Ian Beardsley
10-06-2004, 09:47 PM
Here is a project I have undertaken to do art with scientific tools, and knowlege.
An Anthropic Solar System and Planet by Ian Beardsley
Copyright by Ian Beardsley October 5, 2004
Abstract: It is shown here that there is a connection between the microworld, the world of elements and compounds, and the macroworld, the world of our solar system, especially where humans are concerned. Such a connection is evidence for an anthropic universe.
Furthermore, 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. This correlation between the microworld and the macroworld, where our solar system is concerned, may define some of the parameters necessary to achieve this, and may be related to why the solar system is life bearing.
Part 1
An interesting family of substances is 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, and are all in amino acids, the hydrogen as well. 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. Is what I mean by structural systems is that there are only three structural systems, the tetrahedron, the octahedron, and the icosohedron. They are the only stable solids, that is non-collapsing flex corners whose faces are
triangles. Most compounds are something other than these, like pentagons with linear off shoots for example, that comprise the wrong number of atoms to make a "solid" unit, and I mean solid as in the pythagorean solids, the geometric term. Both methane and ammonia make different variations on the tetrahedron, a pythagorean solid.
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)=32.00/16.05=1.992~2
(NH_3)/(CH_4)=17.04/16.06=1.061~1
(O_2)/(N_2)=32.00/28.02=1.142~sqrt(2)
(CO_2)/(N_2)=44.01/28.02~1.6=(sqrt(5)+1)/2=phi
(O_2)/(H_2O)=32.00/18.02=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=phi=|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) is 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.
Part 2
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/M_s)r=(5.976E27/1.989E33)(1.495979E13)=(4.495E7)cm=449.5km
We now divide that result by the radius of the earth, R_e:
(449.5km)/(6378.5km)=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))
Having showed the last equation, where hydrogen is the most abundant element in the universe and nitrogen is the most abundant element in the earth atmosphere, then since Mars is a terrestrial planet upon which we can set foot as opposed to
Venus and Mercury, let’s apply the same idea to mars. The most abundant gas in the Mars atmosphere is carbon dioxide, or CO_2. It in fact comprises 95.3% of its atmosphere. We have:
H/(CO_2)=1.01/44.01=0.02
Now let M_m= mass of mars, M_s = mass of the sun,r = the distance between them, and R_m= the radius of mars. We have
(M_m)(r)/(M_s)(R_m)=(6.418E26)(2.279409E13)/(1.989E33)(3.393096E8)=0.02
And therefore,
H/(CO_2)=(M_m)(r)/(M_s)(R_m)
Keep in mind these equations, both for the earth and mars, hold for a solar system at its peak as an orderly arrangement of parts. Eventually it will begin to degenerate. The sun is losing mass every day and therefore r, for any of the planets, will grow.
Thus we say in general:
H/A=(M_p)r/(M_s)(R_p)
where H is the molar mass of hydrogen, A is the molar mass of the most abundant element or gas in the planet’s atmosphere, (M_p) is the mass of the planet, (M_s) is the mass of the star, r is the distance between the planet and the star and (R_p) is the radius of the planet. Lets look at the quantity (M_p)r/(M_s). It is equal to (d_1)/(d_2)(d_1+d_2), the ratio of the distances between the balancing point of a cosmic teeter totter and the planet and the star balanced on it, times its length. We then compare such a distance to the radius of the planet.
Part 3
The relative equatorial surface gravities uncorrected for centrifugal force of the earth and mars respectively are 1.000 and 0.380. Their proportions are
1.000/0.380=2.63
The ratio of the molar mass of oxygen gas to that of carbon is
(O_2)/C=32.00/12.01
Thus, (g_e)/(g_m)~(O_2)/C
where g_e is the equatorial surface gravity of the earth and g_m is the equatorial surface gravity of mars. The centrifugal forces being nominal, this 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 if the atmospheric pressures are excluded. Carbon is the basis of life and oxygen gas its necessity (for human life).
The data for this study came from the Handbook Of Space Astronomy And Astrophysics, by Martin V. Zombeck, Cambridge University Press, 1982.
