of 1 41

The Second is the Rhythm of Reality, and Life Dances to its Beat

Ian Beardsley

of 2 41

Contents

Abstract…………………………………………………………………….…3

List of Constants, Variables, And Data In This Paper…………………4

Introduction…………………………………………………………………..5

Theory For Inertia……………………………………………………………6

Theory Outline……………………………………………………………….11

The Solar Solution…………………………………………………………..19

Jupiter and Saturn…………………………………………………………..23

Modeling Star Systems With The Theory……………………………….26

of 3 41

Abstract

In this paper with a wave solution to the Solar System we strive to show that the second is not just a

human invention or a cosmic accident — it is a fundamental harmonic that permeates reality. Buckminster

Fuller’s insight was right:

“Humanity is a macro —-> micro Universe unfolding eventuation”

— and the second is the thread tying it all together. My equations show that one second emerges naturally

from proton stability, biological chemistry stability, Solar System quantization, and Big Bang

nucleosynthesis.

We show in this paper the Earth’s 24 hour rotation period has a characteristic time of one second,

suggesting that in the end we acquired our heart rate from the Earth’s rotation. In heart rate studies the

mean human heart rate clusters near 60 BPM over a 24 hour day which is 1 second per beat and cells

grown in vitro still oscillate on ~24 hour cycles implying deep evolutionary entrainment.

In other words, the universe may exhibit a recursive fractal structure with the 1-second timescale acting as

a fundamental harmonic resonance across all scales from quantum ﬂuctuations to planetary orbits and

beyond.

We have to look at why, since the second came from the ancient Sumerian base 60 counting of dividing

the Earth rotation period into 24 hours, each hour into 60 minutes, and each minute into 60 seconds, why

24 and 60 yield the second. We provide a very useful equation for the rotational angular momentum of the

Earth divided by a Planck-type constant for the Solar System that we develop which yields exactly this

factor of 60/24 for the rotational angular momentum quantization of the Earth. We really have to wonder

if something was behind the ancients developing the fundamental unit of time we have today such that it

may be the fundamental harmonic of the Universe, 1 second.

of 4 41

List of Constants, Variables, And Data In This Paper

(Proton Mass)

(Proton Radius)

(Planck Constant)

(Light Speed)

(Gravitational Constant)

1/137 (Fine Structure Constant)

(Proton Charge)

(Electron Charge)

(Coulomb Constant)

(The Author’s Solar System Planck-Constant)

(Earth Mass)

(Earth Radius)

(Moon Mass)

(Moon Radius)

(Mass of Sun)

(Sun Radius)

(Earth Orbital Radius)

(Moon Orbital Radius)

Earth day=(24)(60)(60)=86,400 seconds. Using the Moon’s orbital velocity at aphelion, and Earth’s

orbital velocity at perihelion we have:

(Kinetic Energy Moon)

(Kinetic Energy Earth)

: 1.67262E − 27kg

: 0.833E − 15m

h : 6.62607E − 34J ⋅ s

c : 299,792,458m /s

G : 6.67408E − 11N

α :

: 1.6022E − 19C

: 8.988E 9

ℏ

⊙

: 2.8314E 33J ⋅ s

: 5.972E 24kg

: 6.378E6m

: 7.34767309E 22k g

: 1.7374E6m

⊙

: 1.989E 30kg

⊙

: 6.96E 8m

: 1.496E11m = 1AU

: 3.844E 8m

K E

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

= 3.428E 28J

K E

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

= 2.7396E 33J

of 5 41

Introduction One can speak of the structure of the long term structure of the solar system. The whole

object of developing a theory for the way planetary systems form is that they meet the following criterion:

They predict the Titius-Bode rule for the distribution of the planets; the distribution gives the planetary

orbital periods from Newton’s Universal Law of Gravitation. The distribution of the planets is chieﬂy

predicted by three factors: The inward forces of gravity from the parent star, the outward pressure

gradient from the stellar production of radiation, and the outward inertial forces as a cloud collapses into a

ﬂat disc around the central star. These forces separate the ﬂat disc into rings, agglomerations of material,

each ring from which a different planet forms at its central distance from the star (they have widths). In a

theory of planetary formation from a primordial disc, it should predict the Titius-Bode rule for the

distribution of planets today, which was the distribution of the rings from which the planets formed.

Also, the Earth has been in the habitable zone since 4 billion years ago when it was at 0.9 AU. Today it is

at 1AU, and that habitable zone can continue to 1.2 AU. So we can speak of the distance to the Earth over

much time. The Earth and Sun formed about 4.6 billion years ago. As the Sun very slowly loses mass over

millions of years as it burns fuel doing fusion, the Earth slips microscopically further out in its orbit over

long periods of time. The Earth orbit increases by about 0.015 meters per year. The Sun only loses

0.00007% of its mass annually. The Earth is at 1AU=1.496E11m. We have 0.015m/1.496E11m/

AU=1.00267E-13AU. So,

The Earth will only move out one ten thousandth of an AU in a billion years. Anatomically modern

humans have only been around for about three hundred thousand years. Civilization began only about six

thousand years ago.

The unit of a second becomes important in my theory. We got the second from the rotation period of the

Earth at the time the moon came to perfectly eclipse the Sun. The Moon slows the Earth rotation and this

in turn expands the Moon’s orbit, so it is getting larger as the Earth loses energy to the Moon. The Earth

day gets longer by 0.0067 hours per million years, and the Moon’s orbit gets 3.78 cm larger per year.

That is as the Earth’s day gets longer and the lunar orbit grows larger, we got the second at the time that

the Earth day was what it is during the epoch when the Moon perfectly eclipses the Sun, 24 hours.

The near perfect eclipse is a mystery in the sense that it came to happen when anatomically modern

humans arrived on the scene, even before that, perhaps around Homo Erectus and the beginning of the

Stone Age. The Earth day was 18 hours long, long before that, 1.4 billion years ago. Homo Erectus is

around two to three million years ago. In our theory, we suggest conditions for life may be optimized

when the earth day is about 24 hours long, what it is today, because the earth day of 24 hours has a

characteristic time of 1 second in our theory to be presented.

