of 1 38
Life As A Natural Property of the Universe
By
Ian Beardsley
Copyright © 2024 by Ian Beardsley
of 2 38
Contents
The Theory…………………………………………………………….4
The Value of the Planetary Planck Constant……………..10
The Strange Duration of a Second……………………………10
Yellow G2V Stars With Habitable Planets
May Be A Property Of The Universe…………………………11
Arriving At The Planetary Planck Constant
From The Proton Radius………………………………………..16
Finding The Delocalization Time For Earth……………..19
The Computations For All of the Planets………………….21
The TOI 700 Planetary System……………………………….28
of 3 38
Here I present my findings that the solar system mirrors atomic systems and has a quantum
mechanical solution to the Schrödinger wave equation based on a base unit of 1 second. I further
suggest that this solution for our spectral class G2V star the Sun, may indicate that life is a
uniform cosmic unfolding throughout the Universe. In order to test whether this works for many
G2V stars like our Sun we would need to detect Earth sized planets around other such stars, to
see if my Planck constant works for other planetary systems, which I show is connected to my
theory for the radius of a proton in terms the Natural constants such as the gravitational
constant for gravity and the Planck constant for atoms. Though we measure time with the
duration of a second and it came from no idea in Nature but rather how we divided up the day
into units since ancient times, it turns out it is a natural duration in this theory and I suggest it is
because of how we divided up the day and year from observing the apparent motions of the Sun
and stars due to the Earth’s motion, that we may have inadvertently had an idea in this that was
Natural in terms of the Nature pf atoms and the Earth/Moon/Sun system. Our theory also
seems to based on a sixfold symmetry based around the skeletons of life chemistry, the
hydrocarbons.
I further develop an equation of state for the periodic table of the elements and suggest an
intermediary mass between that of the proton and the stars. This results in a constant k that is
characteristic of the universe and is in inverse velocity. With these we show it may be that yellow
G2V stars with a planet in the habitable zone may be a property of the Universe. I find that the
Earth/Moon/Sun system gives the eigenvalue or basis vector for Natural Systems predicting the
life span of the Universe in the standard cosmological model.
At the end of the paper we apply our theory to another star system, one that is the only one we
have detected so far with an Earth-sized planet in it habitable zone. Unfortunately this star is
not a yellow G2V star like our Sun, but a red dwarf M star. But we get a feeling for how to apply
this theory.
The central theme of this paper is that life may be a Natural property of the Universe.
of 4 38
The Theory My findings are that a G2V main sequence star, like our Sun has a habitable planet
in the orbit given by
1.
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
2.
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
3.
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
4.
5.
Where
6.
Equation 1 is the solution to Schrödinger’s wave equation
7.
Which put in spherical coordinates is
n = 3
K E
p
= n
R
s
R
m
⋅
G
2
M
2
p
M
3
m
2h
2
s
K E
p
M
p
M
m
h
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
h
s
h
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 5 38
8.
Just as
9.
is the solution to the Schrödinger wave, equations 7 and 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
10.
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 and I present how I arrived at it starting page 16 Arriving At The
Planetary Planck Constant From The Proton Radius. 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 and 9 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
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, 459m /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
2h
2
s
E = −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
of 6 38
Since for our solar system
That is, the Sun is 400 times the size of the Moon, we see we can write equation 1, that is the
equation for the Earth, in terms of the atomic number which is in the the equation for the
atom, as
11.
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 3, 5, and 6, yield the radius of a proton accurately
12.
Interestingly we can write
13.
14.
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 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. Some of the output looks like
R
s
R
m
= 400
Z
K E
p
= n Z
2
Ca
⋅
G
2
M
2
p
M
3
m
2h
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
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 d s = hydr ogen(H )
m
p
of 7 38
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
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=299792459,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 8 38
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
15.
So, for Jupiter we have (And again using the maximum orbital velocity which is at perihelion):
Jupiter is mostly composed of hydrogen gas, and secondly helium gas, so it is appropriate that
.
Our equation for Jupiter is
16.
Where is the atomic number of hydrogen which is 1 proton, and for the orbital
number of Jupiter, . Now we move on to Saturn…
E =
Z
n
G
2
M
2
m
3
2h
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
E =
Z
H
5
G
2
M
2
j
M
3
m
2h
2
s
Z
H
n = 5
n = 5
K E
S
=
1
2
(5.683E 26kg)(10140m /s)
2
= 2.92E 34J
of 9 38
=
The equation for Saturn is then
17.
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
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:
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
2h
2
s
K E
moon
K E
earth
(Ear th Da y) = 1.08secon d s
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon ds
r
planet
r
moon
≈
R
star
R
moon
of 10 38
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.
The Value Of The Planetary Planck Constant
=!
!
= !
= !
!
!
!
= !
!
