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The Earth A Habitable Planet As A Solution To The Wave Equation

By

Ian Beardsley

of 2 51

Contents

1.0 The Theory………………………………………………………5

2.0 Arriving At The Planetary Planck Constant

From The Proton Radius………………………………………..7

3.0 The Strange Duration of a Second……………………..10

4.0 The Value Of The Planetary Planck Constant………11

5.0 Finding The Delocalization Time For Earth………..11

6.0 The Solar Formulation……………………………………..13

7.0 Analysis Of Our Equation………………………………….16

8.0 Applying The Solution to Jupiter and Saturn……..18

9.0 Hydrocarbons The Backbones of Life Chemistry………22

10.0 Modeling The Asteroid Belt……………………………….24

11.0 Equating The Solutions…………………………………..25

12.0 Why The Historical Second And

Physical Second Coincide And Why They

Are The Base Unit…………………………………………………28

13.0 Solution For the Earth As An Orbital……………………30

14.0 Solving Habitable Planetary Systems……………….33

15.0 Bohr Radius For Planets…………………………………38

16.0. Explicitly The Quantization In Terms

Of The Moon and 1 Second…………………………………….42

17.0 Electron Energy Levels, Planetary Orbitals,

And Spin……………………………………………………………..44

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Abstract

It would seem the Earth/Moon/Sun system and the atom with its proton can be conveniently

formulated in terms of a base unit of one second, which results in a solution to the quantum

mechanical wave equation of Schrödinger of atoms as orbital systems for the planets and the

Sun as an orbital system. It would seem the Moon is a perfect yardstick for formulating reality,

which is curious because the Moon makes life on Earth possible because its mysterious size and

near perfectly circular orbit hold the Earth at its inclination to the Sun allowing for the seasons

preventing extreme hot and cold. The Moon is a mystery in astronomy, it is almost as if it should

not be there. One example of this is its low mass for its size, which has lead some to say it is

hollow. If it is, then it has to have an artificial structure beneath its surface keeping it from

collapsing. This has lead to some suggesting it is a spacecraft that came here a long time ago. It

is as if thought went into its design, and this might explain why it allows for the wave solution of

the solar system presented in this paper. Aside from my formulation in terms of the Moon, I

present a solar solution as well. The base unit of one second is intriguing as well; its origins are

historical, not physical yet it becomes the basis of physical dynamics in this paper. I suggest why

that historical development of such a unit of time, which ultimately came to us from the Ancient

Sumerians of Mesopotamia, who were of the first to settle down from wandering and hunting to

invent agriculture, mathematics, and writing, may have aligned with physics because one can

find a sixfold property to it evolution, and Nature seems to be conveniently formulated in terms

of the sixfold. One of the advantages to our quantum mechanical solution of star systems from

the Schrödinger wave equation may be that in its application to the asteroid belt, it may be able

to find the probability of when and where an asteroid will be ejected from the belt and put on a

possible collision course with the Earth. One could easily be compelled to suggest whatever

might be behind the Moon as a hollow craft, might be behind the historical development of the

unit of second as the base unit to our calendar. I further show that the base unit of a second

found in the atom’s proton is in the hydrocarbons, the chemical skeletons of life chemistry.

Near the end of the paper in section 16.0 we show explicitly that the Solar System is quantized in

terms of the Moon and 1 second. At the end in section 17.0 we model the analog to spin in

quantum mechanics for the planets with success. Spin in quantum mechanics for the atoms is

messy and could only be modeled in an abstract sense where spin integers represent something

that is not concrete. What is actually going on as far as spin of electrons goes is not known

because they are too small to look at, but we can see the planets, so with a model for their spin

perhaps we can get an idea of what is going on with spin of electrons and other particles.

of 4 51

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

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

distances.

The whole object of developing a theory for the way planetary systems form is that they meet the

following criterion: They predict the Titius-Bode rule for the distribution of the planets; the

distribution gives the planetary orbital periods from Newton’s Universal Law of Gravitation. The

distribution of the planets is chiefly predicted by three factors: The inward forces of gravity from

the parent star, the outward pressure gradient from the stellar production of radiation, and the

outward inertial forces as a cloud collapses into a flat disc around the central star. These forces

separate the flat disc into rings, agglomerations of material, each ring from which a different

planets form at its central distance from the star (it has a thickness). In a theory of planetary

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

planets.

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1.0 The Theory My findings are that a G2V main sequence star, like our Sun, has a habitable

planet in the orbit given by

1.0

Where is its orbital kinetic energy, is the mass of the planet, and is the mass of its

moon. is a Planck constant for such a planetary orbital system that plays a similar role as the

Planck constant for atoms. is the radius of the star orbited and is the radius of the moon

orbiting the planet. If the planet has a moon, then it can be habitable because it needs one for

the planet to have seasons, because it could hold the planet at an inclination to its orbit that

would prevent extreme hot and cold. This energy determines a base unit of 1 second such

that

1.1

Where is the orbital kinetic energy of the moon around the planet in an approximately

circular orbit, and is the day of the planet or, its rotation period in other words.

This base unit of one second is given in terms of the atom and natural constants by

1.2

is the fine structure constant, is the mass of a proton, is its radius, is the universal

constant of gravitation, is the speed of light, and is Planck’s constant. The Planck constant

for the planetary system (Planck star) denoted is given by

1.3

1.4

Where

1.5

Equation 1 is the solution to Schrödinger’s wave equation

1.6

Which put in spherical coordinates is

n = 3

K E

p

= n

R

s

R

m

⋅

G

2

M

2

p

M

3

m

2ℏ

2

s

K E

p

M

p

M

m

ℏ

s

h

R

s

R

m

K E

p

K E

m

K E

p

(Pl a netDa y) = 1.0secon d s

K E

m

Pl a netDa y

1

6α

2

m

p

h 4π r

2

p

Gc

= 1.004996352secon d s

α

m

p

r

p

G

c

h

ℏ

s

ℏ

s

= (hC )K E

p

hC = 1secon d

C =

1

3

⋅

1

α

2

c

1

3

⋅

2π r

p

G m

3

p

(

−

ℏ

2

2m

∇

2

+ V

)

ψ = E ψ

of 6 51

1.7

Just as

1.8

is the solution to the Schrödinger wave, equations 1.7 and 1.8, for the atom as given by the Bohr

model of the atom.

It may be that for the planet to be an advanced habitable system that we have the following

condition

1.9.

Where is the orbital radius of the planet, is the orbital radius of the planet’s moon,

is the radius of the star, and is the radius of the planet’s moon. All of this holds for

the Earth/Moon/Sun system. This can all be verified with following data:

(Proton Mass)

(Planck Constant)

(Proton Radius)

(Gravitational Constant)

(light speed)

(Fine Structure Constant)

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

Earth’s orbital velocity at perihelion we have:

Looking at equations 1.0 and 1.8 for planets and the atom respectively, which are

,

−

ℏ

2

2m

[

1

r

2

∂

∂r

(

r

2

∂

∂r

)

+

1

r

2

sinθ

∂

∂θ

(

sin θ

∂

∂θ

)

+

1

r

2

sin

2

θ

∂

2

∂ϕ

2

]

ψ + V(r)ψ = E ψ

E = −

Z

2

(k

e

2

)

2

m

e

2ℏ

2

n

2

r

planet

r

moon

≈

R

star

R

moon

r

planet

r

moon

R

star

R

moon

m

P

: 1.67262 × 10

−27

kg

h :6.62607 × 10

−34

J ⋅ s

r

p

: 0.833 × 10

−15

m

G: 6.67408 × 10

−11

N

m

2

kg

2

c : 299,792,458m /s

α : 1/137

q

p

= q

e

= 1.6022E − 19coulom bs

k

e

= 8.988E 9

Nm

2

C

2

K E

moon

=

1

2

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

2

= 3.428E 28J

K E

earth

=

1

2

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

2

= 2.7396E 33J

K E

p

= n

R

s

R

m

⋅

G

2

M

2

p

M

3

m

2ℏ

2

s

E = −

Z

2

(k

e

2

)

2

m

e

2ℏ

2

n

2

of 7 51

Since for our solar system

That is, the Sun is 400 times the size of the Moon, we see we can write equation 1.0, that is the

equation for the Earth, in terms of the atomic number which is in the equation for the atom, as

1.10

That is, we can write it in terms of the atomic number of calcium because it is 20, and square

root 400 is 20. A G2 star like the sun has strong Ca (II) spectral lines and this makes their study

used more than anything else in astronomy in the study of such G2 stars.

Combining equations 1.2, 1.4, and 1.5, yield the radius of a proton accurately

1.11

2.0 Arriving At The Planetary Planck Constant From The Proton Radius

Here is how we arrive at the planetary Planck Constant from the radius of a proton. Energy is

given by Planck’s constant and frequency

We have

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

The frequency of a proton is

This gives

R

s

R

m

= 400

Z

K E

p

= n Z

2

Ca

⋅

G

2

M

2

p

M

3

m

2ℏ

2

s

1

6α

2

m

p

h 4π r

2

p

Gc

=

1

3

⋅

h

α

2

c

1

3

⋅

2π r

p

G m

3

p

r

p

=

2

3

⋅

h

cm

p

E = h f

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

E = J = Joules = energ y

m

p

E = m

p

c

2

f

p

=

m

p

c

2

h

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The radius of a proton is then

Equation 2.1.