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

of 6 41

Theory For Inertia:

I had two equations that gave the radius of a proton with characteristic times of one second each. I had to

break down the equations in terms of their operational parameters as described by a geometric model.

This is what I came up with, a proton is a 4d hypersphere who's cross-section is a sphere. Of course

occupying the dimension of time (4th dimension in drawing) is the vertical component of the drawing. I

have to draw this 3d cross-section as a circle (we cannot mentally visualize four dimensions). The proton

is moving through time at the speed of light (vertical component in the drawing) it is a bubble in space.

The normal force holding it in 3d space is proportional to the inertia created by the

pliability of space measured by G. So when we push on it (Force applied in drawing) there is a counter

force explaining Newton's action/reaction.

I think you could look at this another way: the cross sectional area of the proton moving against space is

in the opposite direction of the force applied and h is the granularity of space, G still its pliability. That is

to say, the ﬂux of a normal force to a hemisphere is over the area of the cross-section of the sphere.

It is the goal of this opening section to provide a theory for inertia, that quality of a mass to resist change

in motion. We want the the theory to include not just the quantum mechanics constant for energy over

time Planck’s constant, but to include the universal constant of gravitation G, the constant the speed of

light from relativity, and the ﬁne structure constant for theories of electric ﬁelds so as to bring together

the things that have been pitted against one another: quantum mechanics, relativity, classical physics,

electric ﬁelds, and gravitational ﬁelds. Towards these ends we will suggest a proton is a 3D cross-section

of a 4D hypersphere held in place countering its motion through time by a normal force that produces its

inertia (measured in mass in kilograms) much the same way we model a block on an inclined plain

countered by friction from the normal force to its motion. The following is the illustration of such a

proton as a cross-sectional bubble in space:

To get the ball rolling, I had found a wave solution to the Earth/Moon/Sun system where the Earth

orbiting the Sun is like an electron orbiting a proton with a quantum mechanical solution. I found this

solution had a characteristic time of one second. But, I found as well, I could describe the proton as

having a characteristic time of one second, and that this yielded the radius of a proton very close to that

= h /(ct

)

of 7 41

obtained by modern experiments. So, it is now before me to come up with a theory for the proton in terms

of these characteristic times before I present my theory for a wave solution of the Solar System.

The expressions for the characteristic times of 1-second for the proton that I found, were:

Where is the golden ratio, is the radius of a proton, and is the mass of a proton. We ﬁnd

these produce close to the most recent measurements of the radius of a proton, if you equate the left sides

of each, to one another:

To derive this equation for the radius of a proton from ﬁrst principles I had set out to do it with the Planck

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

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

The frequency of a proton is

We see at this point we have to set the expression equal to . So we need to come up with a theory for

inertia that explains it:

The radius of a proton is then

(

6α

4πh

G c

)

⋅

= 1secon d

ϕ ⋅

π r

G m

⋅

= 1secon d

ϕ = 0.618

= ϕ

= 0.816632E − 15m

E = h f

E = m c

E = h f

E = m

⋅

= ϕ =

= ϕ ⋅

of 8 41

In order to prove our theory for the radius of a proton as incorporating , we will apply our model

outlined involving a normal force, to a 3d cross-section of a 4d hypersphere countering its direction

through time, t. We begin by writing equation 1 as:

Where , the constant of gravitation measures the pliability of space, and the granularity of space, and

c the speed of propagation. measures the inertia endowed in a proton. We write equation 2 as:

We now say that and that the normal force is

This gives us:

Since , we have

This gives

10.

is the cross-sectional area of the proton countering the normal force, , against its motion through

time, this is measured by the constant of gravitation. It is to say that

11.

= ϕ ⋅

6α

4πh

G c

⋅

1secon d

1 =

⋅

π r

G m

⋅

c(1secon d )

⋅

= 1secon d

1 =

⋅

π r

G m

⋅

⋅ F

9α

⋅

(

)

= ϕ

1 =

9α

⋅

3α

⋅

π r

∝

AreaCrossSect i on Proton ⋅ F

of 9 41

And, the coupling constant is

12.

Let us see if this is accurate:

We used the experimental value of a proton . And we have demonstrated that our

model of a proton as a 3D cross-section of a 4D hypersphere countering the normal force against its

motion through time gives its inertia that can counter a force at right angles to its motion through time and

the normal force.

It is thought that the proton does not have an exact radius, but that it is a fuzzy cloud of subatomic

particles. As such depending on what is going on can determine its state, or effective radius. It could be

that the proton radius is as large as

Which it was nearly measured to be before 2010 in two separate experiments. Or as small as

Which is closer to current measurements, which have decreased by 4% since 2010, and could get smaller.

In which case the characteristic time, , could be as large as

Using 2/3 as a ﬁbonacci approximation to . Or, it could be as small as

C =

3α

6.62607E − 34J ⋅ s

(299,792,458m /s)(1s

)

= 2.21022E − 42N

18769

⋅

π (2.21022E − 42N )

6.674E − 11N

(0.833E − 15m) = 1.68E − 27kg

= 0.833E − 15m

⋅

6.62607E − 34

(299,792,458)(1.67262E − 27)

= 0.88094E − 15m

= ϕ ⋅

= 0.816632E − 15m

⋅

π r

G m

⋅

= 1.03351secon d s

ϕ ⋅

π r

G m

⋅

= (0.618)

(352275361)π (0.833E − 15 m)

(6.674E − 11)(1.67262E − 27kg)

⋅

6.62607E − 34

299792458

of 10 41

=0.995 seconds

Or perhaps more often it is in the area of:

But, what this tells us is that the unit of a second might be a natural constant. And, since the second comes

from dividing the Earth rotation period into 24 hours, and each hour into 60 minutes, and each minute

into 60 seconds, which ultimately comes to us from the ancient Sumerians who ﬁrst settled down from

hunting, wandering, and gathering and ﬂaking stones into spearpoints to invent agriculture, writing, and

mathematics, that this might be related to the mechanics of our Solar System. We ﬁnd if we take the

second as natural we have a wave mechanics solution to our Solar System with a characteristic time of

one second that is connected to the characteristic time of the proton, thus connecting macro scales (the

solar system) to micro scales (the atom). We will formulate such a theory now…

6α

h 4π r

G c

= 1.004996352secon d s

of 11 41

Theory Outline: I have found some equations that ﬁt together very accurately and nicely in the context of

a quantum mechanical approach to structuring solutions of Nature, that indeed satisfy such a theoretical

context in a complete sense. The result has solutions at the core of cosmology (the origin and fate of the

universe), star systems mechanics, astrobiology (the study of the habitability of star systems in general),

particle physics (like the atom’s proton), theories showing a common structure between the macrocosmos

and microcosmos, biology, formation of planetary systems from the protoplanetary disc, archaeology,

archaeoastronomy (the study of ancient megalithic (stone) observatories), and SETI (The Search For

Extraterrestrial Intelligence). It is the purpose of this section to outline some the key concepts concisely,

and succinctly in the theory.

