## Thursday, June 20, 2013

### 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:

PT = 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)