This is close to the CODATA value for the proton radius around 2018 (0.842E-15m). Let us now

suggest there is some Planck-type constant for the Earth-Sun system that is connected to the

Planck constant h for quanta we will call it and denote h as . We will say is connected to

by some constant and is given by the kinetic energy of the Earth . That is, we have

Equation 2.2 If Equation 2.3

We find

Equation 2.4

Satisfies this condition.

We write

Or,

Equation 2.5

We multiply equation 2.1 by to get

=

m

p

c

2

h

⋅

r

p

c

=

2

3

≈ ϕ =

m

p

c

h

r

p

m

p

r

p

=

2

3

⋅

h

c

r

p

=

2

3

⋅

h

cm

p

h

⊙

h

p

h

⊙

h

p

C

K E

e

h

⊙

= (h

p

C )K E

e

h

p

C = 1secon d

C =

1

3

⋅

1

α

2

c

1

3

⋅

2π r

p

G m

3

p

h

p

C = t

1

t

1

=

1

3

⋅

h

p

α

2

c

1

3

⋅

2π r

p

G m

3

p

r

p

/r

p

r

2

p

=

2

3

⋅

hr

p

cm

p

r

p

=

2

3

⋅

hr

p

cm

p

2

hr

p

cm

p

⋅

1

6

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Multiply the radicand by to get

=

Multiply outside the radical by to get

This factors into

Equation 2.6

We have

Equation 2.7

Because of equation 2.5:

Where .

Thus the our solar system is quantized like an atom with a base unit of 1 second in terms of the

Earth’s Moon because

Equation 2.8

h

⋅

m

2

p

m

2

p

⋅

c

r

p

= 2

h

2

m

2

p

c

2

⋅

r

p

c

m

3

p

h

⋅

1

6

2π G /2π G

r

p

= 2

h

2

m

2

p

c

2

⋅

r

p

c

m

3

p

h

⋅

2π G

12π G

2h

m

p

c

r

p

c

m

3

p

h

⋅

2π

3

⋅

1

4π

⋅

G

α

2

/α

2

r

p

=

2h

α

2

c

⋅ α

2

m

p

1

3

⋅

2π r

p

G m

3

p

⋅

Gc

4πh

r

p

=

1

3

⋅

h

α

2

c

1

3

⋅

2π r

p

G m

3

p

(

18

3

α

2

m

p

Gc

4πh

)

r

p

= 6α

2

m

p

Gc

4πh

⋅ t

1

t

1

=

1

3

⋅

h

p

α

2

c

1

3

⋅

2π r

p

G m

3

p

t

1

= 1secon d

K E

moon

K E

earth

(Ear th Da y) = 1.08secon d s

of 10 51

Where is the kinetic energy of the moon, is the kinetic energy of the earth, and

is the rotation period of the Earth, which has little difference in these results if we

consider the Earth Day with respect to the Sun, or with respect to the stars. We have the second

is a natural unit, and a base unit

Equation 2.9

Using a more recent value of the proton radius 0.833E-15m.

3.0 The Strange Duration of a Second

The interesting thing is that we have the duration of a second is what it is not because we were

attempting to find some base unit of time for Nature, but rather we have the duration of a

second from ancient times that came to us over many years of dividing the rotation of the Earth

into hours, minutes, and seconds in such a way that we ended up with it. However we may have

had some sort of a thought going into it that was six-fold in Nature that lead it to being natural if

we consider:

3.1

And that thought may be was in a six-fold pattern that the motions of the Earth/Moon/Sun

system had in it approximately which we inadvertently characterized in the calendar:

3.2

Which can be written

3.3

Where there are 360 degrees in a circle and as such the Earth approximately moves through one

degree per day around the Sun. We achieved this because the ancients of Sumeria and Babylonia

from which the ancient Greeks received their counting system used base 60, sexagesimal

because 60 is evenly divisible by so much, that is by

1,2,3,4,5,6, 10, 12, 15, 20, 30,…

We have

K E

moon

K E

earth

Ear th Da y

1

6α

2

m

p

h 4π r

2

p

Gc

= 1.004996352secon ds

1secon d =

1

365.25

⋅

1

24

⋅

1

60

⋅

1

60

= 0.0000000316881year

(

360d a ys

year

)(

24h ours

d a y

)(

60min

h our

)(

60sec

min

)

= 31104000secon ds /year

12

2

(

60d a ys

year

)(

1h our

d a y

)(

60min

h our

)(

60sec

min

)

K E

m

K E

p

(Pl a netDa y) = 1.0secon d s

1

6α

2

m

p

h 4π r

2

p

Gc

= 1.004996352secon ds

of 11 51

4.0 The Value Of The Planetary Planck Constant

=!

!

= !

!

= !

!

5.0 Finding The Delocalization Time For Earth

If we want to prove that our planetary Planck constant is correct, the delocalization time for the

the Earth should be 6 months using it, the time for the Earth to travel the width of its orbit. We

want to solve the Schrödinger wave equation for a wave packet and use the most basic thing we

can which is a Gaussian distribution. We want to then substitute for Planck’s constant h that is

used for quanta and atoms our Planck-type constant (h bar star) for the Earth/Moon/Sun

system then apply it to predict the delocalization time for the Moon in its orbit with the Earth

around the Sun.

We consider a Gaussian wave-packet at t=0:

hC =

1

3

⋅

h

α

2

c

1

3

⋅

2π r

p

G m

3

p

= 1.0secon ds

C =

1

3

⋅

1

α

2

c

1

3

⋅

2π r

p

Gm

3

p

1

3

⋅

18769

299792458

1

3

⋅

2π (0.833E − 15)

(6.67408E − 11)(1.67262E − 27)

3

1.55976565E33

s

m

kg

3

⋅

s

2

kg

m

3

=

s

m

s

2

kg

2

m

2

=

s

m

⋅

s

kg ⋅ m

=

1

kg

⋅

s

2

m

2

1

C

= kg

m

2

s

2

=

1

2

mv

2

= energy

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

hC =

(

kg

m

s

2

⋅ m ⋅ s

)

(

1

kg

⋅

s

2

m

2

)

(

kg

m

2

s

)(

1

kg

⋅

s

2

m

2

)

= secon d s

ℏ

s

= (hC )K E

earth

= (1.03351s)(2.7396E33J ) = 2.8314E 33J ⋅ s

ℏ

s

of 12 51

We say that is the delocalization length and decompose the wave packet with a Fourier

transform:

is the harmonics of the wave function. We use the identity that gives the integral of a

quadratic:

Solve the equation

With the initial condition

A plane wave is the solution:

Where,

The wave-packet evolves with time as

Calculate the Gaussian integral of

and

The solution is:

where

ψ (x,0) = Ae

−

x

2

2d

2

d

ψ (x,0) = Ae

−

x

2

2d

2

=

∫

dp

2π ℏ

ϕ

p

e

i

ℏ

px

ϕ

p

∫

∞

−∞

e

−α

2

x+βx

d x =

π

α

e

β

2

4α

iℏ∂

t

ψ (x, t) =



p

2m

ψ (x, t)

ψ (x,0) =

∫

dp ⋅ e

p

2

d

2

2ℏ

2

⋅ e

−

i

ℏ

px

e

i

ℏ

( px−ϵ( p)t)

ϵ( p) =

p

2

2m

ψ (x, t) =

∫

dp ⋅ e

p

2

d

2

2ℏ

2

⋅ e

−

i

ℏ

( px−

p

2

2m

t)

dp

α

2

=

d

2

2ℏ

2

+

it

2mℏ

β =

i x

ℏ

ψ

2

= ex p

[

−

x

2

d

2

⋅

1

1 + t

2

/τ

2

]

τ =

m d

2

ℏ

of 13 51

is the delocalization distance, which for instance could be the width of an atom. is the

delocalization time, the average time for say an electron to traverse the diameter of the atom and

even leave it, to delocalize. If we substitute for our , and say that the delocalization distance

is for the Moon, the width of the Earth orbit, we should get a half a year for the delocalization

time, the time for the Moon and Earth to traverse the diameter of their orbit around the Sun. We

have

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

Now let’s compute a half a year…

(1/2)(365.25)(24)(60)(60)=15778800 seconds

So we see our delocalization time is very close to the half year over which the Earth and Moon

travel from one position to the opposite side of the Sun. The closeness is

Thus we know our is accurate, it continues to function in a theoretical framework. The thing

about this is that it means we can predict the mass of the Moon from the Earth year. In terms of

what we said earlier that the Moon allows for life by creating the seasons, holding the Earth at

its tilt to the Sun, so we don’t go through extreme heat and cold, this suggests the Moon has a

mass that follows from Earth orbit, which is the habitable zone of the Sun, the right distance for

water to exist as liquid, and thus could be as it is for a reason, which means life might be part of

a physical process throughout the Universe, that it unfolds naturally in the evolution of star

systems.

6.0 The Solar Formulation Our solution of the wave equation for the planets gives the

kinetic energy of the Earth from the mass of the Moon orbiting the Earth, but you could

formulate based on the Earth orbiting the Sun. In our lunar formulation we had:

6.1.

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

6.2.

