, which can be thought of as

surface

= 4π (r

) = 3.52989E − 20m

surface

= 4π (r

) = 5.64104E − 20m

4π(53pm)

4π(67pm)

= 0.62575 ≈ 0.618... = ϕ

= ϕ

6pr oton s

⋅

4πh

G c

= 1secon d

1pr oton

⋅

4πh

G c

= 6secon d s

G c

∫

dt = r

G c

∫

dt = (1.67262E − 27kg)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec

(9.2E − 12m /s)(6secon d s) = 5.5E − 11m

G c

∫

dt = (1.67262E − 27kg)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec+1sec

(9.2E − 12m /s)(7secon d s) = 6.44E − 11m

of 40 93

16).

our, equations give

17).

which are the same thing. Thus we have the basis for a theory of everything in that it includes the macro

scale, the Earth/Moon/Sun System, because we had:

and it includes the radius of the proton which is the radius and mass of a proton and gives 1

second

and we have the hydrocarbons the skeleton of the chemistry of life give one second

1). is carbon (C)

2). is hydrogen (H)

and because they predict the radii of the carbon and hydrogen atoms at the core of life

10).

11).

+ r

= 5.3E − 11m + 6.7E − 11m = 1.2E − 11m

+ r

= 5.5E − 11m + 6.44E − 11m = 1.2E − 11m

1second =

K E

(Ear th Da y)

1second =

6α

⋅

4πh

6pr oton s

⋅

4πh

G c

= 1secon d

1pr oton

⋅

4πh

G c

= 6secon d s

G c

∫

dt = (1.67262E − 27kg)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec

(9.2E − 12m /s)(6secon d s) = 5.5E − 11m

G c

∫

dt = (1.67262E − 27kg)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec+1sec

(9.2E − 12m /s)(7secon d s) = 6.44E − 11m

of 41 93

9.0 The Radius of a Planet One of the things we should notice here is that in our theory we

have determined everything about G2V star systems except one thing: the radius of the

habitable planet itself. If we had that, there would be a lot more that we could determine that

would be of great interest. We may be able to do that. The answer may come from ancient

Indian Vedic astrology. To the Hindus, 108 is an important number and, as we talked about in

section 6.0, there are in their system of measuring time 108,000 vispalas in a day, where a

vispala is their base unit of time equal to 0.4 seconds. How does this help? In India they have

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 (the Earth orbital radius) is about 108 solar

diameters. Is it possible that for F, G, and K stars that the following conclusion holds:

In other words is it true that,

Which can be written

9.1.

We have

The actual radius of the Earth is , so this is an accuracy of about

We had found (section 6.0) that the angular momentum of the the earth to the Solar System

Planck constant was

This was related to our time measuring system ultimately given to us by the ancient Sumerians

We can now with the radius of the habitable planet around the star compute its angular

momentum. We have

⊙

= 108

⊙

= 2

(6.96265E8m)

1.496E11m

= 6,481.082m ≈ 6,481k m

= 6,378k m

6,378

6,481

100 = 98.4 %

earth

ℏ

⊙

= 2.5

earth

ℏ

⊙

24 = 60

of 42 93

This gives

9.2.

We see (section 7.0) that for the pressure gradient for the protoplanetary disc from which our

Solar System formed was

7.10.

Which was accurate because the exponent is indeed thought to be 2.5.

planet

π M

planet

ℏ

⋆

= p

P(R) = P

(

)

ex p

[

−

earth

ℏ

⊙

]

of 43 93

10.0 Intersection Of Optimal Life Conditions At Earth-like Star Systems

Our assumption that the moons of planets of stars like F0V stars have about 16 new moons in

their night sky per their year going down to the next spectral class of G0V stars having about

12 new moons in their night sky per their year then down to the spectral class after that of K0V

stars having about 9 new moons per their year in their night sky by way of pivotal matrices,

was reasonable. Because planets in the habitable zones of K and M stars are closer in to those

of G stars due to decreased luminosity. !

The Hill radius, the region where the planet’s gravity dominates that of the star, constrains how

far out a moon can orbit its planet.!

If a K0V star had a planet in the habitable zone with a moon having an orbit of 22.65 days

which gave 9.6 moons in its year which would be about 9 new moons in its night sky over its

year then the planet’s Hill radius must be large enough to allow this orbit. The Hill radius is

given by!

If we model for a K0V star with a PlanetDay characteristic time of almost exactly 2 seconds

(2.042286s) which has an orbital radius of 101463662592m ~ 1.0146366E11m=0.678AU. The

star is 0.88 solar masses and the mass of the planet is 5.699199E24kg. The planet’s day would

be 2 Earth Days.!

The moon’s orbit is well within this limit. It is!

Further, it is reasonable that we suggest a 2 Earth day, day for its day (for its rotation period).

The rotation speed of a planet is set by the accretion material and object impacts and we know

a fast spinning proto-planet could form with a short rotation period. Tidal forces from stars act

to slow planetary rotation, but planets around K0V stars in the the habitable zone are further

from the star than in M-type stars where the tidal forces are weaker (they are at about 0.5-0.8

AU). This reduces the tidal breaking rate allowing the planet to retain a faster rotation for a

= a

(

⋆

)

1/3

⋆

= (0.88M

⊙

)(1.989E30kg/M

⊙

) = 1.75E 30kg

(5.699199E 24kg)

(5.972E 24kg)

= 0.9543Ear th Masses

= (1E11m)

(

5.699E 24kg

3(1.75E30kg)

)

1/3

= 1E9m = 0.00687AU

= a

(

22.65

27.3

)

= (a)

a = 0.88295Lun arRadii(3.84E8m) = 3.4E8m

of 44 93

longer time. An Earth-mass planet has a signiﬁcant moment of inertia so it is less aﬀected by

tidal acceleration than in less massive planets.!

A moon can stabilize the planet’s rotation, but it can also slow its rotation over billions of years.

For a planet to maintain a short rotation period over the 20 to 40 billion year lifespan of a K0V

star seems unlikely, but is plausible over perhaps a few billion years. The time scale for tidal

breaking is given by!

, , ,

(linked to internal heat friction and heat dissipation)!

Tidal breaking is much longer for a planet in the habitable zone of K0V stars than it is for one in

an M0V star. As you can see the braking time is given by the distance of the planet to the sixth

power, so it goes up quick with distance. The habitable zone around 0.7 AU in a K0V star is far

enough to reduce locking timescales signiﬁcantly. For an Earth-like planet (Q~100) it would be

tens of billions of years for tidal breaking to occur, it would take longer than the lifespan of the

star. Thus K0V stars are exciting candidates for hosts to planets with habitable conditions.!

This is because they have masses of to , which means they burn their fuel more

slowly than for GV stars like our Sun and thereby remain in the main sequence for 20-40 billion

years, much longer than our Sun which lasts on the main sequence for about 10 billion years

allowing for more time for life develop and evolve complexity. While K0V stars have more solar

ﬂares and radiation bursts than our Sun does, they have less of this than in M-type stars

making them more suitable for complex life than M-stars in this regard, but less, though, than

stars like our Sun, in this regard. Also, unlike M stars, they don’t emit such powerful, high

frequency solar ﬂares that can strip away a planets atmosphere. And, of course, as we

explained, because their habitable zones are further out than in M stars, tidal locking is

reduced, allowing for more stable climates. Tidally locked planets are common in M stars,

which result in extreme heat on one side of the planet and extreme cold on the other side. K0V

stars emit less ultraviolet light than our Sun does making them better for nurturing the

development of organic molecules on their habitable planets, yet have enough UV to sustain

other necessities life has for it.!

In general a planet must have the right combination of mass and composition for retaining an

atmosphere, the presence of tectonic activity for carbon cycling, and magnetic ﬁelds for

radiation protection. In short, K0V stars would be good hosts for supporting complex life.!

Let’s derive the Hill radius. The gravitational potential due to the planet is!

And, the gravitational potential due to the star is!

breaking

∝

a = Semi Ma jorA xis

= PlanetMass

= PlanetRadiu s

Q = Pla netDissipation Factor

0.8M

⊙

0.9M

⊙

= −

⋆

= −

⋆

of 45 93

We introduce a test particle at the Hill radius where the gravitational potential inﬂuence of the

planet equals that of the star. The magnitude of the tidal force from the star is given by!

This comes from the gradient of the potential in the radial direction ( ). It is !

This is the gravitational ﬁeld of the star. Now we introduce the test particle, in this case the

moon orbiting the planet. It is at a distance from the planet. The separation between the star

and the planet is!

