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A Partial Theory Of Everything And The Hierarchy Of Life In The Universe

Ian Beardsley

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Contents

Important To Point Out…………………………………………..3

Introduction…………………………………………………………..4

1.0. The Theory………………………………………………….……8

2.0 The Proton………………………………………………………..11

3.0 The Solar Formulation……………………………………….15

4.0 Equating The Lunar And Solar Formulations

Yield Our 1 Second Base Unit………………………………….18

5.0 The Solutions For Jupiter and Saturn………………….20

6.0 The Origin of a Second………………………………………24

7.0 Integrating Analytic and Wave

Solutions of the Solar System………………………………….30

8.0 A Theory For Biological Hydrocarbons……………….35

9.0 Solving The F Class Star…………………………………….40

10.0 The Radius of a Planet……………………………………..44

11.0 Solving The K Class Star…………………………………..47

12.0 Computer Modeling The Planetary Systems……….55

13.0 Further Modeling Star Systems…………………………61

14.0 Modeling Red Dwarf Star Systems…………………….66

Appendix 1…………………………………………………………….71

Appendix 2 (Data Used In This Paper)……………………..73

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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 the 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 a 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

OurMoon

F05star

OurMoon

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) there 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 brown 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 =

⊙

OurMoon

6.96265E8m

3.84E8m

= 1.8132 =

F0Vmoon

OurMoon

2.060

1.74

= 1.184 ≈ 1.2

F0Vmoon

OurMoon

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

6315615secon d 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 solution

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

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

= en erg y

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

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

1secon d =

K E

(Ear th Da y)

K E

planet

moon

≈

star

moon

ℏ

⊙

1secon d

⋅

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

1secon d =

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 = en erg 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 ds

= 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)

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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 17 74

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 18 74

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

⊙

= 2πℏ

⊙

= 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 19 74

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

⊙

= 2πℏ

⊙

= 5.7365E 39J ⋅ s

K E

= n

⊙

⋅

2ℏ

⊙

K E

⊙

⋅

⊙

2ℏ

⊙

⋅

2ℏ

⊙

⋅

⊙

⋅ ℏ

⊙

ℏ

⊙

= (hC )K E

of 20 74

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

⋅

⊙

(1secon d )

⊙

(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 %

1secon d = 2r

⊙

of 21 74

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

1secon d = 2r

⋆

E =

2ℏ

⊙

K E

(1.89813E 27kg)(13720 m /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

= 1.006proton s ≈ 1.0proton s = hydrogen(H )

Z = Z

of 22 74

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)(10140 m /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 = Heliu m(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 23 74

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 24 74

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)(60m in /hour) = 1440min /d a y

1440minutes /60 = 24m inutes /h our

L = 2π M f r

L =

π M f r

ℏ

⊙

= 2.8314E 33J ⋅ s

of 25 74

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 26 74

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

1secon d =

(

)

earth

(

)

(7.05E 33)

(6.67408E − 11)((7.347673E 22kg)

(299792458m /s)

(

)

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

2π

of 27 74

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

1vispala = 0.4secon d s

(0.4s)(2.5) = 1secon d

of 28 74

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 − 5r a d ) = 464m = 0.464k m = 0.2883mi ≈ 0.3mi

of 29 74

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 30 74

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 31 74

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.

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

dρ

= −

2ηGM

⋆

∫

dρ

= −

2ηGM

⋆

∫

(

)

2ηGM

⋆

(

−

)

ρ(r) = ρ

exp

2ηGM

⋆

(

−

)

= c

exp

2ηGM

⋆

(

−

)

2ηGM

⋆

(

−

)

≈ 1 + x

≈ P

1 +

2ηGM

⋆

(

−

)

= c

η = −

(

)

dln P

dln R

μm

= 1.38E − 23J/K

μ ≈ 2.3

of 32 74

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…

= 1k m /s

P(R) ∝ R

−q

q ≈ 2.5

dln P

dln R

≈ − 2.5

η = −

(

1k m /s

30k m /s

)

(−2.5) = 1.5E − 3

dln P

dln R

dln P

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 33 74

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

ln P = −

earth

ℏ

⊙

ln R + C

P(R) = P

(

)

−

earth

ℏ

⊙

P(R) = P

(

)

exp

[

−

earth

ℏ

⊙

]

earth

π M

ℏ

⊙

= GM

c(1secon d )

moon

ℏ

⊙

⋅

= 1.0secon d s

of 34 74

(

6α

4πh

)

⋅

= 1secon d

of 35 74

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:

1secon d =

6α

⋅

4πh

1secon d =

K E

(Ear th Da y)

α = 1/137

K E

Ear th Da y

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6secon d s = hydrogen(H )

of 36 74

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 37 74

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 ds

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 38 74

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 39 74

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

1secon d =

K E

(Ear th Da y)

1secon d =

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 40 74

9.0 Solving The F Class Star We now want see how our theory for the wave solution of our

solar system (section 1.0) applies to our theory about F, K, and G stars in terms of being

associated with copper-silver civilizations and silver-gold civilizations (Introduction in this

paper). We have equation 1.17

1.17.

Which can be written

9.1

In our theory, outlined in the introduction, we surmised for a 16 month year of the F0V star that

the mass of the planet would have to be a little less than the Earth’s, about 5E24kg. Thus we see

the basis unit of time for our F class star would be different than for ours, which is 1 second. We

find it is:

Nicely it is about 2 seconds, or very closely exactly double our 1 second so, we see our system of

measuring time may very well have given the second as something Natural in the Universe. With

a characteristic time of 2 seconds we have the Planck constant for this spectral class F star

system. We have from the following equations

1.6

1.7

1.8

That,

9.2

⊙

⋅

= (1secon d )c

1sec

⋅ c ⋅ M

⋆

⋅

1sec

1.732

(299,792,458)(3.2E 30kg)

(6.96265E8m)

(4E11m)

= 2.410E 24kg

(2.410E 24)t = 5E 24

t = 2secon d s

ℏ

⊙

= (hC )K E

hC = 1secon d

C =

⋅

π r

ℏ

⋆

⋅

π r

K E

= (2sec)KE

K E

⋆

(6.674E − 11)(3.2E 30kg

4E11m

= 23,107m /s

K E

(5E 24kg)(23,107m /s)

= 1.33483E 33J

of 41 74

9.3

We have the Planck constant for this F star copper-silver civilization. Now we find the mass of

its moon. We use the equation for the ground state of our solar system, equation 2.6:

2.6.

We write it for the spectral class F star system

9.4

Thus the density of this moon is

These are very good, and realistic, results. This is, in fact, the density of Saturn’s Moon,

Enceladus. Enceladus is considered one of the most promising candidates for life in our Solar

System. It has a liquid water ocean beneath its icy crust confirmed by the Cassini spacecraft. The

presence of geysers at its South Pole suggest hydrothermal vents at it ocean floor that could

provide the energy and chemistry to support life. Cassini detected complex organic molecules in

the plumes of water vapor erupting from its surface that could potentially be the building blocks

for life. We may be able to sample this material from the subsurface ocean because the plumes

eject material into space.

