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mystic math



 
 
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  #1  
Old July 2nd 04, 05:25 AM
Ian Beardsley
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Default mystic math

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)

Ian Beardsley


  #2  
Old July 2nd 04, 05:39 AM
Sam Wormley
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Posts: n/a
Default mystic math

Ian Beardsley wrote:

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)

Ian Beardsley


Crank Information
http://www.google.com/search?q=numer...Awww.crank.net
http://www.csicop.org/cgi-bin/search...i?q=numerology
  #3  
Old July 2nd 04, 05:39 AM
Sam Wormley
external usenet poster
 
Posts: n/a
Default mystic math

Ian Beardsley wrote:

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)

Ian Beardsley


Crank Information
http://www.google.com/search?q=numer...Awww.crank.net
http://www.csicop.org/cgi-bin/search...i?q=numerology
  #4  
Old July 2nd 04, 06:02 AM
Ian Beardsley
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Posts: n/a
Default mystic math (now it becomes mystic)


mystic math

Group: sci.astro.amateur Date: Thu, Jul 1, 2004, 9:25pm From:
(Ian=A0Beardsley)
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)
Ian Beardsley

-----------------------------------
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.

Ian Beardsley


  #5  
Old July 2nd 04, 06:02 AM
Ian Beardsley
external usenet poster
 
Posts: n/a
Default mystic math (now it becomes mystic)


mystic math

Group: sci.astro.amateur Date: Thu, Jul 1, 2004, 9:25pm From:
(Ian=A0Beardsley)
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)
Ian Beardsley

-----------------------------------
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.

Ian Beardsley


  #6  
Old July 2nd 04, 06:17 AM
Sam Wormley
external usenet poster
 
Posts: n/a
Default mystic math (now it becomes mystic)

Crank Information
http://www.google.com/search?q=numer...Awww.crank.net
http://www.csicop.org/cgi-bin/search...i?q=numerology
  #7  
Old July 2nd 04, 06:17 AM
Sam Wormley
external usenet poster
 
Posts: n/a
Default mystic math (now it becomes mystic)

Crank Information
http://www.google.com/search?q=numer...Awww.crank.net
http://www.csicop.org/cgi-bin/search...i?q=numerology
  #8  
Old July 2nd 04, 10:53 AM
Ian Beardsley
external usenet poster
 
Posts: n/a
Default mystic math (now it becomes mystic)


mystic math (now it becomes mystic)

Group: sci.astro.amateur Date: Fri, Jul 2, 2004, 5:17am (PDT+7) From:
(Sam=A0Wormley)
Crank Information
=A0=A0=A0=A0http://www.google.com/search?q=3Dnumerology+site%3Awww.crank.n=
et
http://www.csicop.org/cgi-bin/search...q=3Dnumerology
--------------------------------------------
Por dios, there is nothing cranky about it!!!
Ian

  #9  
Old July 2nd 04, 10:53 AM
Ian Beardsley
external usenet poster
 
Posts: n/a
Default mystic math (now it becomes mystic)


mystic math (now it becomes mystic)

Group: sci.astro.amateur Date: Fri, Jul 2, 2004, 5:17am (PDT+7) From:
(Sam=A0Wormley)
Crank Information
=A0=A0=A0=A0http://www.google.com/search?q=3Dnumerology+site%3Awww.crank.n=
et
http://www.csicop.org/cgi-bin/search...q=3Dnumerology
--------------------------------------------
Por dios, there is nothing cranky about it!!!
Ian

  #10  
Old July 2nd 04, 11:11 AM
DT
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Posts: n/a
Default mystic math (now it becomes mystic)

We have read your post and we have marked you uninteresting.

Denis
--
DT
Replace nospam with the antithesis of hills
 




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