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
K E
p
= n
R
s
R
m
⋅
G
2
M
2
p
M
3
m
2h
2
s
R
star
R
moon
=
R
s
R
m
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
Gm
3
p
1
3
⋅
18769
299792459
1
3
⋅
2π (0.833E − 15)
(6.67408E − 11)(1.67262E − 27)
3
1.5997656E33
s
m
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)(1.5997656E 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
h
s
= (hC )K E
earth
= (1.03351s)(2.7396E33J ) = 2.8314E 33J ⋅ s
of 11 38
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:
18.
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:
19.
Which can be written
20.
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
Yellow G2V Stars With Habitable Planets May Be A Property Of The Universe
It may be that yellow G2V stars with a planet in the habitable zone may be a property of the
Universe. I find that The Earth/Moon/Sun system gives the eigenvalue or basis vector for
Natural Systems, where the eigenvalue is the value that maps something into itself (here 1
second to 1 second): !
21. !
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
hC =
1
3
⋅
h
α
2
c
1
3
⋅
2π r
p
G m
3
p
= 1.0secon ds
h
G
⋅
c
3
m
p
2 3M
m
(
1
k
)
2
(1secon d ) = (1second )
of 12 38
Where !
22. !
is a basic energy characteristic of the Universe. is the mass of the Earth’s moon, is a
constant characteristic of the Universe in terms of velocity, and 1 second is given by!
23. !
where is the fine structure constant, is the mass of a proton, is the radius of a proton,
is the universal constant of gravitation, is Planck’s constant, and is the speed of light.
The following holds for the Earth, and may be characteristic of other habitable star systems:!
24. !
These habitable systems would be characterized by the integer 6, and the orbital velocity of
the habitable planet in terms of the constant , which is true of Earth:!
25. !
But the constant is can also be derived by hypothesizing an intermediate mass characteristic
of the Universe that is the geometric mean between the mass of a proton and the
Chandrasekhar limit for a white dwarf star to not collapse into a blackhole:!
26. !
The Chandrasekhar limit is given by!
27. !
28.
Where we have approximated 0.77~3/4 in equation 27.
29.
h
G
⋅
c
3
m
p
= 1.599298E 29J
M
m
k
1
6α
2
m
p
h4πr
2
p
Gc
= 1.004996352secon ds
α
m
p
r
p
G
h
c
KE
moon
KE
earth
(Ear th Day) = 1.08secon d s
k
k v
e
= 6
k
m
i
= Mm
p
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼 =
1
773.5
s
m
of 13 38
30.
31.
32
is any element and is the number of protons in times Avogadro’s number per that
number of grams. would be carbon is 6 grams per 6 protons.
We can hone this by reintroducing 0.77 for 3/4
Thus precisely:
We have
We have honing our
33.
Thus since we said with our estimate
we have a honed value of
34.
To derive this we introduce Giordano’s relationship:
N
A
= 6E 23
proton s
gr a m
ℍ = 1
gr a m
proton
N
A
⋅ 𝔼 = 6E 23
𝔼
N
A
𝔼
ℂ
3
2
= 0.8660
0.77 = 0.8775
0.8775
0.8660
= 1.01328
m
i
m
i
= (67.9943)(1.01328) = 68.897kg ≈ 69kg ≈ 70kg
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼 =
1
773.5
⋅
s
m
k =
1
(68.897kg)
2
(6.62607E − 34) ⋅
(1.007299)
6.67408E − 11
⋅ 6.02E 23 = 0.001268291s /m
1
k
= 788.4626
m
s
h(1 + α) ⋅ 10
23
= G
of 14 38
The number on the left is the number on the right, but with different units. I found I could
eliminate the and gain six-fold symmetry by creating an equation of state for the periodic
table of the elements using Avogadro’s Number ( ). We say that
And,
Which is basically true, hydrogen is one proton and the mass of the electron is nominal. For
every 6E23 protons of hydrogen, there is one gram. Then we always have
We can say for any element
where is the number of protons in the element, so for carbon
Thus we have
35.
And we have introduced our 6 of six-fold symmetry. If we take equation 21 for the Earth/Moon/
Sun system as the eigenvalue, which is given by the mass of the moon:
21.
10
23
6.02E 23
N
A
= 6E 23
proton s
gr a m
ℍ = 1
gr a m
proton
N
A
⋅ 𝔼 = 6E 23
𝔼
N
A
=
Z ⋅ 6E 23pr oton s
Z ⋅ gr a m s
𝔼 =
Z ⋅ gr a m s
Z ⋅ proton s
Z
ℂ
ℂ =
6gr a m s
6proton s
N
A
=
6(6E 23proton s)
6gr a m s
N
A
ℂ = 6E 23
h
(1 + α)
G
⋅ N
A
𝔼 = 6.00kg
2
⋅
s
m
h
G
⋅
c
3
m
p
2 3M
m
(
1
k
)
2
(1secon d ) = (1secon d )
of 15 38
and we substitute for the mass of the Moon our intermediary mass , we get the life span of the
universe in the Friedmann model of the Universe, which is 1E14 years:
36.