We now present the solution to the Schrödinger wave equation for this copper-silver civilization

spectral class F star:

9.5

For this system, the condition of the eclipse gives

, where as for ours it is

We have

ℏ

⋆

= (2sec)(1.33483E33J ) = 2.67E 33J ⋅ s

ℏ

⊙

⋅

= 1secon d

ℏ

⋆

⋅

= 2secon d s

ℏ

⋆

2 sec

(2.67E 33)

(6.674E − 11)(299,792459)2sec

= 5.6268E 22kg

π R

π (2.060E6m)

= 3.66E19m

5.6268E 22 kg

3.66E19m

= 1,537.377kg /m

= 1.537g /cm

K E

= 3

⋆

⋅

2ℏ

⋆

= 574.5

⊙

= 400

of 42 74

We had found that . So this is an accuracy of

This is good. Let’s see what the length of the day of this planet would be (its rotation period). We

call upon equation 1.10

1.10

And write

9.6

=27.498hours

As we will get into in the next section, the planet day characteristic time is not necessarily the

same as the characteristic time. It could be one second here as well. Their day is about 27.5

hours compared to ours which is 24 hours. It is good for astronomy, which would have a 13.75

hour night during its spring and fall, and about 15 hour night during its winter. Since this moon

has a month of 6315615 seconds, then that is month of 73 day month. In terms of F0V days that

is (24/25.5)73~64 F0V day month, which is 8 squared. If in their ancient times if they

formulated a calendar and wanted to break up their month into smaller segments, like we did

when we made weeks, their week might be 8 days long. Interestingly (as we have 4 months in

our year) that results in 8 times 4 = 32 days, which is about our calendar month for some

months. The year of its planet is (3.44664 yr)(365.25 day/yr)=1,259 days. In terms of its own

days this is 1,098.76 days. However, in this paper we will not model possible calendars of FOV

copper-silver civilizations. Keep in mind, while modeling calendars, you might make some

months 28 days long, others 31 days long, introduce leap years, and much more to make things

work conveniently with mathematical relationships, like we did.

We will recall in section 4.0 that we equated the lunar formulation with the solar formulation to

get a characteristics time of 1 second for our planetary system. We had equation 4.5

K E

= (1.732)(574.5)

(6.674E − 11)

(5E 24kg)

(5.6268E 22 kg)

2(2.67E 33J ⋅ s)

= 1.38446E 33Joules

K E

= 1.33483E 33J

1.33483E 33

1.38446E 33

100 = 96 %

1secon d =

K E

(Ear th Da y)

2 secon d s =

K E

(Pl anetDa y)

(6.674E − 11)(5E 24kg

6.96265E8m)

= 692.3m /s

K E

= =

(5.6268E 22 kg)(692.3m /s)

= 1.3484E 28J

Pl anetDa y = (1secon d )

K E

= 1s

(1.33483E 33J )

(1.3484E 28J )

= 98,993.6secon d s

of 43 74

4.5

In terms of the F0V star system we should have

9.7

We have

We see it works great and the characteristic time for the F0V star system is indeed 2 seconds. So

the whole system solves nicely in terms of our theory. There is something further to consider

here, The period of the planet (It’s year) and period of its moon (its’s month)…

Periods in Earth Days

There is some room for play here and we chose the mass of the planet so it was close to 16 new

moons per year. The moons per year have to be a little larger than 16 to have 16 new moons per

year because the sidereal month is shorter than the lunar month. Thus there should be about 76

days in a lunar month if there are about 73 days in a sidereal month so we have close to 16 new

moons per year of the planet. The Earth is close to twelve new moons per year but has 13.3

moons per year. We wanted the relationship between this F star and our G star to be close to

12/16~3/4. We found this works quite well.

Periods in F0V Days

The planet day we found was 27.5 hours is approximately 1.146 days. In F0V days where a day is

one rotation for the habitable planet that is for the year

1secon d = 2r

⊙

2secon ds = 2r

⋆

(6.674E − 11)((3.2E 30kg)

4E11m

= 23,106.71m /s

⋆

= 2(4E11)

692.3m /s

(23,106.71m /s)

(5.6268E 22kg)

(1.732)

(5E 24kg)

(3.2E 30kg)

= 2.0372secon d s ≈ 2secon d s

F0Vplanet

= 3.44664years

F0Vmoon

= 6315615 = 73.0974d ays

(3.44664d ays)(365.25d a ys /year) = 1,258.885d a ys ≈ 1,259d a ys

1,259d a ys

73.1d a ys

= 17.223m oons /year

of 44 74

For the days of in the year of the planet in terms of the F0V planet day. The days for its moon’s

orbit, its lunar month, are

Thus for inhabitants of this planet we could say they would see their moon orbit their planet in

66.31 days and they might make this their month. However, each special class has a range of

mass for the star that makes it an F0V star, and depending on its peculiarities, and indeed we

can adjust that a little and as well the mass of the planet, so we could be seeing that the month

for their moon is close to 60 days even, because the month our moon is close to 30 days. Such

approximations are good for mathematics, and we have made them for our mathematics by

dividing the circle into 360 degrees, which is a good number with which to do as such because it

is very divisible by numbers used a lot in formulating Nature. We have 365.5 days in our year,

which means the Earth moves through about one degree a day in its orbit about the Sun. We

may be talking about a civilization that approximates things according to

to reconcile base 10 ( with base 60. Both are good bases for mathematics. Base 16 is

actually used a lot in software engineering. We use base 10 in general probably because we have

10 fingers to count on. Ten fingers are natural because with five fingers on each hand, five is a

number characteristic of the mathematics of life, its square root is used in obtaining the golden

ratio, which has mathematical properties used by life. That is why many animals have five

appendages on each hand and foot. Physical things in Nature are more characteristic of 6-fold

symmetry, like the points on a snowflake, or the sides on regular hexagons like in honeycombs.

1,259d a ys

1.146d a ys

= 1,098.6F 0Vd a ys /year

76d a ys

1.146d a ys

= 66.31Da ys

1,000d a ys /year

60d a ys /m onth

≈ 16m oons /year

= 1000)

of 45 74

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

have determined everything about G2V and F0V 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

10.1.

We have

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

Which is pretty good. Thus the radius of the F0V habitable planet is

10.2.

We computed the mass of this habitable planet to be a little less than that of the Earth, to be

5E24kg. Thus, the average density of the planet would be

⊙

= 108

⊙

= 2

(6.96265E8m)

1.496E11m

= 6,481.082m ≈ 6,481k m

= 6,378k m

6,378

6,481

100 = 98.4 %

= 2

⋆

2(1.18365E 9m)

(4E11m)

= 7,005,136.5m ≈ 7000k m

π R

π (7.005E6m)

= 1.4398E 21m

5E 24kg

1.4398E 21m

= 3.4727kg /m

= 3.4727g /cm

of 46 74

The density of the Earth is 5.51g/cm3. This result for the planet around the F0V star is actually

close to the density of Mars, which is 3.93 g/cm3. The planet could be rocky with oceans, it

would just have a smaller iron core. Or it could have more water to land than the Earth.

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 F0V star compute its angular

momentum. We have

This gives

10.3.

This is actually very good, which means the length of our planet day is correct which means our

characteristic time of 1.00 seconds is correct. We see We had (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. We have for F0V stars

10.4.

This works well because while the exact value of the exponent, q, depends on the individual star

and disc characteristics, observations suggest F0V stars have a slightly shallower pressure

gradient compared to G-type stars that are in fact to . This seems to be consistent

with both the observational data (photos of such discs) and theoretical models adjusted for the

hotter and more luminous nature of F-type stars.

earth

ℏ

⊙

= 2.5

earth

ℏ

⊙

24 = 60

planet

π M

π (5E 24kg)(1/(98,993.6s)(7.00E6m)

= 6.22E 33J ⋅ s

planet

ℏ

⋆

6.33E 33J ⋅ s

2.67E 33J ⋅ s

= 2.33

P(R) = P

(

)

exp

[

−

earth

ℏ

⊙

]

P(R) = P

(

)

exp

[

−

planet

ℏ

⋆

]

p ≈ 2.3

2.4

of 47 74

11.0 Solving The K Class Star While we said F-type stars are characteristic of copper-silver

civilizations because they have lower metallicities, this is not always true; it depends on the type

of F star. The older F-type stars have lower metallicity than G-type stars because they formed in

an environment that was not metal-rich because the environment has not been enriched by

previous generations of stars. Younger F-type stars (that formed more recently) formed in an

environment where previous stellar generations enriched the environment from which the F-

star formed. Also, F-type stars are more luminous meaning they have higher radiation pressure,

thus may be blowing the heavier metallic elements further out, perhaps beyond the habitable

zone of the planet. in general you can say F-type stars have lower metallicity than G-type stars.

In general, younger G-type stars like our Sun tend to have higher metallicities compared to older

stars, because they formed in a part of the galaxy that has undergone more stellar generations

enriching the interstellar medium. Also they have lower radiation pressure than F stars,

meaning the heavier elements needed to form terrestrial planets have not been so heavily blown

out beyond the habitable zone.

