of 1 30
Sentient Beings In The Universe: Here And Beyond Earth
Ian Beardsley!
(University of Oregon, Department of Physics, 2022)!
!
Genesis Project California 2022!
⋅
⋅
of 2 30
This is two works, one dealing with humans and their relationship to Nature, and one the
archaeology of other star systems, thus the relationship of life beyond Earth to Nature. It
is the author’s hope that we will be able to find the common ground between the two.
Contents
Paper 1………………………3
Paper 2………………………12
Closing Remarks…………..29!
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The Human Scenario
Ian Beardsley!
(University of Oregon, Department of Physics, 2022)!
!
Genesis Project California 2022!
⋅
⋅
of 4 30
Life is intricately complex and a lot of perfect circumstances come together to make it
possible. It may be that someone or something is responsible for it. It may be as well that
we can find something hidden that can be measured here on Earth and from Earth that
can tell us something about the human scenario, why we are here, and where we are
headed. In the course of my research I have found some clues that may be precisely this.
They are outlined here.!
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The Theory
It started with a theory I had and a constant I found. In order to present the elements as
mathematical structures we need to explain the matter from which they are made as
mathematical constructs. We need a theory for Inertia. I had found (Beardsley Essays In
Cosmic Archaeology. 2021) where I suggested the idea of proton seconds, that is six proton-
seconds, which is carbon the core element of biological life if we can figure out a reason to
divide out the seconds. I found!
!
Where h is Planck’s constant 6.62607E—34 Js, is the radius of a proton 0.833E-15m, G is
the universal constant of gravitation 6.67408E-11 (Nm2)/(kg2), and c is the speed of light
299,792,459 m/s. And is t=1 second. is the Sommerfeld constant (or fine structure
constant) is 1/137. The mass of a proton is !
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed
of light squared: !
!
Thus we see the duration of a second might be important, and as it turns out it may very well
be because through this constant it predicts the radius of a proton, one of the fundamental
particles from which matter is made.!
The Unit of One Second And The Proton Radius
I formulated the constant in the form of an integral and found as such it solved a lot of
problems across a large amount of fields. I wrote: In that we have!
!
And the periodic table of the elements is cyclical with 18 groups and!
!
Then perhaps we are supposed to write!
!
1
t
1
α
2
m
p
h4π r
2
p
Gc
= 6protons
r
p
t
1
α
m
p
= 1.67262E − 27kg
α
2
=
U
e
m
e
c
2
1
α
2
m
p
h4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
6 =
1
α
2
m
p
h4π r
2
p
Gc
3
α
2
m
p
h4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = 18
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In fact, what if the 3 is supposed to be pi, then!
!
Then we would say that!
k=18/pi=5.7229577951!
The parameter in our constant with the most uncertainty is the radius of a proton . If the 3 is
supposed to be pi, then the radius of a proton becomes:!
!
Which gives!
!
About 95% raw most recent value measured. But, if !
!
Is supposed to be 6 and it is supposed to be multiplied by three to give 18 even which we
need for chemistry so we have 18 protons in the last group of the periodic table which is
important because we need argon with 18 protons for predicting valence numbers of elements
in terms of their need to attain noble gas electron configuration. Then we get!
!
This is in very close agreement with the most recent value measured which is!
!
Which is good for chemistry and we see that the unit of one seconds determines the radius of
a proton.!
Origins of the Second
In that the constant predicts the radius of a proton with the unit of a second we have to ask
why the second has the duration it has. The second comes to us from the Ancient Greeks
dividing the degree of a circle into 60 minutes and that into 60 seconds because they used a
sexagesimal (base 60) counting system, and this might be the answer because 60 was
π
α
2
m
p
h4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = 18
r
p
r
p
= kα
2
m
p
Gc
4πh
r
p
= 8.790587E − 16m
1
α
2
m
p
h4π r
2
p
Gc
r
p
= 8.288587E − 16m = 0.829f m
r
p
= 0.833 + / − 0.014
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probably chosen because it is evenly divisible by 1,2,3,4,5,6..12, 15, 20, 30… and the Ancient
Greeks got that from the Babylonians who got it from the Sumerians. Consider then that the
second comes from 365 days and year, 12 hours a day, 60 minutes in and hour, 60 seconds in
a minute. Thus the calendar formed like this (sexagesimal) could be related to the nature of
solar system formation because the 12 moons per year (earth orbital period) divided up by
sexagesimal reconcile these periods.!