Where,
23.
Data And Computations
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
Earth day=(24)(60)(60)=86,400 seconds.
The mass of the Moon is
Using the Moon’s orbital velocity at aphelion, and Earth’s orbital velocity at perihelion we have:
We can even use not the solar day used here (rotation of Earth with respect to the Sun) but the
sidereal day (rotation of Earth with respect to the stars) and the equation has even slightly more
accurate results. Our computation are:
m
i
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
(1secon d ) = Li feSpanUniverse
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
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, 459m /s
α : 1/137
q
p
= q
e
= 1.6022E − 19coulom bs
k
e
= 8.988E 9
Nm
2
C
2
7.34767309E 22kg
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
h
G
⋅
c
3
m
p
2 3M
m
(
1
k
)
2
(1sec) =
1.599298E 29J
(3.464)(7.34767E 22kg)(788.4626
m
s
)
2
(1sec) =
(3.50secon d s)
3.464
= 1secon d
of 16 38
(365.25)(24)(60)(60)=31557600 seconds/year
This agrees with the life span of the Universe as modeled in the standard cosmological model,
the Freedman model based on the mass in the universe and from that when the last stars formed
will burn out. would come in because 3 is the earth orbital number and the 2 because when
solving for end states we have kinetic energy equals potential energy and since the two are
added for the kinds of energy of the system at different times we have twice the energy given in
the equation.
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
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
(1sec) =
1.599298E 29J
(68.897kg)(788.4626
m
s
)
2
(1sec) = 3.734E 21secon ds
3.734E 21secon d s
31557600sec /yr
= 1.182E14years
2 3
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
m
p
c
2
h
⋅
r
p
c
=
2
3
≈ ϕ =
m
p
c
h
r
p
of 17 38
The radius of a proton is then
Equation 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. If Equation 3.
We find
Equation 4.
Satisfies this condition.
We write
Or,
Equation 5.
We multiply equation 1 by to get
m
p
r
p
=
2
3
⋅
h
c
r
p
=
2
3
⋅
h
cm
p
h
⊙
h
p
h
⊙
h
p
C
KE
e
h
⊙
= (h
p
C )KE
e
h
p
C = 1second
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
Gm
3
p
h
p
C = t
1
t
1
=
1
3
⋅
h
p
α
2
c
1
3
⋅
2π r
p
Gm
3
p
r
p
/r
p
r
2
p
=
2
3
⋅
hr
p
cm
p
r
p
=
2
3
⋅
hr
p
cm
p
of 18 38
=
Multiply the radicand by to get
Multiply the radicand by to get
=
Multiply outside the radical by to get
This factors into
Equation 6.
We have
Equation 7.
Where . Since this views the Earth-Sun system as a quantum mechanical
system analogous to that which describes an atom, we include the Moon and find its
orbit with respect to the Earth orbit is . I find that the following holds
2
hr
p
cm
p
⋅
1
6
h
h
⋅
m
2
p
m
2
p
⋅
c
c
r
p
= 2
h
2
m
2
p
c
2
⋅
r
p
c
m
3
p
h
⋅
1
6
2πG /2πG
r
p
= 2
h
2
m
2
p
c
2
⋅
r
p
c
m
3
p
h
⋅
2πG
12πG
2h
m
p
c
r
p
c
m
3
p
h
⋅
2π
3
⋅
1
4π
⋅
G
G
α
2
/α
2
r
p
=
2h
α
2
c
⋅ α
2
m
p
1
3
⋅
2π r
p
Gm
3
p
⋅
Gc
4πh
r
p
=
1
3
⋅
h
α
2
c
1
3
⋅
2π r
p
Gm
3
p
(
18
3
α
2
m
p
Gc
4πh
)
r
p
= 6α
2
m
p
Gc
4πh
⋅ t
1
t
1
= 1secon d
t
1
= 1secon d
of 19 38
Equation 8.
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 9.
Using a more recent value of the proton radius 0.833E-15m.
Finding The Delocalization Time For Earth
The delocalization time for the the Earth should be 6 months, 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 solar) for
the Earth/Moon/Sun system then apply it to predict the delocalization time for the Moon in its
orbit with the Earth around the Sun.
We consider a Gaussian wave-packet at t=0:
We say that is the delocalization length and decompose the wave packet with a Fourier
transform:
is the harmonics of the wave function. We use the identity that gives the integral of a
quadratic:
Solve the equation
With the initial condition
KE
moon
KE
earth
(Ear th Day) = 1.08secon ds
KE
moon
KE
earth
Ear th Day
1
6α
2
m
p
h4πr
2
p
Gc
= 1.004996352 seconds
h
⊙
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)
of 20 38
A plane wave is the solution:
Where,
The wave-packet evolves with time as
Calculate the Gaussian integral of
and
The solution is:
where
is the delocalization distance, which for instance could be the width of an atom. is the
delocalization time, the average time for say an electron to traverse the diameter of the atom and
even leave it, to delocalize. If we substitute for our , and say that the delocalization distance
is for the Moon, the width of the Earth orbit, we should get a half a year for the delocalization
time, the time for the 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
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
d
τ
ℏ
h
s
τ =
m
moon
(2r
moon
)
2
h
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 %
of 21 38
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.