Thus equation 6.1 becomes

6.3.

d

τ

ℏ

s

τ =

m

moon

(2r

moon

)

2

ℏ

s

m

moon

r

moon

τ = 4

(7.34767E 22kg)(3.844E 8m)

2

2.8314E 33J ⋅ s

= 15338227secon ds

15338227

15778800

100 = 97.2 %

ℏ

s

K E

e

= 3

R

s

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

s

R

s

R

m

=

r

e

r

m

K E

e

= 3

r

e

r

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

s

of 14 51

The kinetic energy of the Earth is

6.4.

Putting this in equation 6.3 gives the mass of the Sun:

6.5.

We recognize that the orbital velocity of the Moon is

6.6.

So equation 6.5 becomes

6.7.

This gives the mass of the Moon is

6.8.

Putting this in equation 6.1 yields

6.9.

We now multiply through by and we have

6.10.

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

momentum quantized as Bohr discovered for the atom by The angular

momentum we will call , the subscript for Planck. We have

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

K E

e

=

1

2

⋅

GM

⊙

M

e

r

e

M

⊙

= 3r

2

e

⋅

GM

e

r

m

⋅

M

3

m

ℏ

2

s

v

2

m

=

GM

e

r

m

M

⊙

= 3r

2

e

⋅ v

2

m

⋅

M

3

m

ℏ

2

s

M

3

m

=

M

⊙

ℏ

2

s

3r

2

e

v

2

m

K E

e

=

R

s

R

m

⋅

G

2

M

2

e

M

⊙

2r

2

e

v

2

m

M

2

e

/M

2

e

K E

e

=

R

s

R

m

⋅

G

2

M

4

e

M

⊙

2r

2

e

v

2

m

M

2

e

ℏ

s

h /2,2h /2,3h /2,...

L

p

L

p

= r

e

v

m

M

e

= r

e

v

m

M

e

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

m

2

s

L

2

p

= r

2

e

v

2

m

M

2

e

= 7.4483E 77J ⋅ m

2

⋅ kg = 8.3367E 77kg

2

⋅

m

4

s

2

of 15 51

6.11.

Let us compare this to that of an atom:

6.12.

We notice that in equation 6.11

; ; ; ;

is really . We can write 6.11 as

6.13.

We say . That is

Let us see how accurate our equation is:

=

We have that the kinetic energy of the Earth is

K E

e

=

R

s

R

m

⋅

G

2

M

4

e

M

⊙

2L

2

p

E = −

Z

2

n

2

⋅

k

2

e

4

m

e

2ℏ

2

Z

2

n

2

→

R

s

R

m

k

2

e

→ G

2

e

4

→ M

4

e

m

e

→ M

⊙

ℏ

2

→ L

2

p

L

p

ℏ

s

K E

e

=

R

s

R

m

⋅

G

2

M

4

e

M

⊙

2ℏ

2

s

ℏ

s

= h

s

/2π

ℏ

s

= 9.13E 38J ⋅ s

h

s

= 2π ℏ

s

= 5.7365E 39J ⋅ s

K E

e

=

R

s

R

m

⋅

G

2

M

4

e

M

⊙

2ℏ

2

s

R

s

R

m

(6.67408E − 11)

2

(5.972E 24kg)

4

(1.9891E 30kg)

2(8.3367E 77kg

2

⋅

m

4

s

2

)

R

s

R

m

(6.759E 30J )

R

s

R

m

=

6.957E8m

1737400m

= 400.426

K E

e

= 2.70655E 33J

K E

earth

=

1

2

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

2

= 2.7396E 33J

of 16 51

Our equation has an accuracy of

Which is very good.

7.0 Analysis Of Our Equation

How we arrived at our equation for the solution of the planets to the Schrödinger wave equation

is as follows: The Bohr solution of the wave equation for the atom is

Let us say

Then,…

and

Where e is charge and is the mass of the planet and is the mass of the Moon. We write

, ,

Where is the Planck constant for the Sun and is the mass of an electron, or star. We have

our solution for the solar system is

But we need to include the orbital number, n, and the atomic number, Z. However the planets

don’t orbit a proton, Z for the Bohr equation is the atomic number, or number of protons orbited

by an electron or by electrons. The Earth orbits the Sun, it would turn out Z squared is the Sun

as given by its radius. But we must divide it by the radius of the Moon to normalize it. As such

the Moon is a natural yardstick and says the radius of the Sun is 400:

,

Our solution to the Schrödinger equation for the planets is:

2.70655E 33J

2.7396E 33J

= 98.79 %

E = −

Z

2

(k

e

2

)

2

m

e

2ℏ

2

n

2

e

2

= e

1

e

2

M

p

= e

1

M

m

= e

2

M

p

M

m

e

2

= M

p

M

m

= M

2

k

e

= G

h = h

s

M

m

= m

e

ℏ

s

m

e

E =

(GM

2

)

2

M

m

2ℏ

2

s

1

n

2

→ n

Z

2

→

R

s

R

m

K E

p

= n

R

s

R

m

G

2

M

2

p

M

3

m

2ℏ

2

s

of 17 51

Let us check if the units work:

,

Let us see what the accuracy of our equation is.

=

The kinetic energy of the Earth using its orbital velocity at perihelion, the velocity at its closest

approach to the Sun, which is when it is going fastest is

Thus the accuracy of our equation is

Our theory for the kinetic energy of the Earth predicts it very accurately. The value I use here for

G is the CODATA value in 2014. We only know G to three places after the decimal so it doesn’t

matter that the fourth digit after the decimal can go up or down from year to year depending on

how we measure G.

In determining we have

=

G = N

m

2

kg

2

= kg

m

s

2

⋅

m

2

kg

2

=

m

3

s

2

kg

h = J ⋅ s = kg

m

s

2

⋅ m ⋅ s = kg

m

2

s

G

2

M

2

p

M

3

m

=

m

6

s

4

kg

2

⋅ kg

5

=

m

6

s

4

kg

3

h

2

= kg

2

⋅

m

4

s

2

(

m

6

s

4

kg

3

)(

s

2

kg

2

m

4

)

=

m

2

s

2

kg =

1

2

mv

2

= K E

K E

p

= n

R

s

R

m

G

2

M

2

p

M

3

m

2ℏ

2

s

3

6.96E8m

1737400m

⋅

(6.67408E − 11

2

)(5.972E 24 kg)

2

(7.34763E 22kg)

3

2(2.8314E 33

2

)

3

6.96E8

1737400

(3.930355E 30) = 2.7271E 33J

K E

earth

=

1

2

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

2

= 2.7396E 33J

2.7271E 33J

2.7396E 33J

100 = 99.54 %

ℏ

s

C =

1

3

⋅

1

α

2

c

1

3

⋅

2π r

p

Gm

3

p

of 18 51

=

Also we have

All this using the following data

(Proton Mass)

(Planck Constant)

(Proton Radius)

(Gravitational Constant)

(light speed)

(Fine Structure Constant)

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

Earth’s orbital velocity at perihelion we have:

8.0 Applying The Solution to Jupiter and Saturn

We have the solution for the Bohr hydrogen atom is

8.1

1

3

⋅

18769

299792458

1

3

⋅

2π (0.833E − 15)

(6.67408E − 11)(1.67262E − 27)

3

1.55976565E 33

hC = (6.62607E − 34)(1.55976565E 33) = 1.03351secon ds

ℏ

s

= (hC )K E

earth

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

1

6α

2

m

p

h 4π r

2

p

Gc

= 1.004996352secon ds

K E

moon

K E

earth

(Ear th Da y) = 1.08secon d s

m

P

: 1.67262 × 10

−27

kg

h:6.62607 × 10

−34

J ⋅ s

r

p

: 0.833 × 10

−15

m

G: 6.67408 × 10

−11

N

m

2

kg

2

c : 299,792,458m /s

α : 1/137

q

p

= q

e

= 1.6022E − 19coulom bs

k

e

= 8.988E 9

Nm

2

C

2

K E

moon

=

1

2

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

2

= 3.428E 28J

K E

earth

=

1

2

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

2

= 2.7396E 33J

E = −

Z

2

(k

e

2

)

2

m

e

2ℏ

2

n

2

of 19 51

is the atomic number, the number of protons orbited, but for our system it is 1, because one

body is orbited. is the orbit number, which we will deal with later. We have our solution for the

Earth/Moon/Sun system is

8.2

Let is the earth mass, is the lunar mass, and is our

Planck constant for the Earth/Moon/Sun system (Another way of writing ) We compute

equation 8.2:

=

The kinetic energy of the Earth is

This is

The Moon perfectly eclipses the Sun which means as seen from the Earth the Moon appears to

be the same size as the Sun. This is because

8.3

Where is the Earth orbital distance, is the Moon’s orbital distance, is the Sun’s radius,

and is the Moon’s radius. This is because though the Moon is 400 times further from the Sun

than it is from the Earth, the Sun is 400 times larger than the Moon. That is

Thus

8.4

We find the quantum mechanical solution to the Earth/Moon/System is

8.5 !

for Earth orbit (Third Planet).!