Generating A Table
]I made a program that looks for close to whole number solutions so we can create a table of
values for problem solving. I set it at decimal part equal to 0.25. You can choose how may
values for t you want to try, and by what to increment them. Here are the results for
incrementing by 0.25 seconds then 0.05 seconds. Constant to all of this is hydrogen and
carbon. The smaller integer value of seconds gives carbon (6 protons at 1 second) and the
largest integer value of seconds gives hydrogen (1 proton at six seconds) and outside of that
for the other integer values of protons you get are at t>0 and t<1. Equation 1 really has some
interesting properties. Here are two runs of the program( decpart is just me verifying that my
boolean test was working right to sort out whole number solutions):!
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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 !
0.2436 protons 24.750000 seconds 0.243635 decpart !
By what value would you like to increment?: 0.05!
How many values would you like to calculate for t in equation 1 (no more than 100?): 100!
40.1998 protons 0.150000 seconds 0.199837 decpart !
30.1499 protons 0.200000 seconds 0.149879 decpart !
24.1199 protons 0.250000 seconds 0.119904 decpart !
20.0999 protons 0.300000 seconds 0.099918 decpart !
17.2285 protons 0.350000 seconds 0.228500 decpart !
15.0749 protons 0.400000 seconds 0.074938 decpart !
12.0599 protons 0.500000 seconds 0.059950 decpart !
10.0500 protons 0.600000 seconds 0.049958 decpart !
8.0400 protons 0.750000 seconds 0.039966 decpart !
7.0941 protons 0.850000 seconds 0.094088 decpart !
6.0300 protons 1.000000 seconds 0.029975 decpart !
5.2435 protons 1.150000 seconds 0.243457 decpart !
5.0250 protons 1.200000 seconds 0.024980 decpart !
4.1586 protons 1.450000 seconds 0.158605 decpart !
4.0200 protons 1.500000 seconds 0.019985 decpart !
3.1737 protons 1.899999 seconds 0.173673 decpart !
3.0923 protons 1.949999 seconds 0.092296 decpart !
3.0150 protons 1.999999 seconds 0.014989 decpart !
2.2333 protons 2.699999 seconds 0.233325 decpart !
2.1927 protons 2.749999 seconds 0.192719 decpart !
2.1536 protons 2.799999 seconds 0.153564 decpart !
2.1158 protons 2.849998 seconds 0.115782 decpart !
2.0793 protons 2.899998 seconds 0.079303 decpart !
2.0441 protons 2.949998 seconds 0.044061 decpart !
2.0100 protons 2.999998 seconds 0.009993 decpart !
1.2433 protons 4.850000 seconds 0.243294 decpart !
1.2306 protons 4.900001 seconds 0.230607 decpart !
1.2182 protons 4.950001 seconds 0.218177 decpart !
Here is the code for the program:!
#include <stdio.h>!
#include <math.h>!
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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);!
}}}}!
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 values 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 biological chemistry.!
!
!
Hominids
At this point we ask, what are Humans? A Hominid is a member of the family Hominidae which
are the great apes which are the orangutans, gorillas, chimpanzees and humans. A hominine is
a member of the subfamily of this that includes gorillas, chimpanzees, and humans but
excludes orangutans. A hominin is a member of the tribe Hominini which are the chimpanzees
and humans. !
1
α
2
m
p
h4π r
2
p
Gc
= (6protons)(1second )
1
α
2
m
p
h4π r
2
p
Gc
= (1proton)(6secon ds)
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Lucy is 3.2 million years old. Hominins appeared on the scene about 6 million years ago in the
Miocene, with our evolutionary path taking us after this ended 5.3 million years ago through the
Pliocene, the Pleistocene, and into the Holocene which started about 12,000 years ago when
civilization began with the Sumerians in Mesopotamia. !
The Miocene
We look at our time unit and notice it is a little more than a second, this predicts the Miocene,
the epoch where the human trajectory begins. Protons gives a little more than a second. This
makes a shorter day. We have!
!
H=1.00784 g/mol, carbon = 6 protons!
h=6.62607E-34, r_p=0.833E-15, G=6.67408E-11, c=299,792,459!
(24hours)/(1.004996352)=23.8806837hours/day!
Over the past several billion years the length of the year has not deviated much from 365.25
days because of Kepler’s period for the orbit of a planet but, because the earth loses energy to
the moon, its days becomes longer over time by 0.0067 hours per million years which, we can
see from examining sedimentation band growth, which follows the lunar month. To get our
23.8806837 hours per day we have to go back about 18 million years:!
24-23.8806837=0.0067t!
t=17.80840299 million years!
This is the Miocene, a time which the Earth cooled slowly towards a series of ice ages,
between the Oligocene and the Pliocene defined by the boundaries of the cooler Oligocene
and the warmer Pliocene. While we say hominins began about 6 billion years ago in the latter
part of the Miocene 6 million years ago the Miocene begins about 23.03 million years ago and
goes to 5.33 million years ago. Our figure of 18 million years makes sense because hominids
did not just appear over night, evolution is very gradual. It would make sense to multiply our 6
million years by 3 to get our 18 million years, which we could estimate as when the precursor
to hominins really started to take shape.!