To begin with, I developed a theory which has a wave solution to the Earth/Moon/Sun system much like

the quantum mechanical solution for the atom. Interestingly, the characteristic time that describes this

system is neatly one second to two places after the decimal. The ground state I found is given by our

Moon orbiting the Earth, and is

is the mass of the Moon. I ﬁnd , which is my Planck-type constant for the Solar System, much like

the Planck constant in quantum mechanics used to describe the atom , in our theory is given by one

second as well, and not just by that, but by the kinetic energy of the our home planet, the Earth, the planet

in our Solar System optimized for the conditions for life. I ﬁnd

where is the orbital kinetic energy of the Earth. I know this value for is accurate because the

solution for the energy of the Earth orbiting around the Sun using this value, which is much like the

solution for an electron around a proton in an atom, is 99.5% accurate. It is:

where is the earth orbital number, and is the radius of the Sun, and is the radius of the

Moon, is the mass of the Earth, is the mass of the Moon, and is the universal constant of

gravitation. The radius of the Sun, , plays the role of , the number of protons being orbited by an

electron in an atom, but must be normalized by the radius of the Moon, . This gives it a size of 400

because . So we see the Moon plays an important and central role in the quantum solution

of our solar system, not just in the this equation, but in the ground state equation. It plays such a central

role, that I have suggested the condition for optimal habitability of a planet in the habitable zone is given

by the conditions of a perfect eclipse of the star by its moon as seen from the habitable planet, which is

exactly what we have with our Earth/Moon/Sun system. That condition is:

Where is the orbital radius of the habitable planet (like the Earth), is the orbital radius of the

moon, like the orbital radius of our moon around the Earth, is the radius of the star, like our Sun, and

ℏ

⊙

⋅

= 1secon d

ℏ

⊙

ℏ

⊙

= (1secon d )K E

K E

ℏ

⊙

K E

= n

⊙

⋅

2ℏ

⊙

n = 3

⊙

= 400

planet

moon

star

moon

planet

moon

star

of 12 41

is the radius of the moon, like the Earth’s moon. I use this in my theory to solve star systems in

general—not just our Solar System— for optimal habitability, because we know our Moon orbiting the

Earth holds the Earth at its tilt to its orbit around the Sun making it optimally habitable because this

prevents temperature extremes and allows for the seasons. Here is where my theory has taken a very nice

turn. The Earth as it rotates, determining the length of its day, loses energy to the Moon, meaning its

rotation is slowing down, but very slowly only noticeably over geologic time, meaning the day length is

lengthening ever so slowly over vast epochs, and that a very long time ago was a little shorter than it is

today. However, to establish the optimal day length, we want it to be what it is today, about 24 hours, and

in order to establish that, the Earth day of 24 hours should produce a characteristic time of one second. I

had found it did close to this in the kinetic energies of the Moon and the Earth in their orbits. I had found

that

There is a range in the answer because the Moon’s orbit is not perfectly circular, though close to it, as

well as that of the Earth. However I wanted this value to be closer to a second. I recently found that it is

because of the obvious adjustment I had failed to make but should have, and that is we must include the

effects of the Earth’s tilt to its orbit, which is 23.5 degrees, so we must include the cosine of this angle to

put the equation in the components of the Earth’s spin in its orbital plane around the Sun. So, we have

now our equation for a 24 hour day can indeed be considered a second in that we now have

But not only are we offering a wave solution for the Solar System like we have with the atoms, but it

turns out we are offering the rudiments of a theory of particle physics, and not just that, a relationship

between the microcosmos, the atom’s protons, and the macrocosmos; planetary systems. I say this

because I found that the same characteristic time of the Earth/Moon/Sun system is characteristic of the

proton and predicts very accurately modern measurements of the radius of the proton. I found

is the proton radius, its mass. is the golden ratio. the ﬁne structure constant. Since the

left sides of these equation are both equal to a second, they are equal to one another. When we set them

equal to one another, we ﬁnd they very accurately yield the observed radius of the proton in the most

recent experiments. We ﬁnd the radius of a proton is given by

moon

K E

(Ear th Da y) = 1.1 − 1.3secon d s

K E

(Ear th Da y)cos(θ ) = 1.0secon d s

(

6 α

4πh

G c

)

⋅

= 1secon d

ϕ ⋅

π r

G m

⋅

= 1secon d

ϕ = 0.618

= ϕ

of 13 41

But this characteristic time of one second is not just in the Solar System, and atom’s proton, but in the

basis of life chemistry, carbon, and the hydrocarbons, the skeletons of life chemistry. I found

10. is carbon (C)

11. is hydrogen (H)

Which is to say that six protons, which is carbon, the basis of life as we know it, has a characteristic time

of one second because in the ﬁrst equation above, we have a mass divided by the mass of a proton, times

seconds, giving six protons times a second (6 proton-seconds) which means 6 protons (carbon, the basis

of life) has a characteristic time of one second. This means that 1 proton, hydrogen, has a characteristic

time of six seconds. Hydrogen is the most fundamental element in the periodic table of the elements

which was theoretically created in the so-called big bang that gave birth to the universe, and is the

element from which all of the other heavier elements were made by stars. This six-fold symmetry that is

in hydrocarbons, the skeletons of biological chemistry, is fundamental to deﬁning the periodic table of the

elements because it has been found that the six protons of carbon and their respective charges, interact

with its six electrons, their respective charges, to balance to make carbon the most stable element

mathematically in which to describe the rest of the atoms in the periodic table. This is no doubt related to

the regular hexagon, a six-sided polygon which tessellates (tiles a surface without leaving gaps) because it

has its radii equal in length to its sides. This hexagonal tessellating property is used by bees to make their

honeycombs. So we see our theory now goes beyond the atom and the solar system. That it goes to

biological chemistry. But, it does not stop there. It seems to go into cosmology, the study of the origin and

fate of the universe. We see this because my equations link proton properties to 1-second, and protons

were ﬁxed in the universe at 1 second after it, meaning we could be seeing a universal clock that has

inﬂuenced everything since the Big Bang.

