Homework 4 Solutions
Problem 1
We start with
L(r)
L
r
x=
R
f=
Then, we can write
dL(r)
= 4πr2 ρ
dr
R
df
= 4πr2 ρ
dx
L
Next, we substitute = 0 ρT ν and use the ideal gas law in the form ρ =
to write
2
df
R
µ
2
= 4πr 0
P Tν
dx
L
Rg T
µ
Rg T
P
Switching to dimensionless variables yields
R3
df
µ2
=
4πx2 0 2 p2 P02 tν−2 T0ν−2
dx
L
Rg
Finally, we can use P0 =
GM 2
4πR4
and T0 =
µGM
RRg
to write
df
0 µν Gν M ν+2 2 2 ν−2
=
x p t
dx
4πLRgν Rν+3
= Dx2 p2 tν−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.
NH =
Msol
= 1.2 × 1057
MH
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 × 1058
= 5 × 101 2 s = 160, 000 years
T =
Lsol
Which is far too short of a time.
Problem 10.5
We use equations 10.9 and 8.1 to write
Z
1 ∞
P =
mnv v 2 dv
3 0
Z
m 3/2
2
4π ∞
=
e−mv /2kT v 4 dv
mn
3 0
2πkT
This integral can be evaluated to yield the answer.
√ πnm m 3/2
P = π
2
2πkT
2kT
m
5/2
= nkT
Problem 10.7
The virial theorem in the form of equation 2.46 is
−2hKi = hU i
According to equation 10.22 the gravitational potential energy of the sun is
U ≈−
3 GM 2
= −2.28 × 1041 J
5 R
The thermal kinetic energy is given by 23 N kT , where N is twice the number of
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
3N k
= 2.29 × 106
T =−
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 D0 ) = −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
1
ξ 2 D00 = − ξ 3 + C
3
C
1
0
D0 = − ξ + 2
3
ξ
Now, the boundary condition Dn0 (0) = 0 implies that C = 0. From here, we
integrate once more to obtain
1
D0 (ξ) = − ξ 2 + C 0
6
Then, Dn (0) = 1 implies that C 0 = 1 so we finally have
1
D0 (ξ) = 1 − ξ 2
6
√
which of course means that ξ1 = 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 = mc2 as
Lsol /c2 = 4.27 × 109 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 101 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 H − ions. So, the Saha
equation will give the ratio of neutral hydrogen to H − . As usual, we take
ZH = 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
NH
= 5 × 107
NH −
NH −
= 2 × 10−8
NH
Problem 8
The Luminosity is given by the Stefan-Boltzmann equation
L = 4πR2 σT 4 = 1.27 × 1027 W = 4.02 × 103 4 J/yr
Now we use the formula for the gravitational potential energy to find the radius
after 1 year
1
3GM 2
1
= 4.02 × 1034 J
−
10
Rf
3Rsol
Rf = 2.08652 × 109 m
= .9999998 × 3Rsol
Combining the Stefan-Boltzmann equation with the inverse square law yields
the following for the flux.
4πR2 σT 4
R2
=
σT 4
4πr2
r2
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 =
Ff
= 100−.001/5
Fi
2
Rf
= .999079
Ri
Rf
= .99954
Ri
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
= 230 years
1 − .999998
4