The Dinosaurs
At this point it seems humans are here for a reason. It would seem the dinosaurs going extinct
gave small mammals the opportunity to evolve paving the way for humans. When the
dinosaurs roamed the Earth 70 million years ago the day was 23.7 hours long, about a half an
hour shorter. They seemed to have gone extinct about 65 million years ago when an asteroid
hit in the Yucatan. We seem to understand that this asteroid came from the outer reaches of
the asteroid belt, and that is why we catalog their trajectories. The asteroid belt is about 1AU
thick at 2.2 to 3.2 AU from the sun, the earth-sun separation being 1 AU. Carbon-12 the basis
of life as we know it is 12.01 g/mol. It is made in stars from Beryllium-8 is 8.0053051 g/mol.
Thus we have!
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
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!
!
Delta x equal to 1 AU is both Earth and The Asteroids. Mars is at 1.52AU. Delta X cancels with
the Earth leaving Mars equal to carbon to beryllium, which is life. Does this say we need to be
able to colonize Mars to succeed as a species? To put things in perspective 65
million-18million=47 million years as the time from the dinosaurs going extinct to mammals
becoming the precursor to hominins or something like hominins themselves.!
Conclusion
The Ancients divided up the motions of the Earth and Moon to reconcile with one another in
such a way that it created the duration of a second. This very unit of time in terms of the
Natural constants that govern the universe predict the radius of a proton, one of fundamental
constituents of matter that make the elements and the most primary one. In turn this predicts
the hydrocarbons, the backbones of biological chemistry that makes life possible. The exact
value of time for our constant when reconciled with the the exact duration of a second predicts
the Miocene, the epoch—from now where we discovered the proton radius— where the human
story began with Hominins. It may be the earth was hit by an asteroid to bring about the
extinction of dinosaurs so that small mammals could evolve into humans. There is some
indication that Mars is there for us to colonize for a reason."
12
C
8
Be
=
Mars
Ear th
Δx
Δx = 1AU
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Archaeology For Other Star Systems
Ian Beardsley!
(University of Oregon, Department of Physics, 2022)!
!
Genesis Project California 2022!
⋅
⋅
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Abstract:
A lot comes together for the Earth orbiting the sun to let us know there is a mystery before us if
we look at the archaeology of Earth’s astronomy. Thus, does a lot come together considering
the archaeology of other star systems as well that can indicate to its intelligent life that they are
part of something larger than themselves, as well. We can infer the existence of a planet
around a star we cannot see, but can we say something about the archaeology of its people as
well. The cycles of the Moon, Sun, and Earth follow patterns that are elegant mathematically
and that result in a dynamic calendar, it may have been the formulation of the calendar that had
much of a hand in the development of the system of mathematics we have today. But could
other civilizations of other star systems be on planets that have different patterns that are just
as elegant mathematically, and would this result in other systems of mathematics? "
of 14 30
Star System Archaeology
Kepler’s Law of planetary motion is !
Equation 1. !
For the Sun with T the orbital period of the planet and its distance from the Sun in
astronomical units, for circular orbits. For other stars we have to include a constant k involving
the masses of the bodies:!
Equation 2. !
If the mass of the body orbiting the Star m is is small compared to the mass of the star it is
orbiting we have!
Equation 3. !
The Earth is the third planet, is in the habitable zone, and its distance from the Sun defines 1
AU. Thus we ask: What is k for other star systems? For stars on the main sequence their
luminosity is proportional to their masses raised to the power of 3.5 as an estimate. We have:!
Equation 4: !
Where L is in solar luminosities and M is in solar masses:!
!
!
Further if we say since the Earth is in the right zone to be habitable ( ) then if a star is 100
times brighter than the Sun by the inverse square law its habitable zone is is
10 times further from the star it orbits than the Earth is from the Sun. We have!
Equation 5: !
Combining equations 4 and 5:!
Equation 6: !
T
2
= a
3
a
1
k
= G
M + m
4π
2
≈ G
M
4π
2
a
3
=
GM
4π
2
T
2
L =
(
M
M
⊙
)
3.5
L
⊙
= 3.9E26J/s
M
⊙
= 1.98847E30kg
ℍ
100 = 10AU
ℍ =
(
L
L
⊙
)
ℍ =
(
M
M
⊙
)
3.5
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For another star system we can write equation 3!
Equation 7. !
Where n is the number of solar masses of the star and k has M in solar masses. Combining this
with equation 6 we have for the habitable zone of a star:!
