Problem 1
We start with
L ( r ) f = x =
L r
R
Then, we can write dL ( r )
= 4 πr
2
ρ dr df dx
=
R
L
4 πr
2
ρ
0
ρT ν and use the ideal gas law in the form ρ =
µ
R g
T
P Next, we substitute = to write df dx
=
R
L
4 πr
2
0
µ
R g
T
P
Switching to dimensionless variables yields
2
T
ν df dx
=
R
3
4 πx
2
L
µ
2
0
R 2 g p
2
P
0
2 t
ν − 2
T
0
ν − 2
Finally, we can use P
0
= GM
2
4 πR 4 and T
0
=
µGM
RR g to write df dx
=
0
µ ν G ν M ν +2 x
2 p
2 t
ν − 2
4 πLR ν g
R ν +3
= Dx
2 p
2 t
ν − 2
Problem 10.3
First we calculate the number of atoms in the sun by dividing the solar mass by the mass of a hydrogen atom.
N
H
=
M sol
M
H
= 1 .
2 × 10
57
1

If a chemical reaction releases 10 eV per atom then we can get the total chemical energy contained in the sun by multiplying the number of hydrogen atoms above by 10 eV. Then, we divide by the solar luminosity(in eV per second) to get the time.
1 .
2 × 10 58
T = = 5 × 10
1
2 s = 160 , 000 years
L sol
Which is far too short of a time.
Problem 10.5
We use equations 10.9 and 8.1 to write
P =
=
1
3
4
Z
∞
π mn v v
2
0
Z
∞ mn
3
0 dv m
2 πkT
3 / 2 e
− mv
2
/ 2 kT v
4 dv
This integral can be evaluated to yield the answer.
P =
√
π
πnm
2
= nkT m
2 πkT
3 / 2 2 kT m
5 / 2
Problem 10.7
The virial theorem in the form of equation 2.46 is
− 2 h K i = h U i
According to equation 10.22 the gravitational potential energy of the sun is
U ≈ −
3
5
GM 2
R
= − 2 .
28 × 10
41
J
The thermal kinetic energy is given by 3 N kT , where N is twice the number of
2 hydrogen atoms that we calculated in problem 10.3 (if all of the hydrogen is ionized each atom contributes a proton and an electron). So, we can plug in to get the temperature
U
T = −
3 N k
= 2 .
29 × 10
6
2

Problem 10.17
For n = 0 the left hand side of the Lane-Emden equation is just − 1 and we get
∂
ξ
( ξ
2
∂
ξ
D
0
) = − ξ
2
We can solve this differential equation by successive integrations while applying the boundary conditions along the way to fix the integration constants. Integrating once gives
ξ
2
D
0
0
D
0
0
= −
1
3
ξ
3
+
= −
1
3
ξ +
C
ξ
C
2
Now, the boundary condition D
0 n
(0) = 0 implies that C = 0. From here, we integrate once more to obtain
D
0
( ξ ) = −
1
6
ξ
2
+ C
0
Then, D n
(0) = 1 implies that C
0
= 1 so we finally have which of course means that ξ
1
D
0
( ξ ) = 1 −
1
6
ξ
2
=
√
6.
Problem 11.2
A)
If all of the sun’s energy comes from its mass then the rate at which its mass diminishes is given by E = mc 2 as
L sol
/c
2
= 4 .
27 × 10
9 kg/s = 6 .
77 × 10
− 14 solar masses per year
B)
The mass loss rate due to the solar wind was estimated in the text as 3 × 10
− 14 solar masses per year, about half of the answer in part a.
C)
Given that the sun’s main sequence lifetime is about 10 1 0 years both mass loss processes combined would only decrease the sun’s mass by less than a thousandth of a solar mass in this time.
3

11.3
We can consider the neutral hydrogen atoms as ionized equation will give the ratio of neutral hydrogen to H
H
− ions. So, the Saha
− . As usual, we take
Z
H
= 2, but for the ion, as the problem points out, the ground state is non degenerate so we set the partition function to unity. Then, the last ingredient is the ionization energy, which is .75 eV. So, now we can just plug in to get
N
H
N
H −
N
H −
N
H
= 5 × 10
7
= 2 × 10
− 8
Problem 8
The Luminosity is given by the Stefan-Boltzmann equation
L = 4 πR
2
σT
4
= 1 .
27 × 10
27
W = 4 .
02 × 10
3
4 J/yr
Now we use the formula for the gravitational potential energy to find the radius after 1 year
3 GM
10
2 1
R f
1
−
3 R sol
R f
= 4 .
02 × 10
34
J
= 2 .
08652 × 10
9 m
= .
9999998 × 3 R sol
Combining the Stefan-Boltzmann equation with the inverse square law yields the following for the flux.
F =
4 πR 2 σT 4
4 πr 2
=
R 2
σT
4 r 2
Now, we can calculate the flux ratio for a difference in magnitude of .001 and set it equal to the ratio of final and initial flux.
F f
R f
F i
2
= 100
− .
001 / 5
= .
999079
R i
R f
R i
= .
99954
Now we can use our previous result for the change in the radius over the course of a year to figure out how many years it would take for the radius to shrink by the factor of .99954 needed to see the change in brightness.
1 − .
99954
1 − .
999998
= 230 years
4