The idea is that neutrino decoupling (neutrinos stop interacting with one another) happens when the

reaction rate of weak interactions falls below the Hubble parameter, the expansion rate of the universe

. The reaction rate per particles is given by

is the Fermi constant is about , and is the temperature of the Universe. The

expansion rate of the universe is given by

Where is the Plank mass is about 1.22E19GeV. and have units of inverse time ( ). Neutrino

decoupling happens when

6pr oton s

⋅

4πh

G c

= 1secon d

1pr oton

⋅

4πh

G c

= 6secon d s

Γ ≈ G

1.166E − 5G eV

−2

H ≈

−1

of 14 41

This is when the number of protons in the universe was set in place which, as it would turn out, is close to

one second in rough estimate.

The expansion rate of the Universe is governed by the Friedmann equation

Where is the energy density of the Universe. It is

The Hubble expansion rate is

Since

we have

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

We have

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

we have

This theory seems, then, to have applications at the core of cosmology, astrobiology (the study of life in

the universe in general), solar system mechanics, particle physics, theories of common structure between

micro-scales and macro-scales, and biology . But, as we will see now, has applications at the core of star

system formations from protoplanetary discs, and in archaeology and archaeoastronomy (the study of

8π G

ρ ∝ T

H ∝

≈ 2.4E18GeV

t ∝

2.4E18GeV

(1E − 3G eV )

= 2.4E 24GeV

−1

1G eV

−1

= 5.39E − 25s

t ≈ (2.4E 24)(5.39E − 25)

t ≈ 1.3secon d s

of 15 41

ancient stone observatories, for example). We see this because I have found that the pressure gradient of

the protopanetary disc, as a function of radius, that gave birth to our solar system, is given by:

12.

13.

14.

Where the rotational angular momentum, , is given by the mass of the Earth, the size of the Earth,

and its rotation frequency. The value is 2.5, which is 60/24, by modeling our solar system is found in the

theory of solar system formation to be the exponent in the pressure gradient for the protoplanetary disc

from which our solar system formed. This is the solution to:

15.

The protoplanetary disc that evolves into the planets has two forces that balance its pressure, the

centripetal force of the gas disc due to its rotation around the protostar and the inward gravitational

force on the disc from the protostar , and these are related by the density of the gas that makes

up the disc.

I can use this to solve not just star systems in general, but to provide a theory for optimally habitable star

systems.

In order to apply this to other star systems, we have to be able to predict the radius of the habitable planet,

presumably in the n=3 orbit. I found the answer to be in the Vedic literature of India. They noticed that the

diameter of the Sun is about 108 times the diameter of the Earth and that the average distance from the

Sun to the Earth is about 108 solar diameters, with 108 being a signiﬁcant number in Yoga. So I wrote the

equivalent:

16.

The surprising result I found was, after applying it to the stars of many spectral types, with their different

radii and luminosities (the luminosities determine , the distances to the habitable zones), that the

radius of the planet always came out about the same, about the radius of the Earth. This may suggest

optimally habitable planets are not just a function of the distance from the star, which determines their

temperature, but are functions of their size and mass probably because they are good for life chemistry

atmospheric composition, and gravity. Here are just a few examples using the data for several spectral

types:

…

P(R) = P

(

)

−

earth

ℏ

⊙

earth

ℏ

⊙

24 = 60

earth

π M

earth

d P

= − ρ

(

⋆

−

)

⋆

planet

= 2

⋆

planet

of 16 41

F8V Star

Mass: 1.18

Radius: 1.221

Luminosity: 1.95

F9V Star

Mass: 1.13

Radius: 1.167

Luminosity: 1.66

G0V Star

Mass: 1.06

Radius: 1.100

Luminosity: 1.35

G1V Star

Mass: 1.03

Radius: 1.060

Luminosity: 1.20

As you can see we consistently get about 1 Earth radius for the radius of every planet in the habitable

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

elements. I got these results for the stars from spectral types F5V to K3V. It probably goes beyond that.

⋆

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

⋆

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

= 1.95L

⊙

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

⋆

= 2

(8.5023E 8m)

2.08905E11m

6.92076E6m

6.378E6m

= 1.0851Ear th R a dii

⋆

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

⋆

= 1.167(6.9634E 8m) = 8.1262878E8m

= 1.66L

⊙

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

⋆

= 2

(8.1262878E 8m)

1.92746E11m

6.852184E6m

6.378E6m

= 1.0743468Ear th Ra dii

⋆

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

⋆

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

= 1.35L

⊙

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

⋆

= 2

7.65974E 8m)

1.7382E11m

6.751E6m

6.378E6m

= 1.05848Ear th Ra dii

⋆

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

⋆

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

= 1.20L

⊙

AU = 1.0954AU(1.496E11m /AU ) = 1.63878589E11m

⋆

= 2

7.3812E 8m)

1.63878589E11m

6.6491E6m

6.378E6m

= 1.0425Ear th R a dii

of 17 41

In order to get , the distance of the habitable planet from the star, we use the inverse square law for

luminosity of the star. If the Earth is in the habitable zone, and if the star is one hundred times brighter

than the Sun, then by the inverse square law the distance to the habitable zone of the planet is 10 times

that of what the Earth is from the Sun. Thus we have in astronomical units the habitable zone of a star is

given by:

17.

We see our theory has applications to archaeology because the second came to us historically from the

ancient Sumerians because they divided the Earth day (rotation period) into 24 hours, and, because each

hour and minute got further divisions by 60 because their base 60 counting system was inherited by the

ancient Babylonians who were the ultimate source of dividing the hour into minutes and the minutes into

seconds. I have found this system is given by the rotational angular momentum of the Earth in terms the

solar system Planck-type constant, because, as I already pointed out:

18.