Like with F-stars and G stars, younger K stars have higher metallicities than older K stars

because they formed after other stars enriched the environment where they formed. Also, the

radiation pressure is lower in K-stars than in G stars, meaning perhaps not so much metal got

blown out beyond the habitable zone.

It in fact may be that we want to say that the stars of the silver-copper type civilizations are the

Population II stars which is to say they are older and did not form from gas clouds enriched by

earlier generations of stars, and that the the stars of the silver-gold type civilizations are the

Population I stars, like our Sun, which is to say they are younger and formed from clouds

enriched with heavier metals from pervious generations of stars. However we do have to say it

may depend on spectral class as well because the hotter F-stars have more radiation pressure

and whether they formed from metal-rich environments or not, they may blow more heavier

elements out of their habitable zones.

However, there is more to it. K-type stars are generally older than G-type stars, they have longer

life-spans, meaning they formed in environments that were less enriched by previous

generations of stars, that is to say they are usually Population II stars. Thus, they in general have

lower metallicities than G-type stars. F-stars, in general, do have lower metallicities than G-type

stars as well, because they are usually older than G-stars. Many F-type stars are part of the thick

disc or halo populations in our galaxy the Milky Way. G-type stars are generally found in the

thin disc region of our galaxy, which tend to have higher metallicities because they are younger

and closer to regions of recent star formation. Thus stars like the Sun, G-type stars, in general do

have higher metallicities than F or K stars, even thought they are in-between the two in spectral

class. So, we were right to make F-type stars copper-silver type civilizations, and G-type stars

silver-gold type civilizations because in general this would be true, though there can be

exceptions. We suggested G-star civilizations can become more sophisticated than F-star

civilizations because they have longer life-spans giving more time for intelligent life to evolve.

However, K stars have longer life-spans than G stars, so we suggested they can have more

sophisticated life than found around G-stars. However, we did not associate any metals in

particular with these stars because we started at F-type stars with copper and silver, went to G-

type stars with silver and gold, and after gold in this group of elements there is no suitable metal

left to utilize in this approach to the theory, Indeed most F-type stars would have copper-silver

type civilizations because they are in general older than G-type stars. So, as we begin to solve K-

type stars we will start with them as copper-silver civilizations as in general they have lower

metallicities.

However it may be, regardless of the metallicity of the star, if it is more luminous, having more

radiation pressure, it may blow more heavier elements out of its habitable zone. However, it may

of 48 74

not, because the habitable zone of a more luminous star is further away from it. How that

balances out has to be computed. ChatGpt says that in less luminous stars (like G and K-type

stars) the habitable zone is closer in, and radiation pressure is lower, allowing rocky planet

formation to occur more easily in the inner system (as opposed to easier formation of gas giants

from lighter elements). That, these G and K stars are therefore often prime candidates for

finding terrestrial planets in stable habitable zones.

We now begin to solve the K-class star. We know their typical properties. For a K0V star we have

in solar mass, solar radii, and solar luminosities:

K0v:

Thus, the habitable zone of the K0V star is at

This is the orbital radius of the planet we are considering, . We have the mass of the star is

We now want to figure out the characteristic time of this K0V system, and from that, its . We

call upon equation 9.1 for the the Earth, a G2V system:

9.1

11.1

We theorize this system has 9 new moons per year because the F0V star we guessed had 16 from

Nature’s 4x4 array, we know the Earth has about 12 completed new moons per year from

Nature’s 3x4 array, so we surmise the K0V system is Nature’s 3x3=9 array from the cross

product in three dimensional space. We have

0.88M

⊙

0.813R

⊙

0.46L

⊙

habi table

= 0.46L

⊙

= 0.678AU = (0.678AU )(1.496E11m /AU ) = 1.014288E11m ≈ 1E11m

⋆

= (0.88M

⊙

)(1.989E 30kg/M

⊙

) = 1.75E 30kg

K0Vstar

= R

⋆

= (0.813)R

⊙

= (0.813)(6.96265E 8m) = 5.66E8m

⋆

K0Vmoon

= (1.7)r

K0Vmoon

= r

5.66E8m

1.7

= 3.33E 8m

ℏ

⋆

1sec

⋅ c ⋅ M

⋆

⋅

1sec

(299,7092458m /s)(1.75E 30kg)

(3.33E8m)

(1E11m)

= 1.00865E 24kg ≈ 1E 25kg

KOVplanet

4π

K0Vstar

K0Vplanet

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Its planet has year of a little more than one half of an Earth year. 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 9 moons per the planet’s year. I think this is what we want. I

find that value is 5.4E24 kg.

Returning to equation 11.1, gives for the characteristic time of the K0V system

11.2.

Now we can determine , the Planck-type constant for the K0V system. We have the orbital

velocity of the planet is

So, the kinetic energy of the planet is

So,

11.3.

Which means the mass of the moon of this planet is

4π

K0Vstar

4π

(6.674E − 11)(1.75E 30kg)

= 3.380E − 19

K0Vplanet

= (3.380E − 19)(1E11m)

= 18384776secon d s = 212.8 ≈ 213da ys

= 0.5825784years ≈ 0.58years

K0Vmoon

4π

K0Vplanet

F0Vmoon

K0Vmoon

4π

(6.674E − 11)(5.4E 24kg)

(3.33E8m)

= 2011204.5sec = 23.2779d a ys

K0Vplanet

K0Vmoon

212.8d ays

23.278d ays

= 9.14m oons ≈ 9m oons

= 1.0E 25kg

(1.0E 25kg)t

= 5.4E 24kg

= 0.54secon d s

ℏ

⋆

(6.674E − 11)(1.75E 30kg

1E11m)

= 34,175.28m /s

K E

(5.4E 24kg)(34,175.28m /s)

= 3.15E 33J

ℏ

⋆

= (0.54secon ds)(3.15E 33J ) = 1.7E 33J ⋅ s

of 50 74

11.4.

Thus, the density of this moon is from a perfect eclipse

11.5.

11.6.

Thus, the density of the moon orbiting the habitable planet of the K0V star is 2.3g/cm3. This

density is closest to the moon Ganymede in our solar system, Jupiter’s largest moon and, the

largest moon in our solar system, which is even larger than our smallest planet, Mercury. It is

the third major moon orbiting Jupiter, that is it is the third largest Galilean satellite. It has a

density of 1.936g/cm3. The orbital velocity of this K0V moon is

11.7.

And, its kinetic energy is, then

11.8.

Thus, we would compute the K0V planet’s day to be (its rotation period)

Which is quite a rapid rotation, nearly twice as quick as the the Earth’s rotation, which is about

24 hours. But, as we will see this is not correct, and the the actual value works out quite nice, as

well as is the reason for it. But let us first verify that our characteristic time of 0.54 seconds is

correct on all fronts according to the theory. We start with equation 9.5, the solution for the

Schrödinger wave equation:

9.5

The equation for a perfect eclipse is

ℏ

⋆

⋅

= 0.54secon d s

(1.7E 33)

(6.674E − 11)(299,792,458m /s)(0.54sec)

= 6.443E 22kg

K0Vmoon

= R

K0Vstar

K0Vmoon

K0Vplanet

= 5.66E 8

3.33E8m

1E11m

= 1.88478E6m

π R

π (1.88478E6)

6.443E 22

2.8E19

= 2.3E 3kg /m

= 2.3g /cm

(6.674E − 11)(5.4E 24kg)

(3.33E8m)

= 1040.322m /s

K E

(6.443E 33kg)(1040.322m /s)

= 3.4865E 28J

Pl an etDa y = (0.54sec)

3.15E 33

3.4865E 28 = 48788.183secon d s

= 13.55h ours

K E

= 3

⋆

⋅

2ℏ

⋆

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We have for our K0V system

And we see the eclipse ratio is nearly perfectly 300. Four our solar system it is nearly perfectly

400. We have

11.9

We found earlier in this section

Thus, that is an accuracy of

We want to verify equations 4.5 and 9.7. For the Earth (G2V star) we have

4.5

And, for the F0V star system we have

9.7

Thus we want for the K0V system

11.10.

