Text 74, 199 rader
Skriven 2004-09-16 13:02:00 av Ian Beardsley (1:278/230)
Ärende: Project Genesis
=======================


 
Project Genesis, by Ian Beardsley 
Since other stellar systems may not even exist as we need them, and the
distances between them are so immense, it might be better to unlock the
mysteries of making them, and find the structure in ours that allows for
so much life. There is, I have found, a correlation between the
microworld and the macroworld, where our solar system is concerned. It
may be related to why it is life bearing. 
Abundant in the early earth atmosphere 
were methane (CH 4), ammonia (NH 3) and water vapor (H 2O). Methane is 
tetrahedral in structure, a carbon atom sourounded by 4 hydrogens. 
Ammonia is trigonal pyramidal, a nitrogen atom surrounded by 3 hydrogen 
atoms, and water vapor is triangular, or bent, an oxygen atom surrounded 
by two hydrogens. These represent stable structural systems as they are 
all systems of triangles, which are the only stable polygons. These 
substances combined under energy with hydrogen gas form amino acids, the 
building blocks of life. The core atoms of these molecules, carbon, 
nitrogen, and oxygen, are all in period two of the periodic table and 
follow directly one after the other. 
When plants perform 
photosynthesis, they combine carbon dioxide with water and release 
oxygen. The reaction is: 
CO 2+2H 2O----->CH 2O+O 2+H 2O 
As can be seen a sugar is made. Important to most plants to do this is 
nitrogen. Nitrogen (N 2) is the most abundant gas in the earth 
atmosphere, comprising about 78.03% of it. We 
now calculate the molecular masses of these special gases: 
CH 4=(12.01+4(1.01))=16.05 
NH 3=(14.01+3(1.01))=17.04 
CO 2=(12.01+2(16.00))=44.01 
H 2O=(2(1.01)+16.00)=18.02 
N 2=(14.01+14.01)=28.02 
O 2=(16.00+16.00)=32.00 
We now form some ratios between these molecular masses: 
O 2/CH 4=1.992~2 
NH 3/CH 4=1.061~1 
O 2/N 2=1.142~sqrt(2) 
CO 2/N 2~1.6=(sqrt(5)+1)/2=phi 
O 2/H 2O=1.776~sqrt(3) 
Notice that these values are given by the sequence: |2cos(pi/n)| 
n=1,2,3,4,5,6 pi/n radians 
Observe: 
2=|2cos(pi)| 
0=|2cos(pi/2)| 
1=|2cos(pi/3)| 
sqrt(2)=|2cos(pi/4)| 
(sqrt(5)+1)/2=|2cos(pi/5)| 
sqrt(3)=|2cos(pi/6)| 
Geometrically sqrt(2) is the ratio of the side of a square to its 
radius. Phi is the ratio of the chord of a regular pentagon to its side. 
sqrt(3) the ratio of the side of an equilateral triangle to its radius, 
and 1 is the ratio of the side of a regular hexagon to its radius. The 
square, the regular hexagon and the equilateral triangle are the 
tessellating regular polygons. The regular pentagon is one of the 
archemedian tessellators. 
Note: All of the compounds considered above are combinations of any of
the two 
elements that always occur in amino acids, the building blocks of life.
They are hydrogen, carbon, nitrogen and oxygen. 
We compare the mass of the earth to the mass of the sun, and multiply
that ratio by the distance between them. Let the mass of the earth be M
e and the mass of the sun be M s. Let the distance between them be r. (M
e)r/(M s)= 
(5.976E27)(1.495979E13cm)/(1.989E33)
=4.495E7cm=449.5km 
We now divide that result by the radius of the earth, R e:
449.5km/6378.5k=0.07 
Hydrogen is the most abundant element in the universe and nitrogen is
the most abundant element in the earth atmosphere. We now compare their
molar masses: 
H/N=1.01/14.01=0.07
And we see that
H/N=(M e)r/(M s)(R e) 
Now it becomes mystic. We apply the same concept to mars and get the
same result. The most abundant gas in the mars atmophere is CO 2. 
(H)/(CO 2)= 1.01/44.01=0.02 
M m=mass of mars, M s=mass of sun, r = the distance between them and R m 
= the radius of mars. We have 
(M m)r/(M s)(R m)=0.02 
M m=6.418E26g
r=2.279409E13cm
M s=1.989E33g
R m=3.393096E8cm 
(H)/(CO 2)=(M m)(r)/(M s)(R m) 
Keep in mind these equations hold for a solar system that is at its peak
as an orderly arrangement of parts. Eventually the order will
degenerate. The sun is losing mass every day and therefore r for any of
the planets will grow. 