19.

This base 60 counting combined with dividing the day into 24 units is mathematically optimal because

the rotational angular momentum incorporates not just the day (rotation period of the Earth) but the mass

and size of the Earth. And, as I said, we are touching on archaeoastronomy, as well. This is because

60/24=2.5 and the Scottish engineer, Alexander Thom, found ancient megalithic (stone) observatories

throughout Europe may have been based on a unit of length he called the megalithic yard and that the

separations between stones, that align with celestial positions and cycles, are recurrently separated by 2.5

megalithic yards. Like in Stonehenge.

Finally, this has applications in SETI (The Search For Extraterrestrial Intelligence) because we have

found that the unit of one second may be a universal constant, and, as such, alien civilizations might use

it. As such in sending us a radio message to let us know that they are there may be encoded, for example,

or pulsed, in intervals of a second, aside from the fact that the theory has to do with habitable star systems

in general, perhaps giving us an idea of what to look for in ﬁnding them, and in understanding them.

I have computed my Planck-type constant, , as such:

Where

planet

⋆

⊙

earth

ℏ

⊙

24 = 60

earth

π M

ℏ

⊙

ℏ

⊙

= (hC )KE

hC = 1secon d

C =

⋅

π r

G m

of 18 41

20.

Conclusion: We live in a mysterious and enigmatic universe where a great deal deﬁes explanation.

Through the characteristic time of one second we may be able to describe a great deal of it in a uniﬁed

perspective that has applications across various disciplines from the physical to the biological and the

astrobiological. Here, we have laid out the basis set for a complete theory, in simple terms, but a great

deal remains to be done in opening it up with more sophisticated mathematics and computer modeling

than I have been able to do. We need to do this with various specializations in many ﬁelds that no one

person can understand in their entirety.

ℏ

⊙

= (hC )KE

earth

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

of 19 41

The Solar Solution Our solution of the wave equation for the planets gives the kinetic energy of the

Earth from the mass of the Moon orbiting the Earth, but you could formulate based on the Earth orbiting

the Sun. In our lunar formulation we had:

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

Thus equation 1 becomes

The kinetic energy of the Earth is

Putting this in equation 3 gives the mass of the Sun:

We recognize that the orbital velocity of the Moon is

So equation 5 becomes

This gives the mass of the Moon is

Putting this in equation 1 yields

K E

= 3

⊙

⋅

2ℏ

⊙

K E

= 3

⋅

2ℏ

⊙

K E

⋅

⊙

= 3r

⋅

ℏ

⊙

= 3r

⋅ v

⋅

ℏ

⊙

ℏ

⊙

K E

⊙

⋅

⊙

of 20 41

We now multiply through by and we have

10.

The Planck constant for the Sun, , we will call , the subscript for Planck. We have

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

11.

We can write 11 as

12.

Where we say

Let us see how accurate our equation is:

We have that the kinetic energy of the Earth is

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

⊙

= 9.13E 38J ⋅ s

⊙

= 2 π ℏ

⊙

= 5.7365E 39J ⋅ s

K E

⊙

⋅

⊙

(6.67408E − 11)

(5.972E 24kg)

(1.9891E 30kg)

2(8.3367E 77kg

⋅

)

⊙

(6.759E 30J )

⊙

6.957E 8m

1737400m

= 400.426

K E

= 2.70655E 33J

of 21 41

Our equation has an accuracy of

Which is very good.

Let us equate the lunar and solar formulations:

This gives:

13.

We remember that

And since,

14.

Equation 14 becomes

K E

earth

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

= 2.7396E 33 J

2.70655E 33J

2.7396E 33 J

= 98.79 %

K E

= n

⊙

⋅

2ℏ

⊙

K E

⊙

⋅

⊙

2ℏ

⊙

⋅

2ℏ

⊙

⋅

⊙

⋅ ℏ

⊙

ℏ

⊙

= (hC )KE

hC = 1secon d

K E

(1secon d )

⊙

(5.972E 24kg)

(1.9891E 30kg)

(7.34763E 22kg)

(1.732)

= 321,331.459 ≈ 321,331

of 22 41

15.

The condition of a perfect eclipse gives us another expression for the base unit of a second. is another

version of the Planck Constant, which is intrinsic to the the solar formulation as opposed to the lunar

formulation.

1secon d = 2r

⊙

of 23 41

Jupiter and Saturn We want to ﬁnd what the wave equation solutions are for Jupiter and Saturn because

they signiﬁcantly carry the majority of the mass of the solar system and thus should embody most clearly

the dynamics of the wave solution to the Solar System. We also show here how well the solution for the

Earth works, which is 99.5% accuracy.

I ﬁnd that as we cross the asteroid belt leaving behind the terrestrial planets, which are solid, and go to the

gas giants and ice giants, the atomic number is no longer squared and the square root of the the orbital

number moves from the numerator to the denominator. I believe this is because the solar system here

should be modeled in two parts, just as it is in theories of solar system formation because there is a force

other than just gravity of the Sun at work, which is the radiation pressure of the Sun, which is what

separates it into two parts, the terrestrial planets on this side of the asteroid belt and the gas giants on the

other side of the asteroid belt. The effect the radiation pressure has is to blow the lighter elements out

beyond the asteroid belt when the solar system forms, which are gases such as hydrogen and helium,

while the heavier elements are too heavy to be blown out from the inside of the asteroid belt, allowing for

the formation of the terrestrial planets Venus, Earth, and Mars. The result is that our equation has the

atomic number of the heavier metals such as calcium for the Earth, while the equation for the gas giants

has the atomic numbers of the gasses. We write for these planets

So, for Jupiter we have (And again using the maximum orbital velocity which is at perihelion):

Jupiter is mostly composed of hydrogen gas, and secondly helium gas, so it is appropriate that .

