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The Sun-like Star KOI-4878 With Earth-like Planet Satisﬁes A Theory

Showing 1-second Time Invariance Across Vastly Diﬀerent Scales,

From The Microcosmos to the Macrocosmos

October, 2025!

By Ian Beardsley!

of 2 46

Abstract

Having found that a solution to the Solar System exists that is similar in form to the solution for the

hydrogen atom with Schrödinger wave equation we apply it to a star system with a candidate Earth-like

planet around a Sun-like star (G-type, main sequence). Earth-like in that it could be on the order of Earth-

size and mass, is in the star’s habitable zone, and could be terrestrial and rocky with water. The star is

KOI-4878 in the constellation Draco, about 1,120 light years distant. We ﬁnd our solution has a

characteristic time of about 1 second. Also developed is a theory for protons, electrons, and neutrons that

has a characteristic time of 1-second as well. As such we say there is a 1-second time invariance across

scales from the microcosmos to the macrocosmos. We will see that star systems from larger, more

luminous stars like spectral type FV, to medium luminosity stars (GV stars) like the Sun (G2V) to less

luminous stars, KV stars, come in line with the characteristic time of one second for the proton, electron,

and neutron, around GV ﬁve stars, like the Sun. This equivalence may be a condition for optimal

habitability of a star system.

I had tried to apply the Schrödinger wave equation to the protoplanetary disc to see if it would predict the

orbits of the planets, but it occurs to me, since the solutions are analogous to those of the electron around

the proton in the hydrogen atom, and the electron and proton did not form from a protoatomic cloud, that

really I shouldn't pursue that. That since gravity is an inverse square ﬁeld like electric ﬁelds of the proton

and electron, that the analogous equations for the planets are just nodes where planets can exist from the

quantization of gravity. Here we ﬁnd the nodes match with the solar system if gravity is quantized by the

Earth's moon, and a basis unit of one second.

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Contents

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

Part A: The Solar System Theory…………………………………………………….5

We put forward a quantum wave solution for the Solar System with a basis unit of one second.

Part B: The Atomic Theory……………………………………………………………6

We put forward a theory for the proton, electron, and neutron with a basis unit of one second.

Part C: The Schimmelpfennig Formulation………………………………………….12

Here we talk about Schimmelpfennig’s alternate solution and its connection to mine.

Part D: The Delocalization Time………………………………………………………15

We see the delocalization time as given by the wave solution for a gaussian wave packet

distribution supports the determination of a solar system Planck-type constant.

Part E: The Equations Bridging Across Scales……………………………………….18

We show 1-second time invariance across enormously different scales, the scale of the atom to the scale

of the Solar System.

Part F: Jupiter And Saturn……………………………………………………………..19

Proposed solutions for Jupiter and Saturn and putting in the number for the Earth solution to show its

accuracy

Part G: Applying Theory to KOI-4878………………………………………………..22

Modeling the one star like the Sun for which we have found an Earth-like planet in the habitable zone. We

ﬁnd it is Sun-like stars that come in line with the proton, electron, and neutron characteristic time of 1-

second. This may be a condition of optimal habitability.

Appendix 1: Pressure Gradient of Protoplanetary Disk……………………………..35

Appendix 2: Deriving the Delocalization Time Form a

Gaussian Wave Packet…………………………………………………………………38

Appendix 3: The Program For Modeling Starsystems……………………………….43

of 4 46

List of Constants, Variables, And Data In This Paper

(Proton Mass)

(Proton Radius)

(Planck Constant)

(Light Speed)

(Gravitational Constant)

1/137 (Fine Structure Constant)

(Proton Charge)

(Electron Charge)

(Coulomb Constant)

(The Author’s Solar System Planck-Constant)

(Earth Mass)

(Earth Radius)

(Moon Mass)

(Moon Radius)

(Mass of Sun)

(Sun Radius)

(Earth Orbital Radius)

(Moon Orbital Radius)

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

orbital velocity at perihelion we have:

(Kinetic Energy Moon)

(Kinetic Energy Earth)

m

p

: 1.67262E − 27kg

r

p

: 0.833E − 15m

h : 6.62607E − 34J ⋅ s

c : 299,792,458m /s

G : 6.67408E − 11N

m

2

s

2

α :

q

p

: 1.6022E − 19C

q

e

: 1.6022E − 19C

k

e

: 8.988E 9

Nm

2

C

2

ℏ

⊙

: 2.8314E 33J ⋅ s

M

e

: 5.972E 24kg

R

e

: 6.378E6m

M

m

: 7.34767309E 22k g

R

m

: 1.7374E6m

M

⊙

: 1.989E 30kg

R

⊙

: 6.96E 8m

r

e

: 1.496E11m = 1AU

r

m

: 3.844E 8m

K E

m

=

1

2

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

2

= 3.428E 28J

K E

e

=

1

2

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

2

= 2.7396E 33J

of 5 46

Part A: The Solar System Theory

The idea: We can model the Solar System and the atom such that across these scales, which are immense,

they can be shown as structured according the same physical concepts.

How: To do this we ﬁnd we can model the Solar System as having a quantum Shrödinger wave solution,

and the atom in terms of its proton, electron, and neutron components as having a solution as being

particles whose inertia is due to their rest motions through time.

The connection: The connection in scales is found to be through a basis unit of one second — making

this unit of time being not just a cultural development handed to us from the ancient Sumerians when they

divided the Earth day into 24 hours, with each hour being 60 minutes, each minute being 60 seconds —

but to be a structural one.

We start with the Schrödinger wave equation in spherical coordinates:

Whose solutions for the hydrogen atom are!

A1.

A2.

Instead of solving the wave equation we suggest since gravity from the Sun is an inverse square ﬁeld

governing the motion of planets around it and the proton is an inverse square ﬁeld as well with the

electron, the solutions would be of the same form except , the atomic number, would not be 1 proton but

the radius of the Sun. However, we ﬁnd for this to work we have to suggest its size is normalized by the

radius of the Moon, giving it a radius of 400. As such we see the Moon is the metric for measuring size,

and as well will see distance and mass as well. This comes to us from the condition for a perfect eclipse of

the Sun by the Moon, which is:

A3.

is the the orbital radius of the Earth, is the lunar orbital radius, is the solar

radius, and is the lunar radius. Thus the solutions are for energy and orbital radius of the

Earth (The energy can be taken as the orbital kinetic energy of the Earth, which matches better

than 99%) :!

A4. , A5.

−

ℏ

2

2m

[

1

r

2

∂

∂r

(

r

2

∂

∂r

)

+

1

r

2

sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

+

1

r

2

sin

2

θ

∂

2

∂ϕ

2

]

ψ + V(r)ψ = E ψ

E

n

= −

Z

2

(k

e

2

)

2

m

e

2ℏ

2

n

2

r

n

=

n

2

ℏ

2

Z k

e

2

m

e

Z

r

earth

r

moon

=

R

⊙

R

moon

r

earth

r

moon

R

⊙

R

moon

E

3

= n

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

r

e

=

1

n

⋅

R

m

R

⊙

⋅

ℏ

2

⊙

G

⋅

M

⊙

M

e

M

m

of 6 46

Where n=3 is the Earth orbital number. We notice it is in terms of the the masses of the Earth and the

Moon. Later, I will convert this to a solar solution in terms of the Sun and Earth. We get that the ground

state equation giving the unit of a second is (introducing to measure distance in time units):

A6.

