## Wednesday, December 21, 2016

### Soloutions to Problems 2, 3 for Suggested Comprehension Ex. ('Constructing A Star', Pt. 2)

2) Do the first two radiative envelope computations for temperature, density and pressure, i.e. at radial fractions r/R = 0.183 and r/R = 0.200.

Soln.  (A)  At r/ R = 0.183   And we have: U = 2.562 and V = 1.389

r =   9.800  x 1010  cm (0.183) =  1.793  x 1010  cm

Mr/M = 0.172   so

Mr   =  3.970  x 1033  g (0.172) = 6.83  x 1032  g

density

r   =     UMr/4p r3

=  [(2.562) (6.83  x 1032  g)]/ 4p (1.793  x 1010  cm) 3

r  =   24.16  g/ cm 3        log   r  = 1. 3831

Pressure:

P   =      G  r Mr/ V r

=  [(6.67 x  10-8 dyne-cm 2/ g.2)  (24.16  g/ cm 3 )(6.83  x 1032  g)/   (1.389)(1.793 x 1010  cm)

P   =  4.417  x 1016    dynecm 2

log P   =   16.645

Temperature:

T = t  m  (H/R) (GM/r)

log t = - 0.15695  so t (dimensionless T) =  0.6969

T =  (0.6969) [m  (H/R) (GM/r) ]      (for  m  = = 0.64,   H  = 1.672  x 10-24)

So:

T =   1.417   x 107  K  and log T  = 7.1516

Soln: (B)

At r/ R = 0.200   And we have: U = 2.450  and V = 1.684
r =   9.800  x 1010  cm (0.200) =  1.960  x 1010  cm

Mr/M = 0.217   so

Mr   =  3.970  x 1033  g (0.217) = 8.61  x 1032  g

density

r   =     UMr/4p r3

=
[(2.450) (8.61  x 1032  g)]/ 4p (1.960  x 1010  cm) 3

r  =   22.31  g/ cm 3        log   r  = 1. 3485

Pressure:

P   =      G  r Mr/ V r

=  [(6.67 x  10-8 dyne-cm 2/ g.2)  (22.31  g/ cm 3 )(6.83  x 1032  g)/   (1.684)(1.96 x 1010  cm)

P   =  3.884  x 1016    dynecm 2

log P   =   16.5893

Temperature:

T = t  m  (H/R) (GM/r)

log t = - 0.19200  so t (dimensionless T) =  0.6426

T =  (0.6426) [m  (H/R) (GM/r) ]      (for  m  = = 0.64,   H  = 1.672  x 10-24)

So:

T =   1.346   x 107  K  and log T  = 7.1291

3) Estimate the uncertainty in the core radius as a consequence of the 'fuzziness' of the interface parameters.

The percentage uncertainty can be computed as:

D  (  r c  )   =  100%(0.172 - 0.148) / 0.172   =  0.024(100%)=   2.4 %