⋆

5.66E5m

1.88478E6m

= 300.313

1E11m

3.33E8m

= 300.300

K E

= (1.732)(300.3)

(6.674E − 11)

(5.4E 24kg)

(6.443E 22)

2(1.7E 33J ⋅ s)

= 3.126E 33J

K E

= 3.15E 33J

3.126E 33

3.15E 33

= 99.24 %

1secon d = 2r

⊙

2secon ds = 2r

⋆

0.54secon d s = 2r

⋆

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So, it check out correctly. As we will remember, in the protoplanetary disc of a F0V star system

the pressure gradient is

10.4.

10.3.

Where 2.33 is called p, and this is correct from what we know about protoplanetary discs of

these stars. And, for a protoplanetary disc of a star like our Sun

7.10

Where p=2.5 and this is correct from what we know about protoplanetary discs of these stars.

So, let us compute p for our K0V star system from 10.2:

10.2

The K0V planet has a radius about that of our Earth. We have its rotational angular momentum

11.11.

We have

From observation and data for K0V protoplanetary disc p is about 2.0 to 2.2. We are way off.

But as you will remember, we said the rotation of the planet was unusually fast. The interesting

thing is as we go go from F stars to G stars p increases (2.3 to 2.5) but then when we go from G

stars to K stars, it decreases to 2.0. This is because of the the relationship of mass to luminosity

of a K star, the relationship for p is not linear (going up with increase in star class) but is

sinusoidal (starts down, then goes up, then down, with increase in star class). Further, our

equation

2(1E11)

(1040.322m /s)

(34,175.28m /s)

(6.443E 22kg)

(1.732)

(5.4E 24)

(1.75E 30)

= 0.53674 ≈ 0.54secon d s

P(R) = P

(

)

exp

[

−

planet

ℏ

⋆

]

planet

ℏ

⋆

6.33E 33J ⋅ s

2.67E 33J ⋅ s

= 2.33

P(R) = P

(

)

exp

[

−

earth

ℏ

⊙

]

earth

ℏ

⊙

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

= 2

⋆

= 2

(5.66E8m)

(1E11m)

= 6407120m = 6,407k m

planet

π M

π (5.4E 24kg)

48788.183s

(6,407,120)

= 1.142E 34J ⋅ s

planet

ℏ

⋆

1.142E 34

1.7E 33

= 6.7176 = p

of 53 74

Exists independent of the theory. Characteristic times from ground states and all other

equations are derived in terms of one another. But this equation is not, it was discovered. And

we see here, its characteristic times are not necessarily the same as the characteristic times of

the star systems. In fact, for the protplanetary discs to have the right q in terms of our theory, it

has to start up, go down, then back up. Where the characteristic time in the equation was the

same as the characteristic time for the F0V star system and close to the same for characteristic

time for the Sun, it now has to go back up to be close to the F0V system for the K0V system.

That is, that short rotation period for the K0V planet has to go up. We find for the PlanetDay

equation the characteristic time is around 1.81 seconds. We have

11.12.

This gives

11.13.

We have

11.14.

Thus we have

11.15.

Which is correct for the pressure gradient of the protoplanetary disc of a K0V star system. For

our Sun the characteristic time is 1.00 seconds but for the PlanetDay it is closer to 1.2 seconds.

But our Sun is a G2V system, not a G0V system. But if we go down to a G0V system, the

characteristic time could become 1.00 seconds, depending on planet mass, radius, mass of star

and so forth. Interestingly if we go up from a K0V system to a K2V system, the PlanetDay

characteristic time could go up from 1.8 seconds to 2.00 seconds in general. We have F0V stars

have a PlanetDay characteristic time of 1.00 seconds but there can be some play in the factors.

We might want to calibrate our factors, planet size, mass,…and so forth so that F0V is 1 second,

G0V is 1 second, K0V is 2 seconds in the diagrams. That is clear if we look at the following

diagrams (next page)…

Ear th Da y = (1secon d )

K E

earth

K E

moon

Pl an etDa y = (1.8secon d s)

K E

planet

K E

moon

Pl an etDa y = 1.81secon d s

3.15E 33

3.4865E 28

= 163530.76s = 45.425h ours

planet

π M

π (5.4E 24kg)

163530.7s

(6,407,120)

= 3.4072E 33J ⋅ s

planet

ℏ

⋆

3.4072E 33J ⋅ s

1.7E 33J ⋅ s

= 2.00 = p

of 54 74

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12.0 Computer Modeling The Planetary Systems

I wrote a program to model our theory for star systems. Keep in mind Mars is further from the

Sun, next planet after the Earth, but it has a day of 24 hours like the Earth Does. That would be a

good length of day for intelligent life to develop in. With our program we can model any good

data on extrasolar planets the comes in. Here is the program. We run it for several types of star

systems.

#include <stdio.h>

#include <math.h>

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

{

float c=299792458,M_star,M_s,r_planet, M_p,r_moon,L_star, ratio,R_s,R_star,n,

r_p,T_p,moons,T_m,M_planet, t_c,KE_planet, hbar_star, v_p, M_moon, R_moon,

KE_p,eclipse, v_m, t_characteristic,R_planet,PlanetDay,KE_moon, seconds, L_planet, p,

V_p, V_m, rho_p, rho_m;

printf ("What is the mass of the star in solar masses? ");

scanf ("%f", &M_s);

printf ("What is the radius of the star in solar radii? ");

scanf ("%f", &R_s);

printf ("What is the luminosity of the star in solar luminosities? ");

scanf ("%f", &L_star);

printf ("What is the ratio of element 2 to element 1, ie Au/Ag=1.823, Ag/

Cu=1.6975? ");

scanf ("%f", &ratio);

printf ("What is the orbital number of the planet? ");

scanf ("%f", &n);

M_star=M_s*(1.989E30);

R_star=R_s*(6.96265E8);

r_p=sqrt(L_star);

r_planet=r_p*(1.496E11);

r_moon=R_star/ratio;

M_p=((1/(sqrt(n))*c*M_star*(r_moon/(r_planet*r_planet)))/1E24);

printf("The habitable zone which is the orbital radius of the planet is: %f m \n",

r_planet);

printf("The mass of the planet is: %f E24 kg \n", M_p);

printf("Which is: %f Earth masses \n", M_p/5.972);

T_p=sqrt((4*3.14159*3.14159*r_planet*r_planet*r_planet)/(6.674E-11*M_star));

printf("The orbital period, or year, of the planet is: %f years \n", T_p/

31557600);

printf ("How many moons do you want for the moon's orbital period?\n");

printf ("For example there are 13.379 moons of our moon per Earth year. \n");

printf ("This gives 12.38 New Moons is approximately 12 New Moons per year. \n");

scanf ("%f", &moons);

T_m=T_p/moons;

printf ("This gives the orbital period of the moon is %f days. \n", T_m/86400);

printf ("We will use this value to determine the suggested mass of the planet it

orbits. \n");

M_planet=(4*3.141*3.141*r_moon*r_moon*r_moon)/(6.674E-11*T_m*T_m);

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

of 56 74

printf ("Now that we know what the mass the planet should be, we \n");

printf ("multiply the original mass M_p by the characteristic time t_c \n");

printf ("to get the mass of the planet M_planet, to find the characteristic time.

