of 1 8
The Theory In the Bohr atom the radius of the electron’s orbit in the ground state (n=1) is given by
1.
2.
I have extended this to our solar system and found that the ground state of our solar system is based on
the Moon of our Earth. I have written
3.
Yes, interestingly it is equal to 1 second, our base unit of time that ultimately came from the way the
Ancient Sumerians divided up the period of rotation of the Earth (1 Day) into 24 hours, each hour into 60
minutes, each minute into 60 seconds. They used based 60 counting which was handed down to the
Babylonians. is the mass of the Moon, and is the Planck-type constant I have proposed for the
Solar System. I found it is given by
4.
Where is the kinetic energy of the Earth in its orbit around the Sun. We see the the third planet from
the Sun, the Earth, determines its Planck constant in terms of the same unit of a second. I also find that the
kinetic energy of the Moon to the kinetic energy of the Earth, that quantity times the Earth rotation period
(24 hours, its day) is close to a second, about 1.2 to 1.3 seconds depending whether you use aphelion or
perihelions, and in what combination. That is
5.
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:
r
1
=
ℏ
2
k e
2
m
e
r
1
≈ 0.529E − 10m
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1secon d
λ
moon
=
ℏ
2
⊙
GM
3
m
=
(2.8314E 33)
2
(6.67408E − 11)(7.34763E 22k g)
3
= 3.0281E 8m
λ
moon
c
=
3.0281E 8m
299,792,458m /s
= 1.010secon d s
λ
moon
c
= 1secon d
M
m
ℏ
⊙
ℏ
⊙
= (1secon d )KE
e
K E
e
K E
m
K E
e
(Ear th D a y) = 1.2 − 1.3secon d s
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
of 2 8
If the second were purely arbitrary, we should expect this ratio to not be so close to 1 second. I also find
that the proton has a characteristic time of 1 second in two forms that when equated give about the radius
of the proton:
6.
7.
Equating these two yield the radius of a proton, in terms of its mass :
8.
I find if I replace 2/3 with , we may have exactly the radius of a proton
9.
This may show an underlying optimization principle where occurs in biological and cosmic structures. I
find I can arrive at this radius of a proton another way, With the Planck energy, , given by
frequency of a particle, and from mass-energy equivalence, :
We take the rest energy of the mass of a proton :
The frequency of a proton is
Since our theory gave us the factor of 2/3 for the radius of a proton we have:
(
1
6 α
2
4πh
G c
)
⋅
r
p
m
p
= 1secon d
(18769)((0.833E − 15m)
6(1.67262E − 27kg)
(6.62607E − 34J ⋅ s)(4π)
(6.674E − 11)(299,792,458m /s)
= 1.00500secon d s
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
r
p
m
p
r
p
=
2
3
⋅
h
cm
p
ϕ = 1.618
r
p
= ϕ
h
cm
p
ϕ
E = h f
E = m c
2
E = h f
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 3 8
The radius of a proton is then
This is very close to the value upon which the proton radius converged historically by two independent
methods which was 0.877E-15m. The result from our theory is
The 0.877fm was challenged in 2010 by a third experiment making it 4% smaller and was 0.842E-15m.
We find it may be that the radius of a proton is actually
Where is the golden ratio. The most recent value is 0.833E-15m. I guessed since one second
comes from the ancient Sumerians dividing the Earth day (rotation period) into 24 hours, and those into
60 minutes, and those into 60 seconds, that this has to do with the rotational angular momentum of the
Earth, . I found
10.
Where
11.
The value is 2.5 which by modeling our Solar System is found to be the exponent in the pressure gradient
for the protoplanetary disc from which our Solar System formed. That is I found
12.
the pressure of the disc as a function of radius. Which suggests that the structure of the protoplanetary
disc could be governed by the same fundamental time of one second in the Earth’s rotation and that the
Earth’s formation process may be encoded in the same number we developed since ancient times to
describe time (24, 60). This is the solution to:
13.