Z

n

E =

G

2

M

2

m

3

2h

2

⊙

n

2

M = M

e

m = M

m

h

⊙

= 2.8314E 33J ⋅ s

ℏ

s

E =

(6.67408E − 11)

2

(5.972E 24kg)

2

(7.347673E 22kg)

3

2(2.8314E 33)

2

n

2

3.93E 30Joules

n

2

K E

e

=

1

2

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

2

= 2.7396E 33J

1/n

2

= 697

r

e

r

m

≈

R

⊙

R

m

r

e

r

m

R

⊙

R

m

6.96E8m

1737400m

= 400.5986

R

⊙

R

m

= 400

E = n

R

⊙

R

m

⋅

G

2

M

2

m

3

2h

2

⊙

n = 3

of 20 51

!

We see the solar Planck constant we developed works in solving the Schrödinger wave equation

for the Earth/Moon/Sun system. My belief as to why this was never done is that it had to be

realized that the solution of the Earth kinetic energy around the Sun as a quantum mechanical

state is based on the Moon around the Earth.

Now we want to find what the wave equation solutions are for Jupiter and Saturn because they

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

clearly the dynamics of the wave solution to the Solar System.

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

and go to the gas giants and ice giants, the atomic number is no longer squared and the square

root of the the orbital number moves from the numerator to the denominator. I believe this is

because the solar system here should be modeled in two parts, just as it is in theories of solar

system formation because there is a force other than just gravity of the Sun at work, which is the

radiation pressure of the Sun, which is what separates it into two parts, the terrestrial planets on

this side of the asteroid belt and the giants on the other side of the asteroid belt. The effect the

radiation pressure has is to blow the lighter elements out beyond the asteroid belt when the

solar system forms, which are gases such as hydrogen and helium, while the heavier elements

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

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

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

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

8.6

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

.

3(400.5986)(3.93E 30J ) = 2.7268585E33J

2.7268585E33

2.7396E33

= 99.5 %

E =

Z

n

G

2

M

2

m

3

2ℏ

2

s

K E

j

=

1

2

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

2

= 1.7865E 35J

E =

Z

H

5

(6.67408E − 11)

2

(1.89813E 27kg)

2

(7.347673E 22kg)

3

2(2.8314E 33)

2

E =

Z

H

5

(3.971E 35J ) = Z

H

(1.776E 35J )

Z

H

=

1.7865E 35J

1.776E 35J

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

Z = Z

H

of 21 51

Our equation for Jupiter is

8.7

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

8.8

It makes sense that Saturn would use Helium in the equation because Saturn is the next planet

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

just like Jupiter, Saturn is primarily composed of hydrogen and helium gas.

Our equation in this paper for the Earth orbit does not depend on the Moon’s distance from the

Earth, only its mass. The Moon slows the Earth rotation and this in turn expands the Moon’s

orbit, so it is getting larger, the Earth loses energy to the Moon. The Earth day gets longer by

0.0067 hours per million years, and the Moon’s orbit gets 3.78 cm larger per year. It is believed

the Moon came from a chunk knocked off the Earth from a collision with a Mars sized

protoplanet. At the time that the Moon formed it was about 4 Earth radii distant 4.5 billion

years ago. After 100 million years the Moon became tidally locked making its rotation period

equal to its orbital period keeping one face always towards the Earth. After 500 million years the

Moon was orbiting at about 20 Earth radii. It is believed that during the period of heavy

bombardment in a chaotic early solar system about 4.1 to 3.8 billion years ago a large object did

a close pass pulling the Moon further changing its orbit giving it its 5 degree offset from the

Earth’s equator. Our wave equation solution may only use the Moon’s mass but the equation for

kinetic energy of the Moon to kinetic energy of the Earth times the Earth Day equal to about one

second:

Which we connect with the equation where the proton gives one second

E =

Z

H

5

G

2

M

2

j

M

3

m

2ℏ

2

s

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 =

Z

He

6

G

2

M

2

s

M

3

m

2ℏ

2

s

K E

moon

K E

earth

(Ear th Da y) = 1.08secon d s

of 22 51

This holds for when the Moon was at a distance from the Earth such that it appears the same

size as the Sun, which means:

Which is when the two equations above for one second connect to our wave equation solution to

the Earth

Because .

The Moon at its inclination to the Earth in its orbit makes life possible here because it holds the

Earth at its tilt to its orbit around the Sun allowing for the seasons so the Earth doesn’t get too

extremely hot or too extremely cold. We see the Moon may be there for a reason.

9.0 Hydrocarbons The Backbones of Life Chemistry Interestingly we can write!

9.1

9.2

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

proton-seconds by not letting cancel with the bodies of these equations on the left, but rather

divide into them, which are in units of mass, giving us a number of protons. I say this is the

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

they display sixfold symmetry. I can generate the time values from these equations for all of the

elements for integer values in a program. Some of the output looks like

1

6α

2

m

p

h 4π r

2

p

Gc

= 1.004996352secon ds

r

planet

r

moon

≈

R

star

R

moon

K E

p

= n

R

s

R

m

⋅

G

2

M

2

p

M

3

m

2ℏ

2

s

R

star

R

moon

=

R

s

R

m

1

α

2

m

p

h 4π r

2

p

Gc

= 6proton ⋅ secon ds = carbon(C )

1

α

2

m

p

h 4π r

2

p

Gc

= 1proton ⋅ 6secon ds = h ydr ogen(H )

m

p

of 23 51

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

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

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

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

chemistry. And carbon is the core element of life. Here is the code for the program:

#include <stdio.h>

#include <math.h>

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

int n;

float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,

c=299792458,protons[100],r=0.833E-15;

do

{

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

scanf("%f", &increment);

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

scanf("%i", &n);

}

while (n>=101);

{

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

{

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

int intpart=(int)protons[i];

float decpart=protons[i]-intpart;

t=t+increment;

if (decpart<0.25)

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

}}}}

of 24 51

10.0 Modeling The Asteroid Belt We want to model the asteroid belt in terms of our wave

solution for the planets, which is important because we want to have a way to predict the

probability of from when and where an asteroid can be ejected from it and go on a collision

course with the Earth.!

The total mass of the asteroid belt is estimated to be 2.39E21 kg. The four largest objects are

Ceres, Vesta, Pallas, and Hygiea, which contain an estimated 62% of the belt’s total mass with

39% of that for Ceres alone. Ceres is at 2.77 AU and the Titius-Bode Rule places it at 2.8 AU,

which carries most of the mass of the Asteroids. The mass of Ceres is 9.1E20 kg. The Titus-

Bode Rule, which gives the distribution of the planets is!

!

The asteroid belt is between 2.2 and 3.2 AU (1 AU thick).!

The Asteroid Belt separates the solid, terrestrial planets, from the gaseous, gas giants. It is not

a single body but a debris ﬁeld, so it does not have an orbital number n in our theory. It is

between the n=4 (Mars) orbit and the n=5 (Jupiter) orbit. We will say it represents an

interference pattern between the gas giants wave and the terrestrial planets wave.!

We have that it is as a solution of the wave equation ( !

!

We have!

!

= !

a = 0.4 + 0.3 × 2

n

n = − ∞,0,1,2,...

−∞ = m ercur y,0 = venus,1 = ear th, . . .

M

A

= Mass AsteroidBelt)

E =

G

2

M

2

A

M

3

m

2h

2

⊙

⋅

M

J

M

S

M

U

M

N

M

E

M

V

M

⋅ N

A

M

J

= MassJupiter = 1.898E 27kg

M

S

= MassSat ur n = 5.683E 26kg

M

U

= MassU ranus = 8.681E 25kg

M

N

= MassNept u ne = 1.024E 26kg

M

= MassMars = 6.39E 23kg

M

E

= MassEar th = 5.972E 24kg

M

V

= MassVenu s = 4.867E24kg

M

= MassMercur y = 3.185E 23kg

E =

(6.67408E − 11)

2

(2.39E 21kg)

2

(7.34767E 22kg)

3

2(2.8314E33J ⋅ s)

⋅

(9.588E105)

(6.10E 96)

⋅ N

A

(6.295E 23J )(1.572E 9)N

A

of 25 51

This gives the asteroids are objects that make up the debris ﬁeld, which is true.

There are an estimated 100,000 objects in the asteroid belt with diameters from 1 km to

500km, however the shattering process produces a size distribution strongly weighted towards

small diameters, the vast majority of which are around 1 km in diameter.!

We have this value because the average orbital velocity of the asteroid belt is 17.9 km/s, the

orbital velocity of Ceres which makes up 1/3 of the Belt by mass. That is!

!

What needs to be done is to ﬁnd the wave equation of the gas giants, and the wave equation

of the terrestrial planets, and then the interference pattern that would be the Asteroid Belt,

which would not be a trivial undertaking, though the above we guess is part of the solution.!

11.0 Equating The Solutions We had both the lunar and solar formulations:

1.0

6.13.

However has a different value in each. They are respectively:

That is in the solar formulation . This 6.13 can be written:

6.11.