\n");

t_c=M_planet/(M_p*1E24);

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

printf ("Now we compute the orbital velocity of the planet. It is: \n");

v_p=sqrt((6.674E-11)*M_star/r_planet);

printf ("%f m/s \n", v_p);

printf ("Mass of planet is %f kg \n", M_planet);

KE_planet=0.5*M_planet*v_p*v_p;

printf ("Its kinetic energy is: \n");

printf ("%f E33 J \n", KE_planet/1E33);

hbar_star=KE_planet*t_c;

printf ("Since we know this and the characteristic time, we have \n");

printf ("the star system Planck constant. It is: \n");

printf ("%f E33 Js \n", hbar_star/1E33);

printf ("Knowing this we can determine the mass of the moon. \n");

M_moon=cbrt(hbar_star)*cbrt(hbar_star)/cbrt(6.674E-11*c*t_c);

printf ("It is: %f E22 kg \n", M_moon/1E22);

printf ("The radius of the moon is given by a perfect eclipse. It is: \n");

R_moon=R_star*(r_moon/r_planet);

printf ("%f E6 m \n", R_moon/1E6);

printf ("We need only verify that our star system Planck constant \n");

printf ("works, by trying it in our KE of the planet solution to the wave

equation. \n");

printf (“R_star over R_moon is: \n");

eclipse=R_star/R_moon;

printf ("%f \n", eclipse);

KE_p=sqrt(n)*eclipse*M_planet*(6.674E-11)*M_planet*(6.674E-11)*(M_moon/

(2*hbar_star))*(M_moon/hbar_star)*(M_moon);

printf ("KE_p is: %f J \n", KE_p/1E33);

printf ("We need to verify that the characteristic time is correct \n");

printf ("from the equation for it from equating solar and lunar solutions. \n");

v_m=sqrt((6.674E-11*M_planet)/r_moon);

printf ("Orbital velocity of moon is: %f m/s \n", v_m);

of 57 74

t_characteristic=2*r_planet*(v_m/(v_p*v_p)*sqrt((M_moon/M_planet)*(M_moon/

M_planet)*(M_moon/M_star)*sqrt(n)));

printf ("The characteristic time is: %f seconds \n", t_characteristic);

printf ("Now we want to compute the characteristic time for the PlanetDay \n");

printf ("We need to determine the rotational angular momentum of the planet. \n");

printf ("And, the planet radius to do this. \n");

printf ("The planet radius is: \n");

R_planet=2*(R_star*R_star)/r_planet;

printf ("%f km \n", R_planet/1000);

printf ("What is the rotation period, or day, in EarthDays? \n");

scanf ("%f", &PlanetDay);

PlanetDay=PlanetDay*86400;

printf ("PlanetDay is: %f seconds \n", PlanetDay);

printf ("The PlanetDay characteristic time is: \n");

KE_moon=0.5*M_moon*v_m*v_m;

seconds=(KE_moon/KE_planet)*PlanetDay;

printf ("%f seconds \n", seconds);

printf ("The planetary angular momentum is: \n");

L_planet=0.8*3.14159*M_planet*(1/PlanetDay)*R_planet*R_planet;

printf ("%f E33 Js \n", L_planet/1E33);

printf ("Thus the pressure gradient for the protoplanetary disc \n");

printf ("has an exponent of p is: ");

p=L_planet/hbar_star;

printf ("p=%f \n", p);

printf ("All that is left to do is to compute the densities of the planets and the

moons. \n");

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

V_m=(1.33333)*(3.14159)*R_moon*R_moon*R_moon;

rho_p=M_planet/V_p;

rho_m=M_moon/V_m;

printf ("The density of the planet is: %f g/cm3 \n", rho_p*1000/(100*100*100));

printf ("The density of the moon is: %f g/cm3 \n", rho_m*1000/(100*100*100));

}

We now run the program for our solar system and we find it works well…

of 58 74

G2V System (Our Solar System)

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

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

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

What is the ratio of element 2 to element 1, ie Au/Ag=1.823, Ag/Cu=1.6975? 1.8

What is the orbital number of the planet? 3

The habitable zone, which is the orbital radius of the planet is: 149599993856.000000m

The mass of the planet is: 5.950231 E24 kg

Which is: 0.996355 Earth masses

The orbital period, or year, of the planet is: 0.999913 years

How many moons do you want for the moon's orbital period?

For example there are 13.379 moons of our moon per Earth year.

This gives 12.38 New Moons is approximately 12 New Moons per year.

13.379

This gives the orbital period of the moon is 27.297884 days.

We will use this value to determine the suggested mass of the planet it orbits.

The mass of the planet is 6.152200 E24 kg

Now that we know what the mass the planet should be, we

multiply the original mass M_p by the characteristic time t_c

to get the mass of the planet M_planet, to find the characteristic time.

The characteristic time is: 1.033943 seconds

Now we compute the orbital velocity of the planet. It is:

29788.230469 m/s

Mass of planet is 6152200133217478516932608.000000 kg

Its kinetic energy is:

2.729543 E33 J

Since we know this and the characteristic time, we have

the star system Planck constant. It is:

2.822192 E33 Js

Knowing this we can determine the mass of the moon.

It is: 7.274836 E22 kg

The radius of the moon is given by a perfect eclipse. It is:

1.80030 E6 m

We need only verify that our star system Planck constant

works, by trying it in our KE of the planet solution to the wave equation.

R_star over R_moon is:

386.749268

KE_p is: 2.729543 J

We need to verify that the characteristic time is correct

from the equation for it from equating solar and lunar solutions

Orbital velocity of moon is: 1030.284790 m/s

The characteristic time is: 1.033943 seconds

Now we want to compute the characteristic time for the PlanetDay

We need to determine the rotational angular momentum of the planet.

And, the planet radius to do this

The planet radius is:

6481.083008 km

What is the rotation period, or day, in EarthDays?

PlanetDay is: 86400.000000 seconds

The PlanetDay characteristic time is:

1.222170 seconds

The planetary angular momentum is:

7.517118 E33 Js

Thus the pressure gradient for the protoplanetary cloud

has an exponent of p is: p=2.663574

All that is left to do is to compute the densities of the planets and the moons.

The density of the planet is: 5.395113 g/cm3

The density of the moon is: 2.976465 g/cm3

Program ended with exit code: 0

of 59 74

F0V System

What is the mass of the star in solar masses? 1.61

What is the radius of the star in solar radii? 1.7

What is the luminosity of the star in solar luminosities? 7.24

What is the ratio of element 2 to element 1, ie Au/Ag=1.823, Ag/Cu=1.6975? 1.7

What is the orbital number of the planet? 3

The habitable zone, which is the orbital radius of the planet is: 402532433920.000000m

The mass of the planet is: 2.381736 E24 kg

Which is: 0.398817 Earth masses

The orbital period, or year, of the planet is: 3.478191 years

How many moons do you want for the moon's orbital period?

For example there are 13.379 moons of our moon per Earth year.

This gives 12.38 New Moons is approximately 12 New Moons per year.

17.223

This gives the orbital period of the moon is 73.762367 days.

We will use this value to determine the suggested mass of the planet it orbits.

The mass of the planet is 4.914012 E24 kg

Now that we know what the mass the planet should be, we

multiply the original mass M_p by the characteristic time t_c

to get the mass of the planet M_planet, to find the characteristic time.

The characteristic time is: 2.063206 seconds

Now we compute the orbital velocity of the planet. It is:

23042.150391 m/s

Mass of planet is 4914012012894422959128576.000000 kg

Its kinetic energy is:

1.304525 E33 J

Since we know this and the characteristic time, we have

the star system Planck constant. It is:

2.691503 E33 Js

Knowing this we can determine the mass of the moon.

It is: 5.598608 E22 kg

The radius of the moon is given by a perfect eclipse. It is:

2.047374 E6 m

We need only verify that our star system Planck constant

works, by trying it in our KE of the planet solution to the wave equation.

R_star over R_moon is:

578.131042

KE_p is: 1.304525 J

We need to verify that the characteristic time is correct

from the equation for it from equating solar and lunar solutions

Orbital velocity of moon is: 686.315674 m/s

The characteristic time is: 2.063206 seconds

Now we want to compute the characteristic time for the PlanetDay

We need to determine the rotational angular momentum of the planet.

And, the planet radius to do this.

The planet radius is:

6961.071777 km

What is the rotation period, or day, in EarthDays?

1.146

PlanetDay is: 99014.406250 seconds

The PlanetDay characteristic time is:

1.000793 seconds

The planetary angular momentum is:

6.044071 E33 Js

Thus the pressure gradient for the protoplanetary cloud

has an exponent of p is: p=2.245612

All that is left to do is to compute the densities of the planets and the moons.