An Anthropic Solar System and Planet by Ian Beardsley
Copyright by Ian Beardsley October 5, 2004
Abstract: It is shown here that there is a connection between the microworld, the world of elements and compounds, and the macroworld, the world of our solar system, especially where humans are concerned. Such a connection is evidence for an anthropic universe.
Furthermore, 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. This correlation between the microworld and the macroworld, where our solar system is concerned, may define some of the parameters necessary to achieve this, and may be related to why the solar system is life bearing.
Part 1
An interesting family of substances is 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, and are all in amino acids, the hydrogen as well. 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. Is what I mean by structural systems is that there are only three structural systems, the tetrahedron, the octahedron, and the icosohedron. They are the only stable solids, that is non-collapsing flex corners whose faces are
triangles. Most compounds are something other than these, like pentagons with linear off shoots for example, that comprise the wrong number of atoms to make a "solid" unit, and I mean solid as in the pythagorean solids, the geometric term. Both methane and ammonia make different variations on the tetrahedron, a pythagorean solid.
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)=32.00/16.05=1.992~2
(NH_3)/(CH_4)=17.04/16.06=1.061~1
(O_2)/(N_2)=32.00/28.02=1.142~sqrt(2)
(CO_2)/(N_2)=44.01/28.02~1.6=(sqrt(5)+1)/2=phi
(O_2)/(H_2O)=32.00/18.02=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=phi=|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) is 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.
Part 2
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/M_s)r=(5.976E27/1.989E33)(1.495979E13)=(4.495E7)cm=449.5km
We now divide that result by the radius of the earth, R_e:
(449.5km)/(6378.5km)=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))
Having showed the last equation, where hydrogen is the most abundant element in the universe and nitrogen is the most abundant element in the earth atmosphere, then since Mars is a terrestrial planet upon which we can set foot as opposed to
Venus and Mercury, let’s apply the same idea to mars. The most abundant gas in the Mars atmosphere is carbon dioxide, or CO_2. It in fact comprises 95.3% of its atmosphere. We have:
H/(CO_2)=1.01/44.01=0.02
Now let M_m= mass of mars, M_s = mass of the sun,r = the distance between them, and R_m= the radius of mars. We have
(M_m)(r)/(M_s)(R_m)=(6.418E26)(2.279409E13)/(1.989E33)(3.393096E8)=0.02
And therefore,
H/(CO_2)=(M_m)(r)/(M_s)(R_m)
Keep in mind these equations, both for the earth and mars, hold for a solar system at its peak as an orderly arrangement of parts. Eventually it will begin to degenerate. The sun is losing mass every day and therefore r, for any of the planets, will grow.
Thus we say in general:
H/A=(M_p)r/(M_s)(R_p)
where H is the molar mass of hydrogen, A is the molar mass of the most abundant element or gas in the planet’s atmosphere, (M_p) is the mass of the planet, (M_s) is the mass of the star, r is the distance between the planet and the star and (R_p) is the radius of the planet. Lets look at the quantity (M_p)r/(M_s). It is equal to (d_1)/(d_2)(d_1+d_2), the ratio of the distances between the balancing point of a cosmic teeter totter and the planet and the star balanced on it, times its length. We then compare such a distance to the radius of the planet.
Part 3
The relative equatorial surface gravities uncorrected for centrifugal force of the earth and mars respectively are 1.000 and 0.380. Their proportions are
1.000/0.380=2.63
The ratio of the molar mass of oxygen gas to that of carbon is
(O_2)/C=32.00/12.01
Thus, (g_e)/(g_m)~(O_2)/C
where g_e is the equatorial surface gravity of the earth and g_m is the equatorial surface gravity of mars. The centrifugal forces being nominal, this 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 if the atmospheric pressures are excluded. Carbon is the basis of life and oxygen gas its necessity (for human life).
The data for this study came from the Handbook Of Space Astronomy And Astrophysics, by Martin V. Zombeck, Cambridge University Press, 1982.