The gravitational potential at these points is!

The tidal force is the diﬀerence between between the gradients at and . For small

compared to expand in a Taylor series around !

Thus it is proportional to:!

We use this latter from the ﬁrst order term in expansion. The centrifugal force of the rotating

planet-star system is!

tidal

∝

⋆



∇Φ

⋆

= −

∂Φ

⋆

∂R



R =

⋆



= a

⋆

= −

⋆

−

⋆

≈

⋆

tidal

≈

2GM

⋆

tidal

≈

⋆

centrifigal

= ω

⋆

centrifugal

⋆

of 46 93

For the Hill radius the moon is at the boundary where the planet’s and star’s gravitational forces

and centripetal forces, balance…!

This is the Hill radius where we have introduced a factor of 3 in the denominator of the cube

root. This is because the factor of 3 arises from a rigorous analysis of the three-body problem,

from the Lagrange points and the volume of stability. It accounts for the geometry and stability

constraints.!

Now that we have made the case for decreasing lunar orbital periods — F0V moon 74 days,

around Earth moon (G0V) 33 days, K0V 23 days — we can proceed to show that the

characteristic time and PlanetDay characteristic time intersects around 1.00 seconds for Earth-

like star systems. This coincides with the characteristic time of the proton, which we found to

be one second as well.#

⋆

2GM

⋆

= a

(

⋆

)

1/3

of 47 93

11.0 Lunar Eclipses Since we have said it may be the optimization of life occurs under the

condition where the moon of a planet perfectly eclipses the star that the planet in the habitable

zone orbits as seen from from the planet, we will look at that condition a bit, which is equation

8 in the introduction!

To look at this we only need to know two things: The Earth loses energy to the Moon such that

the Moon’s orbit grows (recedes) by 0.0000382 kilometers per year, and the Earth day becomes

longer because of this by 0.00002 seconds per year. Before we look at this we want to consider

equation 1.10

1.10

Which is really more accurately represented if we consider the average orbital velocity of the

Moon and the average orbital velocity of the Earth:

Where the EarthDay is 86,400 seconds. The sidereal day is a little less than this, but by so little

there is no point in using it. A variation on 1.10 is 2.2, is

2.2.

We can write this

Or,…

We want it like this to see the base 60 of the ancient Sumerians interacting with their 24 for the

hours in a day. But, we will remember we estimated 1.252 seconds to be 1 second. For the

equation to work perfectly we want to write

planet

moon

star

moon

1second =

K E

(Ear th Da y)

K E

(7.3476731kg)(1,022m /s)

= 3.837E 28J

K E

(5.972E 24kg)(29,780m /s)

= 2.648E 33J

K E

(Ear th Da y) =

3.837E 28J

2.648E 33J

(86,400s) = 1.252secon d s

(

(Ear th Da y)

)

⋅

⊙

= 1secon d

⊙

⋅

1second

Ear th Da y

⊙

⋅

1second

86400sec

⋅

(24)(60)(60)

of 48 93

The Appearance of Photosynthesizers And An Oxygen Rich Atmosphere

We ask: How many years ago was the Earth day 18.794 hours long? Convert it to seconds

The Earth’s rotation slow by 0.00002 seconds per year. We have

The Earth day, then, was 18.794 hours long about 937 million years ago. Thus, how much closer

was the Moon to the Earth at this time? It was

Putting it at

It’s angular size in the sky was then

Thus, the Moon appeared to be about 0.57 degrees wide in the Earth sky. How big did the Sun

appear in the Earth sky?

So, 937 million years ago, when the Earth day was about 18 hours long, the Moon was larger in

the sky than the Sun, on the average. By how much?

The Sun was then about 94% the size of the Moon in the sky back then. At his time we could

write

⊙

⋅

(18.794)(60)(60)

(18.794hr)(60 m in)(60sec) = 67,658.4secon d s

24hrs − 18.794hrs = 5.206hrs

(5.206hrs)(60min)(60min) = 18,741.6secon d s

(18,741.6s)(yr /0.00002sec) = 937,080,000years

(937,080,000yrs)(0.0000382k m /yr) = 35,796.456k m

384,400k m − 35,796.456k m = 348,603.544k m

moon

= 2arcsin

(

)

= 2arcsin

(

1,737.4k m

348,603.544k m

)

= 0.57

∘

sun

= 2arcsin

(

⊙

)

= 2arcsin

(

696,340k m

149,597,870.7k m

)

= 0.5334

∘

0.57

0.5334

1.0686 ≈ 107 %

(Ear th Da y)

⊙

(1second )

of 49 93

We ask what era of the Earth history was this, and what was characteristic of it? It was the

Tonian Period of the Neoproterozoic Era in the Proterozoic Eon. It was the time of early steps

towards multicellular life with the appearance of multicellular eukaryotes such as red algae. It

was the beginning of abundance in photosynthesizers appearing and converting the primordial

CO2 atmosphere into an atmosphere of oxygen setting the stage for the development of

organisms that breathed oxygen to create their energy from food. It was the beginning of

multicellular life and a rich oxygen atmosphere on Earth, setting the scene for the diversification

and development of more complex life. We see 937 million years ago the Sun was about 94% the

size of the Moon in the Earth sky and we ask when was it the same size as the Moon in the Earth

sky. That is when were perfect eclipses most dominant. Today, the annual diameter of the Moon

The moon has an angular diameter on the average throughout a year of about a half a degree.

Let’s see what the size of the Sun is…

Thus the Moon is 97%

Now, the Moon on the average is 97% the size of the Sun throughout the year meaning annular

eclipses dominate.

The Time of Perfect Eclipses

Thus, to find when the Moon was the same size as the Sun, when there were perfect eclipses

dominating, we write

This is how far the Moon was from the Earth at the time perfect eclipses dominated. We ask how

far it has moved since then?

We ask how long did it take the Moon to move this distance. We have

moon

= 2arcsin

(

)

= 2arcsin

(

1,737.4k m

384,400k m

)

= 0.518

∘

sun

= 2arcsin

(

⊙

)

= 2arcsin

(

696,340k m

149,597,870.7k m

)

= 0.5334

∘

moon

sun

0.518

0.5334

= 97 %

0.5334

∘

= arcsin

(

1,737.4k m

)

x = 373,251k m

384,400k m − 373,251k m = 11,149k m

11,149k m = (0.0000382k m /yr)t

t = 291,858,638.7years

of 50 93

About 292 million years ago we had perfect eclipses dominating. We ask what was the Earth day

at this time? It was

At the time of perfect eclipses of the Sun by the Moon as seen from the Earth dominating the

Earth sky, which was about 292 million years ago, the Earth day was 22.38 hours long. We ask

what era was this in the Earth history and what were its characteristics. This was the Permian

Period which was from 298.9 million years to 251.9 million years ago. The Earth’s continents

were mostly joined into a single massive supercontinent called Pangea, which was surrounded

by a global ocean called Panthalassa. Ecosystems were dominated by lycophytes, ferns, and seed

ferns in wetter regions that were gradually replaced by gymnosperms, including conifers and

ginkgoes which could tolerate the drier conditions. This was a time of the the diversification of

reptiles and large amphibians continued to thrive in wetter areas. The oceans were teeming with

life such as brachiopods, bivalves, and ammonoids. The late Permian period ended around 252

million years ago in the most severe extinction event in the Earth’s history where it lost about

90% of marine species and 70% of terrestrial species, but this period set the stage for the rise of

dinosaurs in the Mesozoic Era.