The Computations For All of the Planets
We have the solution for the Bohr hydrogen atom is
1.
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
2.
Let is the earth mass, is the lunar mass, and is our
Planck constant for the Earth/Moon/Sun system. We compute equation 7:
=
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
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
h
⊙
E = −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
Z
n
E =
G
2
M
2
m
3
2h
2
⊙
n
2
M = M
e
m = M
m
h
⊙
= 2.8314E 33J ⋅ 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
of 22 38
Thus
4.
We find the quantum mechanical solution to the Earth/Moon/System is
5. !
for Earth orbit (Third Planet).!
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.
So let us move onto Venus, and guess its orbital energy is quantized in terms of the Earth’s
moon as well. We have equation
6.
was equal to 3, the earth orbital number, now it is 2 for Venus. From all that has been said we
can write it in terms of calcium:
7.
We will write the equation for Venus in terms of copper (Cu):
8.
The mass of Venus is . Why choose copper? As it would turn out the
condition for the Earth to perfectly eclipse the Sun, which is
This condition holds because , is to say the solar radius to the Earth
orbital radius is the mass of a gold atom to a silver atom, or equivalently the molar mass of gold
to the molar mass of silver. The other one that works for this is indium to copper:
6.96E8m
1737400m
= 400.5986
R
⊙
R
m
= 400
E = n
R
⊙
R
m
⋅
G
2
M
2
m
3
2h
2
⊙
n = 3
E = 3
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2h
2
⊙
n
E = 3Z
2
Ca
⋅
G
2
M
2
e
M
3
m
2h
2
⊙
E = 2Z
2
Cu
⋅
G
2
M
2
v
M
3
m
2h
2
⊙
M
v
= 4.867E 24kg
r
e
r
m
≈
R
⊙
R
m
R
⊙
/r
m
= 1.8 = Au /Ag
of 23 38
Copper is in the same group in the periodic table of the elements as gold (Au) and silver (Ag)
since we are going from Earth to Venus, n=3 to n=2, then the next logical element up the group
is copper (Cu). We want to suggest n=2 (Venus) is actually the first planet because mercury
(n=1) is so small and has such an irregular orbit that it really is not part of the equation. Indeed
in the Titius-Bode rule that predicts the orbits of the planets, it does not take Mercury as an
integer for its orbit, but rather for the equation to work, and the rest are integers. So here,
Mercy is just a placeholder for n=1. We have for Venus
= (2.61E30J)=3.1E33
The kinetic energy of Venus is
Where we have used the kinetic energy at perihelion (the maximum orbital velocity) because
that is what we used with Earth. The accuracy is
The result for Earth was
=
And, the kinetic energy of the Earth is
That’s an accuracy of
Now moving onto Mars we have
R
⊙
r
m
=
In
Cu
= 1.8
−∞
E = 2(29
2
)
(6.67408E − 11)
2
(4.867E 24kg)
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
2(841)
K E
v
=
1
2
(4.867E 24kg)(36260m /s)
2
= 3.1999E 33J ≈ 3.2E 33J
3.1E 33
3.2E 33
100 = = 96.875 %
E = 3(20
2
)
(6.67408E − 11)
2
(5.972E 24kg)
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
(692.82)(3.93E 30J ) = 2.72278E 33J
K E
e
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
2.72278E 33
2.7396E 33J
100 = 99.386 %
K E
mars
=
1
2
(6.39E 23kg)(26500m /s)
2
= 2.2437E 32J
of 24 38
It works great that the element here is tin (Tn) because so far we have had the involvement of
gold (Au), silver (Ag), and copper (Au) all the metals of metallurgy and electronics most used.
Copper was associated with Indium (In) and tin (Tn) comes right after it in the same period of
the periodic table. We have
9.
We have covered the terrestrial planets with reasonably circular orbits and substantials masses.
We have not considered Mercury as it is in the Titus-Bode Rule where the other
planets are integers, though it might actually work here and have the eccentric orbit it has for a
reason that makes it work with our equation. We could also go on to consider the gas giants,
Jupiter, and Saturn, and the ice giants Uranus, and Neptune. Considering the asteroid belt, it
would be difficult. The Titius-Bode Rule includes it (It is between Mars and Jupiter) but then it
does not require mass in its input and we do. So let’s move on to the the gas giants Jupiter and
Saturn, then the ice giants Uranus and Neptune.