Let us equate 6.11 with 1.0:

This gives:

11.1

N

A

= 124,000

1

2

Mv

2

av e

=

1

2

(2.39E 21kg)(17,900m /s)

2

= 1.2268E38J

K E

p

= n

R

s

R

m

⋅

G

2

M

2

p

M

3

m

2ℏ

2

s

K E

e

=

R

s

R

m

⋅

G

2

M

4

e

M

⊙

2ℏ

2

s

ℏ

s

ℏ

s

= (hC )K E

earth

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

L

p

= r

e

v

m

M

e

= r

e

v

m

M

e

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

m

2

s

L

p

= ℏ

s

K E

e

=

R

s

R

m

⋅

G

2

M

4

e

M

⊙

2L

2

p

3

R

s

R

m

⋅

G

2

M

2

p

M

3

m

2ℏ

2

s

=

R

s

R

m

⋅

G

2

M

4

e

M

⊙

2L

2

p

L

p

=

M

2

e

M

⊙

M

3

m

3

⋅ ℏ

s

of 26 51

We remember that

1.3

1.4

1.5

1.2

This gives

11.2

We have

11.3

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

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

and the orbital number of the Earth. Let us compute

11.4

Let us see how well equation 11.3 works. at aphelion is 966 m/s and .

. We have

That is an accuracy of

Equation 11.3 can be written:

ℏ

s

= (hC )K E

p

hC = 1secon d

C =

1

3

⋅

1

α

2

c

1

3

⋅

2π r

p

G m

3

p

1

6α

2

m

p

h 4π r

2

p

Gc

= 1.004996352secon ds

r

e

v

m

M

e

=

1

6α

2

⋅

r

p

m

p

h 4π

Gc

⋅

1

2

M

e

v

2

e

⋅

M

2

e

M

⊙

M

3

m

3

2v

m

=

v

2

e

r

e

(1secon d )

M

2

e

M

⊙

M

3

m

3

M

2

e

M

⊙

M

3

m

3

=

(5.972E 24kg)

2

(1.9891E 30kg)

(7.34763E 22kg)

3

(1.732)

= 321,331.459 ≈ 321,331

v

m

v

e

= 29,800m /s

r

e

= 1AU = 1.496E11m

2(966m /s) =

(29,800m /s)

2

1.496E11m

(1sec)(321,331.459)

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

1907

1932

= 98.7 %

of 27 51

11.5

From equation 1.2:

We have

11.6

Since , the diameter of the Earth orbit, we have

11.7

And we see the Earth/Moon/Sun system determines the radius and mass of the proton and vice

versa. We basically have

11.8.

Where is the Earth orbital number. We have

= 7.83436E4seconds

EarthDay=(24)(60)(60)=86400 seconds,

1secon d = 2r

e

v

m

v

2

e

M

3

m

3

M

2

e

M

⊙

1

6α

2

m

p

h 4π r

2

p

Gc

= 1.004996352secon ds

1

6α

2

⋅

r

p

m

p

h 4π

Gc

= 2r

e

v

m

v

2

e

⋅

M

3

m

3

M

2

e

M

⊙

2r

e

= d

e

1

6α

2

⋅

r

p

m

p

h 4π

Gc

= d

e

v

m

v

2

e

⋅

M

3

m

3

M

2

e

M

⊙

1secon d = 2v

m

r

e

v

2

e

M

3

m

3

M

2

e

M

⊙

1secon d =

K E

moon

K E

earth

Ear th Da y

1secon d =

M

m

v

2

m

M

e

v

2

m

Ear th Da y

Ear th Da y =

2r

e

v

m

M

m

M

⊙

n

n = 3

Ear th Da y =

2(1.496E11m)

966m /s

7.34763E 22kg

1.9891E 30kg

1.732

of 28 51

accuracy

This last equation, equation 11.8, we can use to find the rotation period, or the length of the day,

of an earth-like planet in the habitable zone of any star system, so it would be very useful.

12.0 Why The Historical Second And Physical Second Coincide And Why They Are

The Base Unit

We want to turn our attention to equation 11.5 and write it

12.1

Because it may tell us why 1 second is the basic unit of our quantum mechanical solution for the

Earth/Moon/Sun system, which happens to coincide with the the fundamental unit of time we

developed since ancient times in our calendar we have today.

Let us talk, first, about how we got the unit of a second to measure time. The Sumerians, who

were of the first to settle down from wandering and gathering, from following the herds and

hunting, who fired clay to make the first homes, developed a system of writing and mathematics,

and agriculture, some 12 thousand years ago, gave us our system of measuring time.

They made their system of counting base 60, called sexagesimal, or hexagesimal, where we use

today base 10. They chose base 60 because it is evenly divisible by so much:

1,2,3,4,5,6, 10, 12, 15, 20, 30,…

and, because there are 12 months in a year, approximately the number of new Moons in a year,

and there are 5 fingers on each hand, and 12 times 5 is sixty. Twelve as well is an important

number and they wanted to factor base twelve into their counting (duodecimal) where twelve is

very abundant too, divisible evenly by

1,2,3,4,6,…

So they divided the day into 12 hours and the night into 12 hours, so the time from sunrise to

sunrise and sunset to sunset was 24 hours. So the Earth’s rotation period to this day is divided

into 24 hours. Because they had a base sixty counting system, they divided each hour into 60

minutes, and each minute into 60 seconds. If we divide the Earth’s rotation period into 24

hours, and each hour into 60 minutes, and each minute into sixty seconds, we get the duration

of a second we have today.

Looking at equation 12.1 the velocity of the Moon is given by

12.2

And the velocity of the Earth is given by

7.834E4s

86,400s

100 = 90.675 %

1secon d = d

e

v

m

v

2

e

M

3

m

3

M

2

e

M

⊙

v

m

=

GM

e

r

m

of 29 51

12.3

For the Earth to be in the habitable zone of the star it orbits, must be exactly what it is and

therefore must be exactly what it is. If it is a condition for life to optimize that the Moon must

perfectly eclipse the Sun as seen from the Earth it has to be that:

12.4

And, then must be exactly what it is, and therefore must be exactly what it is. Therefore the

orbital period of the Moon must be

12.5.

And, the orbital period of the Earth must be

12.6.

Thus the orbital period of the Moon has to be

12.7.

And, the orbital period of the Earth has to be

12.8

Also the Sumerians divided the circle into 360 parts giving us the 360 degrees we have in a circle

today because 6 times 60 is 360. Thus since there are 365.25 days in a year, the time for the

Earth to revolve once around the Sun, it moves through about 1 degree per day. The Moon

moves through about . Thus we must have the Earth and lunar

orbits give one second because

Where . Because we have said that in this equation and must be exactly what they

are, which is such that the Moon moves through 12 times more degrees per day than the Earth

and there are then about 12 moons in a year because the the Earth moves through about 1

degree in a day.

v

e

=

GM

⊙

r

e

r

e

v

e

r

e

r

m

≈

R

⊙

R

m

r

m

v

m

T

m

=

2π r

m

v

m

T

e

=

2π r

e

v

e

T

m

= 27.32d ays

T

e

= 365.25d a ys

360deg /30d a ys = 12d eg /d a y

1secon d = d

e

v

m

v

2

e

M

3

m

3

M

2

e

M

⊙

d

e

= 2r

e

v

m

v

e

of 30 51

13.0 Solution For the Earth As An Orbital We had since the Bohr atom was the solution to

the wave equation for the Energy!

13.1. !

That the solution for the Earth/Moon/Sun System was in the lunar solution!

13.2. !

We want to now look at the solution for the orbital of an electron in the atom from the wave

equation in the Bohr atom which is!

13.3. !

Thus the solution for the Earth/Moon/Sun system might be!

13.4. !

This gives!

13.5. !

= !

Where n=3 is the Earth orbit and we see our Planck constant continues to work. The average

distance of the Earth from the Sun is the astronomical unit AU=1.496E11 meters, so this works

with an accuracy of!

!

The units work, as we can see:!

!

E = −

Z

2

(k

e

2

)

2

m

e

2ℏ

2

n

2

KE

p

= n

R

s

R

m

⋅

G

2

M

2

p

M

3

m

2ℏ

2

s

r

n

=

n

2

ℏ

2

Zk

e

2

m

e

r

n

=

2ℏ

2

s

GM

3

m

⋅

R

s

R

m

⋅

1

n

r

3

=

2(2.8314E33J ⋅ s)

2

(6.67408E − 11)(7.34763E 22kg)

3

⋅ (400.5986)

1

1.732

1.40075E11meters

1.40075E11

1.496E11

100 = 93.633 %

h = J ⋅ s = kg

m

s

2

⋅ m ⋅ s = kg

m

2

s

G = kg

m

s

2

⋅

m

2

kg

2

=

m

3

s

2

kg

h

2

= kg

2

m

4

s

2

of 31 51

!

The solar solution was!

13.6. !

Where had a diﬀerent value given by!

13.7. !

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

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

might be!

13.8. !

Where is !

We have

13.9.

This has an accuracy of

Equating the lunar and solar solutions we have

Yields

r

n

=

(

kg

2

m

4

s

2

)(

s

2

kg

m

3

)

(

1

kg

3

)

= meters

KE

e

=

R

s

R

m

⋅

G

2

M

4

e

M

⊙

2ℏ

2

s

ℏ

s

L

p

=

M

2

e

M

⊙

M

3

m

3

⋅ ℏ

s

L

p

ℏ

s

r

n

M

e

M

⊙

r

n

=

ℏ

2

s

GM

3

e

⋅

R

m

R

s

h

s

ℏ

s

= 9.13E 38J ⋅ s

r

3

=

(9.13E 38)

2

(6.67408E − 11)(5.972E 24kg)

3

⋅

1

400.5986

= 1.4638E11m

1.4638E11m

1.496E11m

100 = 97.85 %

2ℏ

2

s

GM

3

m

⋅

R

s

R

m

⋅

1

n

=

L

2

p

GM

3

e

⋅

R

m

R

s

of 32 51

13.10.