Equation 8. !
Thus if the star is 100 times more luminous than the sun!
!
Or,…!
!
Yielding from equation 6!
!
And!
Equation 9. !
Which is 31.623 years for 3.72759. This result is (31.623 yr)(365.25 days)=11,550 days for the
orbital period. 10 AU from this star is about the distance Saturn is from the Sun (9.54AU), with
year equal to 10,759 days. The orbital period is about the same but will be shorter by the right
amount if we include because we have said the star is more massive so it is orbiting
faster to keep from falling in. Turns out to be 16.38 year orbit. (See Appendices 1 and 2).!
If in the planet that has life orbiting a star has an indication to its intelligence that there is a
mystery before it, just as we do on the Earth in that we have a moon that perfectly eclipses the
Sun as seen from the Earth, then…!
This is because:!
T
2
a
3
=
k
n
T
2
=
k
n
(
M
M
⊙
)
21
4
ln(100) = 3.5ln(M /M
⊙
)
M
M
⊙
= 3.72759
ℍ = 3.72759
3.5
= 10AU
T ∝
(
M
M
⊙
)
21/8
k /n
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!
!
Which are approximately equal. As well we can look at it as:!
!
!
Which are about the same as well. The interesting thing is that since our ratios are around
0.0025 and 0.0045, then…!
!
I say this is interesting because this the ratio of the precious metal gold (Au) to that of silver
(Ag) by molar mass these elements being used for religious and ceremonial jewelry:!
!
We have:!
!
!
Taking the average between these:!
Equation 10. !
Where, is the lunar orbit, is the solar radius, is the earth orbit, and is the radius of
the moon. What this means is that the moon perfectly eclipses the sun because the solar
radius is 1.8 times greater than the lunar orbital radius. And interestingly gold is yellow, silver is
silver and the sun is gold, and the moon is silver, the moon perfectly eclipses the sun allowing
us to study its outer atmosphere.!
(lunar − orbit)
(ear th − orbit)
=
384,400k m
149,592,870k m
= 0.00257
(lunar − radius)
solar − radius
=
1,738.1
696,00
= 0.0025
(lun ar − ra diu s)
(lun ar − orbit)
=
(1,738.1)
(384,400)
= 0.00452
solar − ra dius
ear th − orbit
=
696,000
149,597,870
= 0.00465
0.0045
0.0025
=
9
5
= 1.8
Au
Ag
=
196.97
107.87
= 1.8
(lun ar − ra diu s)(ear th − orbit)
(lun ar − orbit)
2
= 1.759577590
(solar − ra dius)
2
(ear th − orbit)(lunar − radius
= 1.863
1
2
⋅
(
r
2
m
⋅ R
2
⊙
+ r
2
e
R
2
m
r
2
m
⋅ r
e
⋅ R
m
)
=
Au
Ag
r
m
R
⊙
r
e
R
m
of 17 30
The Calendar of a People’s Star System
By dividing the day into 24 hours, the hour into 60 minutes, and the minute into 60 seconds,
the second is 1/86400 of day. By doing this we have a twelve-hour daytime at spring and fall
equinox on the equator, 12 being the most divisible number for its size (smallest abundant
number). That is to say that twelve is evenly divisible by 1,2,3,4,6 which precede it and
1+2+3+4+6=16 is greater than twelve. As such there is about one moon per 30 days and about
12 moons per year (per each orbit) giving us a twelve-month calendar. This is all further
convenient in that the moon and earth are in very close to circular orbits and the circle is evenly
divisible by 30, 45, 60, and 120 if we divide the circle into 360 degrees which are special angles
very useful to the workings of physics and geometry. Further, the 360 degrees of a circle are
about the 365 days of a year (period of one earth orbit) so as such the earth moves through
about a degree a day in its journey around the sun. Thus, through these observations down
through the ages since ancient times we have constructed the duration of a second wisely
enough to make a lot work together.!
Essentially as the moon orbits the earth it makes 12 revolutions for each revolution of the Earth
around the sun which is 365.25 days. That is to say!
!
!
These are frequencies of!
!
!
In radians per day these are:!
!
!
Thus the equations of their phases are:!
!
!
Where t is in days and 2.55605E-3=!
(radius lunar orbit)(radius earth orbit)=384,400km/(149,597,876km).!
We can say the frequency of the moon is 0.21/0.0172=12.21 times greater than that of the
earth. Thus we have the following plots of lunar phases to earth phases:!
T
e
= 365.25days
T
m
= 29.53059days
f
e
= 0.002737851s
−1
f
m
= 0.033863191da ys
−1
ω
e
= 0.0172
ω
m
= 0.21
y
e
= cos(ω
e
t)
y
m
= (2.57E − 3)cos(ω
m
t)
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!