(Follows directly from the Bohr atom ground state.) And, we ﬁnd the Earth/Moon/Sun system solution

above has a Solar System Planck-type constant given by:

A7.

Since the Earth day lengthens by small amounts as the lunar orbit grows because the Earth loses energy to

the Moon, we suggest the characteristic length of the Earth Day is of 24 hours because it is given by one

second in:

A8.

the kinetic energy of the Moon, the kinetic energy of the Earth, EarthDay is 24 hours (86,400

seconds, the rotation period of the Earth) and is the Earth’s tilt to its orbit around the Sun.

Part B: The Atomic Theory

Now we need to show that the electron, neutron, and proton have a characteristic time of one second that

we ﬁnd on the much larger scale of the Earth/Moon/Sun system.

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:

We ﬁnd the 1-second time invariant equations are:

c

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

ℏ

⊙

= (1secon d )K E

e

K E

m

K E

e

(Ear th Day)cos(θ ) = 1.0secon d s

K E

m

K E

e

θ = 23.5

∘

F

n

of 7 46

B1. ,

B2. , ,

, and so on…

Proton

B3. !

= !

=1.00500 seconds !

Neutron

B4. !

= !

=1.004779 seconds!

Electron

B5. !

= !

=0.99773 seconds!

B6. !

, , , !

F

n

=

h

ct

2

1

t

1

= 1secon d

m

p

=

1

3α

2

⋅

π r

2

p

F

n

G

m

e

=

π r

2

eClassical

F

n

G

m

n

=

1

3α

2

⋅

π r

2

n

F

n

G

π r

2

p

= Ar eaCr ossSec t ion Pr oton

1second =

r

p

m

p

⋅

πh

Gc

1

3α

2

(0.833E − 15)

(1.67262E − 27)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

(18769)

3

1second =

r

n

m

n

⋅

πh

Gc

1

3α

2

(0.834E − 15)

(1.675E − 27)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

(18769)

3

1second =

r

eClassical

m

e

⋅

πh

Gc

(2.81794E − 15)

(9.11E − 31)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

1second =

r

i

m

i

⋅

πh

Gc

κ

i

κ

p

= 1/3α

2

κ

n

= 1/3α

2

κ

e

= 1

r

e

= r

eClassical

of 8 46

These were derived from the formulations of the time invariant second that are the forms

B7.

B8.

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 (0.831E-15m), if you

equate the left sides of each, to one another:

B9.

B10.

To derive this equation for the radius of a proton from ﬁrst principles (thought we have done it

geometrically, above) I had set out to do it with the Planck energy, , given by frequency of a

particle, and from mass-energy equivalence, :

B11.

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

B12.

The frequency of a proton is

B13.

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:

B14.

B15.

The radius of a proton is then

(

1

6α

2

4πh

G c

)

⋅

r

p

m

p

= 1secon d

ϕ ⋅

π r

p

α

4

G m

3

p

1

3

⋅

h

c

= 1secon d

ϕ = 0.618

r

p

m

p

r

p

= ϕ

h

cm

p

r

p

= 0.816632E − 15m

E = h f

E = m c

2

E = h f

m

p

E = m

p

c

2

f

p

=

m

p

c

2

h

ϕ

m

p

c

2

h

⋅

r

p

c

= ϕ =

m

p

c

h

r

p

m

p

r

p

= ϕ ⋅

h

c

of 9 46

B16.

Something incredible regarding the connection between microscales (the atom’s proton) and macro scales

(the solar system) if you want to get very close to modern measurements of the proton and to exactly a

characteristic time of one second. The radius of a proton is not constant, but depends of the nature of the

experiment, because protons are thought to be a fuzzy cloud of subatomic particles. Therefore, not using

in our equations for protons and the characteristic time of one second, but the right ratio of terms in the

ﬁbonacci sequence, we ﬁnd that ratio is 5/8:

=0.6303866

If

0, 1, 1, 2, 3, 5, 8, 13,…

is the ﬁbonacci sequence whose successive terms converge on , the golden ratio, then the two terms that

come closest to this are 5/8=0.625.

This is a characteristic time from

B17.

that has a value of

B18.

1.0007seconds

Combining

B19.

with

r

p

= ϕ ⋅

h

cm

p

ϕ

r

p

= ϕ

h

cm

p

ϕ =

r

p

m

p

c

h

=

(0.833E − 15)(1.67262E − 27)(299,792,458)

6.62607E − 34

ϕ

ϕ ⋅

π r

p

α

4

G m

3

p

1

3

⋅

h

c

= 1.0335secon d s

5

8

⋅

(352275361)π (0.833E − 15m)

(6.674E − 11)(1.67262E − 27)

3

1

3

⋅

(6.62607E − 34)

299,792,458

=

5

8

⋅

π r

p

α

4

G m

3

p

1

3

⋅

h

c

= 1.0007secon d s

of 10 46

B20.

Gives the radius of a proton to be

B21.

B22.

With this, while we get very close to one second (1.0007 seconds) with the ﬁbonacci ratio of 5/8 we also

get something very much in line with the most recent measurement for the radius of a proton

(0.831E-15m).

We ﬁnd that for our solutions of the Solar System, that is derived from the Fibonnaci approximation

ratio to that is 2/3, where as for the subatomic particles solution they take on the Fibonacci ratio 5/3

approximation to .

The electron has no size, its classical radius is determined by

B23.

This comes from the electron not having mass or size, from the idea that all of its mass is bound up in its

electrostatic energy. We see that from our 1-second time invariant equations that they are all one equation.

The equations for the proton and neutron have and for the electron is , because for the

electron there is no spacial impedance because it has no size or mass. Again, equations B3-B6:

Proton!

!

Neutron!

!

Electron!

!

Our equation for the subatomic particles is (we say )!

(

1

6 α

2

4πh

G c

)

⋅

r

p

m

p

= 1secon d

r

p

=

5

8

h

cm

p

r

p

=

5

8

(6.62607E − 34)

(299,792,458)(1.67262E − 27)

= 0.8258821E − 15m

ℏ

⊙

ϕ

r

eClassical

=

e

2

4π ϵ

0

⋅

1

m

e

c

2

κ

i

= 1/3α

2

κ

i

= 1

1second =

r

p

m

p

⋅

πh

Gc

1

3α

2

1second =

r

n

m

n

⋅

πh

Gc

1

3α

2

1second =

r

eClassical

m

e

⋅

πh

Gc

r

eClassical

= r

e

of 11 46

; , , !

We see, as we said, that the Solar System Planck-Type Constant can nicely be taken as (uses the 2/3

ﬁbonacci approximation to :

B24.

B25. where,

B26.

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:

B27.

radius of the star. The surprising result I found was, after applying it to the stars of all spectral types

from F through K, with their different radii and luminosities (the luminosities determine , the

distances to the habitable zones), that the radius of the planet always came out about the same, about the

radius of the Earth. This may suggest optimally habitable planets are not just a function of the distance

from the star, which determines their temperature, but are functions of their size and mass probably

because they are good for life chemistry, atmospheric composition, and gravity when they are the size and

mass of the Earth.

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

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

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

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

given by:

B28.

the luminosity of the star, and the luminosity of the Sun. We compute the orbital radius of the

Moon…

B29.