Where . The accuracy of this is

Thus the energy equations gave equation 13.7:

13.7. !

And equating the orbital equations gives

13.10.

These last two yields

13.11.

The accuracy is

Is 98% accuracy. is the radius of the star, which is the Sun, , so we write it

13.12.

ℏ

2

s

M

3

e

M

3

m

=

3

2

L

2

p

R

2

m

R

2

s

3/2 = cos(π /6)

(2.8314E 33)

2

(5.972E 24)

3

(7.34763E 22)

3

= (0.866)(9.13E 38)

2

(1737400)

2

(6.96E8)

2

4.29059E 72 = 4.5005E 72

4.29059E 72

4.5005E 72

100 = 95.3 %

L

p

=

M

2

e

M

⊙

M

3

m

3

⋅ ℏ

s

L

2

p

=

2 3

3

⋅

M

3

e

M

3

m

R

2

s

R

2

m

⋅ ℏ

2

s

2

R

s

R

m

M

e

M

⊙

= 1

(400.5986)

5.972E 24kg

1.9891E 30kg

= 1

0.98 = 1

R

s

R

⊙

2

R

⊙

R

m

M

e

M

⊙

= 1

of 33 51

14.0 Solving Habitable Planetary Systems

We want to solve a habitable planetary system in general and apply it to another star system. We

must include the solution of its moon as well, because the Moon of the Earth makes life

optimally possible. However, our Moon is counter-intuitive, it breaks all of the rules of the other

moons in our solar system. One is it has a low mass for its size. It has been said that it could be

hollow. Even suggested that it is a hollow-spacecraft put there to make life as we know it

possible. In our quantum theory for the solar system, we may have discovered the science of the

engineers who may have made the Moon, and we are possibly applying their science here. Our

quantum theory for the solar system also has a solar solution as compared to a lunar solution, so

it works for star systems in general.

The perfect star system to use for our purposes here would be Tau Ceti. Tau Ceti is a star

spectrally similar to the Sun and the closest of such a G-class star, at about 12 light years distant.

Its mass is 78% of the Sun’s. Its mass is

Its radius is:

And its luminosity is

It has several unconfirmed planets with one potentially in the habitable zone. It has been the

subject of science fiction literature. It is in the constellation Cetus. From Tau Ceti the Earth

would be seen to be in the northern hemisphere constellation Bootes with an apparent

magnitude of 2.6. It is spectral class G8 V where our Sun is G2 V.

For the luminosity of a main sequence star we only need to know the mass of the star to get its

luminosity:

1.

This holds in solar masses and solar luminosities. The habitable zone of a star is given by

luminosity. If the star is 100 times brighter than our Sun, then the habitable zone is 10 times

further than the Earth is from the Sun by the inverse square law.

2.

In astronomical units and solar luminosities. From we can get the orbital period in Earth

years:

3.

We can get the radius of the Moon of the planet if we know the mass of the planet, :

M

s

= 0.783 + / − 0.012M

⊙

R

s

= 0.793 + / − 0.004R

⊙

L

s

= 0.488 + / − 0.0010L

⊙

L = M

3.5

r

p

= L

s

r

p

T

2

p

= r

3

p

M

e

of 34 51

4.

We want the mass of the Earth from the mass of the Sun. We would guess we can get it from the

mass of the Sun, the size of the Sun, and the base 60 sexagesimal system upon which we have

found everything is based. I find the relationship exists for the Earth:

5.

(36)(5)=180=360/2 and 360/6=60 is base 60 sexagesimal. This gives

This is an accuracy of

We can get the orbital distance of the planet’s Moon from the planet:

6.

Once we have that we can get the radius of the Moon of the planet

7.

Now we can get the rotation period of the planet, its length of day:

8.

Because the orbital velocity of the planet’s Moon is

9.

We can use the the delocalization time get the moon’s mass, which is 1/2 the planet’s year, with

10.

2

R

⊙

R

m

M

e

M

⊙

= 1

M

e

=

5

6

2

M

⊙

R

2

⊙

r

2

e

M

e

= 0.13889(1.9891E 30kg)

6.96E8m

2

1.496E11m

2

= 5.979748E 24kg

5.972E 24

5.979748E 24

100 = 99.87 %

r

planet

r

moon

≈

R

star

R

moon

2

R

⊙

R

m

M

e

M

⊙

= 1

Ear th Da y =

2r

e

v

m

M

m

M

⊙

3

v

m

=

GM

e

r

m

τ =

m

moon

(2r

moon

)

2

ℏ

s

of 35 51

And the whole system is solved. Let us see how the theoretical luminosity appears with that

measured with equation 1:

It is close to the measured. The equation for predicting is approximate as the actual

values can vary a little with things like metallicity of the stars. Tau Ceti has low metallicity

compared to our Sun.

Let’s find the orbital distance of the habitable planet if it exists with equation 2:

Let’s find the orbital period of the habitable planet (It’s year) with equation 3:

,

=

Let’s get the mass of the habitable planet from equation 5:

=

Which is good because the planet, if it exists, is thought to be larger than the Earth. Let’s get the

radius of its moon from equation 4:

=

Let’s see at what distance it orbits the planet with equation 6:

L

s

= M

3.5

s

= 0.783

3.5

= 0.425L

⊙

0.448L

⊙

r

p

= 0.488 = 0.69857AU = (0.6986AU )(1.496E11m /AU ) = 1.045E11m eters

T

2

p

= r

3

p

T

2

p

= (0.69867AU )

3

T

p

= 0.584years(365.25 d ays)((24hrs)(60min)(60sec) = 1849638secon d s

(0.584)(365.25d a ys) = 213.306d ays

M

p

=

5

36

M

s

R

2

s

r

2

p

M

s

= (0.783)(1.9891E 30kg) = 1.5575E 30kg

R

s

= (0.793)(6.96E8m) = 5.52E8m

M

p

=

5

36

(1.5575E 30kg)

(5.52E8m)

2

(1.045E11m)

2

= 6.036E 24kg

6.036E 24

5.972E 24

= 1.0107Ear th Ma sses

R

m

= 2R

s

M

p

M

s

2(5.52E 8m)

(6.036E 24kg)

(1.5575E 30kg)

= 1.53656E6m =

1.53656E6m

1.7374E6m

= 0.8844Lu n arRa dii

of 36 51

=

Let’s get the orbital velocity of this moon with equation 9:

So the orbital period of the moon is:

=

Compared to that of the Earth which is 27.3 days for the sidereal month and is a 29.53day lunar

month. The Moon of Tau Ceti would have a slightly longer lunar month, too. Now we can

determine the rotation period of the planet which would be its day:

We get the mass of the moon from equation 10:

=

So the rotation period of the planet (Its Day) is from equation 8:

=

If the moon of Tau Ceti is to perfectly eclipse its star, like our Moon does with the Sun, to let its

inhabitants know that they are there for reason, and so as to theoretically be a condition for

optimizing life, then we have the following should hold:

r

moon

= r

planet

R

moon

R

star

= (1.045E11m)

1.53656E6m

5.52E8m

= 2.91E8m

2.91E8m

3.84E8m

= 0.7578L un ar Orbital Ra dii

v

m

=

GM

p

r

m

=

(6.67408E − 11)(6.0636E 24kg

2.91E8m

= 1,176.6m /s

T

m

=

2π r

m

v

m

=

2π (.91E8m)

1,176.6m /s

= 1.554E6secon d s

(1.554E6s)

min

60sec

h our

60min

d a y

24h ours

= 17.987d a ys

M

m

=

ℏ

s

τ

4r

2

m

=

(2.8314E 33J ⋅ s)(0.5)(18429638s)

4(2.91E 8m)

2

= 7.7E 22kg

7.7E 22kg

7.34763E 22kg

= 1.048Ear th Moon s

Pl a netDa y =

2r

p

v

m

M

m

M

s

3

2(1.045E11m)

1,176.6m /s

7.7E 2kg

1.5575E 30kg

= 39495.61secon ds = 658.26min = 10.971h ours

10.971

24h ours

= 0.457Ear th Da ys

r

planet

r

moon

≈

R

star

R

moon

of 37 51

We have

And it does hold. For the Earth/Moon/Sun system we get

For the Tau Ceti system the 359 is close to 360, which is a convenient amount of degrees into

which divide a circle, like we did. We see the Tau Cetians might do the same, because like us

they might choose the base 60 sexagesimal system of counting, and this 360 might influence the

way the inhabitants of a planet orbiting this star would make their calendar. Just like the 365

day year here on Earth corresponds closely to the 360 degrees of a circle. Aside from having here

on Earth the degree, we have gradians, of which there are 400 in a circle. They make it very easy

for computing right angles to a given angle. They are used more in Europe and in surveying and

architecture, though it might have some interesting connection with our 400=400.

The habitable planet of Tau Ceti would orbit it star with a period of 213 days, which would be its

year. It star is about 78% the mass of our Sun, and about 79% its size. Its habitable planet would

be a little larger than the Earth, but not much. Its Moon would be about 88% the size of our

moon, but a little bit more massive, but not much more massive. It would orbit Tau Ceti once

about every 18 days meaning there would be 11.833 months in the Tau Ceti Year, or about 12

months like we have. It would orbit the planet at about three quarters the distance ours does.