When The PlanetDay Characteristic Time Was Exactly One Second

Let us ask when was the Earth day characteristic time exactly one second. That is

Which is the PlanetDay characteristic time today using average orbital velocities and an

EarthDay of 86,400 seconds. The sidereal day is a little shorter, but we don’t use it here because

the difference is too small to be noted. To find when the Earth day is such that the PlanetDay

characteristic time is exactly one second for the the Earth, we write

The Earth day had to be 69,012 seconds long compared to the 86,400 seconds it is today. That is

it was 69,012/86,400~80% what it is today. We write

(0.00002s /yr)(291,858,638.7yr) = 5837sec = 1.62hrs

24hrs − 1.62hrs = 22.38hrs

K E

(Ear th Da y) = 1secon d

K E

(7.3476731E 22kg)(1,022m /s)

= 3.837E 28J

K E

(5.972E 24kg)(29,780m /s)

= 2.648E 33J

K E

(Ear th Da y) =

3.837E 28J

2.648E 33J

(86,400s) = 1.252secon d s

K E

t = 1.00secon d s

t = 69,012secon d s

69,012sec = 19.17hrs

of 51 93

The PlanetDay characteristic time for the Earth was 1 second around about 870 million years

ago. You might think the problem is when we go back in time to about 870 million years, the

Moon is closer to the Earth so its kinetic energy has changed due to the increase in its orbital

velocity and so the PlanetDay characteristic time is then no longer the same. Let’s look at this…

The lunar orbital radius is now

We find the orbital velocity of the Moon was

Since the average orbital velocity of the Moon today is 1,022 m/s, we see it has increased by only

0.69 m/s, which is practically nothing. So we don’t have to solve a differential equation and the

answer is then, 869,400,000 years ago the PlanetDay characteristic time for the Earth was 1

second which it remained close to for a long time, perhaps hundreds of millions of years before

and after that time.

This intersection of carbon, the proton, and planet day characteristic time and star system

characteristic time of one second, was when plants on Earth were converting the atmosphere to

oxygen leading to the more complex organisms that exist today using oxygen to burn food for

energy. Oxygen started entering the atmosphere during the great oxidation event about 2.4

billion years ago, but the turning point where oxygen began to rise more reliably and

consistently was about 900 million years ago. This is about when animals started using oxygen

for respiration. This time also marks the beginning of multicellular life.

When We Will Lose Perfect Eclipses

We ask, when will the Earth lose perfect eclipses of the Sun by the Moon. The Moon is closest to

the Earth at perihelion, which is 360,000 km. We have

So, on the average the Moon is smaller than the Sun (97%) and because the orbit is elliptical at

its closest approach it is 103.67% larger than the Sun. So we live in a time when we have annular

eclipses ( ) perfect eclipses ( ) and total eclipses ( ). But when

the closest approach of the Moon no longer has there will no longer be perfect or

86,400sec − 69,012sec = 17,388secon ds

(17,388sec)(yr /0.00002sec) = 869,400,000years

(869,400,000years)(0.0000382k m /year) = 3,321k m

= 384,400k m − 3,321k m = 381,079k m

(6.674E − 11)(5.972E 24kg)

381,079,000m

= 1,022.69m /s

moon

= 2arcsin

(

)

= 2arcsin

(

1,737.4k m

360,000k m

)

= 0.553

∘

moon

sun

0.553

0.5334

= 103.67454 %

moon

< θ

sun

moon

= θ

sun

moon

> θ

sun

moon

> θ

sun

of 52 93

total eclipses, only annular eclipses. Let’s find in how long that will be…A perfect eclipse we

found occurs when the Moon is at 373,251km. We have

In about 347 million years we will no longer have perfect eclipses. This is a long time from now,

perhaps longer after the Earth can reasonably sustain life and we have moved on to other

realms.

373,251k m − 360,000k m = 13,251k m

(13,251k m)(yr /0.0000382 k m) = 346,884,817years

of 53 93

12.0 Ideal Way to Model Habitable Star Systems 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, which we will

model shortly. 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!

ℏ

⊙

⋅

= 1secon d

earth

πM

P(R) = P

(

)

−L

ear th

ℏ

⊙

ℏ

⊙

= t

= 1secon d

earth

ℏ

⊙

= p

ℏ

⊙

ℏ

⊙

= t

earth

ℏ

⊙

⋅

= 1secon d

earth

⊙

of 54 93

9.0. !

Where I suggest that we introduce where for , (F2V,G2V,K2V)=(2,1, 0). For a K2V star

is tidally locked, there is no solution. The radius of the planet’s moon we suggested is

given by a perfect eclipse:!

10.0. !

Modeling Our Solar System***************************************************************************!

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…!

⊙

⋆

(1.7)

i = 0

⋆

earth

πM

π(5.972E 24kg)

(86400secon d s)

(6.378E6m)

7.07866672E 33J ⋅ s

⋆

(6.674E − 11)(1.989E30kg)

(1.496E11m)

= 29,788.24m /s

(5.972E 24kg)(29,788.24m /s)

= 2.65E33J

ℏ

⋆

ℏ

⋆

(7.07866672E 33J ⋅ s)

2.5

= 2.831467E33J ⋅ s

ℏ

⋆

(2.831467E33J ⋅ s)

2.65E33J

= 1.068secon ds ≈ 1second

of 55 93

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!

So this gives the correct characteristic time for the Earth/Moon/Sun system. Let’s compute the

planet day characteristic time…!

(2.831467E33J ⋅ s)

(6.674E − 11)(299,729,458m /s)(1.068seconds)

= 7.213E 22kg

= R

⋆

= R

⋆

/(1.8) =

6.957E8m

1.8

= 3.865EE 8m

= R

⋆

= (6.957E8m)

3.865E8m

1.496E11m

= 1.79738E6m

πR

π(1.79738E6m)

= 2.432E19m

7.213E 22 kg

2.432E19m

= 2.96587g/cm

≈ 3g/cm

1secon d = 2r

⊙

(6.674E − 11)(5.972E 24kg)

(3.865EE8m)

= 1,015.5m /s

2(1.496E11m)

1,015.5m /s

(29,788.24m /s)

(7.213E 22kg)

(1.732)

(5.972E 24kg)

(1.989E30kg)

= 1.03648sec ≈ 1secon d

1secon d ≈

(Ear th Day)

of 56 93

We see the system for modeling star system works. The thing is while M class stars are the

most abundant in the galaxy and have longer life spans than the Sun, their planets are thought

to be tidally locked, their day is equal to their year, leaving them perpetually night on one side

of the planet and perpetually day on the other, only being cool enough for life in the twilight

region between night and day. K class stars show a lot of promise to host life on planets in their

habitable zones because they are far enough away from their star that they might not always

become tidally locked, while being more stable than the Sun and longer lived. It is easiest to

detect planets and get data for these M2V and K2V stars, but when you get to our Sun a G2V

star, they are so bright they wash out the light of their planets in the habitable zone. However,

we have gotten data for a planet in the habitable zone of a G2V star, the same size, radius,

mass, and luminosity as our Sun. And the planet in its habitable zone has the same radius, and

mass as our Earth. All we need to do is to detect a moon around its planet, and we will have

veriﬁed my theory for G2V stars. This star is called KOI-4878 and its planet is KOI-4878.01.

Here is the information on it…!

Comment The characteristic time of the star system is given by the mass of the moon alone. It

is inversely proportional to the mass of the moon cubed:!

(7.213E 22 kg)(1,015.5m /s)

= 3.719E 28J

(PlanetDay) =

(3.719E 28J )

(2.65E33J )

(86,400sec) = 1.2secon ds

∝

of 57 93

We know the mass of our moon, but have not yet had the technology to detect the moon

orbiting a planet in another star system, just the planet orbiting the star. However, eﬀorts are

being made to do that very thing. We suggest for optimal life conditions this characteristic time

becomes close to the PlanetDay characteristic time when the system is in the epoch of perfect

eclipses of the star by the moon which happens with our Earth/Moon/Sun system. That is!

And, the condition of perfect eclipse are!

, !

For the Earth/Moon/Sun system this ratio is 400. The constant of proportionality for the

characteristic time inversely proportional to the mass of the moon is given by!

F2V stars********************************************************************************************

We actually have fairly accurate data for M2V stars and K2V stars with planets in their habitable

zones. But these spectral classes are to the right of our Sun, a G2V, star, and we want data for

a star to the left of our Sun, as well, preferably the ﬁrst one, the F stars. But such stars are

much brighter than our Sun, really washing out the light of the planet. However, if we want the

data for one so we can plot characteristic times for star systems against spectral class we will

want to model a habitable F2V system with our theory. !

We have our planet radius from what we use that holds for the Earth which is!

We start with the radius of the planet…!

In the tables we have for F2V stars!

Mass: 1.46 solar!

Radius: 1.622 solar!

Luminosity: 5.13 solar!

The habitable zone of this star would be by the inverse square law:!

1secon d ≈

(Ear th Day)

⋆

⊙

= 400

ℏ

⊙

⋅

= 1secon d

ℏ

⊙

(2.8314E33J ⋅ s)

(6.674E − 11)(299,792,458)

= 4.00678E68kg

⋅ s

⋆

= 5.13 = 2.265AU = 2.265(1.496E11m) = 3.388E11m

of 58 93

We want that the density of the planet is the same as the Earth’s to ensure a rocky planet with

oceans…!

