Text 82, 206 rader
Skriven 2004-09-17 06:09:00 av John Morrison (1:278/230)
Ärende: Re: Project Genesis
===========================


Ian Beardsley wrote:
> 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.
>

    I'll grant you (at most) /some/ of that - if and _only_ if you can
justify all the various postings of Archimedes Plutonium as well ...

John
johnDOTmorrisonATtescoDOTnet
-- 
"Gegen [die] Dummheit kämpfen [die] Götter selbst vergebens." - Friedrich
von Schiller ("Die Jungfrau von Orléans", Act iii, Scene 6.)
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