The day from sunrise to sunrise, or sunset to sunset would be about 11 hours, or close to half the

length of our day. This is if Tau Ceti was engineered for life, but it doesn’t have to have a planet

so optimally designed for life as ours. Our 24 hour day would be better for when entering the

realm of doing astronomy, because longer nights mean more complete observations of the

universe surrounding us than for the Tau Cetians. It may mean for the conditions for life to be

optimal, like here on Earth, the star would have to be as massive as the Sun, to provide a longer

year and a longer day. But we see here how the science of engineering planetary systems optimal

for life, might be done.

1.045E11m

2.91E8m

=

5.52E8m

1.53656E6m

359 = 359

400 = 400

of 38 51

15.0 Bohr Radius For Planets

The wave function for the hydrogen atom that is the solution of the Schrödinger wave equation

15.1.

is for the ground state, , that is angular momentum (l=0) is zero, only

depends on the radial component (r) is well established is:

15.2.

Where

15.3.

is the Bohr radius, which is the most probable distance for the electron in a hydrogen atom, that

is orbiting a single proton, in the ground state. Where is the permittivity of free space which is

Where the Farad is

Where Ampere (A) is 1 Coulomb (C) per second in SI base units. Since

the Bohr radius is

We want the ground state for the planetary system and we have found it quantizes according to

the mass of the Moon, because it is some kind of a Natural Yardstick, So we put for and use

our planetary Planck constant, and we find the ground state for the solar system is

15.4.

where we put the mass of Mercury because it is the planet n=1, and should be the ground state.

We should get is somewhere in the Mercury orbit. Mercury has a very elliptical orbit, so it is

between

−

ℏ

2m

e

(

∂

2

ψ

∂ x

2

+

∂

2

ψ

∂y

2

+

∂

2

ψ

∂z

2

)

− k

e

2

r

ψ = E ψ

n = 1,l = 0,m = 0

ψ

100

=

1

π

⋅

1

a

3/2

0

e

−r/a

0

a

0

=

4πϵ

0

ℏ

2

e

2

m

e

ϵ

0

8.854E − 12F/m

F =

1Coulom b

Volt

=

1

kg

1

m

2

s

4

⋅ A

2

k

e

=

1

4πϵ

0

a

0

=

ℏ

2

k

e

2

m

e

G

k

e

a

0

=

1000ℏ

2

s

GM

mercur y

M

2

moon

a

0

of 39 51

So let’s see if we get that:

=

Like this it tends towards aphelion at . But if we find the ground state in the solar

formulation we get:

15.5.

=

So in AU we have

The Mercury orbit is for its average . So our result has an accuracy of

The wave equation for the n=2 orbit is well established ( Would be Venus) would be

15.6.

Where and, for Earth, the n=3 orbit, would be

15.7.

The probability that an electron will be found in a given volume at a given time is

15.8.

If that volume is a sphere with radius r, then it is given by

per ih elion = 46,000,000k m

aph elion = 70,000,000k m

a

0

=

(1000)(2.8314E 33J ⋅ s)

2

(6.67408E − 11)(3.285E 23kg)(7.34763E 22kg)

2

6.77300E10m

7.00E 7k m

a

0

=

L

2

p

GM

2

mercur y

M

⊙

(9.13E 38J ⋅ s)

2

(6.67408E − 11)(3.285E 23kg)

2

(1.9891E 30kg)

= 5.818657E10m

1AU = 1.496E11m

a

0

=

5.818657E10m

1.496E11m /AU

= 0.389AU

r

m

= 0.387AU

0.387

0.389

100 = 99.5 %

ψ

200

=

1

4 2π

1

a

3/2

0

(2 −

r

a

0

)e

−r/2a

0

n = 2,l = 0,m = 0

ψ

300

=

1

81 3π

1

a

3/2

0

(27 − 18

r

a

0

+ 2

r

2

a

2

0

)e

−r/3a

0

P(r)dr = ∣ ψ ∣

2

dV

of 40 51

15.9

If it is the ground state, then since

We have

15.10.

This is

15.11.

Set this equal to 0 and you get . That is if you plot P(r) versus r you get the probability of

finding an electron at any r as shown in the graph below. The most probable radial position

occurs at the maxima when dP/dr=0. If you integrate P(r)dr from r1 to r2 to get the area under

the curve you get the probability of finding the electron in the region from r1 to r2.

You can put this integral in the Wolfram computational engine and integrate from the center of

the Sun to the orbit of Venus and you find the probability of finding Mercury in that region. If

you do, you get 100%, which is true:

dV = 4π r

2

dr

ψ

100

=

1

π

⋅

1

a

3/2

0

e

−r/a

0

P(r)dr =

4

a

3

0

r

2

e

−2r/a

0

dr

d P

dr

=

8re

−2r/a

0

a

4

0

(a

0

− r)

r = a

0

of 41 51

of 42 51

16.0. Explicitly The Quantization In Terms Of The Moon and 1 Second

Thus if the ground state of the Solar System is

16.1.

Where

.

Then if we put in the mass of the Moon for , we have

16.2

Then wavelength associated with the Moon divided by the speed of light should be 1 second if

our planetary system is quantized in terms of the Moon and one second. We have

16.3.

And we see it is, so we have

16.4.

The energy of a photon in terms of its wavelength is

16.5.

For the wavelength of the Moon we have

16.6.

We notice here we don’t use the reduced Planck constant , but the regular Planck constant . It

is given by

16.7.

We have

a

0

=

ℏ

2

s

GM

1

M

2

m

M

1

= Ma ssMer cur y

M

2

= Ma ssVenu s

M

3

= Ma ssEar th

M

1

λ

moon

=

ℏ

2

s

GM

3

m

=

(2.8314E 33)

2

(6.67408E − 11)(7.34763E 22kg)

3

= 3.0281E8m

λ

moon

c

=

3.0281E8m

299,792,458m /s

= 1.010secon ds

λ

moon

c

= 1secon d

E =

hc

λ

E

moon

=

h

s

c

λ

moon

ℏ

h

s

= 2π ℏ

s

= 1.779021E 34J ⋅ s

E

moon

=

(1.779021E 34J ⋅ s)(299,792,458m /s)

3.0281E8m

= 1.7612928E 34J

of 43 51

16.8.

We then have further in terms of the frequency associated with the Moon:

16.9.

Where is the frequency associated with the Moon is

16.10.

And we see the Solar System is quantized in terms of the Moon and 1 second. Another thing

where wavelength and frequency are associated with particles is the deBroglie Wave Length:

16.11.

Where is momentum given by velocity times mass. We could write in terms of the Moon:

16.12.

We see

16.13.

If we were to write equation 16.6 with instead of we would have

16.14.

We already had in equation 16.7. We have

We divide this by 2 and get

That is only because

λ

moon

c

=

h

s

E

moon

=

1.779021E 34J ⋅ s

1.761293E 34J

= 1.010secon ds

E

moon

= h

s

f

moon

f

moon

f

moon

=

c

λ

moon

= 1secon d

−1

λ =

h

p

=

h

mv

p

λ =

h

s

M

m

v

m

=

1.779021E 34J ⋅ s

(7.34767E 22kg)(1,022m /s)

= 2.3691E8m

λ

moon

λ

=

3.0281E8m

2.3691E8m

= 1.278

ℏ

s

h

s

E

moon

=

ℏ

s

c

λ

moon

=

(2.8314E 33Js)(299,792,458m /s)

3.0281E8m

= 2.803185E 33J

E

moon

= 1.7612983E 34J

1.7612983E 34J

2.803185E 33J

= 6.283204

6.283204

2

= 3.1416 = π

h

s

= 2π ℏ

s

= 1.779021E 34J ⋅ s

of 44 51

17.0 Electron Energy Levels, Planetary Orbitals, And Spin

The Bohr model of the atoms was formulated before the Schrödinger wave equation, but it is the

solution to the Schrödinger wave equation for the atom. Let’s see how Niels Bohr arrived at it,

then arrive at it for the planets. In the Bohr Model of the atom Niels Bohr made three postulates:

1. Electrons exist in stable orbits at discrete distances from the nucleus.

2. The angular momentum of the electron is quantized according integer multiples of the

reduced Planck constant.

3. Electrons only gain and lose energy by jumping from one orbit to another.

Thus to begin we equate the centripetal force of the electron to the force between two charges

given by Coulomb’s law:

1.

This gives a velocity for the electron of

2.

The angular momentum of the electron is quantized by integer multiples of the reduced Planck

constant:

3.

Putting the velocity from equation 2 into equation 3:

4.

Z=1 is the smallest radius possible for r and is the Bohr radius:

Putting the equation for v in the equation for kinetic energy:

5.

And, into that the equation for r, we get

6.

m

e

v

2

r

= Z

k

e

2

r

2

v = Z

k

e

2

m

e

r

m

e

vr = nℏ

r

n

=

n

2

ℏ

2

Z k

e

2

m

e

r

1

=

ℏ

2

Z k

e

2

m

e

= 5.29E − 11m = 52.9pm

E =

1

2

Mv

2

E = −

Z

2

(k

e

2

)

2

m

e

2ℏ

2

n

2

= −

13.6Z

2

n

2

eV

of 45 51

We now write equation 1 but not using Coulomb’s law but rather Newton’s universal law of

gravitation, We have

7.