Which works for our Solar System, Ag and Au the relative masses of silver and gold atoms.

1second =

r

i

m

i

⋅

πh

Gc

κ

i

κ

p

= 1/3α

2

κ

n

= 1/3α

2

κ

e

= 1

ϕ

ℏ

⊙

= (hC )KE

e

hC = 1secon d

C =

1

3

⋅

1

α

2

c

2

3

⋅

π r

p

G m

3

p

ℏ

⊙

= (hC )KE

earth

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

R

planet

= 2

R

2

⋆

r

planet

R

⋆

r

planet

r

planet

r

planet

=

L

⋆

L

⊙

AU

L

⋆

L

⊙

r

m

= R

⋆

Ag

Au

= R

⋆

/(1.8) =

6.957E 8m

1.8

= 3.865EE8m

of 12 46

Part C: The Schimmelpfennig Formulation

Schimmelpfennig made a big development that should really open up the theory. He started out by writing

the equation of the potential for an orbit:

C1.

He then deﬁned the effective action constant as the orbital angular momentum:

C2.

Then said the Bohr-like orbital radius is

C3.

He then observed by substituting into this last equation (equation 3) we have

C4.

He then observes that is the exact orbital equation, the identity. That, in other word his solution collapses

to the classical solution. I then show close to a second appears in his equation for the Moon:

C5.

( ) That would be the distance to the Moon. I then suggested we divide that by the speed of

light to measure distance in time:

C6.

It is close to one second. We will see my ground state equations are:

C7.

Which is a little less than the Moon’s orbital distance (3.844E8m). We then divide this by the speed of

light and we get exactly 1-second for all practical purposes.

C8.

V(r) = −

GMm

r

h * := L = mvr

r =

h *

2

GMm

2

h * = mvr

v

2

=

GM

r

h *

2

GM

e

M

2

m

=

(2.866E 34)

2

(6674E − 11)(5.972E 24)(7.347E 22)

2

= 3.8179E 8m

h* = M

m

v

m

r

m

h *

2

GM

e

M

2

m

⋅

1

c

=

(2.866E 34)

2

(6674E − 11)(5.972E 24)(7.347E 22)

2

1

299,792,458m /s

= 1.29secon d s

ℏ

2

⊙

GM

3

m

=

(2.8314E 33)

2

(6.67408E − 11)(7.34763E 22kg)

3

= 3.0281E 8m

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

of 13 46

We saw my solution was the equation:

C9.

This would mean my equation for the orbit is, using and :

C10.

Where n=3 is the Earth orbital number. Thus Schimmelpfennig has

C12.

Where for my equation the Planck-type constant is:

C13.

And his is

C14.

This has me thinking if the moon is close to a light-second away from the Earth several things: ﬁrst, if we

are to measure distance with time, that is why we introduce the speed of light c into the equation for the

ground state. Second, in my solution the energy equation uses the radius of the Sun in place of the atomic

number, number of protons, but its size is normalized by the size of the moon giving it a radius of 400. In

the same way the Moon being close to a light-second away from the Earth, means its orbital radius

normalizes distance. This works great because the moon has a nearly perfect circular orbit, it is a near

perfect sphere, and its mass is very accurately known. Thirdly, the moon as seen from the Earth near

perfectly eclipses the Sun, this gives .This allowed me to make a Earth/Sun

solution as opposed to the Moon/Earth solution (In this paper) with a characteristic time of 1 second as

well. Fourthly, It might be good to ﬁnd how far the moon was from the Earth and when, when its distance

was exactly a light second from the Earth as opposed to the 1.3 light-seconds of today.

Equating my with his yields the relationship between our solar system Planck-type constants;

C15.

This can be written

C16.

We then consider my solar solution for energy from using the eclipse condition:

E

3

= n

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

r

e

K E = (1/2)mv

2

v

2

= GM

⊙

/r

e

r

e

=

1

n

⋅

R

m

R

⊙

⋅

ℏ

2

⊙

G

⋅

M

⊙

M

e

M

m

r

e

=

h *

2

GM

⊙

M

2

e

ℏ

⊙

= (1secon d )(K E

e

)

h* = M

e

v

e

r

e

R

⊙

/R

moon

= r

earth

/r

moon

r

e

h*

2

ℏ

⊙

=

1

n

⋅

R

m

R

⊙

⋅

M

2

⊙

M

e

M

3

m

M

3

m

M

2

⊙

M

e

=

ℏ

2

⊙

h*

2

⋅

1

n

⋅

R

⊙

R

m

of 14 46

C17. , and , ,

Also using , and using:

C18.

Which is:

C19.

And consider the equation for a second that followed from equating this solar formulation with the

original lunar formulation :

C20.

Putting the terms from equation C16

C21.

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

C22.

C23.

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

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

and size of the Earth. We can say we are touching on archaeoastronomy. This is because 60/24=2.5 and

the Scottish engineer, Alexander Thom, found ancient megalithic (stone) observatories throughout Europe

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

R

⊙

R

m

=

r

e

r

m

K E

e

=

1

2

⋅

GM

⊙

M

e

r

e

v

2

m

=

GM

e

r

m

E

3

= n

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

K E

e

= E

3

ℏ

⊙

: L

p

= r

e

v

m

M

e

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

m

2

s

E

3

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2ℏ

2

⊙

1secon d = 2r

e

v

m

v

2

e

M

3

m

3

M

2

e

M

⊙

M

3

m

M

2

⊙

M

e

M

3

m

M

2

⊙

M

e

=

ℏ

2

⊙

h*

2

⋅

1

n

⋅

R

⊙

R

m

1secon d = 2r

e

⋅

v

m

v

2

e

⋅

ℏ

⊙

h*

R

m

R

⊙

⋅

M

⊙

M

e

L

earth

L

earth

ℏ

⊙

24 = 60

L

earth

=

4

5

π M

e

f

e

R

2

e

of 15 46

stones, that align with celestial positions and cycles, are recurrently separated by 2.5 megalithic yards.

Like in Stonehenge.

You will notice the 4/5 in the above equation can be written as 2/5, meaning the equation becomes:

C24.

the rotational frequency of the Earth, Since , where is the angular velocity of the Earth, we

have:

C25.

We notice , the ancient Sumerian factors deﬁning the second from the rotation period of the

Earth, We see the rotation, mass and radius (size) of the Earth are optimally formulated with ancient

Sumerian base 12 and base 60 mathematics (they divided the day into 12 hours, and the night into 12

hours).

I have found that the pressure gradient of the protopanetary disc (Appendix 1) , as a function of radius,

that gave birth to our solar system, is given by:

C26.

Because we can know the exponent in C26 for the protocloud from which a star system forms, given by

where the radius of the disc where pressure is , we can determine for the star

system. This is the solution to:

C27.

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. This becomes useful for modeling other star systems because is the exponent in the

equation for the pressure gradient of the protoplanetary disc.

Part D: The Delocalization Time

In order to show that our hypothesis is right, we solve the wave equation for a Gaussian wave packet and

determine the delocalization time (Appendix 2), . If it is about six months, the time it takes the Earth to

delocalize (travel its orbital diameter), using the Moon playing the role of the mass of an electron and our

as above to describe the Earth, then the hypothesis can be taken as correct, and we can solve the whole

system for the Earth/Moon/Sun system from the rest of the equations in the hydrogen atom solution to the

Schrodinger wave equation.