There are 12 moons in a year and 24 hours in a day. Divide twelve by 2 and we have 6, divide
24 by 2 and we have 12. We have:!
!
In that days=1, moons-12, hours=24.!
The Geometrical Explanation of Seconds
We have suggested the second is important as well in terms of the phases of the moon and the
earth and that these determined the calendar system we use. We further suggested there is a
connection of this to the structure found in geometry, and this is what we want to explore
further, here. We have all of this can be compactly written as:!
!
Where n=3, 4, 5, ,6 ,…!
We could evaluate this for n equal to other integers, or even the numbers, but these produce
the special triangles, and geometries we are most interested. Thus we will begin by pointing
out that!
Equation 11. 3 X 4 X 5 X 6 = 360!
The amount of degrees into which we divide a circle and that, as such it approximates the
number of days in a year (1 revolution of the earth around the sun) and thus we see that!
Equation 12. !
Represents days as well (The earth moves through about 1 degree a day in its orbit around the
sun) by solving it for n this and the following are what might be characteristic equations of
habitable star systems:!
moons
2
hours ⋅ d ays
= 6
2 cos(π /n) = 1, 2, Φ, 3, . . .
2 cos(π /n) = 1, 2, Φ, 3, . . .
of 19 30
Equation 13. !
Where,!
!
Which correspond respectively to:!
!
Which are the unit triangle, unit the square, the unit regular pentagon, and the unit regular
hexagon.!
The Arrival of Mammalian Life
The asteroid belt seems to have been the source of an asteroid that hit Earth in the Yucatan 65
million years ago leading to the extinction of the dinosaurs and giving mammalian life a chance
to evolve into intelligence. Is this part of a natural process that takes place in other star
systems?!
The asteroid belt is about 1AU thick at 2.2 to 3.2 AU from the sun, the earth-sun separation
being 1 AU. Carbon-12 the basis of life as we know it is 12.01 g/mol. It is made in stars from
Beryllium-8 is 8.0053051 g/mol. Thus we have!
Equation 14. !
Equation 15. !
Delta x equal to 1 AU is both Earth and The Asteroids. Mars is at 1.52AU. Delta X cancels with
the Earth leaving Mars equal to carbon to beryllium, which is life, Does this say we Mars was
made available to Humans to colonize?!
!
Or, does it mean we need to put bases on the moon to mine Helium-3 as a clean, renewable
energy source.!
Application of the Theory
Kepler-442b is a larger than earth rocky planet orbiting a K-type main sequence star with a
mass of 0.61 solar masses, luminosity 12% that of the sun and 0.60 solar radius. It is about 2.9
billion years old compared to thus Sun at 4.6 billion years. It is at a distance of about 1,206
light years in the constellation Lyra. The planet orbits at 0.409 AU with an orbital period of
112.3 days (112 days 9 hours 10 minutes 24 seconds). It is 2.3 earth masses and 1.34 earth
radius. It receives 70% the light the earth receives from the sun.!
As it would turn-out this exoplanet is tidally locked with the star it orbits being closer in to its
star than the earth making its day perhaps weeks to months long. In so far as we don’t know
da ys = cos
−1
(y/2)
y = 1, 2,
5 + 1
2
, 3, . . .
n = 3,4,5,6,...
12
C
8
Be
=
Mars
Ear th
Δx
Δx = 1AU
12
C =
4
He +
8
Be
of 20 30
the length of its day we cannot determine the orbital period of any moon it might have in terms
of such a day. Nor do we even know if it has a moon to define a month from which we can infer
the number of months in a year on the planet. All we really know is it has a year 112.3 earth
days.!
However, let us speculate upon a scenario from what we have developed so far. Equation 9
was derived with the condition:!
Equation 16. !
We might formulate another scenario, While gold (Au) and silver (Ag) are metal used for making
religious and ceremonial jewelry so are silver (Ag) and copper (Cu). The thing all of these have
in common is that they are the finest malleable and ductile metals used fro the purpose of
making electrical wire as they are the finest conductors. Copper is among one of the first
metals used for the purpose of adornments and tools because it is soft enough to pound-out
without heating it, which was the method first discovered by our ancient ancestors. It was used
before smelting (separating from its ores) in its naturally occurring form. We thus have:!
Equation 17. !
Putting in the radius of the star as 0.6 the radius of the sun which is 696,000km and the molar
mass of silver is 107.87 and copper is 63.55 we have as the orbital
distance for the moon of Kepler-44b. Thus with the astronomical unit of the the moon (LAU)
being one corresponds to a period of 29.3 days. We have!
!
!
!
