a problem in star formation

[josephus]

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