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 copper and tin, while the equation for the giants has the
atomic numbers of the gasses. We write for these planets
10.
So, for Jupiter we have (And again using the maximum orbital velocity which is at perihelion):
E = 4(Z
2
)
(6.67408E − 11)
2
(6.39E 23kg)
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
= 4Z
2
= 4.4998863E 28J
2Z
2
(4.4998863E 28) = 2.2437E 32J
Z = 49.93 = 50proton s = t in(T n)
E = 4(Z
2
Tn
)
G
2
M
2
mars
M
3
moon
2h
2
⊙
n = − ∞
E =
Z
n
G
2
M
2
m
3
2h
2
⊙
K E
j
=
1
2
(1.89813E 27kg)(13720m /s)
2
= 1.7865E 35J
of 25 38
Jupiter is mostly composed of hydrogen gas, and secondly helium gas, so it is appropriate that
.
Our equation for Jupiter is
11.
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
12.
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.
Now we move on to the ice giants, Uranus and Neptune. They are composed of mainly water,
methane, and ammonia, so they should have the elements nitrogen (N) from ammonia (NH3)
and carbon (C) from methane (CH4). We have for Uranus (n=7)
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
E =
Z
H
5
G
2
M
2
j
M
3
m
2h
2
⊙
Z
H
n = 5
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
2h
2
⊙
K E
U
=
1
2
(8.681E 25kg)(7130m /s)
2
= 2.21E 33J
of 26 38
=
So Z is nitrogen for ammonia (NH3) just as we guessed. The equation for Uranus is then
13.
Now we conclude with Neptune (n=8)…
=
=
We have to multiply this by , the golden ratio, to get a whole number of protons. We
have
This is what I guessed because Neptune is nitrogen (N) of ammonia (NH3) so this should be
carbon (C) of methane (CH4). Neptune like Uranus, is an ice giant primarily composed of water,
methane, and ammonia.
We see that the terrestrial planets in our equations are characterized by the heavier elements, by
the metals, and that the gas giants and ice giants by the lighter elements like hydrogen and
helium. This makes sense because our equation needs to take into account their compositions to
account for their different masses, and thus sizes. Which it does nicely, and exactly how we
would guess it would. We see the Moon of the Earth is the key factor in terms of which the orbits
of the planets satisfy a Schrödinger type wave equation for the atom. This is the Moon of the
Earth, which is the n=3 orbit and habitable zone, it is this habitable zone for a G2 spectral class
star that is the basis for a Schrödinger type equation for the planets, which is characterized by
calcium (Ca) which is prominent in all G2 stars giving them calcium (II) spectral lines.
E =
Z
7
(6.67408E − 11)
2
)(8.681E 25kg)
2
(7.347673E 22)
3
2(2.8314E 33)
2
Z(3.139E 32)
Z = 7.038proton s ≈ 7proton s = Nitrogen(N )
E =
Z
N
7
G
2
M
2
u
M
3
m
2h
2
⊙
K E
N
=
1
2
(1.024E 26kg)(5470m /s)
2
= 1.532E 33J
E =
Z
8
(6.67408E − 11)
2
(1.024E 26kg)
2
(7.347673E 22)
3
2(2.8314E 33)
2
Z
8
(1.15558E 33J )
Z(4.0856E 32J )
Z(4.0856E 32J ) = (1.532E 33J )
Z = 3.75proton s
Φ = 1.618
Z = (3.75)Φ = 6proton s = carbon(C )
of 27 38
Our equation for Neptune is
14.
In conclusion the solution to the Schrödinger wave equation for atoms is
For the Earth, which is the basis set of the rest of the planets is
In General for the inner, terrestrial planets
Where Z is the atomic numbers of the most prominent metals for metallurgy and electronics.
For the outer, gas giants, and ice giants the equation is in general:
Where Z is the atomic numbers of the lighter gas elements from which these planets are
primarily made.
Now we apply our theory to an Earth sized planet in another star system. It is the first such
planet we have detected, but it orbits and M-type red dwarf star, and different kind of a star than
our Sun. M-type stars are the coolest and smallest, so the Planck constant for it we shall see may
have to be different…
E =
Z
C
8
G
2
M
2
n
M
3
m
2h
2
⊙
E = −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
E = 3
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2h
2
⊙
E = n Z
2
⋅
G
2
M
2
M
3
m
2h
2
⊙
E =
Z
n
G
2
M
2
M
3
m
2h
2
⊙
of 28 38
The TOI 700 Planetary System
Now we turn our attention to the TOI 700 Planetary System…It is an M type (red dwarf) star
101.4 ly from Earth (It is close) and harbors the first Earth sized planet in a habitable zone
(There are two) with 4 planets all together in its habitable zone. The star is in the Southern
Hemisphere constellation Dorado. Two of the planets are in the optimistic habitable zone (TOI
700 d and e). TOI 700 is about 40% the solar mass, about 40% the solar radius, and 55% the
solar temperature. The planets were detected by the Transiting Survey Satellite Object of
Interest (TESS).