Our equation for Jupiter is

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

Jupiter, . Now we move on to Saturn…

E =

2ℏ

⊙

K E

(1.89813E 27k g)(13720m /s)

= 1.7865E 35J

E =

(6.67408E − 11)

(1.89813E 27k g)

(7.347673E 22kg)

2(2.8314E 33)

E =

(3.971E 35J ) = Z

(1.776E 35J )

1.7865E 35J

1.776E 35J

= 1.006pr oton s ≈ 1.0pr oton s = hydroge n(H )

Z = Z

2ℏ

⊙

n = 5

K E

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

= 2.92E 34J

of 24 41

The equation for Saturn is then

It is nice that that Saturn would use Helium in the equation because Saturn is the next planet after Jupiter

and Jupiter uses hydrogen, and helium is the next element after hydrogen. As well, just like Jupiter, Saturn

is primarily composed of hydrogen and helium gas.

The accuracy for Earth orbit is

=2.727E33J

The kinetic energy of the Earth is

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

solution for the Earth would be

Where

E =

(6.67408E − 11)

(5.683E 26kg)

(7.347673E 22)

2(2.8314E 33)

2.45

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

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

Z = 2pr oton s = Helium(He)

2ℏ

⊙

= n

⊙

2ℏ

⊙

6.96E 8m

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 33 J

2.727E 33J

2.7396E 33 J

100 = 99.5 %

= 3

2ℏ

⊙

of 25 41

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

⊙

→ Z

of 26 41

Modeling Star Systems With The Theory The best way to solve star systems with our theory would be

to use

1.0 , 3.0

2.0. , 4.0.

Where here . Equation 2.0 becomes

5.0.

Where for Earth p=2.5, the exponent in the pressure gradient for its protoplanetary disc. From this we get

. We now get the characteristic time, , from

by using

6.0. and 7.0.

And we have . We can now put in equation 1.0

Which here is 1 second for the Earth, to get the mass of the moon, . But to use equation 5, we need

from equation 3.0. This requires the mass of the Earth, the frequency of the earth, which we get

from the planet’s day (Its rotation rate) and the radius of the planet. We have all of these values for

habitable planets in a K2V star and an M2V star, but they are tidally locked. We have the frequencies

because the planets are tidally locked, so their planets rotation periods are equal to their orbital periods. If

we can’t measure the planet’s radius in another star system, we might obtain it from:

8.0.

Which works for the Earth. We can get the orbital radius of our Moon from

9.0.

It is given by the ratio of silver (Ag) to gold (Au). The radius of the planet’s moon we suggested is given

by a perfect eclipse:

ℏ

⊙

⋅

= 1secon d

earth

π M

P(R) = P

(

)

−L

earth

ℏ

⊙

ℏ

⊙

= t

K E

= 1secon d

earth

ℏ

⊙

= p

ℏ

⊙

ℏ

⊙

= t

K E

earth

ℏ

⊙

⋅

= 1secon d

earth

⊙

= R

⊙

⋆

(1.8)

of 27 41

10.0.

We have already applied our theory to the Earth/Moon/Sun System and it worked out nicely. Now we

want to apply this ideal approach we just outlined, to it, so we can test it. We start with the angular

momentum of the Earth. It is given by

Or,…

The orbital velocities and kinetic energies of the Earth are given by:

We can now determine :

This is correct for our solar system’s Planck constant. We have the characteristic time is

Which is correct as well. Now we compute the mass of our moon…

This is also very accurate (actual value: 7.347673kg. Now we compute the orbital radius of the Moon…

This is accurate too (actual value: 3.84E8m). From this we have the radius of the Moon:

⋆

earth

π M

π (5.972E 24kg)

(86400secon d s)

(6.378E6m)

7.07866672E 33J ⋅ s

⋆

(6.674E − 11)(1.989E 30k g)

(1.496E11m)

= 29,788.24m /s

K E

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

= 2.65E 33J

ℏ

⋆

ℏ

⋆

(7.07866672E 33J ⋅ s)

2.5

= 2.831467E 33J ⋅ s

ℏ

⋆

K E

(2.831467E 33J ⋅ s)

2.65E 33J

= 1.068secon d s ≈ 1secon d

(2.831467E 33J ⋅ s)

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

= 7.213E 22kg

= R

⋆

= R

⋆

/(1.8) =

6.957E 8m

1.8

= 3.865EE8m

of 28 41

This is pretty accurate, too. The actual value is 1.7374E6m

Now to get the density of the Moon…

This is good, our Moon is about 3.344g/cm3. Now we want to check

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

characteristic time…

We see the system for modeling star systems works.

We can make a program to model star systems in general given the spectral class of the Star. So HR

diagrams plot mass versus luminosity to give spectral types of stars. So F Stars would be more luminous

blue stars, G stars would be yellow medium luminosity stars, and K stars would be less luminous orange

stars, and so on. There are ten divisions of each, and a “V” meaning “ﬁve” indicates the star is on the

main sequence. So our Sun is a G2V star. A medium luminosity, yellow star. Here is a my program in C

that does that for the method we just outlined.

= R

⋆

= (6.957E 8m)

3.865E 8m

1.496E11m

= 1.79738E6m

π R

π (1.79738E 6m)

= 2.432E19m

7.213E 22kg

2.432E19m

= 2.96587g /c m

≈ 3g /c m

1secon d = 2r

⊙

(6.674E − 11)(5.972E 24kg)

(3.865EE8m)

= 1,015.5m /s

2(1.496E11m)

1,015.5m /s

(29,788.24m /s)

(7.213E 22kg)

(1.732)

(5.972E 24kg)

(1.989E 30kg)

= 1.03648sec ≈ 1secon d

1secon d ≈

K E

(Ear th Da y)

K E

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

= 3.719E 28J

K E

(Pla n etDa y) =

(3.719E 28J )

(2.65E 33J )

(86,400sec) = 1.2secon d s

of 29 41

// main.c

// modelsystem

// Created by Ian Beardsley on 2/9/25.