L

earth

=

2

5

M

e

2 π f

e

R

2

e

f

e

2 π f

e

= ω

e

ω

e

L

earth

=

2

5

M

e

ω

e

R

2

e

2/5 = 24/60

P(R) = P

0

(

R

0

)

−

L

earth

ℏ

⊙

p

p = L

earth

/ℏ

⊙

R

0

P

0

ℏ

⊙

d P

dr

= − ρ

(

GM

⋆

r

2

−

v

2

ϕ

r

)

v

2

ϕ

/r

GM

⋆

/r

2

ρ

L

earth

ℏ

⊙

τ

ℏ

⊙

of 16 46

The delocalization time of a particle, molecule, or mass in general, , is the time it takes a particle to

delocalize. If we want to apply our wave equation theory of the Solar System to this concept, then the

delocalization time should be the time for the Earth to travel the diameter of it’s orbit, which would be

half a year (about six months). In order to derive the delocalization time we must consider a Gaussian

wave packet…

The Gaussian wavefunction in position space is

D1.

It’s Fourier wave decomposition is

D2.

We use the Gaussian integral identity (integral of quadratic)

D3.

τ

ψ (x,0) = Ae

−

x

2

2d

2

ψ (x,0) = Ae

−

x

2

2d

2

=

∫

dp

2 π ℏ

ϕ( p)e

i

ℏ

px

∫

∞

−∞

e

− a x

2

+bx

d x =

π

a

e

b

2

4a

of 17 46

We ﬁnd via the inverse Fourier transform. It is

D4.

Substitue :

D5.

The solution is standard and is:

D6.

D7.

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 uses 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

D8.

Where is the mass of the Moon, and is the orbital radius of the Moon. We have

D9.

Now let’s compute a half a year…

D10. (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

D11.

Conclusion We see we the quantization of the Earth/Moon/Sun system is with a characteristic time of one

second and the mass of the Moon.

ϕ( p)

ϕ( p) =

∫

∞

−∞

d x ψ (x,0)e

−

i

ℏ

px

ψ (x,0)

ϕ( p) = A

∫

∞

−∞

e

−

x

2

2d

2

e

−

i

ℏ

[ px]

d x

|

ψ (x, t)

|

2

=

[

−

x

2

d

2

⋅

1

(1 + t

2

/τ

2

)

]

τ =

m d

2

ℏ

d

τ

ℏ

⊙

τ =

m

moon

(2r

moon

)

2

ℏ

⊙

m

moon

r

moon

τ = 4

(7.34767E 22k g)(3.844E 8m)

2

2.8314E 33J ⋅ s

= 15338227secon d s

15338227

15778800

100 = 97.2 %

of 18 46

Part E: The Equations Bridging Across Scales

In Schimmelpfennigs Formulation

C5.

That would be the distance to the Moon. In units of time, this is close to a second:

C6.

Beardsley’s ground state equations are:

C7.

C8.

Which is a little less than the Moon’s orbital distance (3.844E8m), but gives the light-second accurately.

The 1-second time invariant equations are

Proton

B3. !

= !

=1.00500 seconds !

Neutron

B4. !

= !

=1.004779 seconds!

h *

2

GM

e

M

2

m

=

(2.866E 34)

2

(6674E − 11)(5.972E 24)(7.347E 22)

2

= 3.8179E 8m

h *

2

GM

e

M

2

m

⋅

1

c

=

(2.866E 34)

2

(6674E − 11)(5.972E 24)(7.347E 22)

2

1

299,792,458m /s

= 1.29secon d s

ℏ

2

⊙

GM

3

m

=

(2.8314E 33)

2

(6.67408E − 11)(7.34763E 22kg)

3

= 3.0281E 8m

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

1second =

r

p

m

p

⋅

πh

Gc

1

3α

2

(0.833E − 15)

(1.67262E − 27)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

(18769)

3

1second =

r

n

m

n

⋅

πh

Gc

1

3α

2

(0.834E − 15)

(1.675E − 27)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

(18769)

3

of 19 46

Electron

B5. !

= !

=0.99773 seconds!

B6. !

, ,

Part F: 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):

1second =

r

eClassical

m

e

⋅

πh

Gc

(2.81794E − 15)

(9.11E − 31)

π(6.62607E − 34)

(6.674E − 11)(299,792,458)

1second =

r

i

m

i

⋅

πh

Gc

κ

i

κ

p

= 1/3α

2

κ

n

= 1/3α

2

κ

e

= 1

E =

Z

n

G

2

M

2

m

3

2ℏ

2

⊙

K E

j

=

1

2

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

2

= 1.7865E 35J

E =

Z

H

5

(6.67408E − 11)

2

(1.89813E 27kg)

2

(7.347673E 22k g)

3

2(2.8314E 33)

2

E =

Z

H

5

(3.971E 35J ) = Z

H

(1.776E 35J )

of 20 46

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…

=

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

=

Z

H

=

1.7865E 35J

1.776E 35J

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

Z = Z

H

E

5

=

Z

H

5

G

2

M

2

j

M

3

m

2ℏ

2

⊙

Z

H

n = 5

K E

S

=

1

2

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

2

= 2.92E 34J

E =

Z

6

(6.67408E − 11)

2

(5.683E 26kg)

2

(7.347673E 22)

3

2(2.8314E 33)

2

Z

2.45

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

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

Z = 2pr oton s = Heliu m(He)

E

6

=

Z

He

6

G

2

M

2

s

M

3

m

2ℏ

2

⊙

E

n

= n

R

⊙

R

m

G

2

M

2

e

M

3

m

2ℏ

2

⊙

R

⊙

R

m

=

6.96E 8m

1737400m

= 400.5986

E

3

= (1.732)(400.5986)

(6.67408E − 11)

2

(5.972E 24kg)

2

(7.347673E 22k g)

3

2(2.8314E 33)

2

of 21 46

=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

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

K E

e

=

1

2

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

2

= 2.7396E 33J

2.727E 33J

2.7396E 33J

100 = 99.5 %

E

3

= 3

Z

2

Ca

G

2

M

2

e

M

3

m

2ℏ

2

⊙

R

⊙

R

m

→ Z

2

of 22 46

Part G: Applying Theory to KOI-4878 We want to apply our wave theory for the planets to a star

system other than that of the Earth-Sun system, but that are similar, and fortunately we have discovered

one such a candidate star system. The G4V spectral type star KOI-4878 is in the constellation Draco with

location coordinates

RA: 19h 04m 54.7s

Dec: +50deg 00min 48.70sec

While M class stars are the most abundant in the galaxy and have longer life spans than the Sun, their

planets are thought to be tidally locked, their day is equal to their year, leaving them perpetually night on

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

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

zones because they are far enough away from their star that they might not always become tidally locked,

while being more stable than the Sun and longer lived. It is easiest to detect planets and get data for these

M2V and K2V stars, but when you get to our Sun a G2V star, they are so bright they wash out the light of

their planets in the habitable zone a great deal. However, we have gotten data for a planet in the habitable

zone of a G4V star, about the same size, radius, mass, and luminosity as our Sun. And the planet in its

habitable zone has about the same radius, and perhaps mass as our Earth. All we need to do is to detect a

moon around its planet, and we will have veriﬁed my theory for habitable GV stars. We have yet

developed the technology to detect moons around a planet, but we are getting close to it, and we are going

to try to with the James Webb Space Telescope. This star is called KOI-4878 and its planet is

KOI-4878.01. Here is the information on it…

I have written a program in C that models star systems with our theory (Appendix 3). First we will do the

computation by hand so as to see how the program works. For KOI-4878 we will use some of the lowest

possible values, and ﬁnd we can get in the ball park. We will ﬁnd after running the program for higher

values of the data within the errors of the measurements for this star system, we can get close to this star

system. It becomes clear that with variations of parameters, this theory accounts for this star system if it

has a moon. And the moon can be similar to that of the Earth.