The volume of the planet is:!

We will choose the planet day equal to (24hrs)=86,400s. Now let’s compute the planet’s

rotational angular momentum:!

From this we have :!

We will need the planet’s orbital velocity and kinetic energy:!

Now we can ﬁnd the characteristic time:!

From this we have the mass of the moon:!

⋆

= (1.622)(6.957E8m) = 1.1284254E 9m

2(1.1284254E 9m)

3.388E11m

= 7.516788E6m = 1.17855Ear th Ra dii

πR

π(7.516788E6m)

= 1.78E 21m

1.78E 21m

= 5,510kg/m

= 5.51g/cm

= 9.8078E 24kg = 1.6423Ear th Masses

π(9.8078E 24kg)

(86,400s)

(7.516788E6m)

= 1.612E 34J ⋅ s

ℏ

⋆

ℏ

⋆

(1.612E 34J ⋅ s)

2.0

= 8.06E33J ⋅ s

⋆

(6.674E − 11)(2.9E30kg)

(3.388E11m)

= 23,901.24m /s

(9.8078E 24kg)(23,901.24m /s)

= 2.8E33J

ℏ

⋆

(8.06E33J ⋅ s)

2.8E33J

= 2.87857secon ds

of 59 93

Now we can ﬁnd the density of the moon:!

This is good. The density of the Earth is 5.51g/cm3, We want it to be about the same, but a

little less. The density or our Moon is 3.34 g/cm3, but it is thought to be unusually low, so much

so some have suggested it is hollow, so this seems really good. We want to check if the

equation for the characteristic time we obtained by equating the lunar and solar formulas,

works…!

We needed a characteristic time of 2.87857seconds seconds. When we got to this phase of

modeling the Earth, where everything worked, this worked too. So, we suggest the equations

of the theory work for spectral class G2V stars. And we can actually conﬁrm this because we

have detected a habitable planet around another G2V star that has all the same characteristics

as our Earth and Sun; radii, mass, luminosity, all the same. Here we see, since this equation

has the masses of the moon, the planet, and the Sun - in the radical - the the thing that needs

to be diﬀerent is the orbital velocity of the moon. The orbital velocity of the planet cannot

(8.06E33J ⋅ s)

(6.674E − 11)(299,792,458m /s)(2.87857seconds)

= 1.041E 23kg

⋆

= 1.622(6.957E8m) = 1.1284E 9m

⋆

= R

⋆

/((1.414)(1.7)) =

1.1284E 9m

2.4038

= 4.6942E8m

= R

⋆

= (1.1284E9m)

4.6942E 8m

3.388E11m

= 1.56344E6m

πR

π(1.56344E6m)

= 1.60E19m

1.041E 23kg

1.60E19m

= 6.5g/cm

1secon d = 2r

⊙

(6.674E − 11)(9.8078E 24kg)

(4.6942E 8m)

= 1180.859m /s

2(3.388E11m)

1180.859m /s

(23,901.24m /s)

(1.041E 23kg)

(1.732)

(9.8078E 24kg)

(2.9E30kg)

= 3.7067secon ds

of 60 93

change, it is given by the mass of the star it orbits and its distance from the star. That distance

cannot be changed much without taking the planet out of its habitable zone. If the orbital

velocity of the moon is to change, we need to change its orbital distance. If we change its

orbital distance, we no longer have a perfect eclipse. We would conclude that perfect eclipses

are a condition applying to habitable planets around G2V stars, like our Sun. Namely,!

So, we want to look at equation 1.4, the solution to the wave equation for the kinetic energy of

the planet. It is!

= !

Which is (2.3594E33)(1.732). We found the kinetic energy was!

That is 2.685E33(1.732)=4.65E33 introduces a factor of . !

=3.00E33!

Which works. But the litmus test is does it work in!

And it works because we need 2.87857 seconds. We have, then,…!

⋆

= 3

⋆

2ℏ

⋆

(1.732)

1.1284E 9m

1.56344E6m

⋅

(6.674E − 11)

(9.8078E 24kg)

(1.041E 23kg)

2(8.06E33J ⋅ s)

4.65E33J

(9.8078E 24kg)(23,901.24m /s)

= 2.8E33J

5/2 = 1.118

1.1284E 9m

1.56344E6m

⋅

(6.674E − 11)

(9.8078E 24kg)

(1.041E 23kg)

2(8.06E33J ⋅ s)

2(3.388E11m)

1180.859m /s

(23,901.24m /s)

(1.041E 23kg)

(1.118)

(9.8078E 24kg)

(2.9E30kg)

= 2.978secon ds

of 61 93

For F2V Habitable!

For G2V Habitable, !

For K2V Habitable; is tidally locked, has no solution.!

Thus quantizing as we move right in the HR diagram by spectral class starting at!

Thus means for F2V stars the habitable planet is planet 1 because such system shouldn’t

be as tightly packed as G2V systems, means G2V stars, because their habitable planets

are planet 3. We see the equation for G2V stays the same because .!

Let us now compute the planet day characteristic time:

We now want to apply our modeling method to K2V stars and M2V stars. We have data for

these stars with planet’s in their habitable zones.!

Modeling Actual K2V Star System*******************************************************************!

We have all of the data for the equations for our Solar System, but it is often diﬃcult to obtain

for other star systems because they are so far away. However, we do have it with reasonable

margins of errors for the spectral class K2V star Kepler 62f and its planet in the habitable zone

Kepler 62f. We also have it for the spectral class M2V star TOI 700 and its planet in the

habitable zone TOI 700e. It is easier to detect and measure planets in these stars than with

G2V stars like our Sun because, as we said, they are not so bright, thus they don’t wash out

⋆

⊙

⋅

⋆

2ℏ

⋆

⊙

⋆

⋅

⋆

2ℏ

⋆

= L

⊙

⋆

⋅

⋆

2ℏ

⋆

(F 2V, G2V, K 2V ) = ( 1/2, 3/1, 5 /0)

⋆

= L

⊙

1secon d ≈

(Ear th Day)

(1.041E 23kg)(1180.859m /s)

= 7.258E 28J

(9.8078E 24kg)(23,901.24m /s)

= 2.8E33J

(PlanetDay) =

(7.258E 28J )

(2.8E33J )

(86,400s) = 2.2396secon ds

of 62 93

the light of the planets as much. K stars show a lot of promise to be good hosts for habitable

planets because they are even more stable than our Sun, and have even longer life spans. They

are so-called orange dwarf stars. M stars are the most abundant in the galaxy, so-called red

dwarfs, and if they can sustain complex life, this would mean there would be a lot of life in the

Universe. Let us perform the computations for these systems starting with the K2V star. The

star is called Kepler 62, and its planet in the habitable zone is called Kepler 62f. The data is!

We will use the converted values to kilograms/meters/seconds. We have!

= !

This is the rotational angular momentum of the habitable planet in the K2V star system. Its

orbital velocity is!

Its kinetic energy is!

We use the typical value for the the pressure gradient exponent. It is p=2.0!

Thus, the characteristic time is!

π(1.672E 25kg)

(267.291da ys)(86400sec /d a y)

(9.318E6m)

1.58E32J ⋅ s

⋆

(6.674E − 11)(1.52E 30kg)

(1.074E11m)

= 30,733.55m /s

(1.672E 25kg)(30,733.55m /s)

= 7.896E33J

ℏ

⋆

(1.58E32J ⋅ s)

2.0

= 7.9E31J ⋅ s

ℏ

⋆

⋅

= t

of 63 93

We have

The mass of its moon is then!

There is no solution for the moon as we said for K2V stars because . This does not mean

it can’t have a moon, it just means it doesn’t have one as a stabilizing factor. That factor is its

star because it is tidally locked. But we will compute it using to see what we get.!

The radius of the moon is given be a perfect eclipse:!

The volume of the moon is!

The density of the moon is!

This is a very good result. The density of the Earth is 5.51 g/cm3. The density of our moon is

3.34 g/cm3. It is mysteriously low, so much so that people have theorized that it is hollow.

There are so far 5 planets detected in the Kepler 62f system. This one is the ﬁfth (n=5) orbit. I

think n=3 is the scaling factor for the most optimally habitable planet that is in an interior orbit

of solid planets, not exterior orbits of gas giants. This manner of modeling with angular

momentum of the planet and the pressure gradient exponent to get does not use orbit

number. We use it in the following where we get the characteristic time from the intersections

of the lunar and solar formulations, but we didn’t use this to get the characteristic time. We can

put in our values and we get something in the right area: !