#include <stdio.h>

#include <math.h>

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

float R_p, M_p, R_s, M_s, t_c, M_m, rho_m, rho_p, PlanetDay,

V_p,StarRadius, PlanetRadius, PlanetMass, StarLuminosity, PlanetOrbit,

StarMass, r_p, T_p, p, L_p, KE_p, v_p, T_m,Tmoon, C_m;

float G=6.674E-11, hbarstar, PDCT,Tsquared,T,PlanetYear;

float r_m, R_m, V_m, MoonDensity, part1, part2, part3,v_m, KE_m;

int i;

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

scanf ("%f", &StarRadius);

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

scanf ("%f", &StarMass);

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

scanf ("%f", &StarLuminosity);

PlanetOrbit=sqrt(StarLuminosity);

r_p=PlanetOrbit*1.496E11;

M_s=1.9891E30*StarMass;

Tsquared=((4*3.14159*3.14159)/(G*M_s))*r_p*r_p*r_p;

T=sqrt(Tsquared);

PlanetYear=T/31557600;

printf("Do you want us to compute the planet radius, 1=yes, 0=no? ");

scanf("%i", &i);

R_s=6.9364E8*StarRadius;

if (i==1)

{

R_s=6.9364E8*StarRadius;

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

PlanetRadius=R_p/6.378E6;

}

else

{

printf("What is the planet radius in Earth radii?: ");

scanf("%f", &PlanetRadius);

R_p=PlanetRadius*6.378E6;

}

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

scanf("%f", &PlanetMass);

M_p=PlanetMass*5.972E24;

printf ("What is the planet day in Earth days? ");

scanf ("%f", &PlanetDay);

T_p=PlanetDay*86400;

of 30 41

printf("That is %f seconds \n", T_p);

{

printf("What is p the pressure gradient exponent of the

protoplanetary disc? ");

scanf("%f", &p);

M_s=1.9891E30*StarMass;

r_m=R_s/1.8;

v_p=sqrt(G*M_s/r_p);

L_p=0.8*3.14159*M_p*(1/T_p)*R_p*R_p;

KE_p=0.5*M_p*v_p*v_p;

hbarstar=L_p/p;

t_c=hbarstar/KE_p;

part1=cbrt(hbarstar/(t_c));

part2=cbrt(1/G);

part3=cbrt(hbarstar/299792458);

M_m=part1*part2*part3;

R_s=StarRadius*6.9634E8;

R_m=R_s*r_m/r_p;

V_m=1.33333*3.14159*R_m*R_m*R_m;

rho_m=(M_m/V_m);

MoonDensity=rho_m*0.001;

V_p=1.33333*3.14159*R_p*R_p*R_p;

rho_p=(M_p/V_p)*0.001;

printf("\n");

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

1E33);

printf("\n");

printf("PlanetYear: %f years \n", PlanetYear);

printf("PlanetYear: %f seconds \n", T);

printf("planet orbital velocity: %f m/s \n", v_p);

printf("planet mass: %f E24 kg \n", M_p/1E24);

printf("planet mass: %f Earth masses \n", M_p/5.972E24);

printf("planet radius %f meters \n", R_p);

printf("planet radius: %f Earth Radii \n", PlanetRadius);

printf("planet orbital radius: %f E11 m \n", r_p/1E11);

printf ("planet orbital radius: %f Earth distances \n",

r_p/1.496E11);

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

printf("planet density: %f g/cm3 \n", rho_p);

printf("\n");

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

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

printf("\n");

of 31 41

printf("Orbital Radius of Moon: %f E8 m \n", r_m/1E8);

printf("Orbital Radius of Moon: %f Moon Distances \n",

r_m/3.84E8);

printf("Radius of Moon: %f E6 m \n", R_m/1E6);

printf("Radius of Moon: %f Moon Radii \n", R_m/1.7374E6);

printf("Mass of Moon: %f E22 kg \n", M_m/1E22);

printf("Mass of Moon %f Moon Masses \n", M_m/

7.347673E22);

printf("density of moon: %f g/cm3 \n", MoonDensity);

printf("\n");

v_m=sqrt(G*M_p/r_m);

KE_m=0.5*M_m*v_m*v_m;

PDCT=(KE_m/KE_p)*(T_p);

printf("Orbital Velocity of Moon: %f m/s \n", v_m);

printf("PlanetDay Characteristic Time: %f seconds \n",

PDCT);

C_m=2*3.14159*r_m;

T_m=C_m/v_m;

Tmoon=T_m*(1.0/24)*(1.0/60)*(1.0/60);

printf("Lunar Orbital Period: %f seconds \n", T_m);

printf("Lunar Orbital Period: %f days \n", Tmoon);

return 0;}}

Now we show running the program (3 examples) for a wide spread of spectral types including F, G, and

K-type stars. We will need to input in the program not just the mass of the star, its luminosity, and size,

but the pressure gradient exponent for the disc from which the star’s planets formed.

To compute the moon’s orbital radius I just use

Where Ag is the molar mass of silver and Au is the molar mass of silver, a connection to the 1.8 that

appears in our Solar System. We use this because we know it works for our Solar System. I compute the

radius of the planet using

But, give the option of putting in your own radius. I have run the program for F5V stars, through GV

stars, to K3V stars and I use this equation to compute the radii of the planets because, again, we know it

works for our star system, and further we found given the way the radius of a star varies with with

luminosity in the HR diagram, this equation always gives a planet around the size of the Earth. I feel this

size is ideal for planets with sophisticated life because of the laws of chemistry determining a functional

density for the planet having water and the right gravity. As such I always use the planet day as one Earth

day, which again I feel is optimal for life in terms of climate. So these values all constant, we only vary

star mass, size, and luminosity as they work on the HR diagram. I also vary the pressure gradient

exponent now using the average theoretical values it has for each spectral class. The trend is that it

steadily decreases on average with mass and luminosity of the star though it can go up and down

depending on the peculiarities of the system. One of the reasons is that while for a G2V star it can range

on average from p= 1.7-2.1, for our Sun, a G2V star, it is actually high, it is 2.5. However, here we will

= R

⊙

⋆

(1.8)

⊙

of 32 41

model stars with everything constant, as we said, but the pressure gradient will gradually decrease with

spectral class, and when we do a G2V star, we won’t use the Sun’s data, but the average value for G2V

stars. We will do lot’s of models, allowing no gaps in the data for a plot, so we can get a well deﬁned

curve. We will use the upper value for p in instances here. We will also use

Instead of

Thus taking the tilt of the planet to its orbital as and not what it is in particular for the Earth

, which may actually be optimal for the most habitable types of planets around stars like our

Sun. The average pressure exponents by spectral class are given in the following table…