m
p
r
p
=
2
3
⋅
h
c
r
p
=
2
3
⋅
h
cm
p
r
p
=
2
3
⋅
6.62607E − 34
(299,792,458)(1.67262E − 27)
= 0.88094E − 15m
r
p
= ϕ ⋅
h
cm
p
= 0.816632E − 15m
ϕ = 0.618
L
earth
L
earth
ℏ
⊙
24 = 60
L
earth
=
4
5
π M
e
f
e
R
2
e
P(R) = P
0
(
R
R
0
)
−
L
ear th
ℏ
⊙
d P
dr
= − ρ
(
GM
⋆
r
2
−
v
2
ϕ
r
)
of 4 8
The protoplanetary disc that evolves into the planets has two forces that balance its pressure, the
centripetal force of the gas disc due to its rotation around the protostar and the inward gravitational
force on the disc from the protostar , and these are related by the density of the gas that makes
up the disc. It is the pressure gradient of the disc in radial equilibrium balancing the inward gravity and
outward centripetal force. In order to apply this to other star systems, we have to be able to predict the
radius of the habitable planet, presumably in the n=3 orbit. I found the answer to be in the Vedic literature
of India. They noticed that the diameter of the Sun is about 108 times the diameter of the Earth and that
the average distance from the Sun to the Earth is about 108 solar diameters, with 108 being a significant
number in Yoga. So I wrote the equivalent:
14.
The surprising result I found was, after applying it to the stars of many spectral types, with their different
radii and luminosities (the luminosities determine , the distances to the habitable zones) that the
radius of the planet always came out about the same, about the radius of the Earth. This may suggest
optimally habitable planets are not just a function of the distance from the star, which determines their
temperature, but are functions of their size and gravity probably because it is good for life chemistry. Here
are just a few examples using the data for several spectral types:
F8V Star
Mass: 1.18
Radius: 1.221
Luminosity: 1.95
v
2
ϕ
/r
GM
⋆
/r
2
ρ
R
planet
= 2
R
2
⋆
r
planet
r
planet
M
⋆
= 1.18(1.9891E 30kg) = 2.347E 30kg
R
⋆
= 1.221(6.9634E 8m) = 8.5023E 8m
r
p
= 1.95L
⊙
AU = 1.3964AU(1.496E11m /AU ) = 2.08905E11m
R
p
=
2R
2
⋆
r
p
= 2
(8.5023E 8m)
2
2.08905E11m
=
6.92076E6m
6.378E6m
= 1.0851Ear th Ra d ii
of 5 8
F9V Star
Mass: 1.13
Radius: 1.167
Luminosity: 1.66
G0V Star
Mass: 1.06
Radius: 1.100
Luminosity: 1.35
G1V Star
Mass: 1.03
Radius: 1.060
Luminosity: 1.20
As you can see we consistently get about 1 Earth radius for the radius of every planet in the habitable
zone of each type of star. It might be that radius is right for life in terms of gravity and densities for the
elements. I got these results for the stars from spectral types F5V to K3V.
In order to get , the distance of the habitable planet from the star, we use the inverse square law for
luminosity of the star. If the Earth is in the habitable zone, and if the star is one hundred times brighter
than the Sun, then by the inverse square law the distance to the habitable zone of the planet is 10 times
that of what the Earth is from the Sun. Thus we have in astronomical units the habitable zone of a star is
given by:
15.
Also, the theory utilizes the fact that the Moon as seen from the Earth perfectly eclipses the Sun as a
possible condition for optimal habitability of the planet, which is
M
⋆
= 1.13(1.9891E 30kg) = 2.247683E 30kg
R
⋆
= 1.167(6.9634E 8m) = 8.1262878E 8m
r
p
= 1.66L
⊙
AU = 1.28841AU(1.496E11m /AU ) = 1.92746E11m
R
p
=
2R
2
⋆
r
p
= 2
(8.1262878E 8m)
2
1.92746E11m
=
6.852184E6m
6.378E6m
= 1.0743468Ear th Ra dii
M
⋆
= 1.06(1.9891E 30kg) = 2.108446E 30kg
R
⋆
= 1.100(6.9634E 8m) = 7.65974E 8m
r
p
= 1.35L
⊙
AU = 1.161895AU(1.496E11m /AU ) = 1.7382E11m
R
p
=
2R
2
⋆
r
p
= 2
7.65974E 8m)
2
1.7382E11m
=
6.751E6m
6.378E6m
= 1.05848Ear th Ra dii
M
⋆
= 1.03(1.9891E 30kg) = 2.11E 30kg
R
⋆
= 1.060(6.9634E 8m) = 7.381E 8m
r
p
= 1.20L
⊙
AU = 1.0954AU(1.496E11m /AU ) = 1.63878589E11m
R
p
=
2R
2
⋆
r
p
= 2
7.3812E 8m)
2
1.63878589E11m
=
6.6491E6m
6.378E6m
= 1.0425Ear th Ra dii
r
planet
r
planet
=
L
⋆
L
⊙
AU
of 6 8
16.