ℏ

⋆

(7.9E31J ⋅ s)

7.896E33J

= 0.010005secon ds

ℏ

⋆

⋅

(7.9E31J ⋅ s)

(6.674E − 11)(299,792,458m /s)(0.01s)

= 3.14786E 22kg

i = 0

i = 1

⋆

(1.7)

4.5936E8m

1.7

= 2.7E8m

= R

⋆

= (4.5936E8m)

2.7E8m

1.074E11m

= 1.155E6m

πR

π(1.155E6m)

= 6.4541E18m

3.14786E 22 kg

6.4541E18m

= 4.8773g/cm

ℏ

⋆

of 64 93

Let’s compute the PlanetDay characteristic time, even though it does not make sense to do so

because the planet’s day has become very large due to becoming tidally locked. Keep in mind

the planet day characteristic time isn’t necessarily the same as the star system characteristic

time, except where they intersect, which is at star systems like our Earth/Moon/Sun system in

the G2V region.!

Because the planet has become tidally locked with the star it orbits, its rotation has slowed

down to that of its orbital period, making the same face always towards the star, the same

away, and giving an absurdly long planet day, and absurdly high planet day characteristic time.!

1secon d = 2r

⊙

= 2r

⋆

⊙

⋅

⊙

(6.674E − 11)(1.672E 25kg)

(2.7E8m)

= 1,017m /s

2(1.074E11m)

1,017m /s

(30,733.55m /s)

(0.2565)

(3.147331E 22kg)

(1.732)

(1.672E 25kg)

(1.52E 30kg)

= 0.04175secon d s

1secon d ≈

(Ear th Day)

(3.147331E 22 kg)(1,017m /s)

= 1.6276E 28J

(PlanetDay) =

(1.6276E 28J )

(7.896E33J )

(23,093,942.4sec) = 47.6second s

1secon d ≈

(Ear th Day)

of 65 93

Modeling Actual M2V Star System********************************************************************!

Now we can move on to an M2V star that has a few planets in its habitable zone. It is called

TOI 700. We are interested in exoplanet TOI 700e that is an Earth-like planet in the habitable

zone.!

The TOI 700e Planet values converted!

The TOI 700 star values converted!

= !

= (0.818)(5.972E 24kg) = 4.885E 24kg

= 0.1340 = 2.00464E10m

= 0.953 = (0.953)(6.378E6m) = 6.078E6m

= 27.80978 = 2.402765E6secon ds

⋆

= 0.416(1.989E30kg) = 8.27424E 29kg

⋆

= 0.420(6.9634E8m) = 2.924628E 8m

⋆

= 0.0233

π(4.885E 24kg)

(2.402765E6secon ds)

(6.078E6m)

1.8876E32J ⋅ s

of 66 93

That density is very low, below that of water. We see that the moon vanishes as a stabilizing

factor and the stabilizing factor becomes the star making the planet tidally locked. We can

have moon’s with a density this low, we have them in our solar system, like Saturn’s moon

Hyperion, which is mostly porous water-ice. However, Saturn is out further from thus Sun

where it is colder and this moon is in the habitable zone making it much warmer, you would

think such a low density would be vaporized by the star. !

⋆

(6.674E − 11)(8.27424E 29kg)

(2.00464E10m)

= 52,485.45m /s

(4.885E 24kg)(52,485.45m /s)

= 6.7284E33J

ℏ

⋆

(1.8876E32J ⋅ s)

2.0

= 9.438E31J ⋅ s

ℏ

⋆

(9.438E31J ⋅ s)

6.7284E33J

= 0.014second s

(9.438E31J ⋅ s)

(6.674E − 11)(299,729,458m /s)(0.014secon ds)

= 3.168E 22kg

= R

⋆

= R

⋆

/(1.7) =

2.924628E8m

1.7

= 1.72E8m

= R

⋆

= (2.924628E8m)

1.72E 8m

2.00464E10m

= 2.51E6m

πR

π(2.51E6m)

= 6.624E19m

3.168E 22 kg

6.624E19m

= 0.478g/cm

(6.674E − 11)(4.885E 24kg)

(1.72E 8m)

= 1,376.77m /s

⋆

(6.674E − 11)(8.27424E 29kg)

(2.00464E10m)

= 52,485.45m /s

of 67 93

Now we want to plot where the Planet day characteristic time intersects with the star system

characteristic time to show visually that they intersect at around spectral class G2V stars like

our Sun. We suggest the possibility that this intersection makes them optimally habitable. We

have all the data we need for our star system for our star, the Sun, for the planets, and for their

moons. We have the data for F2V stars, but need to model the planets and the moons because

it is still too diﬃcult to detect planets around such bright stars. We have the data for the stars

and habitable planets of a K2V star, and a M2V star, but we don’t have the data for their moons

if they have them, they are still too diﬃcult to detect with current technology if they exist. Also

the habitable planets of these stars are tidally locked meaning they have lost their

characteristic times. Their PlanetDay characteristic times are high, absurdly high. We will have

to scale down the results of the K2V and M2V for them to ﬁt on the chart.!

We see we have the intersection of an atomic and planetary science with habitable star

systems at around G2V like our star, the Sun….see plot next page.#

(3.168E 22 kg)(1,376.77m /s)

= 3.00E 28J

(4.885E 24kg)(52,485.45m /s)

= 6.72841E33J

(PlanetDay) =

(3.00E 28J )

(6.72841E33J )

(2.402765E6secon ds) = 10.71secon d s

of 68 93

We see we have the intersection of PlanetDay characteristic times and planetary system

characteristic time at around G2V stars like our Sun . PlanetDay characteristic times of K2V

and M2G stars go oﬀ the charts so they are scaled down. We have the intersection of the

planetary and atomic as such!

, !

And it all happens under a perfect eclipse of the Sun by the Moon: !

ℏ

⊙

⋅

= 1secon d

(

6α

4πh

)

⋅

= 1secon d

⊙

moon

earth

moon

of 69 93

13.0 Modeling With A Computer For easy of modeling and checking my computations I

have made a program in C that models according to this ideal method. The round function has

not been incorporated so the program truncates after 8 digits. I only work to about 4 digits after

the decimal so it doesn’t seem to be much of a problem. You will see all of the program output

verifies my computations, and as well in much of the in-between work, not just the final results.

Here is the program and running it for the F2V, G2V, K2V, and M2V stars we did here…

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// main.c

// Intersections

// Created by Ian Beardsley on 1/7/25.

#include <stdio.h>

#include <math.h>

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

float R_p, M_p, R_s, M_s, L_s, h_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, Root;

float G=6.674E-11, hbarstar, PDCT;

float Mcubed, r_m, root, civilization, 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;

printf ("What is the radius of the planet in Earth radii? May we suggest

it is given by 2StarRadius^2/PlanetOrbitalRadius? 1=yes, 0=no ");

scanf ("%i", &i);

R_p=0;

if (i==1)

{

R_s=6.9634E8*StarRadius;

R_p=(2*R_s*R_s)/r_p;

printf ("The radius of the planet is: %fE6m\n", R_p/1E6);

printf ("that is %f Earth Radii \n", 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? May we

suggest that it is such that it has the density of the Earth? 1=yes, 0=no ");

scanf("%i", &i);

if (i==1)

{

V_p=1.33333*(3.14159)*R_p*R_p*R_p;

M_p=V_p*5510;

PlanetMass=M_p/5.972E24;

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printf("The mass of the planet is %f E24kg \n", M_p/1E24);

printf("That is %f Earth masses \n", PlanetMass);

}

else

{

printf("What is the mass of the planet in Earth Masses? ");

scanf("%f", &PlanetMass);

M_p=PlanetMass*5.972E24;

printf("The mass of the Planet is %f E24kg \n", M_p/1E24);

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);

printf("What integer do you want to use, F0V=2,G0V=1? ");

scanf("%f", &root);

printf("Choose copper-silver civilization (1.7) or gold silver

(1.8)? ");

scanf("%f", &civilization);

Root=sqrt(root);

M_s=1.9891E30*StarMass;

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/(Root*civilization);

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;

printf("\n");

printf("Angular Momentum of Planet: %f E33 \n", L_p/1E33);

printf("\n");

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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 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("\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);

return 0;