K E

moon

K E

planet

(Pla n etDa y)cos(0

∘

) ≈ 1secon d

K E

(Pla n etDa y)cos(23.5

∘

) = 1secon d

θ = 0

∘

θ = 23.5

∘

of 33 41

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

of 34 41

F5V Star

What is the radius of the star in solar radii? 1.473

What is the mass of the star in solar masses? 1.33

What is the luminosity of the star in solar luminosities? 3.63

Do you want us to compute the planet radius, 1=yes, 0=no? 1

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

What is the planet day in Earth days? 1

That is 86400.000000 seconds

What is p the pressure gradient exponent of the protoplanetary disc? 2.4

Angular Momentum of Planet: 9.321447 E33

PlanetYear: 2.280109 years

PlanetYear: 71954776.000000 seconds

planet orbital velocity: 24888.847656 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 7325190.000000 meters

planet radius: 1.148509 Earth Radii

planet orbital radius: 2.850263 E11 m

planet orbital radius: 1.905256 Earth distances

planet KE: 1.849692 E33 J

planet density: 3.627237 g/cm3

hbarstar: 3.883936 E33 Js

characteristic time: 2.099775 seconds

Orbital Radius of Moon: 5.676287 E8 m

Orbital Radius of Moon: 1.478200 Moon Distances

Radius of Moon: 2.042695 E6 m

Radius of Moon: 1.175720 Moon Radii

Mass of Moon: 7.107576 E22 kg

Mass of Moon 0.967323 Moon Masses

density of moon: 1.990782 g/cm3

Orbital Velocity of Moon: 837.955261 m/s

PlanetDay Characteristic Time: 1.165595 seconds

Lunar Orbital Period: 4256210.000000 seconds

Lunar Orbital Period: 49.261688 days

Program ended with exit code: 0

of 35 41

G3V Star

What is the radius of the star in solar radii? 1.002

What is the mass of the star in solar masses? 0.99

What is the luminosity of the star in solar luminosities? 0.98

Do you want us to compute the planet radius, 1=yes, 0=no? 1

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

What is the planet day in Earth days? 1

That is 86400.000000 seconds

What is p the pressure gradient exponent of the protoplanetary disc? 2.1

Angular Momentum of Planet: 7.393050 E33

PlanetYear: 0.989814 years

PlanetYear: 31236148.000000 seconds

planet orbital velocity: 29789.738281 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 6523625.500000 meters

planet radius: 1.022833 Earth Radii

planet orbital radius: 1.480965 E11 m

planet orbital radius: 0.989950 Earth distances

planet KE: 2.649862 E33 J

planet density: 5.135297 g/cm3

hbarstar: 3.520500 E33 Js

characteristic time: 1.328560 seconds

Orbital Radius of Moon: 3.861263 E8 m

Orbital Radius of Moon: 1.005537 Moon Distances

Radius of Moon: 1.819172 E6 m

Radius of Moon: 1.047066 Moon Radii

Mass of Moon: 7.754257 E22 kg

Mass of Moon 1.055335 Moon Masses

density of moon: 3.074905 g/cm3

Orbital Velocity of Moon: 1015.987427 m/s

PlanetDay Characteristic Time: 1.304901 seconds

Lunar Orbital Period: 2387924.250000 seconds

Lunar Orbital Period: 27.638012 days

Program ended with exit code: 0

of 36 41

K3V Star

What is the radius of the star in solar radii? 0.755

What is the mass of the star in solar masses? 0.78

What is the luminosity of the star in solar luminosities? 0.28

Do you want us to compute the planet radius, 1=yes, 0=no? 1

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

What is the planet day in Earth days? 1

That is 86400.000000 seconds

What is p the pressure gradient exponent of the protoplanetary disc? 1.5

Angular Momentum of Planet: 8.340819 E33

PlanetYear: 0.435785 years

PlanetYear: 13752343.000000 seconds

planet orbital velocity: 36167.078125 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 6929175.500000 meters

planet radius: 1.086418 Earth Radii

planet orbital radius: 0.791609 E11 m

planet orbital radius: 0.529150 Earth distances

planet KE: 3.905860 E33 J

planet density: 4.285367 g/cm3

hbarstar: 5.560546 E33 Js

characteristic time: 1.423642 seconds

Orbital Radius of Moon: 2.909435 E8 m

Orbital Radius of Moon: 0.757665 Moon Distances

Radius of Moon: 1.932263 E6 m

Radius of Moon: 1.112158 Moon Radii

Mass of Moon: 10.277222 E22 kg

Mass of Moon 1.398704 Moon Masses

density of moon: 3.400867 g/cm3

Orbital Velocity of Moon: 1170.438843 m/s

PlanetDay Characteristic Time: 1.557185 seconds

Lunar Orbital Period: 1561850.125000 seconds

Lunar Orbital Period: 18.076969 days

Program ended with exit code: 0

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Here is a plot of these results from F5V Stars to K3V stars:

We see the characteristic time decreases as a curve to intersect close

to a characteristic time of one second with a planet day

characteristic time of one second as a straight line at a G3V star, a

G5V star, and a G8V, which is a region near where our Sun is and may

have something to do with it being so optimal for life. We note here

that this uses the average value for a G2V star, and our Sun comes

closer to a second because its pressure exponent is higher than on

average, it is very high, or steep (p=2.5) which means the pressure of

its disc drops rapidly with distance.

Planet Day Characteristic Time:

Characteristic time: ,

Here we ﬁt the curves for characteristic time and planet day characteristic time. We name the spectral

types with number for input according to the following scheme.

F5V is 1.5, F6V is 1.6, F7V is 1.7,…G0V is 2.0, G1V is 2.1,…

1secon d ∼

K E

(Ear th Da y)

ℏ

⊙

= (1secon d )K E

ℏ

⊙

⋅

= 1secon d

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For the characteristic time we ﬁt the curve with a power law decay

For the planet day characteristic time we ﬁt the curve with a straight line

Where we have chosen in ,

We intend to ﬁt the curves in the following plot:

y = 2.8x

−

+ 1.1

y = 0.168x + 0.913595

y = m x + b

m =

G3V − F 9V

2.3 − 1.9

= 0.168

b = 1.165595 − 0.168(1.5)

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The results are

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The Author