We would begin by computing the kinetic energy of planet TOI 700 e, which is close to the same
size of the Earth, and like the Earth may be the third planet, is in the habitable zone, and more
than likely rocky like the Earth and can have surface water like the Earth, and therefore life.
From that we can get the Planck constant for a red dwarf star planetary system like this one. We
have the orbital velocity of this planet is
v =
2π a
T
a = (0.1340AU )(1.496E11m /AU ) = 2.00E10m
T = (27.80978d a ys)(24hrs)(60min)(60sec) = 2,402,765sec
v = 52,300m /s
of 29 38
Compared to the orbital velocity of the Earth which is 29,780m/s. We need the mass of the
planet which is:
So the kinetic energy of the planet is
Now we compute the Planck constant for such a red dwarf star system with equations 4, 5, and
6:
4.
5.
6.
And hC is exactly 1.03351 seconds. We have
Since the solution to the wave equation for terrestrial planets is equation 1:
1.
We have
We have said that the condition for a very successful habitable planet might be equation 10
10.
Which we should write more compactly
This equation holds for the Earth/Moon/Sun system. The other thing that is true of the Earth/
Moon/Sun system is that
M
p
= (0.818)(5.972E 24kg) = 4.8851E 24kg
K E
p
=
1
2
(4.8851E 24kg)(52,300)
2
= 6.6812E 33J
h
s
= (hC )K E
p
hC = 1secon d
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
G m
3
p
h
s
= (6.6812E 33J )(1.03351s) = 6.9E 33J ⋅ s
K E
p
= n
R
s
R
m
⋅
G
2
M
2
p
M
3
m
2h
2
s
M
3
m
=
K E
p
n
⋅
2h
2
s
r
m
G
2
M
2
p
r
p
r
planet
r
moon
≈
R
star
R
moon
r
p
r
m
≈
R
s
R
m
of 30 38
Where Au is the molar mass of gold, and Ag is the molar mass of silver, the most precious of
metals, used since ancient times to make ceremonial jewelry, and being the best conductors of
electricity are important to electronics engineering. This gives us
We have everything for
=
So the mass of the moon holding this planet to the inclination of its orbit so necessary for
advanced life, would be
Compared to the mass of our moon, which is 7.3E22kg. So it is more massive, but that is good
because it is also bigger to eclipse its star, as does our moon with the Sun. From
We have
Compared to the radius of the Earth’s moon which is 1.74E6m. The density of this moon would
be
Compared to the density of the Earth’s moon of 3.34g/cm3. The density of the Moon for TOI
700 e could then be mostly composed of silicates like the Earth’s Moon, but with a bigger iron
core than that of the Earth’s moon. This is on there order of the density of germanium, which is
right below silicon in the period table, which is in turn right below carbon.
R
s
r
m
≈
9
5
≈
Au
Ag
r
m
=
5
9
R
s
= 1.624E8m
M
3
m
=
K E
p
n
⋅
2h
2
s
r
m
G
2
M
2
p
r
p
(6.68E 33J )2(6.9E 33)
2
(1.624E8m)
3(6.67408E − 11)
2
(4.885E 24kg)
2
(2E10m)
=
4.894085E 70
3
= 2.8256E 70kg
3
M
m
= 3.0458E 23kg
R
m
= R
s
r
m
r
p
R
s
= (0.42)(6.96E8m) = 2.9232E8m
R
m
= (2.9232E8m)
1.624E8m
2E10m
= 2.37E6m
V =
4
3
π R
3
m
= (2.37E6)
3
4
3
π = 5.576E19m
3
ρ
m
=
3.0458E 23kg
5.576E19m
3
= 4.04E 3
kg
m
3
= 5.462
g
cm
3
of 31 38
Discussion
We have predicted the possibility of a moon similar in size, mass, and composition as the Earth’s
moon. The wave equation solution we found for the Earth solar system uses the mass of the
Earth’s moon, and we have further pointed out that it is believed that such a moon is necessary
for a planet to be abundantly and complexly suitable for life. The TOI 700 system is close to
Earth (~100ly) which is crucial for getting accurate physical data on a star system. Changes in
motion of a star due to a planet orbiting a star must be measured to attain orbital data of the
planet, and such changes get increasing difficult to measure with greater and greater distances
to the star system. For this reason the TOI 700 planetary system is perfect for obtaining data on
exoplanetary systems. We can determine the orbital data of planets orbiting this star to good
accuracy and it is the first and only system where we have detected an Earth-sized planet in the
habitable zone. For the theory in this paper we want to detect a moon orbiting the planet, and
that is much more difficult to do than detecting the planet orbiting the star. But, with the advent
of the James Webb telescope I think it might be possible to do, since it is being ramped up to try
and detect the atmospheric composition of exoplanets. We can measure the change in motion of
a star being pulled on by a planet, but the question becomes whether in the near future we can
measure the change in motion of a planet due to a moon orbiting it. Nicely, TOI 700 e is not just
Earth-sized, and in the habitable zone, but is planet n=3 like the Earth so far as we know. If
future planets, or bodies are found, orbiting closer in to TOI 700 e it may not affect the n=3
status of TOI 700 e, because if they are very small, they probably wouldn’t be included as having
an n-value in our theory because below a certain mass probably wouldn’t play a role in the
dynamics of the planetary wave solution. Other possible solutions for the moon of this system
could easily exist. For instance, it may be that
Would use elements other than Au and Ag in a M-type (red dwarf) star. It may be that Au and
Ag correspond to a G2V star like the Sun. I have made a program in C that does the modeling we
have done here and tested it against these results, and it has the same output. With this program
we can run it for different elemental ratios than Au/Ag. Let us turn our attention to this…
Modeling Star Systems
R
s
r
m
≈
9
5
≈
Au
Ag
of 32 38
of 33 38
Let’s run the program for Au/Ag in TOI 700 e to show that it gets the results we just got…
//
// main.c
// Planck Planet
//
// Created by Ian Beardsley on 1/5/24.