This star is the only candidate we have for an Earth-like planet in the habitable zone. We have detected

many around M-type red dwarfs because it is easy to detect planets around such abundant (the most

of 23 46

abundant) stars that are so faint that a transit lowers the light from the star by a high percentage. I say

candidate because there are rigid standards to consider them conﬁrmed. To be conﬁrmed you need to

detect them by methods than transit (as this one was) like by measuring radial velocity (Changes in

velocity of the star due to being pulled on by the orbiting planet, by detecting red and blue shifts in the

star). It is hard to apply our theory for habitable planets to M-type stars because their habitable zones are

so close in that tidal forces from the star tidally lock the planet, so their rotation period gets slowed down

to their orbital period, leaving a gap in the data. Tidal forces weaken very rapidly with distance leaving

the Earth very unaffected by them. The tidal force gradient is proportional to , and tidal heating/

dissipation is proportional to . So at Earth, the effects are very small.

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

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

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

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

wrote the equivalent:

radius of the star. The surprising result I found was, after applying it to the stars of all spectral types

from F through K, with their different radii and luminosities (the luminosities determine , the

distances to the habitable zones), that the radius of the planet always came out about the same, about the

radius of the Earth. This may suggest optimally habitable planets are not just a function of their 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 when they are the size and

mass of the Earth.

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

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

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

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

given by:

the luminosity of the star, and the luminosity of the Sun.

1/r

3

1/r

6

R

planet

= 2

R

2

⋆

r

planet

R

⋆

r

planet

r

planet

r

planet

=

L

⋆

L

⊙

AU

L

⋆

L

⊙

of 24 46

A G4V star on average has a mass of 0.985, a radius of 0.991, a luminosity of 0.91 (Sun=1). Since the

above data has a large margin of error taking it to a range of 0.88-1.138 solar masses (avg. 0.9325) we

will use the average for the G4V spectral type that it is, which is 0.985 solar masses. And since the radius

is in the range is 1.072-1.19 solar radii, (avg. 1.131) we will use the average again for its spectral class

G4V which is 0.991. This gives

The mass of the star being taken to be 0.985 solar masses, we have, if the orbit of the planet is close to

circular:

=

This is to see if the period predicted is close to the period measured, which it is because the measured

value is 449.015 days. That is 97.5%. This is good because we used a circular orbit approximation and

average values for G4V stars. Let us compute the kinetic energy of this planet:

Compared to that of Earth, which is 29,784m/s. The mass of the planet we will take to be 0.92 that of

Earth as recommended by Wikipedia because it has a range of 0.66-1.18 Earth masses. That is

[(0.92)(5.972E24kg)]=5.49424E24kg

The kinetic energy is, then:

We now compute , the Planck-type constant for this star system. We use

Where is the exponent in

R

planet

= 2

R

2

⋆

r

planet

= 2

[(0.991)(6.96E 8m)]

2

(1.496E11m /AU )(1.125)

= 5.6534E 6m =

(6.378E6m)

(5.6534E6m)

= 1.1282Ea r th Ra d ii

r

planet

=

L

⋆

L

⊙

AU =

0.91

1

AU = 0.9539392AU = 1.4271E11m

T

2

=

4π

2

GM

a

3

=

(39.4784)

(6.674E − 11)(0.991)(1.989E 30k g)

[(1.125)(1.496E11m)]

3

(3.001E − 19)(1.683E11m)

3

= 1.430600E15

T = 3.7823E 7secon d s = 437.77d a ys

v =

GM

r

=

(6.674E − 11)(1.971E 30k g)

(1.683E11m)

= 27,957.244m /s

K E =

1

2

Mv

2

=

1

2

(5.4942E 24kg)(27,957.244)

2

= 2.147E 33J

ℏ

⋆

L

earth

ℏ

⊙

= p

p

of 25 46

We have

Since Mars is further out and has a day of close to Earth’s 24 hours, and since Venus doesn’t have this

because it is closer to the Sun and greatly slowed down by tidal forces, we will guess for an Earth-sized

planet like this one, its day is 24hrs=86,400sec because we are computing as if this planet hosts life, and a

fast rotation, keeps the planet cool, but it can’t be so fast that the nights and days are to short for life to

function (hunt, build, etc…):

=

For G4V stars the typical range of is p=1.6-2.0 for the exponent in the pressure gradient. We will choose

2.0 since it is closest to that of Earth, which is 2.5:

Compared to that of Earth, which is . Thus we have the characteristic time of this

planet is

Our theory says that

So the mass of the Moon of this planet is:

=6.4989E22kg~6.5E22kg

P(R) = P

0

(

R

0

)

−L

earth

ℏ

⊙

L

planet

=

4

5

π M

p

f

p

R

2

p

L

p

=

4

5

π (5.4942E 24k g)

1

(86400secon d s)

(5.6534E6m))

2

5.108E 33J ⋅ s

(5.108E 33J ⋅ s)

ℏ

⋆

= 2.0

ℏ

⋆

= 2.554E 33J ⋅ s

ℏ

⊙

: 2.8314E 33J ⋅ s

t

c

=

ℏ

⋆

K E

p

=

(2.554E 33J ⋅ s)

(2.147E 33J )

= 1.1877secon d ≈ 1secon d

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

M

3

m

=

(2.554E 33J ⋅ s)

2

(6.674E − 11)(299,729,458m /s )(1.1877secon d s)

of 26 46

Compared to that of the Earth’s moon, 7.347673E22kg. The orbital radius of the Earth’s moon seems to

be governed by the relative masses of the heavy metallic elements gold (Au) and silver (Ag). We will

guess this holds here, which is a similar type of a star system.

It is given by the ratio of silver (Ag) to gold (Au) by molar mass is equal to . The radius of the

planet’s moon we suggested is given by a perfect eclipse:

Compared to that of the Moon, which is 3.84E8m. From this we have the radius of the Moon:

Compared to that of the Moon, which is 1.7374E6m. Now to get the density of the Moon…

Compared to the Earth moon 3.34 g/cm3. The Earth’s moon is consists of silicates for the surface regolith,

which is porous, with a low density starting at 1.5g/cm3 to solid lunar rock and mantle of 3.17-3.22g/cm3

and 3.22-3.34g/cm3. This moon could exist with a smaller iron core and higher proportions of lighter

silicates. We want to compute the orbital kinetic energy of this moon.

Compared to that of the Earth’s moon, which is 1022m/s

Where that with the Earth’s moon it is 3.428E28J using its orbital velocity at aphelion, which is 966m/s.