}}}

of 73 93

Modeling Earth

What is the radius of the star in solar radii? 1

What is the mass of the star in solar masses? 1

What is the luminosity of the star in solar luminosities? 1

What is the radius of the planet in Earth radii? May we suggest it is given

by 2StarRadius^2/PlanetOrbitalRadius? 1=yes, 0=no 0

What is the planet radius in Earth radii?: 1

What is the mass of the planet in Earth masses? May we suggest that it is

such that it has the density of the Earth? 1=yes, 0=no 0

What is the mass of the planet in Earth Masses? 1

The mass of the Planet is 5.972000 E24kg

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? 2.5

What integer do you want to use, F0V=2,G0V=1? 1

Choose copper-silver civilization (1.7) or gold silver (1.8)? 1.8

Angular Momentum of Planet: 7.066666 E33

planet orbital velocity: 29788.980469 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet orbital radius: 1.496000 E11 m

planet orbital radius: 1.000000 Earth distances

planet KE: 2.649727 E33 J

hbarstar: 2.826666 E33 Js

characteristic time: 1.066777 seconds

Orbital Radius of Moon: 3.868556 E8 m

Orbital Radius of Moon: 1.007436 Moon Distances

Radius of Moon: 1.800688 E6 m

Radius of Moon: 1.036427 Moon Radii

Mass of Moon: 7.207029 E22 kg

Mass of Moon 0.980859 Moon Masses

density of moon: 2.946819 g/cm3

Orbital Velocity of Moon: 1015.029358 m/s

PlanetDay Characteristic Time: 1.210587 seconds

Program ended with exit code: 0

of 74 93

Modeling F2V Star

What is the radius of the star in solar radii? 1.622

What is the mass of the star in solar masses? 1.46

What is the luminosity of the star in solar luminosities? 5.13

What is the radius of the planet in Earth radii? May we suggest it is given

by 2StarRadius^2/PlanetOrbitalRadius? 1=yes, 0=no 0

What is the planet radius in Earth radii?: 1.17855

What is the mass of the planet in Earth masses? May we suggest that it is

such that it has the density of the Earth? 1=yes, 0=no 0

What is the mass of the planet in Earth Masses? 1.6423

The mass of the Planet is 9.807816 E24kg

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? 2.0

What integer do you want to use, F0V=2,G0V=1? 2

Choose copper-silver civilization (1.7) or gold silver (1.8)? 1.7

Angular Momentum of Planet: 16.119928 E33

planet orbital velocity: 23916.781250 m/s

planet mass: 9.807816 E24 kg

planet mass: 1.642300 Earth masses

planet orbital radius: 3.388366 E11 m

planet orbital radius: 2.264950 Earth distances

planet KE: 2.805096 E33 J

hbarstar: 8.059964 E33 Js

characteristic time: 2.873329 seconds

Orbital Radius of Moon: 4.697948 E8 m

Orbital Radius of Moon: 1.223424 Moon Distances

Radius of Moon: 1.565994 E6 m

Radius of Moon: 0.901344 Moon Radii

Mass of Moon: 10.415768 E22 kg

Mass of Moon 1.417560 Moon Masses

density of moon: 6.474909 g/cm3

Orbital Velocity of Moon: 1180.388916 m/s

PlanetDay Characteristic Time: 2.235000 seconds

Program ended with exit code: 0

of 75 93

Modeling K2V star from data for Kepler 62

What is the radius of the star in solar radii? 0.660

What is the mass of the star ins solar masses? 0.764

What is the luminosity of the star in solar luminosities? 0.2565

What is the radius of the planet in Earth radii? May we suggest it is given

by 2StarRadius^2/PlanetOrbitalRadius? 1=yes, 0=no 0

What is the planet radius in Earth radii?: 1.461

What is the mass of the planet in Earth masses? May we suggest that it is

such that it has the density of the Earth? 1=yes, 0=no 0

What is the mass of the planet in Earth Masses? 2.8

The mass of the Planet is 16.721600 E24kg

What is the planet day in Earth days? 267.291

That is 23093942.000000 seconds

What is p the pressure gradient exponent of the protoplanetary disc? 2.0

What integer do you want to use, F0V=2,G0V=1? 1

Choose copper-silver civilization (1.7) or gold silver (1.8)? 1.7

Angular Momentum of Planet: 0.158011 E33

planet orbital velocity: 36587.308594 m/s

planet mass: 16.721600 E24 kg

planet mass: 2.800000 Earth masses

planet orbital radius: 0.757662 E11 m

planet orbital radius: 0.506458 Earth distances

planet KE: 11.192028 E33 J

hbarstar: 0.079006 E33 Js

characteristic time: 0.007059 seconds

Orbital Radius of Moon: 2.703438 E8 m

Orbital Radius of Moon: 0.704020 Moon Distances

Radius of Moon: 1.639859 E6 m

Radius of Moon: 0.943858 Moon Radii

Mass of Moon: 3.535522 E22 kg

Mass of Moon 0.481176 Moon Masses

density of moon: 1.914023 g/cm3

Orbital Velocity of Moon: 2031.766602 m/s

PlanetDay Characteristic Time: 150.577621 seconds

Program ended with exit code: 0

of 76 93

Modeling M2V Star From Data From TOI 700

What is the radius of the star in solar radii? 0.420

What is the mass of the star in solar masses? 0.416

What is the luminosity of the star in solar luminosities? 0.0233

What is the radius of the planet in Earth radii? May we suggest it is given

by 2StarRadius^2/PlanetOrbitalRadius? 1=yes, 0=no 0

What is the planet radius in Earth radii?: 0.953

What is the mass of the planet in Earth masses? May we suggest that it is

such that it has the density of the Earth? 1=yes, 0=no 0

What is the mass of the planet in Earth Masses? 0.818

The mass of the Planet is 4.885096 E24kg

What is the planet day in Earth days? 27.80978

That is 2402765.000000 seconds

What is p the pressure gradient exponent of the protoplanetary disc? 2.0

What integer do you want to use, F0V=2,G0V=1? 1

Choose copper-silver civilization (1.7) or gold silver (1.8)? 1.7

Angular Momentum of Planet: 0.188780 E33

planet orbital velocity: 49177.144531 m/s

planet mass: 4.885096 E24 kg

planet mass: 0.818000 Earth masses

planet orbital radius: 0.228354 E11 m

planet orbital radius: 0.152643 Earth distances

planet KE: 5.907038 E33 J

hbarstar: 0.094390 E33 Js

characteristic time: 0.015979 seconds

Orbital Radius of Moon: 1.720369 E8 m

Orbital Radius of Moon: 0.448013 Moon Distances

Radius of Moon: 2.203346 E6 m

Radius of Moon: 1.268186 Moon Radii

Mass of Moon: 3.031771 E22 kg

Mass of Moon 0.412616 Moon Masses

density of moon: 0.676645 g/cm3

Orbital Velocity of Moon: 1376.635132 m/s

PlanetDay Characteristic Time: 11.685454 seconds

Program ended with exit code: 0

of 77 93

14.0 Perfecting The Ideal Modeling I have reﬁned the program (included output with planet

density). Now I have ﬁgured out how to model the planetary system for solutions all around the

board. That is, PlanetDay characteristic time, planetary characteristic time, proton

characteristic time all intersect closely (1.0-1.5 seconds) and the characteristic time from

equating the solar formulation with lunar formulation intersects closely with these, and the

kinetic energy ( ) is accurately predicted.!

We model here a F8V star, a G0V star, a G2V star (which is our solar system for which we have

the actual data to use) and a G4V star. We assume the habitable zone planet is the third planet

in every case because with an F8V star the pressure gradient for the protoplanetary disc is

more shallow than in the G2V so the distribution of the planets (Which should be geometric by

the Titius-Bode rule) is stretched out putting the third planet further out, which it needs to be

to be in habitable zone because the star is more luminous. The planets are on the order of the

size of the Earth as well as their length of days. It is reasonable to think the planet of an F8V

star is of the same day size as the Earth, after all Mars is further from the Sun than the Earth,

but its length of day is about the same. It is also reasonable to guess that more often the

habitable planet of a more massive star than the Sun could be a little more massive than the

Earth, and that the planet of a less massive star than the Sun more often might be less massive

than the Earth, though this would not necessarily always be the case.!

I ﬁnd the program I wrote, in comparing its output to the actual values computed by hand for a

G0V star, are accurate to four places after the decimal for characteristic times and planet and

moon densities. This is reasonable to be so because even though program doesn’t use a round

function, but rather truncates numbers, it only does so after eight decimal places, and our

results for characteristic times and planet and moon densities are only accurate to 2 to 3

places after the decimal. So we can use the program to make data points for a curve.!