The density of the planet is: 3.477929 g/cm3

The density of the moon is: 1.557404 g/cm3

Program ended with exit code: 0

Now we run it for the K0V star we modeled…

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K0V System

What is the mass of the star in solar masses? 0.88

What is the radius of the star in solar radii? 0.813

What is the luminosity of the star in solar luminosities? 0.46

What is the ratio of element 2 to element 1, ie Au/Ag=1.823, Ag/Cu=1.6975? 1.7

What is the orbital number of the planet? 3

The habitable zone which is the orbital radius of the planet is: 101463662592.000000m

The mass of the planet is: 9.798795 E24 kg

Which is: 1.640789 Earth masses

The orbital period, or year, of the planet is: 0.595374 years

How many moons do you want for the moon's orbital period?

For example there are 13.379 moons of our moon per Earth year.

This gives 12.38 New Moons is approximately 12 New Moons per year.

9.14

This gives the orbital period of the moon is 23.792141 days.

We will use this value to determine the suggested mass of the planet it orbits.

The mass of the planet is 5.166112 E24 kg

Now that we know what the mass the planet should be, we

multiply the original mass M_p by the characteristic time t_c

to get the mass of the planet M_planet, to find the characteristic time.

The characteristic time is: 0.527219 seconds

Now we compute the orbital velocity of the planet. It is:

33930.992188 m/s

Mass of planet is 5166112134934765332594688.000000 kg

Its kinetic energy is:

2.973904 E33 J

Since we know this and the characteristic time, we have

the star system Planck constant. It is:

1.567899 E33 Js

Knowing this we can determine the mass of the moon.

It is: 6.153838 E22 kg

The radius of the moon is given by a perfect eclipse. It is:

1.857680 E6 m

We need only verify that our star system Planck constant

works, by trying it in our KE of the planet solution to the wave equation.

R_star over R_moon is:

304.715332

KE_p is: 2.973904 J

We need to verify that the characteristic time is correct

from the equation for it from equating solar and lunar solutions

Orbital velocity of moon is: 1017.576111 m/s

The characteristic time is: 0.527219 seconds

Now we want to compute the characteristic time for the PlanetDay

We need to determine the rotational angular momentum of the planet.

And, the planet radius to do this

The planet radius is:

6316.110352 km

What is the rotation period, or day, in EarthDays?

1.8927

PlanetDay is: 163529.281250 seconds

The PlanetDay characteristic time is:

1.751937 seconds

The planetary angular momentum is:

3.167431 E33 Js

Thus the pressure gradient for the protoplanetary cloud

has and exponent of p is: p=2.020176

All that is left to do is to compute the densities of the planets and the moons.

The density of the planet is: 4.894716 g/cm3

The density of the moon is: 2.291640 g/cm3

Program ended with exit code: 0

of 61 74

13.0 Further Modeling Star Systems

KOV System Modeled For PlanetDay Characteristic Time of 2 Seconds

What is the mass of the star in solar masses? 0.88

What is the radius of the star in solar radii? 0.813

What is the luminosity of the star in solar luminosities? 0.46

What is the ratio of element 2 to element 1, ie Au/Ag=1.823, Ag/Cu=1.6975?

1.7

What is the orbital number of the planet? 3

The habitable zone which is the orbital radius of the planet is:

101463662592.000000 m

The mass of the planet is: 9.798795 E24 kg

Which is: 1.640789 Earth masses

The orbital period, or year, of the planet is: 0.595374 years

How many moons do you want for the moon's orbital period?

For example there are 13.379 moons of our moon per Earth year.

This gives 12.38 New Moons is approximately 12 New Moons per year.

9.6

This gives the orbital period of the moon is 22.652102 days.

We will use this value to determine the suggested mass of the planet it

orbits.

The mass of the planet is 5.699199 E24 kg

Now that we know what the mass the planet should be, we

multiply the original mass M_p by the characteristic time t_c

to get the mass of the planet M_planet, to find the characteristic time.

The characteristic time is: 0.581622 seconds

Now we compute the orbital velocity of the planet. It is:

33930.992188 m/s

Mass of planet is 5699199403774126933934080.000000 kg

Its kinetic energy is:

3.280779 E33 J

Since we know this and the characteristic time, we have

the star system Planck constant. It is:

1.908175 E33 Js

Knowing this we can determine the mass of the moon.

It is: 6.788848 E22 kg

The radius of the moon is given by a perfect eclipse. It is:

1.857680 E6 m

We need only verify that our star system Planck constant

works, by trying it in our KE of the planet solution to the wave equation.

R_star over R_moon is:

304.715332

KE_p is: 3.280779 J

We need to verify that the characteristic time is correct

from the equation for it from equating solar and lunar solutions

Orbital velocity of moon is: 1068.788818 m/s

The characteristic time is: 0.581622 seconds

Now we want to compute the characteristic time for the PlanetDay

We need to determine the rotational angular momentum of the planet.

And, the planet radius to do this.

The planet radius is:

6316.110352 km

What is the rotation period, or day, in EarthDays?

PlanetDay is: 172800.000000 seconds

The PlanetDay characteristic time is:

2.042286 seconds

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The planetary angular momentum is:

3.306809 E33 Js

Thus the pressure gradient for the protoplanetary disc

has an exponent of p is: p=1.732969

All that is left to do is to compute the densities of the planets and the

moons.

The density of the planet is: 5.399798 g/cm3

The density of the moon is: 2.528113 g/cm3

Program ended with exit code: 0

Modeling K0V Again (02)

What is the mass of the star in solar masses? 0.88

What is the radius of the star in solar radii? 0.813

What is the luminosity of the star in solar luminosities? 0.46

What is the ratio of element 2 to element 1, ie Au/Ag=1.823, Ag/Cu=1.6975?

1.7

What is the orbital number of the planet? 3

The habitable zone which is the orbital radius of the planet is:

101463662592.000000 m

The mass of the planet is: 9.798795 E24 kg

Which is: 1.640789 Earth masses

The orbital period, or year, of the planet is: 0.595374 years

How many moons do you want for the moon's orbital period?

For example there are 13.379 moons of our moon per Earth year.

This gives 12.38 New Moons is approximately 12 New Moons per year.

9.8

This gives the orbital period of the moon is 22.189816 days.

We will use this value to determine the suggested mass of the planet it

orbits.

The mass of the planet is 5.939139 E24 kg

Now that we know what the mass the planet should be, we

multiply the original mass M_p by the characteristic time t_c

to get the mass of the planet M_planet, to find the characteristic time.

The characteristic time is: 0.606109 seconds

Now we compute the orbital velocity of the planet. It is:

33930.992188 m/s

Mass of planet is 5939139086244628588920832.000000 kg

Its kinetic energy is:

3.418902 E33 J

Since we know this and the characteristic time, we have

the star system Planck constant. It is:

2.072228 E33 Js

Knowing this we can determine the mass of the moon.

It is: 7.074662 E22 kg

The radius of the moon is given by a perfect eclipse. It is:

1.857680 E6 m

We need only verify that our star system Planck constant

works, by trying it in our KE of the planet solution to the wave equation.

R_star over R_moon is:

304.715332

KE_p is: 3.418902 J

We need to verify that the characteristic time is correct

from the equation for it from equating solar and lunar solutions

Orbital velocity of moon is: 1091.055176 m/s

The characteristic time is: 0.606109 seconds

Now we want to compute the characteristic time for the PlanetDay

We need to determine the rotational angular momentum of the planet.

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And, the planet radius to do this.

The planet radius is:

6316.110352 km

What is the rotation period, or day, in EarthDays?

1.8

PlanetDay is: 155520.000000 seconds

The PlanetDay characteristic time is:

1.915441 seconds

The planetary angular momentum is:

3.828919 E33 Js

Thus the pressure gradient for the protoplanetary cloud

has an exponent of p is: p=1.847731

All that is left to do is to compute the densities of the planets and the

moons.