The period of the moon would be almost exactly 15 days. This connects the moon of this
planet to the moon of our planet in that the moon goes through about 15 degrees in an hour
(360/24=15). But this is if the planet it orbits is the size of the Earth. We must multiply by the
constant to get a shorter period gives about 10 days. This gives credence to
something I had considered earlier…!
Ethnomusicology is a sub-discipline in archaeology. It studies the history of humans through its
musical instruments and music theory. When we discussed the 12 month year based on twelve
moons per earth orbital period, we noticed it coincided with elegant mathematics because 12
was evenly divisible by 1, 2, 3, 4, 6 making it abundant. 12 in flamenco is a form of six. 6/8 is a
form common to may cultures and a favorite for being lively and dynamic throughout India, the
Middle East, and Africa. But perhaps the most beloved in India is Tin Tal, a cycle of 16 which is
four, four times. Four is often play four times throughout the Earth’s cultures. Thus, if a planet
orbited a star that had sixteen moons per its year, the inhabitants might divide the circle into
R
⊙
r
m
=
Au
Ag
R
⊙
r
m
=
Ag
Cu
r
m
= 246,023.3k m
R
⊙
r
m
= 1.6974
T =
(
246,023.3
384,399
)
3
= 0.512
T = (0.512)(29.53) = 15.1209d ays
k /n ≈ 0.659
of 21 30
400 degrees and have four four month seasons rather than four three month seasons. This
results in the following table:!
12 Month Year
360/30=12
2cos(30)
360/36=10
2cos(36)
360/45=8
2cos(45)
360/60=6
2cos(60)
360/90=4
2cos(0)=
3
2
2
1
Φ
16 Month Year
400/x=12
x=100/3
400/x=10
x=40
400/x=8
x=50
400/x=6
x=200/3
40/x=4
x=100
Φ
3
2
1
2
of 22 30
The luminosity of the sun is:!
!
The separation between the earth and the sun is:!
!
The solar luminosity at the earth is reduced by the inverse square law, so the solar constant is:!
!
That is the effective energy hitting the earth per second per square meter. This radiation is
equal to the temperature, to the fourth power by the steffan-bolzmann constant, sigma ( ),
can be called the temperature entering, the temperature entering the earth.!
intercepts the earth disc, , and distributes itself over the entire earth surface, , while
30% is reflected back into space due to the earth’s albedo, a, which is equal to 0.3, so!
!
!
But, just as the same amount of radiation that enters the system, leaves it, to have radiative
equilibrium, the atmosphere radiates back to the surface so that the radiation from the
atmosphere, plus the radiation entering the earth, is the radiation at the surface of
the earth, . However, !
!
And we have…!
!
!
!
!
!
L
0
= 3.9 × 10
26
J/s
1.5 × 10
11
m
S
0
=
39 × 10
2
4π (1.5 × 10
11
)
= 1,370wat ts/m eter
2
T
e
σ
T
e
S
0
π r
2
4π r
2
σ T
e
4
=
S
0
4
(1 − a)
(1 − a)S
0
(
π r
2
4π r
2
)
σ T
a
4
σ T
e
4
σ T
s
4
σ T
a
4
= σ T
e
4
σ T
s
4
= σ T
a
4
+ σ T
e
4
= 2σ T
e
4
T
s
= 2
1
4
T
e
σ T
e
4
=
S
0
4
(1 − a)
σ = 5.67 × 10
−8
S
0
= 1,370
of 23 30
a=0.3!
!
!
!
So, for the temperature at the surface of the Earth:!
!
Let’s convert that to degrees centigrade:!
Degrees Centigrade = 303 - 273 = 30 degrees centigrade!
And, let’s convert that to Fahrenheit:!
Degrees Fahrenheit = 30(9/5)+32=86 Degrees Fahrenheit!
In reality this is warmer than the average annual temperature at the surface of the earth, but in
this model, we only considered radiative heat transfer and not convective heat transfer. In
other words, there is cooling due to vaporization of water (the formation of clouds) and due to
the condensation of water vapor into rain droplets (precipitation or the formation of rain).!
The program in C is as follows:!
#include <stdio.h>
#include<math.h>
int main(int argc, const char * argv[]) {
{
float s, a, l, b, r, AU, N, root, number, answer, C, F;
printf("We determine the surface temperature of a planet.\n");
printf("What is the luminosity of the star in solar luminosities?
