Advanced Astrophysics (3) - Solutions

We now look at the solutions for the last Astrophysics post:

1) Begin by finding the mean molecular weight of the material given that:

X = 0.9 and we use:

m = 4/ (3 + 5X)  =  4/ (3 + 5(0.9)) = 0.533

The gas pressure, P = (r/ mH) kT =  r RT/ m

=  (20.2 kg m^-3) (8.3 x 10³ JK^-1)(3.14 x 10⁶ K)/ 0.533

 

P =  9.88 x 10 ¹¹ Nm^-2

The radiation pressure is: P _R = aT⁴/ 3

Where a = 7.55 x 10 ^-16 Jm^-3K^-4   

So that P _R = 1/3 (7.55 x 10 ^-16 Jm^-3K^-4 )( 3.14 x 10⁶ K)⁴

And: P _R =  2.44 x 10 ¹⁰ Nm^-2

The total pressure is:

P_T = r RT/ m  +  aT⁴/ 3 =

9.88 x 10 ¹¹ Nm^-2   + 2.44 x 10 ¹⁰ Nm^-2  » 10 ¹² Nm^-2  

Thus, the overall contribution of radiation pressure is only on the order of 0.02 or about 2% of the total pressure.

The thermal energy per kg is: U = P/ (g   - 1) r

 

Or: U = (10 ¹² Nm^-2 )(5/3 – 1) (20.2 kg m^-3)

U = 7.4 x  10 ¹⁰ J kg^-1

2)  We use: dL/dM = e  and dM/dr = 4p r² r

Then by the chain rule for derivatives:

(dL/dM) (dM/dr) =

dL/dr = e  (4p r² r)

3)   We approximate: dP/dr = - G M(r) r dr/ r²

To:   P/R = G M r / r²

P = (6.7 x 10^-11 Nm²kg^-2)( 2 x 10³⁰ kg)( 1400 kgm^-3)/ R

Where R = 7 x 10 ⁸ m

Then: P = 2.6 x 10 ¹⁴ Pa

(Note: The actual central pressure is about an order of magnitude larger.)

For the estimate of the central temperature we use:

T =  m P / r R

Where we know m = 0.57 for a fully ionized H-plasma.

Then: T =

(0.57)( 2.6 x 10 ¹⁴ Pa)/(1400 kgm^-3)( (8.3 x 10³ JK^-1)

T = 1.2 x 10 ⁷ K

Or, 12 million degrees Kelvin.


(Compare to 15 million degrees K, using a more detailed approach)