## Monday, December 19, 2016

### Solutions For No. 1 of Suggested Comprehension Exercises ('Construction of A Star', Part 2)

1) Using the appropriate stellar model parameter relations calculate the core values: T c  ,  r c ,  Pc  .

Soln. We have, for the three core parameters -

T  = Tc      ,    q  = (  rr c  )1/n    And q =   (P/ Pc ) (1+ 1/n)

At the interface (core-radiative boundary)  one finds:  q  = 0. 7839768

Then, for the (dimensionless)  core temperature:   t c     =  t /   q

And, since 0.6969 =    t c   q   =     t c    (0.7839768)  then:   t c   =  0.8876

The dimensional form is:  T  =  Tc   q

Since we have, at r/R = 0.172, log T = 7.1643   then

antilog (7.1643)  =   T   =   1.46  x 107  K

And:   Tc      =   /   q

For which we set up the ratio:

(0.6969)/  / (0.8876) =  (1.46  x 107  )/   Tc

Whence:  Tc     =   (1.46  x 107  ) (0.8876)/  (0.6969)  = 1.86  x 107  K

Take log T = log (1.86  x 107  )  = 7.2695

Which is the core temperature log value we find at r /R  = 0.0, i.e., See Table in Part 1

For core density:

q  = (  rr c )1/n

Where:   The polytropic index n can be defined:   n = 1/ (g   - 1)

where g   is the ratio of specific heats. (g  =  C p / C )

In a non-relativistic limit  one will have g = 5/3 and

n = 1 / (5/3 - 1)  =  1/ (2/3)  =  3/2

Then write the core density as:   r c    =    rq 3/2

Now, take  q  =   0.7839  SO:   q 3/2    =   0.6941

Since we have, at r/R = 0.172, log r = 1.4033   then

antilog (1.4033)  =   r   =   25.31 g/ cm3

Therefore:  r c    =    rq 3/2  =   25.31 g/ cm3 / 0.6941

And:     r c    =   36.46   g/ cm3

Now, take log  (r c  ) =   1.5618

Which is the density log value we find at r /R  = 0.0, i.e. the core (See Table in Part 1)

For core pressure, we have:  q =   (P/ Pc ) (1+ 1/n)

Since:  q    =  (p/ p c ) (1+ 1/n)     and n = 1.5 (assigned polytropic index - see top of Lane-Emden function table)

Then:   p c   =     p / q 5/2      and   q 5/2 =  0.5441

And, since we have, from the model table,  at r/R = 0.172, log P = 16.6770   then

antilog (16.6770 )  =   P = 4.754  x 1016    dynecm 2

Therefore:     P c   =   P /   q 5/2 =  (4.754  x 1016    dynecm2 )/  0.5441

P c   =   8.737  x 1016    dynecm2

and the log is: 16.9414 or the value shown at r/R = 0.0 (the core)