");
scanf("%f", &s);
printf("What is the albedo of the planet (0-1)?" );
scanf("%f", &a);
printf("What is the distance from the star in AU? ");
scanf("%f", &AU);
r=1.5E11*AU;
l=3.9E26*s;
b=l/(4*3.141*r*r);
N=(1-a)*b/(4*(5.67E-8));
1,370
4
(0.7) = 239.75
T
e
4
=
239.75
5.67 × 10
−8
= 4.228 × 10
9
T
e
= 255Ke lvin
T
s
= 2
1
4
T
e
= 1.189(255) = 303Kelvin
of 24 30
root=sqrt(N);
number=sqrt(root);
answer=1.189*(number);
printf("\n");
printf("\n");
printf("The surface temperature of the planet is: %f K\n",
answer);
C=answer-273;
F=(C*1.8)+32;
printf("That is %f C, or %f F", C, F);
printf("\n");
float joules;
joules=(3.9E26*s);
printf("The luminosity of the star in joules per second is:
%.2fE25\n", joules/1E25);
float HZ;
HZ=sqrt(joules/3.9E26);
printf("The habitable zone of the star in AU is: %f\n", HZ);
printf("Flux at planet is %.2f times that at earth.\n", b/1370);
printf("That is %.2f Watts per square meter\n", b);
printf("\n");
printf("\n");
}
return 0;
}
Let us run our program for Kepler-442b!
Here we use a single atmospheric layer with no
convection for the planet to be in an equilibrium
state. That is to say, the temperature stays
steady by heat gain and loss with radiative
heat transfer alone.
The habitable zone is calculated using the idea
that the earth is in the habitable zone for a
star like the Sun. That is, if a star is 100
times brighter than the Sun, then the habitable
zone for that star is ten times further from
it than the Earth is from the Sun because ten
squared is 100.
We determine the surface temperature of a planet.
What is the luminosity of the star in solar luminosities? 0.12
What is the albedo of the planet (0-1)?0.3
What is the distance from the star in AU? 0.409
The surface temperature of the planet is: 279.523163 K
That is 6.523163 C, or 43.741692 F
of 25 30
The luminosity of the star in joules per second is: 4.68E25
The habitable zone of the star in AU is: 0.346410
Flux at planet is 0.72 times that at earth.
That is 989.67 Watts per square meter
Ideally if the nearest star to us was similar to our Sun it would be best to start there. And
indeed the nearest star system is composed of a three stars and two of them are comparable
to the sun, in fact almost identical. But being a triple system, with which one of the stars has
been found to have a planet in its habitable zone, it can only be solved with equations other
than ours which work only for a single star with planets small in comparison to to its mass. The
triple star system I speak of is Alpha Centauri A, B, and C.!
Alpha Centauri is the nearest of the stars to us and is a triple system designated Alpha
Centauri A, B, And C. Alpha Centauri A and B are much like the sun, similar in mass and
luminosity. Alpha Centauri A is is 1.1 solar masses and Alpha Centauri B is 0.907 solar masses.
They orbit a common center, which can be taken as the focus of an ellipse at the center of
gravity between the two. The period is 79.91 years. The eccentricity of the orbit is high and is
given as e=0.519.!
Alpha Centauri A is thought to have a Neptune sized planet in its habitable zone. While A and B
orbit one another their closest approach is 11.2 AU, and furtherest is 35.6 AU, but C (Proxima
Centauri) has its orbit very far from these two at 0.21 light years, a whopping 13,000 AU. The
11.2 and 35.6 AU for A and B are distances like Saturn and Pluto from the Sun, respectively.!
With the launching of the James Webb Space Telescope a lot of new data should be coming in
for planets and their moons, and we will have more to work with.!
The VRMS (Root Mean Velocity)
Ruud Loeffen writes in his paper A fair Tale about Expansion, 2021:
“The original accretion ring [from which the solar system formed] did have one velocity. That
velocity must be the Root Mean Square Velocity (VRMS) of the planets as they are orbiting
today. Why? Because of the conservation laws of energy and momentum. Originally the planets
did have the same velocity, but they changed their velocity based on how far they drove away
from the central mass, or how much closer they got to the central mass…”
“Tufail Abbas and were intrigued by the idea that the “gamma factor” in the Lorentz
Transformation of Mass-Energy could be related to the velocity of this original velocity of the
protoplanetary ring in which the planets from our solar system are born…”
“We decided to calculate the Root Mean Square Velocity (VRMS) of all planets in our solar
system. We calculated the VRMS to be about 12.3 km per second.”
In light of this let us consider the universal gravitational constant. We can write it in
astronomical units, years, and solar masses instead of kg/m/s (See Appendix 2). It is!
!
Let us multiply out the sun which is and convert convert the VRMS to these units.!
G = 39.433
AU
3
M
⊙
⋅ year
2
1M ⊙
of 26 30
!
!
Now we divide by the VRMS and multiply by the orbital period of the earth of 1 year and
divide by the earth orbit of 1 AU:!
!
Divide that by the area swept out by the earth orbit in one year which is where R=1 AU:!