The density of the planet is: 5.627133 g/cm3

The density of the moon is: 2.634548 g/cm3

Program ended with exit code: 0

Modeling K0V 03

What is the mass of the star in solar masses? 0.88

What is the radius of the star in solar radii? 0.813

What is the luminosity of the star in solar luminosities? 0.46

What is the ratio of element 2 to element 1, ie Au/Ag=1.823, Ag/Cu=1.6975?

1.7

What is the orbital number of the planet? 3

The habitable zone which is the orbital radius of the planet is:

101463662592.000000 m

The mass of the planet is: 9.798795 E24 kg

Which is: 1.640789 Earth masses

The orbital period, or year, of the planet is: 0.595374 years

How many moons do you want for the moon's orbital period?

For example there are 13.379 moons of our moon per Earth year.

This gives 12.38 New Moons is approximately 12 New Moons per year.

9.8

This gives the orbital period of the moon is 22.189816 days.

We will use this value to determine the suggested mass of the planet it

orbits.

The mass of the planet is 5.939139 E24 kg

Now that we know what the mass the planet should be, we

multiply the original mass M_p by the characteristic time t_c

to get the mass of the planet M_planet, to find the characteristic time.

The characteristic time is: 0.606109 seconds

Now we compute the orbital velocity of the planet. It is:

33930.992188 m/s

Mass of planet is 5939139086244628588920832.000000 kg

Its kinetic energy is:

3.418902 E33 J

Since we know this and the characteristic time, we have

the star system Planck constant. It is:

2.072228 E33 Js

Knowing this we can determine the mass of the moon.

It is: 7.074662 E22 kg

The radius of the moon is given by a perfect eclipse. It is:

1.857680 E6 m

We need only verify that our star system Planck constant

works, by trying it in our KE of the planet solution to the wave equation.

R_star over R_moon is:

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304.715332

KE_p is: 3.418902 J

We need to verify that the characteristic time is correct

from the equation for it from equating solar and lunar solutions

Orbital velocity of moon is: 1091.055176 m/s

The characteristic time is: 0.606109 seconds

Now we want to compute the characteristic time for the PlanetDay

We need to determine the rotational angular momentum of the planet.

And, the planet radius to do this.

The planet radius is:

6316.110352 km

What is the rotation period, or day, in EarthDays?

1.7

PlanetDay is: 146880.000000 seconds

The PlanetDay characteristic time is:

1.809028 seconds

The planetary angular momentum is:

4.054150 E33 Js

Thus the pressure gradient for the protoplanetary disc

has an exponent of p is: p=1.956421

All that is left to do is to compute the densities of the planets and the

moons.

The density of the planet is: 5.627133 g/cm3

The density of the moon is: 2.634548 g/cm3

Program ended with exit code: 0

Modeling GOV Star System

What is the mass of the star in solar masses? 1.06

What is the radius of the star in solar radii? 1.100

What is the luminosity of the star in solar luminosities? 1.35

What is the ratio of element 2 to element 1, ie Au/Ag=1.823, Ag/Cu=1.6975? 1.8

What is the orbital number of the planet? 3

The habitable zone, which is the orbital radius of the planet is: 173819494400.000000

The mass of the planet is: 5.139235 E24 kg

Which is: 0.860555 Earth masses

The orbital period, or year, of the planet is: 1.216354 years

How many moons do you want for the moon's orbital period?

For example there are 13.379 moons of our moon per Earth year.

This gives 12.38 New Moons is approximately 12 New Moons per year.

13.4

This gives the orbital period of the moon is 33.154713 days.

We will use this value to determine the suggested mass of the planet it orbits.

The mass of the planet is 5.551060 E24 kg

Now that we know what the mass the planet should be, we

multiply the original mass M_p by the characteristic time t_c

to get the mass of the planet M_planet, to find the characteristic time.

The characteristic time is: 1.080134 seconds

Now we compute the orbital velocity of the planet. It is:

28452.089844 m/s

Mass of planet is 5551059943186440862564352.000000 kg

Its kinetic energy is:

2.246851 E33 J

Since we know this and the characteristic time, we have

the star system Planck constant. It is:

2.426899 E33 Js

Knowing this we can determine the mass of the moon.

It is: 6.483453 E22 kg

The radius of the moon is given by a perfect eclipse. It is:

1.874837 E6 m

We need only verify that our star system Planck constant

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works, by trying it in our KE of the planet solution to the wave equation.

R_star over R_moon is:

408.510956

KE_p is: 2.246851 J

We need to verify that the characteristic time is correct

from the equation for it from equating solar and lunar solutions

Orbital velocity of moon is: 933.111816 m/s

The characteristic time is: 1.080134 seconds

Now we want to compute the characteristic time for the PlanetDay

We need to determine the rotational angular momentum of the planet.

And, the planet radius to do this

The planet radius is:

6749.413086 km

What is the rotation period, or day, in EarthDays?

1.2

PlanetDay is: 103680.007812 seconds

The PlanetDay characteristic time is:

1.302460 seconds

The planetary angular momentum is:

6.129886 E33 Js

Thus the pressure gradient for the protoplanetary disc

has an exponent of p is: p=2.525810

All that is left to do is to compute the densities of the planets and the moons.

The density of the planet is: 4.310134 g/cm3

The density of the moon is: 2.348706 g/cm3

Program ended with exit code: 0

P is typically 2.1 to 2.6 for G0V stars where it is 2.0 to 2.5 for G2V stars. We know for our Sun

experimentally G2V has p is about 2.5. Our original model for the F0V star system was perfect, it

gave a characteristic time of 2.00 seconds and 1.00 seconds for the PlanetDay characteristic

time which was exactly what we want for and F0V star because F-type stars is where we might

suggest conditions for intelligent life begin in the HR diagram and we want even multiples of 1

second even to extend the naturalness of 1 second in our theory to beyond the proton and our

Solar System.

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14.0 Modeling Red Dwarf Star Systems

It is currently difficult to detect Earth-sized terrestrial planets around the habitable zones of

bright stars like ours, G2V, and is easiest around M-type brown dwarf stars because they are not

so bright and as such don’t flood-out the planet with light, though we may be able to do as such

soon with the James Webb telescope recently put into orbit. Brown dwarfs are the most

populous stars in our galaxy, but stars like our Sun are optimal for hosting intelligent life.

Planets around red dwarf stars in their habitable zones tend to become tidally locked meaning

their rotation slows to their orbital period because the habitable zone is so close-in to the star

because they are cooler than our Sun. This means such planets always have the same face

towards their star as they orbit it which means it is always day on one side and always night on

the other, extreme hot on one side, extreme cold on the other, so they would only be habitable in

the twilight region between night and day, which we would think is not optimal for intelligent

life to evolve, though they do have longer lives than our Sun, while the Sun has a life longer than

the hotter O, B, A, and F stars. The Sun, is in the middle of the HR diagram on the main

sequence making it stable and perhaps the most optimal for intelligent life, though K stars are

becoming of interest because though between G and M stars, their habitable zone is reasonable

and at the same time because they are cooler than our Sun, they burn much longer and perhaps

more stable. We might suggest the F stars, just before the G stars in the diagram is where good

conditions for intelligent life begin Our first try at a model for such stars worked well because

the characteristic time came out to 2.00 seconds even and the PlanetDay characteristic time

came out to be 1.00 second even and we have as a suggestion in our theory that the second is a

Natural unit that has meaning not just for our Earth-Sun system (the ground state in terms of

the Moon is one second) but for the atom’s proton which a second describes. We modeled F and

K stars from the general theories for their masses, sizes, and luminosities, however we modeled

the G2V star from our Sun for which we have perfect observed data. It turns out we have pretty

good data for the M brown dwarf star TOI 700 which has a few planets in its habitable zone, one

Earth-like, but thought to be a bit larger than the Earth. We will use this data to model the M-

type star system and treat it as though it has at least one moon that plays some kind of

functional role beneficial for life, though it may be our theories where moons are involved with

respect to life may be most functional in the region from A to G to K stars. We will model it as a

copper-silver civilization because they would have lower metallicities than stars like our Sun.