We see that the orbital velocity of the Moon is

8.

The angular momentum of the Moon is

9.

Putting the orbital velocity of the Moon from equation 8 into equation 9 we have

10.

We know planetary systems are not like atomic systems, planets aren’t all the same size like

electrons are and like protons are, which have the same charge but planets don’t have the same

gravity because they don’t have the same mass either. Nor do planets jump from one orbit to

another like electrons do. Z is the atomic number, the number of protons at the nucleus of the

atom, but for planetary systems there is only the Sun at the center. We find that for this to work

with the that we formulated in this theory, that

And,…

Where for the Earth n=3 and further that the Moon near perfectly eclipses the Sun as seen from

the Earth meaning that

M

m

v

2

m

r

m

= G

M

e

M

m

r

2

m

v

m

=

GM

e

r

m

M

m

v

m

r

m

= nℏ

s

M

m

GM

e

r

m

r

m

= nℏ

s

M

2

m

GM

e

r

m

⋅ r

m

= n

2

ℏ

2

s

r

m

=

n

2

Z

⋅

ℏ

2

s

GM

2

m

M

e

ℏ

s

n

2

→

1

n

Z →

R

m

R

⊙

r

e

r

m

=

R

⊙

R

m

of 46 51

We see Z, the proton in the atom is the radius of the Sun, but it is divided by the radius of the

Moon to put it in Moon-units, that the Moon is like a natural yardstick. We further have that

So, our final equation is

11.

Let us go back to equation 5 to get the equation for the energy of the Earth in its orbit and the

equation for the velocity equation 8…

And we get

12.

Putting equation 11 for r3 into this

13.

We find that and we write

14.

And we find we multiply this by and we have

15.

The Bohr model of the atom was later reinterpreted by deBroglie where he said Bohr’s condition

that angular momentum be an integer multiple of is really that it is the condition of a standing

wave where

16.

That electrons behave as waves and that the wavelength of the electron is

M

e

→ M

m

r

3

=

R

⊙

R

m

⋅

ℏ

2

s

GM

3

m

⋅

1

3

E =

1

2

Mv

2

v

m

=

GM

e

r

m

E =

1

2

M

GM

e

r

m

E =

1

2

M

GM

e

⋅ GM

3

m

ℏ

2

s

M → M

e

E =

G

2

M

2

e

M

3

m

2ℏ

2

s

n

R

⊙

R

m

E

3

= 3

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

s

ℏ

n λ = 2π r

of 47 51

17.

Which is to say

18.

Or, that

19.

Where is angular momentum, the same units as which is . So we have

20.

Which is Bohr’s second postulate that angular momentum is quantized in integer multiples of

the reduced Planck constant.

Now, with what we have done up to this point finding the wave length of the Moon, a lot begins

to unfold. We have said in section 1.0 of this theory:

Thus with what we arrived at in section 16.0

And with

We have

21.

λ =

h

mv

nh

mv

= 2π r

nh

2π

= mvr

mvr

h

Joules ⋅ secon d s

L =

nh

2π

ℏ

s

= (hC )K E

p

hC = 1secon d

C =

1

3

⋅

1

α

2

c

1

3

⋅

2π r

p

G m

3

p

K E

m

K E

e

(Pl a netDa y) = 1.0secon d s

λ

moon

c

= 1secon d

K E

moon

1.0secon ds

(Ear th Da y) = K E

earth

ℏ

s

= (1secon d )K E

earth

of 48 51

22.

23.

24.

25.

26.

Thus the planet day, or spin, is given by

27.

Where

28.

This has to do with the concept of spin in quantum mechanics. The Earth day can be thought of

as spin. This is the rotation period of the Earth, which is different than angular momentum of

something orbiting something, it is more the angular momentum of something spinning on its

axis, but this is a messy subject in quantum mechanics. It was found not just the orbit of an

electron was necessary to describe, but as well the spin of an electron on its axis. However, when

you describe the rotation of an electron classically, its spin becomes faster than the speed of

light, so it doesn’t work. It became necessary to introduce spin not as something that is actually

happening, but as an abstract concept that represents something that is not concrete. So they

came up with the concept of spin numbers, numbers that can only be given by spin, s=n/2 where

n is an integer.

Similarly, spin for planets becomes difficult to apply. Planets close into the Sun, like Venus, are

tidally locked in with the Sun, their rotation (or spin) being greatly slowed down by the Sun’s

gravity. The rotation of Venus, closer in than the Earth, has a rotation (day) of 243 days. The

Earth of course is 24 hours. Mars is close to the Earth day. The Earth sidereal day is 23 hours 56

minutes 4.09 seconds, the Mars sidereal day is 24 hours 37 minutes 22 seconds. The Jupiter day

is 9 hours 56 minutes, and the Saturn day is 10 hours 34 minutes. This has to do with the spin

concept because

(K E

moon

)(Ear th Da y) = ℏ

s

ℏ

s

=

λ

moon

c

⋅ K E

earth

K E

earth

=

ℏ

s

1secon d

K E

m

K E

e

(Pl a netDa y) =

λ

moon

c

λ

moon

=

ℏ

2

s

GM

3

m

=

(2.8314E 33)

2

(6.67408E − 11)(7.34763E 22kg)

3

= 3.0281E8m

(Pl a netDa y)

3

=

K E

3

K E

m

(1secon d )

K E

1

= M er cur y, K E

2

= Venu s, K E

3

= Ear th, K E

4

= Mars, . . .

K E

m

K E

e

(Pl a netDa y) =

ℏ

2

s

GM

3

m

⋅

1

c

(Pl a netDa y)

3

=

K E

3

K E

m

(1secon d )

of 49 51

We have to find x for x-seconds, the number of seconds that describe the planet’s day. We can’t

consider the spins of Mercury, and Venus because their orbits are so slowed down by tidal forces

from the Sun, Venus even spins in the opposite direction of its orbit. It starts working at Earth

orbit, where life is successful. Equation 27 is the case for the Earth and works quite well. We

postulate that since the gas giants, Jupiter and Saturn, carry the better part of the solar system’s

mass they embody the dynamics behind the idea behind the solar system. We start with Jupiter,

and its largest Moon, Ganymede. It is the third significant Moon around Jupiter and we

postulate the third orbit is where something significant happens, as the with Earth, the third

planet from the Sun. Ganymede has an orbital velocity of 10.880 km/s and its mass is 1.4819E23

kg. Its eccentricity is very low (0.0013) meaning like our Moon has an almost perfectly circular

orbit. We have its kinetic energy is

We have for Jupiter its day is

(9 hours)(60)(60)+56(60)=35,760 seconds

The kinetic energy of Jupiter is

We have

Spin in quantum mechanics is given by s=n/2 where n is an integer giving 1/2, 1, 3/2, 2, 5/2.

Thus spin of Earth is 1 (1 second) we skip Mars (3/2), go to Jupiter is 2 (2 seconds), so Saturn

should be 5/2 (2.5 seconds).

Titan is the largest moon of Saturn and the second largest in the solar system. It is mass is

1.3452E23 kg and its orbital velocity is 5.57 km/s. We have its kinetic energy is

K E

1

= M er cur y, K E

2

= Venu s, K E

3

= Ear th, K E

4

= Mars, . . .

K E

G

=

1

2

(1.4819E 23kg)(10,880m /s)

2

= 8.77095E 30Joules

1

2

(1.898E 27kg)(13,060m /s)

2

= 1.61865E 35Joules

(Pl a netDa y)

5

=

K E

5

K E

G

(x − secon ds)

35,760s =

1.61865E 35J

8.77095E 30J

(x − secon ds)

35,760s = (18,454.67)x

x = 1.9377secon d s ≈ 2secon d s

(Pl a netDa y)

3

=

K E

3

K E

m

(1secon d )

(Pl a netDa y)

5

=

K E

5

K E

G

(2secon ds)

of 50 51

We have for Saturn its day is

(10 hours)(60)(60)+(34)(60)=35,760 seconds

The kinetic energy of Saturn is

Exactly what we predicted. We have:

We can’t consider Venus, its orbit is so slowed down from tidal forces of the Sun. We have

Mars has no Moons that we can use for the calculation, they are small and irregular, and

perhaps captured passing asteroids. This concept of spin, worked very nicely for the planets.

K E

T

=

1

2

(1.3452E 23kg)(5,570m /s)

2

= 2.08673E 30Joules

1

2

(5.68319E 26kg)(10,180m /s)

2

= 2.945E 34Joules

35,760s =

2.945E 34J

2.08673E 30J

x

35,769s = (14,113)x

x = 2.53447secon d s ≈ 2.5secon d s

(Pl a netDa y)

3

=

K E

3

K E

m

(1secon d )

(Pl a netDa y)

5

=

K E

5

K E

G

(2secon ds)

(Pl a netDa y)

6

=

K E

6

K E

T

(5/2secon d s)

n /2 = 2/2 = 1secon d(Ear th)

n /2 = 3/2 = Mars

n /2 = 4/2 = 2secon d s(Jupiter)

n /2 = 5/2 = 2.5secon d s(Sat ur n)

of 51 51

The Author