!
!
!
This is very close to the average distance to Pluto (39.48 AU) which can be considered a
definition for the radius of the solar system. It seems somehow the VRMS connects the radius
of the solar system to universal gravitational constant G. And it does it through the Earth
because our final result was to multiply 4,840.22 by 1 AU which is the Earth-Sun separation.
The VRMS was determined by considering the orbital velocities of Jupiter and Saturn since
they carry most of the angular momentum of the solar system, their angular momentums being
the remnants of the rotation of the protoplanetary disc from which they formed. It makes sense
that earth would play a pivotal role as well, as Jupiter and Saturn are gas giants separated from
Earth and Mars by the asteroid belt, Earth and Mars being terrestrial planets (solid). Essentially
you have Earth and Mars reflected about the asteroid belt circumference. We know the gas
giants are separated from the terrestrial planets because when the solar system was first
forming pressure from the solar radiation blew the lighter elements out beyond where the
asteroid belt is, and the heavier elements were able to remain close in to the Sun. It might be
worth calculating the VRMS for the inner terrestrial planets and compare it to that of the outer
gas giants to consider perhaps there were two VRMS velocities for the protoplanetary disc one
for the inner protoplanetary disc and one for the outer protoplanetary disc."
GM
⊙
= 39.433
AU
3
year
2
12.3
k m
s
AU
1.495987E11k m
⋅
3.154E 7s
year
= 0.002593218
AU
year
GM
⊙
39.433
0.002593232
= 15,206
AU
2
year
π R
2
15,206
AU
2
year
⋅
year
π AU
2
= 4,840.22
π R
2
= 4,840.22
R = 39.25AU
of 27 30
Appendix 1!
And, !
!
!
Multiply this by one over the mass of the star:!
!
Meaning k is seconds squared over meters cubed, which is right because meters cubed cancel
with a cubed leaving T squared equals seconds squared. Thus for our problem!
!
The factor on the right needs to be multiplied by . We have!
!
!
But this needs to be converted to astronomical units and years. We used G in kg/m/s and M is
the mass of the sun in kg.!
1 astronomical unit is: 1.496E11 meters!
Seconds in a year are: 3.154E7!
Square root 1 astronomical unit cubed is: 5.786E16. Thus,…!
!
(31.633 years)(0.518)=16.38 year orbit"
G = N
m
2
kg
2
N = kg
m
s
2
1
N
=
s
2
kg ⋅ m
1
G
=
s
2
kg ⋅ m
⋅
kg
2
m
2
=
s
2
⋅ kg
m
3
T
2
=
s
2
⋅ kg
m
3
⋅
1
kg
⋅ a
3
T ∝
(
M
M
⊙
)
21/8
k /n
k =
4π
2
GM
4π
2
3.73(6.674E − 11)(1.988E 30)
= 2.824E − 10
s
m
3/2
2.824E − 10 ⋅
5.786E16
3.154E 7s
= 0.518
of 28 30
Appendix 2!
We can express our constant k in solar system units by finding the value of G in astronomical
units, years, and solar masses. !
!
Adding earth to does not change the value of M if taken to 5 places after the decimal.!
!
!
!
!
!
!
!
!
Since!
!
And our equation (equation 9) becomes!
!
If we are working solar masses, years, and AU."
M
⊙
= 1.98847E30kg
AU = 1.49597871E11m
year = 3.154E 7s
G = 6.67408E − 11
m
3
kg ⋅ s
2
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅
AU
3
3.3479E33m
3
⋅
1.98847E30kg
M
⊙
⋅
9.9477E14s
2
year
2
G = 39.433
AU
3
M
⊙
⋅ year
2
G ≈ 40
AU
3
M
⊙
⋅ year
2
GM
⊙
= 40
AU
3
year
2
GM
⊙
4π
2
= 0.99885 ≈ 1
k =
4π
2
GM
4π
2
3.73(40)(1)
≈
1
n
T =
1
n
(
M
M
⊙
)
21/8
of 29 30
Closing Remarks
Our history was forged by the elements and minerals beginning with stones to craft
spearpoints and copper, silver, and gold beginning with using these metals to make tools
and weapons to in the end electrical wire for advanced technology. It would seem other
life on other planets around other stars would have been forged by the same things.
Perhaps they even had a moon that hung in their sky to completely eclipse their star as
ours does with the Sun, letting us know that we are part of something cosmic. Perhaps
there are subtle variations in their history, and they have an ancient history as vast and
intriguing as ours. Perhaps one day we will go to other star systems, even become part
of a galactic, or even intergalactic society and we will learn their histories and perhaps
from them even learn something about ourselves and our histories.
of 30 30
The Author!