The data for this study comes from "Handbook of Space Astronomy and
Astrophysics" by Martin V. Zombeck, 1982 Cambridge University Press. 
The relative surface gravities of the earth and mars respectively are 
1.000 and 0.380. Let that of the earth be g_e and that of mars be g_m. 
Then 
(g_e)/(g_m)=1.000/0.380=2.63 
The mass of a mole of oxygen gas (O_2) is 32.00 grams and a mole of 
carbon (C) 
is 12.01 grams. Therefore 
(O_2)/(C)=32.00/12.01=2.66 
Thus we have 
(g_e)/(g_m)=(O_2)/(C) 
Earth and mars are the two terrestrial planets upon which we can set 
foot and are next to one another. Carbon is the basis of life, and 
Oxygen gas its neccesity (for animal life). Carbon and Oxygen are next
to one another in 
the periodic table of the elements, separated only by nitrogen. The last
equation says it takes the 
same amount of energy to lift a mole of carbon on the earth as it does 
to lift a mole of oxygen gas on mars (the same distance). 
Project Genesis (continued) 
mercury 1,     1 
venus   2,     1.8 
earth   3,     2.5 
mars   4,     3.8 
asteroids       5,     5.3 
jupiter 6,     13.33 
saturn 7,     24.49 
uranus 8,     49.26 
neptune 9,     78.21 
pluto   10,     101.54 
r=0.2(2^n)+0.6 
I have fit the above data to a mean curve given by r=0.2(2^n)+0.6, where
n is the planet number and r is its distance from the sun. 
adrastea       1,     1 
amalthea       2,     1.4 
thebe   3,     1.7 
io     4,     3.3 
europa 5,     5.2 
ganymede       6,     8.3 
callisto       7,     14.6 
r=0.1(2^n)+0.9 
I have fit the above data for the natural satellites of Jupiter to a
mean curve of r=0.1(2^n)+0.9. I have started with andrastea as the first
satellite, even though it is Metis. In any case the equation would
remain the same (the two satellites are so close together, they can be
considered one.) 
janus   1,     1 
mimas   2,     1.22 
enceledus       3,     1.57 
tethys 4,     1.94   r=0.05(2^n)+0.9 
dione   5,     2.49 
Rhea   6,     3.48 
titan   7,     8.06 
hyperion       8,     9.78 
iapetus 9,     23.5 
phoebe 10,     85.5 
Distribution of the satellites of Saturn greater than 100 miles in
diameter. 
Here I have fit a mean curve to the satellites of Saturn. Any satellite
less than 100 miles in diameter was not considered to be a satellite.
The mean curve is r=0.05(2^n)+0.9. 
Thus we see that the equation for the distribution of bodies around a
central body is geometric and of the form a(2^n)+c. This is different in
form than the Titius-Bode rule, because n=1 is the first satellite or
planet, and with the Titius-Bode rule the first planet is n=minus
infinity. The arithmetic mean for a is 0.1 and for c is 0.8. Thus we
write for orbital systems in general: r= 0.1(2^n)+0.8. 
This puts the third planet away from any star at 1.6 times further than
than the first, which is the golden mean, or phi. It would seem that in
the way that orbital bodies distribute themselves around a central body,
the third planet from any star would be the one where the conditions
would be right to support life, in so far as the distribution rate is
related to the mass of the central body and the luminosity of any star
is related to its mass. (It is given by L~M^3.5, where L is in solar
luminosities and M is in solar masses.) 
Let’s assume that d=xm, where d is the mean distance of the first
orbital body from the central body, m is the mass of the central body
and x a constant of proportionality. Then for the sun, with mercury as
the first planet: 
1.989*10^30kg(x)=57809197km 
and 
x=2.91*10^-22km/kg 
and for Saturn and Pan: 
133583km=5.686*10^26kg(x) 
and 
x=2.349*10^-22km/kg 
For Jupiter and Metis I get x=6.74*10^-23km/kg 
Assuming that Jupiter is anomalous, and the sun is the best
representation for this model, we average the Sun and Saturn to get
x=2.63*10^-22km/kg. Hypothesis: The size of a planet is given by its
density, its density by its mass, the density of its atmosphere by its
surface gravity, and it surface gravity by the luminosity of the main
sequence star. This is all part of my & #8220;Project Genesis”. 
Aims: I believe it may be better to learn how to build life bearing
stellar 
systems than to seek out other stellar systems for our survival, in that
there may not be many, if any at all, that exist as we will need them. 


http://community.webtv.net/ianbeardsley/Logica
http://mysite.verizon.net/res8ydyw/
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