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 10^{3}JK^{-1})(3.14 x 10^{6}K)/ 0.533
P = 9.88 x 10

^{11}Nm^{-2}
The radiation pressure is: P

_{R}= aT^{4}/ 3
Where a = 7.55 x 10

^{-16}Jm^{-3}K^{-4 }
So that P

_{R}= 1/3 (7.55 x 10^{-16}Jm^{-3}K^{-4 })( 3.14 x 10^{6}K)^{4}
And: P

_{R}= 2.44 x 10^{10}Nm^{-2}
The total pressure is:

P

_{T}= r RT/ m + aT^{4}/ 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 }r

Then by

*the chain rule for derivatives*:
(dL/dM) (dM/dr) =

dL/dr = e (4p r

dL/dr = e (4p r

^{2 }r)^{2}

^{2}

P = (6.7 x 10

^{-11}Nm^{2}kg^{-2})( 2 x 10^{30}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 10^{3}JK^{-1})
T = 1.2 x 10

^{ 7 }K(Compare to 15 million degrees K, using a more detailed approach)

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