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![]() 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. An interesting family of substances are 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. It is a hypothesis of astrobiology that amino acids formed in the protoplanetary cloud before the earth ever formed. In this sense we may have our origins in deep space. 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. These are another interesting family of substances, deeply connected with life as we know it. We now calculate the molecular masses of these special substances: CH 4=3D(12.01+4(1.01))=3D16.05 NH 3=3D(14.01+3(1.01))=3D17.04 CO 2=3D(12.01+2(16.00))=3D44.01 H 2O=3D(2(1.01)+16.00)=3D18.02 N 2=3D(14.01+14.01)=3D28.02 O 2=3D(16.00+16.00)=3D32.00 We now form some ratios between these molecular masses: O 2/CH 4=3D1.992~2 NH 3/CH 4=3D1.061~1 O 2/N 2=3D1.142~sqrt(2) CO 2/N 2~1.6=3D(sqrt(5)+1)/2=3Dphi O 2/H 2O=3D1.776~sqrt(3) Notice that these values are given by the sequence: |2cos(pi/n)| n=3D1,2,3,4,5,6 pi/n radians Observe: 2=3D|2cos(pi)| 0=3D|2cos(pi/2)| 1=3D|2cos(pi/3)| sqrt(2)=3D|2cos(pi/4)| (sqrt(5)+1)/2=3D|2cos(pi/5)| sqrt(3)=3D|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)=3D (5.976E27)(1.495979E13cm)/(1.989E33) =3D4.495E7cm=3D449.5km We now divide that result by the radius of the earth, R e: 449.5km/6378.5k=3D0.07 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=3D1.01/14.01=3D0.07 And we see that H/N=3D(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)=3D 1.01/44.01=3D0.02 M m=3Dmass of mars, M s=3Dmass of sun, r =3D the distance between them and= R m =3D the radius of mars. We have (M m)r/(M s)(R m)=3D0.02 M m=3D6.418E26g r=3D2.279409E13cm M s=3D1.989E33g R m=3D3.393096E8cm (H)/(CO 2)=3D(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)=3D1.000/0.380=3D2.63 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)=3D32.00/12.01=3D2.66 Thus we have (g_e)/(g_m)=3D(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, =A0 =A0 1 venus =A0 2, =A0 =A0 1.8 earth =A0 3, =A0 =A0 2.5 mars =A0 4, =A0 =A0 3.8 asteroids =A0 =A0 =A0 5, =A0 =A0 5.3 jupiter 6, =A0 =A0 13.33 saturn 7, =A0 =A0 24.49 uranus 8, =A0 =A0 49.26 neptune 9, =A0 =A0 78.21 pluto =A0 10, =A0 =A0 101.54 r=3D0.2(2^n)+0.6 I have fit the above data to a mean curve given by r=3D0.2(2^n)+0.6, where= n is the planet number and r is its distance from the sun. adrastea =A0 =A0 =A0 1, =A0 =A0 1 amalthea =A0 =A0 =A0 2, =A0 =A0 1.4 thebe =A0 3, =A0 =A0 1.7 io =A0 =A0 4, =A0 =A0 3.3 europa 5, =A0 =A0 5.2 ganymede =A0 =A0 =A0 6, =A0 =A0 8.3 callisto =A0 =A0 =A0 7, =A0 =A0 14.6 r=3D0.1(2^n)+0.9 I have fit the above data for the natural satellites of Jupiter to a mean curve of r=3D0.1(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 =A0 1, =A0 =A0 1 mimas =A0 2, =A0 =A0 1.22 enceledus =A0 =A0 =A0 3, =A0 =A0 1.57 tethys 4, =A0 =A0 1.94 =A0 r=3D0.05(2^n)+0.9 dione =A0 5, =A0 =A0 2.49 Rhea =A0 6, =A0 =A0 3.48 titan =A0 7, =A0 =A0 8.06 hyperion =A0 =A0 =A0 8, =A0 =A0 9.78 iapetus 9, =A0 =A0 23.5 phoebe 10, =A0 =A0 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=3D0.05(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=3D1 is the first satellite or planet, and with the Titius-Bode rule the first planet is n=3Dminus infinity. The arithmetic mean for a is 0.1 and for c is 0.8. Thus we write for orbital systems in general: r=3D 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=3Dxm, 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)=3D57809197km and x=3D2.91*10^-22km/kg and for Saturn and Pan: 133583km=3D5.686*10^26kg(x) and x=3D2.349*10^-22km/kg For Jupiter and Metis I get x=3D6.74*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=3D2.63*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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