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Topic: a problem in star formation
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Josephus

Posts: 54
Registered: 12/13/04
a problem in star formation
Posted: May 21, 2013 8:19 AM
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This  is  a little   idle  reasoning.   it  is  coherent.  but   the
data may be false.    so please help me. all the  calculations  are
doable with a  TI30S calculator.    I put  'B' into  A  and  'C'  into
B
 and  'c' into C  and  'M' into  D

then     B^(1/4) * M^(1/2)   ==  Surface temperature
         B *M^2              ==  Operating temperature  --   inner
temperature.
         B*C*M^4             ==   #Core#  temperature
         M for our sun is    ==  15.04664737
      surface   of sun is    ==  5778
      OPT       of sun is    ==  5778^4
     #Core#     of sun 0s    ==  B*C *m^4

 plsease  let me know  if I am  wrong anywhere.


 josephus


stars  should  collapse  in a predictable  manner..     Let   the
maximum  Radius  be    17  light days and    let  N    be such that the
  motion  from  N  to  1  increases  the   pressure and    the
Temperature.          N/(N-1)!    Is  the  expression  for    a ring
 at
 N  to  fall  to  one  light day  position.  This  is a  geometrical
exploration of  star  formation
NOTE:   #CORE#   is the  center of the star.    It is ~~ *E23  in our
sun.

The  calculated temperature  F(N) * MASS2  is  the  ignition
 temperature.

Just one   F(N)   to  define  the   collapse of  rings  of  gas.   F(17)

  Solve  for   T ‹­>1/2  G*T^2 =   C*D
      YR = (365.25* 86400)
T= (2*C*D/G)^(1/2) /YR  = 27922 years  to fall  one light  day.
     *18
=  500000 years

ASSERT  a laminar  homogeneous  gas  cloud.
ASSERT    F(N)   =    (N-1)!/N *4K  = Temperature  in  KELVIN.
        N= 17
        B =  4.932009385 E12 K
ASSERT    without proof  that all  falling  masses  will  rotate.
ASSERT  no other  stars  are  made  just  this one.
                M =  mass in  solar masses.

If the gas  falls  in a regular  sense  with no other  stars  around.
It
 should  fall  N/(N-1)!  down  to  one light  day.     The   T vs. P
formula  is   P^(-1( *4K   =    temperature  Kelvin.   This implies
that
  (N-1)!/N  *4 K  is  a general formula for star formation. (still a
constant)

 F(17) = 4.923009385E12     -A Constant
    B  = F(17)
My reasoning  involves the   SUN -  1.0 solar mass.
Surface              operating temp          core temp
      5778            1.114577188E15     6.179363346E23
Formulas:
      Surface^4 ~~ OPT
          OPT* 12!/13 ~~    #CORE#
        C=   12!/13  == 3.684627692E 7 K  a minor constant  - just a
long  step
         D= 15.04665
  Sun  surface  ~~ 5778K

       s^4~~   1.114577188E15
     OPT* C~~ 36846275,92  *D= #CORE#
  Note:   B^(1/4)*D^(1/2) = 5778.00000
     B^1/4                   B                     B*C
     5778             1.114577188E15      4.106801971E22        SUN
Temperature
    1489.6           4.023009385E12      1.813945671E20  New solar mass
temperatures
     D^1/2                 D2                  D3

           B*M^2   =  5778^4
    5778^4*M^(1/2)     M star      surface temperature
     57784*M^2      M star  operating temperature
      5778^4*C*D*M^3
    B*C*M^3       both calculate  core temperature for an M class star.

  D= 15.04665   =SOLAR MASS ( ??)
       from my  mass  calculations.
        where M  may run from    1  to  350



  this is  the  formula to find  how much  gas is in  a  cubic  meter.

The    general  period  for   a  star  to fall  and  ignite   is
18*27922 =  502596 years.  ‹ 5 E5

Sun  Kg/(4/3*(C*D*18))^3  gives    Kg/M^3

 1.989044385E30/(4/3*C^3*D^3*18^3 ) = 4.685273976E-15  Kg/M^3 * MASS

  Kb/m^3 call this  value  K.*(245)    ->   massive star


Use the above formula    to  calculate  the  gas  in Kg/M3  * mass  of
target in solar masses. Of gas
The correct with my step function.
F(N) = B is  the  constant  that  allows  me  to  create    T for
various  stars.       I  am using   the  temperature  model  of  our
sun
 to   extrapolate  to  the surface,  ignition sequence, and core.
   M^(1/2)       M^2           M^3

C = 12!/13  the heat step for the sun  modified by M^3 and  D

B^(1/4)*D^(1/2)= 5778          (*M1/2)
B*D^2 = 5778^4                 (*M^2)
Core =  5778^4*C*D    (*M^3)
Core =  B*C*D^3    =  5778^4*C*D

the fact that  B = (17!/15)*4 = 4.923000385E12
               D = 15.0466737
               C = (12!/13) =  36846276.92

 B^*(1/4)*M^(1/2) =  surface temp
 B*M^2            =  operating temp
 B*M^3*C          =  core temp.

with these value and  M asserted  near  150 solar masses  one can
calculate  the  temperatures  approximate to  reality.


I  really want someone to look and  tell me what is right or wrong
josephus

--


I go sailing in the summer
  and look at stars in the winter
Its not what you know that gets you in trouble
  Its what you know that aint so.   -- Josh Billings



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