We can now computer the PlanetDay characteristic time:

r

m

= R

⊙

Ag

Au

=

R

⋆

(1.8)

r

m

/R

⊙

R

⋆

R

m

=

r

p

r

m

r

m

= R

⋆

Ag

Au

= R

⋆

/(1.8) =

(0.991)(6.96E 8m)

1.8

= 3.832E 8m

R

m

= R

⋆

r

m

r

p

= (6.957E 8m)

3.832E 8m

1.496E11m

= 1.782E6m

V

m

=

4

3

π R

3

m

=

4

3

π (1.782E6m)

3

= 2.37E19m

3

ρ

m

=

6.5E 22k g

2.37E19m

3

= 2742.62 k g /m

3

≈ 2.74262g /c m

3

v =

GM

r

=

(6.674E − 11)(5.49424E 24kg)

(3.832E 8m)

= 978.21m /s

K E

m

=

1

2

(6.5E 22k g)(978.2m /s)

2

= 3.12E 28J

K E

m

K E

p

(Pla n et Da y)cos(θ ) ≈ 1.0secon d s

of 27 46

=

This is close to the characteristic time for star system, which we found was

Running our program for the Earth to verify its accuracy, we have

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

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

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

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? 2.5

Angular Momentum of Planet: 7.187518 E33

PlanetYear: 0.999888 years

PlanetYear: 31554074.000000 seconds

planet orbital velocity: 29788.980469 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 6432306.000000 meters

planet radius: 1.008515 Earth Radii

planet orbital radius: 1.496000 E11 m

planet orbital radius: 1.000000 Earth distances

planet KE: 2.649727 E33 J

planet density: 5.357135 g/cm3

hbarstar: 2.875007 E33 Js

characteristic time: 1.085020 seconds

Orbital Radius of Moon: 3.853556 E8 m

Orbital Radius of Moon: 1.003530 Moon Distances

Radius of Moon: 1.793707 E6 m

Radius of Moon: 1.032408 Moon Radii

Mass of Moon: 7.247882 E22 kg

Mass of Moon 0.986419 Moon Masses

density of moon: 2.998263 g/cm3

Orbital Velocity of Moon: 1017.002930 m/s

PlanetDay Characteristic Time: 1.120820 seconds

Lunar Orbital Period: 2380778.000000 seconds

Lunar Orbital Period: 27.555302 days

Program ended with exit code: 0

(3.12E 28J )

(2.147E 33J )

(86,400s)cos(23.5

∘

) = 1.15secon d s

t

c

=

ℏ

⋆

K E

p

=

(2.554E 33J ⋅ s)

(2.147E 33J )

= 1.1877secon d ≈ 1secon d

of 28 46

We see it works great. Characteristic time is 1.085 seconds, Lunar mass is

0.98 moons, its density is 2.998g/cm3 close that of the Earth’s moon. The

PlanetDay characteristic time is 1.12 seconds.

We run it for KOI-4878

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

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

What is the luminosity of the star in solar luminosities? 1.3

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? 2.5

Angular Momentum of Planet: 5.528859 E33

PlanetYear: 1.217332 years

PlanetYear: 38416076.000000 seconds

planet orbital velocity: 27897.789062 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 5641505.000000 meters

planet radius: 0.884526 Earth Radii

planet orbital radius: 1.705703 E11 m

planet orbital radius: 1.140175 Earth distances

planet KE: 2.323964 E33 J

planet density: 7.940497 g/cm3

hbarstar: 2.211544 E33 Js

characteristic time: 0.951626 seconds

Orbital Radius of Moon: 3.853556 E8 m

Orbital Radius of Moon: 1.003530 Moon Distances

Radius of Moon: 1.573184 E6 m

Radius of Moon: 0.905482 Moon Radii

Mass of Moon: 6.356811 E22 kg

Mass of Moon 0.865146 Moon Masses

density of moon: 3.897743 g/cm3

Orbital Velocity of Moon: 1017.002930 m/s

PlanetDay Characteristic Time: 1.120820 seconds

Lunar Orbital Period: 2380778.000000 seconds

Lunar Orbital Period: 27.555302 days

Program ended with exit code: 0

We find to get the orbital period the planet has, one solution is to run it

at mass and size of the Sun (which is within the errors for its actual value)

and use p=2.5 like it is for the Sun, but the luminosity 1.3 solar

luminosities.

This gives an orbital period (planet year) of 444.63 days

of 29 46

Running it again for KOI-4878 varying parameters…

What is the radius of the star in solar radii? 1.072

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

What is the luminosity of the star in solar luminosities? 1.3

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? 2.5

Angular Momentum of Planet: 7.301542 E33

PlanetYear: 1.217332 years

PlanetYear: 38416076.000000 seconds

planet orbital velocity: 27897.789062 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 6483127.000000 meters

planet radius: 1.016483 Earth Radii

planet orbital radius: 1.705703 E11 m

planet orbital radius: 1.140175 Earth distances

planet KE: 2.323964 E33 J

planet density: 5.232137 g/cm3

hbarstar: 2.920617 E33 Js

characteristic time: 1.256739 seconds

Orbital Radius of Moon: 4.131012 E8 m

Orbital Radius of Moon: 1.075784 Moon Distances

Radius of Moon: 1.807878 E6 m

Radius of Moon: 1.040566 Moon Radii

Mass of Moon: 6.974273 E22 kg

Mass of Moon 0.949181 Moon Masses

density of moon: 2.817760 g/cm3

Orbital Velocity of Moon: 982.256287 m/s

PlanetDay Characteristic Time: 1.147099 seconds

Lunar Orbital Period: 2642476.250000 seconds

Lunar Orbital Period: 30.584215 days

Program ended with exit code: 0

This gives an orbital period (planet year) of 444 days

Importantly, since the Moon is pivotal to our theory, the important thing is

we get its density so its composition is right and close to its orbital

period of 449 days. We get it is in the range of

Range: 2.818g/cm3-3.898g/cm3. 444.63 days Characteristic time: 0.95s

Average: 3.356g/cm3. 444 days. Characteristic time: 1.25674s

of 30 46

The density of the Earth’s moon is 3.34g/cm3. We get about exactly this in the average. That of the Earth

is 5.52g/cm3, we got 5.23g/cm3 for this planet in the second running or the program, but 7.94g/cm3,

about half that of the planet Mercury (13.6g/cm3). Clearly, with variation of parameters, we can get this

star system. We want the moon to be right because it is believed it is very important to have a moon

orbiting the planet if the planet is to be high functioning in its habitability because it prevents hot and cold

weather extremes. It allows for stable conditions over long periods to give life a chance to evolve into

something sophisticated, like intelligent life. It does this by holding the planet at its inclination to it orbit,

which for the earth is about 1/4 of a right angle (23.5 deg) which is what we used here, the same what it is

for the Earth.

The constellation Draco, which is

Latin for “the Dragon” is a large

winding constellation visible all

year in the Northern Hemisphere.

Since it is near the North Star,

Polaris, it goes around it near

the Little Dipper and Big Dipper

always high in the sky. The

brightest star in it is alpha

Draconis, common name Thuban,

which was the pole star when the

Egyptian Pyramids were being

built, and were thus aligned with

it. It will be the pole star again

in 21000 AD due to the Earth’s

precession. It was the pole star

from 3942 BC to 1793 BC.