Here we don’t introduce the factors of or , ,… in the KE solution from the wave

equation and radius of the planet computation because the solutions work all the way around

the board. There can be many kinds of planets around many stars, but the solutions in the

sense that the Earth is an extraordinarily habitable planet I would think would be more like

these. In that that same kinds of stars can have many kinds of planets at the same distances,

there would be many diﬀerent ways of solving the equations, because they are quantum

mechanical in Nature.!

Here is the program and running it for F8V, G0V, G2V, and G4V stars…#

⋆

⊙

of 78 93

// main.c

// Intersections

// Created by Ian Beardsley on 1/7/25.

#include <stdio.h>

#include <math.h>

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

float R_p, M_p, R_s, M_s, L_s, h_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, Root;

float G=6.674E-11, hbarstar, PDCT;

float Mcubed, r_m, root, civilization, 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;

printf ("What is the radius of the planet in Earth radii? May we suggest

it is given by 2StarRadius^2/PlanetOrbitalRadius? 1=yes, 0=no ");

scanf ("%i", &i);

R_p=0;

if (i==1)

{

R_s=6.9634E8*StarRadius;

R_p=(2*R_s*R_s)/r_p;

printf ("The radius of the planet is: %fE6m\n", R_p/1E6);

printf ("that is %f Earth Radii \n", 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? May we

suggest that it is such that it has the density of the Earth? 1=yes, 0=no ");

scanf("%i", &i);

if (i==1)

{

V_p=1.33333*(3.14159)*R_p*R_p*R_p;

M_p=V_p*5510;

PlanetMass=M_p/5.972E24;

of 79 93

printf("The mass of the planet is %f E24kg \n", M_p/1E24);

printf("That is %f Earth masses \n", PlanetMass);

}

else

{

printf("What is the mass of the planet in Earth Masses? ");

scanf("%f", &PlanetMass);

M_p=PlanetMass*5.972E24;

printf("The mass of the Planet is %f E24kg \n", M_p/1E24);

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);

printf("What integer do you want to use, F0V=2,G0V=1? ");

scanf("%f", &root);

printf("Choose copper-silver civilization (1.7) or gold silver

(1.8)? ");

scanf("%f", &civilization);

Root=sqrt(root);

M_s=1.9891E30*StarMass;

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/(Root*civilization);

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);

of 80 93

printf("\n");

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 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);

return 0;

}}}

of 81 93

Running For Our Solar System (G2V Star System)

What is the radius of the star in solar radii? 1

What is the mass of the star in solar masses? 1

What is the luminosity of the star in solar luminosities? 1

What is the radius of the planet in Earth radii? May we suggest it is

given by 2StarRadius^2/PlanetOrbitalRadius? 1=yes, 0=no 0

What is the planet radius in Earth radii?: 1

What is the mass of the planet in Earth masses? May we suggest that it

is such that it has the density of the Earth? 1=yes, 0=no 0

What is the mass of the planet in Earth Masses? 1

The mass of the Planet is 5.972000 E24kg

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?

2.5

What integer do you want to use, F0V=2,G0V=1? 1

Choose copper-silver civilization (1.7) or gold silver (1.8)? 1.8

Angular Momentum of Planet: 7.066666 E33

planet orbital velocity: 29788.980469 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet orbital radius: 1.496000 E11 m

planet orbital radius: 1.000000 Earth distances

planet KE: 2.649727 E33 J

planet density: 5.495144 g/cm3

hbarstar: 2.826666 E33 Js

characteristic time: 1.066777 seconds

Orbital Radius of Moon: 3.868556 E8 m

Orbital Radius of Moon: 1.007436 Moon Distances

Radius of Moon: 1.800688 E6 m

Radius of Moon: 1.036427 Moon Radii

Mass of Moon: 7.207029 E22 kg

Mass of Moon 0.980859 Moon Masses

density of moon: 2.946819 g/cm3

Orbital Velocity of Moon: 1015.029358 m/s

PlanetDay Characteristic Time: 1.210587 seconds

Program ended with exit code: 0

of 82 93

Running For G0V (We find characteristic times intersect)

What is the radius of the star in solar radii? 1.100

What is the mass of the star in solar masses? 1.06

What is the luminosity of the star in solar luminosities? 1.35

What is the radius of the planet in Earth radii? May we suggest it is

given by 2StarRadius^2/PlanetOrbitalRadius? 1=yes, 0=no 0

What is the planet radius in Earth radii?: 0.95

What is the mass of the planet in Earth masses? May we suggest that it

is such that it has the density of the Earth? 1=yes, 0=no 0

What is the mass of the planet in Earth Masses? 1

The mass of the Planet is 5.972000 E24kg

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?

2.3

What integer do you want to use, F0V=2,G0V=1? 1

Choose copper-silver civilization (1.7) or gold silver (1.8)? 1.8

Angular Momentum of Planet: 6.377667 E33

planet orbital velocity: 28452.804688 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet orbital radius: 1.738195 E11 m

planet orbital radius: 1.161895 Earth distances

planet KE: 2.417352 E33 J

planet density: 6.409266 g/cm3

hbarstar: 2.772898 E33 Js

characteristic time: 1.147081 seconds

Orbital Radius of Moon: 4.255411 E8 m

Orbital Radius of Moon: 1.108180 Moon Distances

Radius of Moon: 1.875241 E6 m

Radius of Moon: 1.079338 Moon Radii

Mass of Moon: 6.945268 E22 kg

Mass of Moon 0.945234 Moon Masses

density of moon: 2.514378 g/cm3

Orbital Velocity of Moon: 967.792480 m/s

PlanetDay Characteristic Time: 1.162511 seconds

Program ended with exit code: 0

of 83 93

Running for F8V star system (characteristic times intersect)

What is the radius of the star in solar radii? 1.221

What is the mass of the star in solar masses? 1.18

What is the luminosity of the star in solar luminosities? 1.95

What is the radius of the planet in Earth radii? May we suggest it is

given by 2StarRadius^2/PlanetOrbitalRadius? 1=yes, 0=no 0

What is the planet radius in Earth radii?: 1.1

What is the mass of the planet in Earth masses? May we suggest that it

is such that it has the density of the Earth? 1=yes, 0=no 0

What is the mass of the planet in Earth Masses? 1.1

The mass of the Planet is 6.569200 E24kg

What is the planet day in Earth days? 1.2

That is 103680.007812 seconds

What is p the pressure gradient exponent of the protoplanetary disc?

2.1

What integer do you want to use, F0V=2,G0V=1? 1

Choose copper-silver civilization (1.7) or gold silver (1.8)? 1.8

Angular Momentum of Planet: 7.838111 E33

planet orbital velocity: 27383.435547 m/s

planet mass: 6.569200 E24 kg

planet mass: 1.100000 Earth masses

planet orbital radius: 2.089050 E11 m

planet orbital radius: 1.396424 Earth distances

planet KE: 2.462966 E33 J

planet density: 4.541442 g/cm3

hbarstar: 3.732434 E33 Js

characteristic time: 1.515423 seconds

Orbital Radius of Moon: 4.723506 E8 m

Orbital Radius of Moon: 1.230080 Moon Distances

Radius of Moon: 1.922439 E6 m

Radius of Moon: 1.106503 Moon Radii

Mass of Moon: 7.716396 E22 kg

Mass of Moon 1.050182 Moon Masses

density of moon: 2.592804 g/cm3

Orbital Velocity of Moon: 963.423218 m/s

PlanetDay Characteristic Time: 1.507493 seconds

Program ended with exit code: 0

of 84 93

Running For G4V Star System (characteristic times intersect

What is the radius of the star in solar radii? 0.991

What is the mass of the star in solar masses? 0.985

What is the luminosity of the star in solar luminosities? 0.91

What is the radius of the planet in Earth radii? May we suggest it is

given by 2StarRadius^2/PlanetOrbitalRadius? 1=yes, 0=no 0

What is the planet radius in Earth radii?: 0.95

What is the mass of the planet in Earth masses? May we suggest that it

is such that it has the density of the Earth? 1=yes, 0=no 0

What is the mass of the planet in Earth Masses? 0.98

The mass of the Planet is 5.852560 E24kg

What is the planet day in Earth days? 0.90

That is 77760.000000 seconds

What is p the pressure gradient exponent of the protoplanetary disc?