Orbital radius of the planet to that of the moon is radius of the star to that of the moon. It is known that
the Moon has a lot to do with the conditions for life on Earth being good because its orbit holds the Earth
at its inclination to Sun its orbit preventing temperature extremes and allowing for the seasons. The
Schrödinger Wave Equation must be solved to determine the energies and orbitals of the electron in the
hydrogen atom. In spherical coordinates it is
17.
It has the solutions
18.
19.
I find the solutions are for the Earth orbiting the Sun are:
20.
21.
is the solar radius, that of the Moon. For Earth , third planet. For a star brighter than the Sun,
more massive, larger, it may be that in many cases the habitable zone, which is further out, still is the n=3
orbit for planets because the distribution of the planets might be stretched out putting the n=3 planet in the
habitable zone. The same may be true of stars that are dimmer, less massive and smaller, because the
planet distribution might be tighter. Since the n=3 habitable zone in such a scenario is closer in, n=3
might be in the habitable zone. While we don’t have complete data for brighter stars, like F stars, we do
for dimmer stars because it is easier to obtain and a good example of this is the M2V star TOI 700. We
have found four exoplanets around this star and TOI 700 e is an Earth-like planet in its habitable zone
which is also the third planet. With the equations so far we can solve planetary systems with a method that
is independent of orbit number, n. To solve the wave equation in the case of a protoplanetary disc you
would want the wave equation to be in cylindrical coordinates, to solve it looks like this, the time
independent Schrödinger equation is
22.
r
planet
r
moon
=
R
star
R
moon
−
ℏ
2
2 m
[
1
r
2
∂
∂r
(
r
2
∂
∂r
)
+
1
r
2
si n θ
∂
∂θ
(
si n θ
∂
∂θ
)
+
1
r
2
si n
2
θ
∂
2
∂ ϕ
2
]
ψ + V(r)ψ = E ψ
E
n
= −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
r
n
=
n
2
ℏ
2
Z k
e
e
2
m
e
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
r
n
=
2ℏ
2
⊙
GM
3
m
⋅
R
⊙
R
m
⋅
1
n
R
⊙
R
m
n = 3
−
ℏ
2
2M
e
∇
2
Ψ(ρ, ϕ, z) + V(ρ, z)Ψ(ρ, ϕ, z) = E Ψ(ρ, ϕ, z)
of 7 8
The Laplacian in cylindrical coordinates is
23.
The gravitational potential is
24.
Where is the mass of the central protostar. We can refer to the variables as such:
We separate the variables according to:
25.
Substituting 23 and 25 into 22 we have
26.
Solving it all the way is beyond the scope of this paper. We have hindsight, so we were able to write out
the solutions by form and dimensional analysis. Let us now compute the value of and show that the
units work…
∇
2
=
1
ρ
∂
∂ρ
(
ρ
∂
∂ρ
)
+
1
ρ
2
∂
2
∂ ϕ
2
+
∂
2
∂z
2
V(ρ, z) = −
GM
s
ρ
2
+ z
2
M
s
Ψ(ρ, ϕ, z) = R(ρ) ⋅ Φ(ϕ) ⋅ Z(z)
−
ℏ
2
2M
e
[
1
ρ
d
dρ
(
ρ
d R
dρ
)
Φ ⋅ Z +
1
ρ
2
R ⋅
d
2
Φ
d ϕ
2
⋅ Z + R ⋅ Φ ⋅
d
2
Z
d z
2
]
+ V (ρ, z)RΦZ = E RΦZ
ℏ
⊙
ℏ
⊙
= (hC )K E
e
of 8 8
Where
Where is the radius of a proton, is the mass of a proton, is the speed of light, and is the fine
structure constant. We found this gives the characteristic time of one second in terms of a proton. We
guess the planetary scale is connected to the proton scale because the planets formed from the
protoplanetary disc and are made of different combinations of protons. We derive the value of our solar
Planck constant:
=
=
=
=
27.
hC = 1secon d
C =
1
3
⋅
1
α
2
c
2
3
⋅
π r
p
G m
3
p
r
p
m
p
c
α
C =
1
3
⋅
1
α
2
c
1
3
⋅
2 π r
p
G m
3
p
1
3
⋅
18769
299792458
1
3
⋅
2 π (0.833E − 15)
(6.67408E − 11)(1.67262E − 27)
3
1.55976565E 33
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
= en erg y
hC = (6.62607E − 34)(1.55976565E 33) = 1.03351secon d s ≈ 1.0secon d s
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
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
ℏ
⊙
= (hC )K E
earth
= (1.03351s)(2.7396E 33J ) = 2.8314E 33J ⋅ s