This star is an M2V star meaning it is two notches cooler than an M0V star. The other good

feature, as far as we know, is that the planet is the third planet from the star. Even if there are

discovered other planets before it, this number may not change because below a certain mass,

smaller planets may not take on an orbital number in the equation.

The TOI 700 Planetary System

Now we turn our attention to the TOI 700 Planetary System…It is an M type (red dwarf) star

101.4 ly from Earth (It is close) and harbors the first Earth sized planet in a habitable zone

(There are two) with 4 planets all together in its habitable zone. The star is in the Southern

Hemisphere constellation Dorado. Two of the planets are in the optimistic habitable zone (TOI

700 d and e). TOI 700 is about 40% the solar mass, about 40% the solar radius, and 55% the

solar temperature. The planets were detected by the Transiting Survey Satellite Object of

Interest (TESS).

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We are interested in TOI 700 e. We want to use our program such that it gives the planets mass

Running The Program

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

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

What is the luminosity of the star in solar luminosities? 0.0233

What is the ratio of element 2 to element 1, ie Au/Ag=1.823, Ag/Cu=1.6975?

16.5

What is the orbital number of the planet? 3

The habitable zone, which is the orbital radius of the planet is:

22835447808.000000 m

The mass of the planet is: 4.867534 E24 kg

Which is: 0.815059 Earth masses

The orbital period, or year, of the planet is: 0.092455 years

How many moons do you want for the moon's orbital period?

For example there are 13.379 moons of our moon per Earth year.

This gives 12.38 New Moons is approximately 12 New Moons per year.

This gives the orbital period of the moon is 0.750430 days.

We will use this value to determine the suggested mass of the planet it

orbits.

The mass of the planet is 0.783035 E24 kg

= (0.818)(5.972E 24kg) = 4.8851E 24kg

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Now that we know what the mass the planet should be, we

multiply the original mass M_p by the characteristic time t_c

to get the mass of the planet M_planet, to find the characteristic time.

The characteristic time is: 0.160869 seconds

Now we compute the orbital velocity of the planet. It is:

49175.910156 m/s

Mass of planet is 783034814373008953573376.000000 kg

Its kinetic energy is:

0.946795 E33 J

Since we know this and the characteristic time, we have

the star system Planck constant. It is:

0.152310 E33 Js

Knowing this we can determine the mass of the moon.

It is: 1.931638 E22 kg

The radius of the moon is given by a perfect eclipse. It is:

0.226963 E6 m

We need only verify that our star system Planck constant

works, by trying it in our KE of the planet solution to the wave equation.

R_star over R_moon is:

1288.456177

KE_p is: 0.946795 J

We need to verify that the characteristic time is correct

from the equation for it from equating solar and lunar solutions

Orbital velocity of moon is: 1717.171631 m/s

The characteristic time is: 0.160869 seconds

Now we want to compute the characteristic time for the PlanetDay

We need to determine the rotational angular momentum of the planet.

And, the planet radius to do this

The planet radius is:

7489.764648 km

What is the rotation period, or day, in EarthDays?

PlanetDay is: 2851200.000000 seconds

The PlanetDay characteristic time is:

85.761986 seconds

The planetary angular momentum is:

0.038719 E33 Js

Thus the pressure gradient for the protoplanetary cloud

has an exponent of p is: p=0.254215

All that is left to do is to compute the densities of the planets and the

moons.

The density of the planet is: 0.444928 g/cm3

The density of the moon is: 394.435394 g/cm3

Program ended with exit code: 0

We see the theory falls apart for stars so cool as M-type dwarfs. But then why should we

compute according to perfect eclipses in such stars when it seems unlikely you could get

intelligent life on their habitable planet, at least as far as we suspect it needs be. In fact they

don’t need a moon to hold the planet at their tilt like the Earth does to have stable weather, to

keep the planet from tilting at 90 degrees then to 180 degrees which could cause temperature

extreme swings over short periods of time, because the planet is tidally locked with the star it

orbits which could prevent this from happening. M_p in the program works for predicting the

mass of the planet, but M_planet doesn’t involving the characteristic time, but it works with the

element ratio around 16 to 16.5. So equation 9.1 works

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9.1

not adjusting the characteristic time for the second and using the ratio 16.5 in the ratio of the

elements. So with no moon, at least that plays a role, we should work out a theory based on

equation 9.1 to solve the system for stars of such low masses and temperatures. The ratio 16.5 is

what determines the r_m, the orbital radius of the moon, but we don’t want to include the moon

in a theory for these stars. But in the program r_moon=R_star/ratio where the ratio here is

16.5. This ratio is much larger than gold to silver (1.8) and silver to copper (1.7). Thus, what

determines it? Right off the top of my head this is around oxygen to hydrogen, 16g/mol to 1 g/

mol. We know our theory predicts hydrocarbons (section 8.0), the skeletons of life chemistry, in

the ratio of 1 proton-second to 6 proton-seconds which is carbon, the crux of life, to hydrogen,

the most fundamental element. Carbon to hydrogen is 12/1=12 is not far from 16.5. But then our

star TOI 700 is a M2V star, maybe carbon to hydrogen is what we need in M0V stars not far

from a M2V stars, and M2V is in the ratio of about oxygen to hydrogen.

In the tables for M-class stars the radius of an M2V star is 0.466. TOI 700 is 0.420. Its mass is

0.416. In the tables it is 0.44. But these are averages from surveys, there is some play, because

the s-shaped curve in the HR diagram has a thickness. But we want now to consider an M0V star

to see how it aligns with carbon/hydrogen=12. The tables give the radius is 0.588, the mass is

0.57, and the luminosity is 0.069 all in solar units. Let’s run that section of the program

What is the mass of the star in solar masses? 0.57

What is the radius of the star in solar radii? 0.588

What is the luminosity of the star in solar luminosities? 0.069

What is the ratio of element 2 to element 1, ie Au/Ag=1.823, Ag/Cu=1.6975?

12.0

What is the orbital number of the planet? 3

The habitable zone, which is the orbital radius of the planet is:

39296704512.000000 m

The mass of the planet is: 4.335389 E24 kg

Which is: 0.725953 Earth masses

The orbital period, or year, of the planet is: 0.178304 years

We would want to run it for 16 as wells oxygen/hydrogen. We find 12 is

better!

What is the mass of the star in solar masses? 0.57

What is the radius of the star in solar radii? 0.588

What is the luminosity of the star in solar luminosities? 0.069

What is the ratio of element 2 to element 1, ie Au/Ag=1.823, Ag/Cu=1.6975? 16

What is the orbital number of the planet? 3

The habitable zone, which is the orbital radius of the planet is:

39296704512.000000 m

The mass of the planet is: 3.251542 E24 kg

Which is: 0.544464 Earth masses

The orbital period, or year, of the planet is: 0.178304 years

We see the year of the planet works as well. We got 0.092455 years for TOI 700 e, is 33.769

days.

What happens is that the star may be small, but the planet is close in giving an absurd eclipse

ratio for producing any kind of a reasonable size for the moon resulting in an absurdly high

density of it. That is the moon has to be so close to the planet to perfectly eclipse the star,

because the star is so close to the planet that it is very large in the sky. Its reduced actual size

1sec

⋅ c ⋅ M

⋆

⋅

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doesn’t help because it is only about halved. The star is so close to the planet that the orbital

period of the planet is only about 30 days, that of our moon around the Earth.

We want to work out a theory where the moon vanishes as a factor stabilizing the planets tilt

being replaced by the star playing that role. Perhaps something like the difference between the

angular momentum of the planet around the M0V star in the habitable zone minus the angular

momentum of our moon around the Earth has to be greater than some critical value. We can

already see something like this, in that the orbital period of our moon is about the orbital period

the planet around the M star. And indeed we want to reference this with our planet and our sun,

because the Sun is midway between the the A0V and K0V stars which define the range over

which we are suggesting stars are optimal for intelligent life to take hold and evolve.

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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−

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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

⊙

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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

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The Author