//
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
float
G=6.67408E-11,EarthMasses,EarthDays,T_p,a,M_p,KE,v_p,AU,h_s,R_s,E1,E2,r_m,Sol
arRadii,MassMoonCubed,n,M_m,R_m,V_m,rho;
double part1, part2, part3, part4, part5;
printf ("What is the orbital radius of the planet in AU? ");
scanf ("%f", &AU);
printf ("What is the orbital period of the planet in EarthDays? ");
scanf ("%f", &EarthDays);
a=AU*(1.496E11);
T_p=EarthDays*24*60*60;
v_p=(2*(3.141592654)*a)/T_p;
printf ("The orbital velocity of the planet is: %f m/s \n", v_p);
printf ("What is the mass of the planet in EarthMasses? ");
scanf ("%f", &EarthMasses);
M_p=EarthMasses*(5.972E24);
KE=0.5*M_p*v_p*v_p;
printf ("The kinetic energy of the planet is %fE33 J \n", KE/(1E33));
h_s=KE*(1.03351);
printf ("The Planck Constant for the system is %fE33 Js \n", h_s/(1E33));
printf ("What is the radius of the star in Solar Radii? ");
scanf ("%f", &SolarRadii);
R_s=SolarRadii*6.96E8;
printf ("What is the molar mass of element 1? ");
scanf ("%f", &E1);
printf ("What is the molar mass of element 2? ");
scanf ("%f", &E2);
r_m=(E1/E2)*R_s;
printf ("The Moon's orbital radius is %fE8 m \n", r_m/(1E8));
printf ("What is the orbital number n? ");
scanf ("%f", &n);
part1=cbrt(KE/M_p);
part2=cbrt(h_s/M_p);
part3=cbrt(h_s/a);
part4=cbrt(r_m/(G*G));
part5=cbrt(2/sqrt(n));
M_m=part1*part2*part3*part4*part5;
printf("The mass the this planet's moon is %fE23 kg \n", M_m/(1E23));
R_m=R_s*(r_m/a);
printf ("The radius of this planet's moon is %fE6 m \n", R_m/(1E6));
V_m=(4)*(3.141592654)*(R_m*R_m*R_m)/3;
rho=(M_m/V_m)*(0.001);
printf ("The density of the moon of this planet is: %f g/cm3 \n", rho);
return 0;
}
of 34 38
What is the orbital radius of the planet in AU? 0.1340
What is the orbital period of the planet in EarthDays? 27.80978
The orbital velocity of the planet is: 52420.957031 m/s
What is the mass of the planet in EarthMasses? 0.818
The kinetic energy of the planet is 6.712016E33 J
The Planck Constant for the system is 6.936936E33 Js
What is the radius of the star in Solar Radii? 0.420
What is the molar mass of element 1? 5
What is the molar mass of element 2? 9
The Moon's orbital radius is 1.624000E8 m
What is the orbital number n? 3
The mass the this planet's moon is 3.051824E23 kg
The radius of this planet's moon is 2.368144E6 m
The density of the moon of this planet is: 5.485880 g/cm3
Program ended with exit code: 0
As we can see the program works well. Let us guess that a G2V star like our sun uses Au/Ag, the
more sophisticated metals, and that a more crude star like a red dwarf, M star, uses more crude
metals. It is my guess that because a more luminous star like the Sun has higher radiation
pressure, that in the formation of the its planetary system, the elements have to be heavier to
stay close in. In a red dwarf M star, this shouldn’t be so much the case. Let us run it for 2/3…
What is the orbital radius of the planet in AU? 0.1340
What is the orbital period of the planet in EarthDays? 27.80978
The orbital velocity of the planet is: 52420.957031 m/s
What is the mass of the planet in EarthMasses? 0.818
The kinetic energy of the planet is 6.712016E33 J
The Planck Constant for the system is 6.936936E33 Js
What is the radius of the star in Solar Radii? 0.420
What is the molar mass of element 1? 2
What is the molar mass of element 2? 3
The Moon's orbital radius is 1.948800E8 m
What is the orbital number n? 3
The mass the this planet's moon is 3.243046E23 kg
The radius of this planet's moon is 2.841773E6 m
The density of the moon of this planet is: 3.373621 g/cm3
Program ended with exit code: 0
We find that in fact it uses metals that are closer together, but this is interesting. We get the
density of the Moon in the ratio of 2/3 for the elements associated with it, the base ratio in the
fibonacci sequence whose ratios of successive terms converge on the golden ratio. This is good
because red dwarf M stars are the coolest stars on the main sequence, so they are the basis stars
in HR diagrams. And this is perfectly iron (Fe) over krypton (Kr) which is
(Fe/Kr)=55.85/83.80=0.67 in that 2/3=0.67.