I have applied this program to a wide range of stars, using their average

values for stellar mass, stellar radius, and stellar luminosities. We see the

characteristic times of about 1 second intersect around spectral class GV

stars like our Sun. Here we show such results for F5V stars down to G3V stars

(which are near to the Sun) down to as low in mass, luminosity, and radius

such as K3V stars. We see using our equation

which is where in the program we give the option to compute the planet’s

radius, that it always returns something close to the Earth radius. We use

the equation for that. We see the characteristic time of 1 second for the

star system intersects with the PlanetDay characteristic time of 1 second

around G-type stars like the Sun, putting them inline with the proton,

electron, and neutron.

The results are…

R

planet

= 2

R

2

⋆

r

planet

of 31 46

F5V Star

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

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

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

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

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

What is the planet day in Earth days? 1

That is 86400.000000 seconds

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

Angular Momentum of Planet: 9.321447 E33

PlanetYear: 2.280109 years

PlanetYear: 71954776.000000 seconds

planet orbital velocity: 24888.847656 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 7325190.000000 meters

planet radius: 1.148509 Earth Radii

planet orbital radius: 2.850263 E11 m

planet orbital radius: 1.905256 Earth distances

planet KE: 1.849692 E33 J

planet density: 3.627237 g/cm3

hbarstar: 3.883936 E33 Js

characteristic time: 2.099775 seconds

Orbital Radius of Moon: 5.676287 E8 m

Orbital Radius of Moon: 1.478200 Moon Distances

Radius of Moon: 2.042695 E6 m

Radius of Moon: 1.175720 Moon Radii

Mass of Moon: 7.107576 E22 kg

Mass of Moon 0.967323 Moon Masses

density of moon: 1.990782 g/cm3

Orbital Velocity of Moon: 837.955261 m/s

PlanetDay Characteristic Time: 1.068920 seconds

Lunar Orbital Period: 4256210.000000 seconds

Lunar Orbital Period: 49.261688 days

Program ended with exit code: 0

of 32 46

G3V Star

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

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

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

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

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

What is the planet day in Earth days? 1

That is 86400.000000 seconds

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

Angular Momentum of Planet: 7.393050 E33

PlanetYear: 0.989814 years

PlanetYear: 31236148.000000 seconds

planet orbital velocity: 29789.738281 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 6523625.500000 meters

planet radius: 1.022833 Earth Radii

planet orbital radius: 1.480965 E11 m

planet orbital radius: 0.989950 Earth distances

planet KE: 2.649862 E33 J

planet density: 5.135297 g/cm3

hbarstar: 3.520500 E33 Js

characteristic time: 1.328560 seconds

Orbital Radius of Moon: 3.861263 E8 m

Orbital Radius of Moon: 1.005537 Moon Distances

Radius of Moon: 1.819172 E6 m

Radius of Moon: 1.047066 Moon Radii

Mass of Moon: 7.754257 E22 kg

Mass of Moon 1.055335 Moon Masses

density of moon: 3.074905 g/cm3

Orbital Velocity of Moon: 1015.987427 m/s

PlanetDay Characteristic Time: 1.196672 seconds

Lunar Orbital Period: 2387924.250000 seconds

Lunar Orbital Period: 27.638012 days

Program ended with exit code: 0

of 33 46

K3V Star

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

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

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

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

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

What is the planet day in Earth days? 1

That is 86400.000000 seconds

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

Angular Momentum of Planet: 8.340819 E33

PlanetYear: 0.435785 years

PlanetYear: 13752343.000000 seconds

planet orbital velocity: 36167.078125 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 6929175.500000 meters

planet radius: 1.086418 Earth Radii

planet orbital radius: 0.791609 E11 m

planet orbital radius: 0.529150 Earth distances

planet KE: 3.905860 E33 J

planet density: 4.285367 g/cm3

hbarstar: 5.560546 E33 Js

characteristic time: 1.423642 seconds

Orbital Radius of Moon: 2.909435 E8 m

Orbital Radius of Moon: 0.757665 Moon Distances

Radius of Moon: 1.932263 E6 m

Radius of Moon: 1.112158 Moon Radii

Mass of Moon: 10.277222 E22 kg

Mass of Moon 1.398704 Moon Masses

density of moon: 3.400867 g/cm3

Orbital Velocity of Moon: 1170.438843 m/s

PlanetDay Characteristic Time: 1.428032 seconds

Lunar Orbital Period: 1561850.125000 seconds

Lunar Orbital Period: 18.076969 days

Program ended with exit code: 0

of 34 46

PlanetDay characteristic time:

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,…

We see the tendency is towards characteristic time and planetary characteristic time intersecting at a

minimum in the area GV stars (G3V=2.3) like our Sun, where star systems come in line with the

characteristic time of the proton, electron, and neutron. This may be the place of optimal habitability. GV

stars come in line with the electron, proton, and neutron characteristic time given by:

K E

m

K E

e

(Pl a n etDay)cos(23.5

∘

) = 1second

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

1secon d =

r

i

m

i

⋅

πh

G c

κ

i

of 35 46

Appendix 1 Pressure Gradient of the Protoplanetary Disk.

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

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 1:

2.

Since , we have that which gives from 2:

v

2

ϕ

/r

GM

⋆

/r

2

ρ

d P

dr

= − ρ

(

GM

⋆

r

2

−

v

2

ϕ

r

)

P = c

2

s

ρ

c

s

v

K

=

GM

⋆

r

η

v

ϕ

= v

K

(1 − η)

ρ(r) = ρ

0

(

r

0

)

−s

ρ

0

r

0

v

2

ϕ

= v

2

K

(1 − η)

2

≈

GM

⋆

r

(1 − 2η)

d P

dr

= − ρ

(

GM

⋆

r

2

⋅ 2η

)

P = c

2

s

ρ

d P/dr = c

2

s

dρ /dr

of 36 46

We integrate both sides:

And, we have

3.

We take

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

have

4.

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

5. and

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

ρ

= −

2ηGM

⋆

c

2

s

r

2

dr

∫

ρ

0

dρ

ρ

= −

2ηGM

⋆

c

2

s

r

2

∫

r

0

dr

ln

(

ρ

0

)

=

2ηGM

⋆

c

2

s

(

1

r

0

−

1

r

)

ρ(r) = ρ

0

exp

2ηGM

⋆

c

2

s

(

1

r

0

−

1

r

)

P

0

= c

2

s

ρ

0

exp

2ηGM

⋆

c

2

s

(

1

r

0

−

1

r

)

2ηGM

⋆

c

2

s

(

1

r

0

−

1

r

)

η

e

x

≈ 1 + x

P

r

≈ P

0

1 +

2ηGM

⋆

c

2

s

(

1

r

0

−

1

r

)

P

0

= c

2

s

ρ

0

η

η = −

1

2

(

c

s

v

K

)

2

dln P

dln R

c

s

=

k

B

T

μm

H

k

B

= 1.38E − 23J/K

μ ≈ 2.3

m

H

of 37 46

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

Where , so

7.