2.4

What integer do you want to use, F0V=2,G0V=1? 1

Choose copper-silver civilization (1.7) or gold silver (1.8)? 1.8

Angular Momentum of Planet: 6.944570 E33

planet orbital velocity: 30270.066406 m/s

planet mass: 5.852560 E24 kg

planet mass: 0.980000 Earth masses

planet orbital radius: 1.427093 E11 m

planet orbital radius: 0.953939 Earth distances

planet KE: 2.681283 E33 J

planet density: 6.281081 g/cm3

hbarstar: 2.893571 E33 Js

characteristic time: 1.079174 seconds

Orbital Radius of Moon: 3.833739 E8 m

Orbital Radius of Moon: 0.998370 Moon Distances

Radius of Moon: 1.853810 E6 m

Radius of Moon: 1.067002 Moon Radii

Mass of Moon: 7.292166 E22 kg

Mass of Moon 0.992446 Moon Masses

density of moon: 2.732585 g/cm3

Orbital Velocity of Moon: 1009.380188 m/s

PlanetDay Characteristic Time: 1.077332 seconds

Program ended with exit code: 0

of 85 93

Now we do the modeling by hand for the G0V star, to show how accurate the program is. We

start by converting the data for the G0V star to kg/m/s…!

GOV Star System

We try a planet radius 0.95 of the Earth’s:!

We try a planet mass the same as the Earth’s!

We try planet day the same as the Earth’s!

We now have the angular momentum of the planet…!

⋆

= (1.1)(6.9634E8m) = 7.65974E8m

⋆

= (1.06)(1,9891E30kg) = 2.108446E 30kg

⋆

= (1.35)(3.846E 26Watts) = 5.192E 26Wat ts

= 1.35 = (1.161895)(1.496E11m) = 1.7382E11m

= (0.95)(6.378E6m) = 6.0591E6m

= (1.0)(5.972E 24kg) = 5.972E 24kg

= (1.0)(86,400s) = 86400s

π(5.972E 24)

86,400s

(6.0591E6m)

= 6.37767E33J ⋅ s

⋆

(6.674E − 11)(2.108446E30kg)

1.7382E11m

= 28,452.76m /s

(5.972E 24kg)(28,452.76)

= 2.41734E33J

ℏ

⋆

6.37767E33J ⋅ s

2.3

= 2.7729E33J

ℏ

⋆

2.7729E33J ⋅ s

2.41734E33J

= 1.147secon ds

(2.7729E33)

(6.674E − 11)(299,792,458)(1.147sec)

= 6.94543E 22kg

of 86 93

It is exactly the same value given by!

Now we have the PlanetDay characteristic time:!

We see for this G0V star everything lines up!

And, all of the values are reasonable, the pressure gradient exponent is really good because it

runs between 2.0 and 2.5 in these stars and here its is p=2.3. We have!

ℏ

⋆

⋅

(2.7729E33J ⋅ s)

(6.674E − 11)(299,792,458)(6.94543E 22kg)

= 1.14700secon ds

ℏ

⋆

2.7729E33J ⋅ s

2.41734E33J

= 1.147secon ds

= R

⋆

⋅

7.65974E8m

1.8

= 4.2554E8m

= R

⋆

⋅

= 7.65974E8m ⋅

4.2554E8m

1.7382E11m

= 1.87523E6m

(6.674E − 11)(5.972E 24kg)

4.2554E8m

= 967.7935m /s

(6.94543E 22 kg)(967.7937)

= 3.25263E 28J

(PlanetDay) =

3.25263E 28

2.41734E33

(86,400s) = 1.1625474second s

ℏ

⋆

⋅

= 1.14700secon d ≈ 1.00second s

ℏ

⋆

= 1.147secon ds ≈ 1.00seconds

(

6α

4πh

)

⋅

= 1secon d

6.94543E 22 kg

7.34767E 22 kg

= 0.945Lu narMasses

1.87523E6m

1.7374E6m

= 1.07933LunarRadii

of 87 93

= !

We had , so it works great. The accuracy is!

4.2554E8m

3.84E8m

= 1.108177LunarOrbitalRa dii

5.972E 24kg

= 1.00Ear th Masses

= 0.95Ear th Ra dii

= 1.161895Ear thOrbital Ra dii

π(1.87523E6)

= 2.76218E19m

6.94543E 22

2.76218E19m

= 2,514.464kg/m

= 2.514

π(6.0591E6m)

= 9.3178E 20m

5.972E 24kg

9.3178E 20m

= 6,409.288kg/m

= 6.4092g/cm

1secon d = 2r

⊙

2(1.7382E11m)

967.7937

28,452.76)

(6.94543E 22)

(1.732)

5.972E 24)

(1.9891E30

1.1880secon d s ≈ 1secon d

= 3

⋆

2ℏ

⋆

1.732

7.65974E8m

1.87523E6

(6.674E − 11)

(5.972E 24)

(6.94543E33)

2(2.7729E33)

= 2.4486E33J

= (1/2)M

= 2.41734E33J

2.41734E33

2.4486E33

= 98.72 %

of 88 93

of 89 93

of 90 93

Appendix 1

If we want to prove that our planetary Planck constant is correct, the delocalization time for the

Earth should be 6 months using it, the time for the Earth to travel the width of its orbit. We want

to solve the Schrödinger wave equation for a wave packet and use the most basic thing we can

which is a Gaussian distribution. We want to then substitute for Planck’s constant that is used

for quanta and atoms our Planck-type constant (h bar solar) for the Earth/Moon/Sun system

then apply it to predict the delocalization time for the Moon in its orbit with the Earth around

the Sun.

We consider a Gaussian wave-packet at t=0:

We say that is the delocalization length and decompose the wave packet with a Fourier

transform:

is the harmonics of the wave function. We use the identity that gives the integral of a

quadratic:

Solve the equation

With the initial condition

A plane wave is the solution:

Where,

The wave-packet evolves with time as

Calculate the Gaussian integral of

ℏ

⊙

ψ (x,0) = Ae

−

ψ (x,0) = Ae

−

∫

2π ℏ

ℏ

∫

∞

−∞

−α

x+βx

d x =

4α

iℏ∂

ψ (x, t) =



ψ (x, t)

ψ (x,0) =

∫

dp ⋅ e

2ℏ

⋅ e

−

ℏ

( px−ϵ( p)t)

ϵ( p) =

ψ (x, t) =

∫

dp ⋅ e

2ℏ

⋅ e

−

ℏ

( px−

of 91 93

and

The solution is:

where

is the delocalization distance, which for instance could be the width of an atom. is the

delocalization time, the average time for say an electron to traverse the diameter of the atom and

even leave it, to delocalize. If we substitute for our , and say that the delocalization distance

is for the Moon, the width of the Earth orbit, we should get a half a year for the delocalization

time, the time for the Moon and Earth to traverse the diameter of their orbit around the Sun. We

have

Where is the mass of the Moon, and is the orbital radius of the Moon. We have

Now let’s compute a half a year…

(1/2)(365.25)(24)(60)(60)=15778800 seconds

So we see our delocalization time is very close to the half year over which the Earth and Moon

travel from one position to the opposite side of the Sun. The closeness is

Thus we know our is accurate, it continues to function in a theoretical framework. The thing

about this is that it means we can predict the mass of the Moon from the Earth year. In terms of

what we said earlier that the Moon allows for life by creating the seasons, holding the Earth at

its tilt to the Sun, so we don’t go through extreme heat and cold, this suggests the Moon has a

mass that follows from Earth orbit, which is the habitable zone of the Sun, the right distance for

water to exist as liquid, and thus could be as it is for a reason, which means life might be part of

a physical process throughout the Universe, that it unfolds naturally in the evolution of star

systems.

2ℏ

2mℏ

β =

i x

ℏ

= ex p

[

−

⋅

1 + t

/τ

]

τ =

m d

ℏ

⊙

τ =

moon

(2r

moon

)

ℏ

⊙

moon

τ = 4

(7.34767E 22kg)(3.844E8m)

2.8314E 33J ⋅ s

= 15338227secon d s

15338227

15778800

100 = 97.2 %

ℏ

⊙

of 92 93

Appendix 2

Some of the 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:

: 1.67262 × 10

−27

h : 6.62607 × 10

−34

J ⋅ s

: 0.833 × 10

−15

G: 6.67408 × 10

−11

c : 299,792,458m /s

α : 1/137

= q

= 1.6022E − 19coulom bs

= 8.988E 9

K E

moon

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

= 3.428E 28J

K E

earth

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

= 2.7396E 33J

of 93 93

The Author