An interesting thing that is happening here is the Moon of the Earth is mostly silicates
(composed of silicon) and the moon of TOI 700 e is using the gold to silver of the Earth equation
produces the density of germanium, the next element down below silicon in the periodic table,
and the orbital period of the habitable planet of TOI 700 e about its star that is about the same
size as the Earth, has almost exactly the orbital period of the Moon around the Earth. It seems to
suggest that the yellow spectral class G stars are connected to the red spectral class M stars in
some kind of an inversion. Indeed, one could suggest there might be a uniformity of life among
the red dwarf stars that is related to the uniformity of life in the yellow stars. The possibility begs
for data on the K stars, F stars, B stars, A stars, and O stars concerning habitable planets for
of 35 38
them, to answer whether there is a pattern in the kind of life from red, to yellow, to blue, to
white stars.
Discussion
The mass of the moon of TOI 700 e would be according to this model:
Its radius would be
It orbital radius would be
The kinetic energy of its planet would be
The orbital velocity of its moon would be
The kinetic energy of this moon would be
The orbital period of its moon would be
=
From our formula
We have that the rotation period of the planet, which is its day, is:
M
m
=
3.0458E 23kg
7.3E 22kg
= 4Ear th Moon Ma sses
R
m
=
2,37E6m
1.74E6m
= 1.362Ear th Moon Ra dii
r
m
=
1.9488E8m
3.844E8m
= 0.5Ear th MoonOr ibtal Ra dii
K E
p
= 6.712E 33J
v
m
=
GM
p
r
m
=
(6.67408E − 11)(4.885E 24kg
1,9488E8m
= 1,293.43m /s
K E
m
=
1
2
(3E 23kg)(1,293.43)
2
= 2.51E 29J
2π r
p
= 2π (1.9488E8) = 9.35E 8m
T
m
=
2π r
m
v
m
=
9.35E8m
1293.43m /s
= 7.22884E5secon d s
200.8h ours = 8.3667d a ys
K E
m
K E
p
(Pl a netDa y) = 1secon d
of 36 38
=
Thus this planet which has a year of 27.81 days, and a lunar month of 8.3667 days, would have
3.324 of its lunar months in its year. Its days is 7.428 hours long or 0.31 Earth days. If the moon
of this planet is like our moon (composed of silicates with and an iron core) its mass is 4
EarthMoonMasses and 1.362 times larger than our moon, and it orbits at about half the distance
as ours. We have for the ratio
Since
Then
The metallic component of salt (NaCl). The planetary Planck constant for this M-type red dwarf
star is
For our G-type yellow star the Sun it is
Thus it is
Times greater.
Let us verify our wave solution:
=
By mass and orbital velocity of the planet we had , and this makes the
accuracy of our modeling
Pl a netDa y =
6.712E 33
2.51E 29
1sec = 26,741secon d s
7.428h ours = 0.31Ear th Da ys
R
s
R
m
=
2.9232E8m
2.37E6m
= 123.34
Z
2
= 123.24
Z = 11.106proton s ≈ 11proton s = Sodiu m(Na)
h
s
= 6.937E 33J ⋅ s
h
s
= 2.8314E 33J ⋅ s
6.937E 33
2.8314E 33
= 2.4500
K E
p
= 3Z
2
Na
G
2
M
2
p
M
3
m
2h
2
s
3(11
2
)
(6.67408E − 11)
2
(4.885E 24)
2
(3.0E 23kg)
3
2(6.937E 33J ⋅ s)
2
= 3(3.608E 33J ) = 6.249E 33J
K E
p
= 6.712E 33J
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6.249E 33
6.712E 33
(100) = 93 %
of 38 38
The Author