So, essentially, by the chain rule

to clarify things. The reason 7 is significant is that equation

Where

c

s

= 1k m /s

P(R) ∝ R

−q

q ≈ 2.5

dln P

dln R

≈ − 2.5

η = −

1

2

(

1k m /s

30k m /s

)

2

(−2.5) = 1.5E − 3

dln P

dln R

=

dln P

d R

⋅

d R

dln R

=

1

P

d P

d R

⋅ R =

R

P

d P

d R

L

earth

ℏ

⊙

=

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

1

2

L

earth

=

4

5

π M

e

f

e

R

2

e

ℏ

⊙

= 2.8314E 33J ⋅ s

λ

moon

c

=

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1.0secon d s

of 38 46

Appendix 2: Deriving the Delocalization Time Form a Gaussian Wave Packet

In order to show that our hypothesis is right, we solve the wave equation for a Gaussian wave

packet and determine the delocalization time, . If it is about six months, the time it takes the

Earth to delocalize (travel its orbital diameter), using the Moon playing the role of the mass of

an electron and our as above to describe the Earth, then the hypothesis can be taken as

correct, and we can solve the whole system for the Earth/Moon/Sun system from the rest of

the equations in the hydrogen atom solution to the Schrodinger wave equation, which is in

spherical coordinates:!

!

The delocalization time of a particle, molecule, or mass in general, , is the time it takes a

particle to delocalize. If we want to apply our wave equation theory of the Solar System to this

concept, then the delocalization time should be the time for the Earth to travel the diameter of

it’s orbit, which would be half a year (about six months). In order to derive the delocalization

time we must consider a Gaussian wave packet…!

!

The Gaussian wavefunction in position space is!

τ

ℏ

⊙

−

ℏ

2

2m

[

1

r

2

∂

∂r

(

r

2

∂

∂r

)

+

1

r

2

sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

+

1

r

2

sin

2

θ

∂

2

∂ϕ

2

]

ψ + V(r)ψ = E ψ

τ

of 39 46

!

It’s Fourier wave decomposition is!

!

We use the Gaussian integral identity (integral of quadratic)!

!

We ﬁnd via the inverse Fourier transform. It is!

!

Substitue :!

!

This is of the form:!

!

, !

Using!

!

ψ (x,0) = Ae

−

x

2

2d

2

ψ (x,0) = Ae

−

x

2

2d

2

=

∫

dp

2π ℏ

ϕ( p)e

i

ℏ

px

∫

∞

−∞

e

−a x

2

+bx

d x =

π

a

e

b

2

4a

ϕ( p)

ϕ( p) =

∫

∞

−∞

d x ψ (x,0)e

−

i

ℏ

px

ψ (x,0)

ϕ( p) = ϕ( p) = A

∫

∞

−∞

e

−

x

2

2d

2

e

−

i

ℏ

[ px]

d x

∫

e

{−a x

2

−bx}

d x

a =

1

2d

2

b =

ip

ℏ

∫

∞

−∞

e

−a x

2

+bx

d x =

π

a

e

b

2

4a

ϕ( p) = A

π

1/(2d

2

)

ex p

−

p

2

ℏ

2

4 ⋅

1

2d

2

b

2

4a

= −

p

2

d

2

2ℏ

2

of 40 46

!

We have to ﬁnd the normalization constant, A, because the probability has to be 1 at its

maximum. We have!

!

We now consider the evolution of a free particle. For a free particle!

!

The time evolution in free space is!

!

Substitute:!

!

Factor out the term and we have!

!

The integral is then,!

ϕ( p) = Ad 2πexp

(

−

p

2

d

2

2ℏ

2

)

∫

∞

−∞

|

ψ (x,0)

|

2

= 1

|

ψ (x,0)

|

2

=

|

A

|

2

e

−x

2

/d

2

∫

∞

−∞

e

{−x

2/

d

2}d x

= d π

A = (πd

2

)

−1/4

H =

p

2

2m

ϕ( p, t) = ϕ(p)e

−i

p

2

2m

t/h

= ϕ(p)e

−i

p

2

t

2mℏ

ψ (x, t) =

∫

dp

2π ℏ

ϕ( p)e

{

i

ℏ

px}

e

{−i

p

2

t

2mℏ

}

ϕ( p) = Ad 2πexp

(

−

p

2

d

2

2ℏ

2

)

ψ (x, t) = A

d 2π

2π ℏ

∫

∞

−∞

dpexp

[

−

p

2

d

2

2ℏ

2

− i

p

2

t

2mℏ

+

ipx

ℏ

]

p

2

α =

d

2

2ℏ

2

+

it

2mℏ

=

1

2ℏ

2

[

d

2

+

itℏ

m

]

of 41 46

, Re(a)>0!

, !

!

, , , , !

!

We determine the probability density . We have!

!

, , !

!

Because we multiplied the top and bottom of by and took its real part. We have!

!

∫

∞

−∞

e

{−αp

2

+bp}

dp =

π

α

e

b

2

/(4a)

a = α

b = i x /ℏ

A

d 2π

2π ℏ

π

α

=

Ad

ℏ

1

2α

ψ (x, t) =

Ad

ℏ 2α

ex p

[

−

x

2

/ℏ

2

4α

]

τ =

m d

2

ℏ

m d

2

=

1

τ

ℏ/m

d

2

=

1

τ

itℏ/m

d

2

=

it

τ

ψ (x, t) =

Ad

ℏ 2α

ex p

[

−

x

2

2d

2

(1 + it /τ)

]

|

ψ (x, t)

|

2

|

ψ (x, t)

|

2

=

|

Ad

ℏ 2α

|

2

ex p

[

−

x

2

2d

2

⋅ 2Re

(

1

1 + it /τ

)

]

|

ex p(−Bx

2

)

|

2

= exp(−2ReBx

2

)

B =

1

2d

2

(1 + it /τ)

ReB =

1

2d

2

⋅

1

1 + t

2

/τ

2

2ReB = 2 ⋅

1

2d

2

(1 + t

2

/τ

2

)

=

1

d

2

(1 + t

2

/τ

2

)

B

1 − it /τ

1

1 + i

t

τ

⋅

1

1 − i

t

τ

=

1

1 +

t

2

τ

2

|

ψ (x, t)

|

2

∝ exp

[

−

x

2

d

2

(1 + t

2

/τ

2

)

]

|

ψ (x, t)

|

2

=

[

−

x

2

d

2

⋅

1

(1 + t

2

/τ

2

)

]

τ =

m d

2

ℏ

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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

uses 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

d

τ

ℏ

⊙

τ =

m

moon

(2r

moon

)

2

ℏ

⊙

m

moon

r

moon

τ = 4

(7.34767E 22kg)(3.844E8m)

2

2.8314E 33J ⋅ s

= 15338227secon d s

15338227

15778800

100 = 97.2 %

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Appendix 3: The Program For Modeling Starsystems

//

// 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? ");

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scanf ("%f", &PlanetDay);

T_p=PlanetDay*86400;

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

{

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

protoplanetary disc? ");

scanf("%f", &p);

M_s=1.9891E30*StarMass;

r_m=R_s/1.8;

v_p=sqrt(G*M_s/r_p);

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

KE_p=0.5*M_p*v_p*v_p;

hbarstar=L_p/p;

t_c=hbarstar/KE_p;

part1=cbrt(hbarstar/(t_c));

part2=cbrt(1/G);

part3=cbrt(hbarstar/299792458);

M_m=part1*part2*part3;

R_s=StarRadius*6.9634E8;

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);

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printf("\n");

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

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

r_m/3.84E8);

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

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

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

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

7.347673E22);

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

printf("\n");

v_m=sqrt(G*M_p/r_m);

KE_m=0.5*M_m*v_m*v_m;

PDCT=(KE_m/KE_p)*(T_p)*(0.91706);

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;

}}

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