Problem 7.3
A)
Referring to figure 7.8 we can see that the threshold for an eclipse is at cos i = r
1
+ r
2 a
B) i = cos
− 1
.
0046(10 + 1)
2
= 88 .
55
◦
Problem 7.4
A)
We can get the sum of the masses from Kepler’s third law if we can figure out the semi major axis.
4 π 2 a 3 m a
+ m b
=
GP 2
The semi major axis can be determined from its given angular extent and the distance. The distance is given by
1
.
37921 00
= 2 .
63 pc = 8 .
12 × 10
16 meters
Then, the semi major axis is given by
π
7 .
61
3600 × 180
= a d a = 3 .
00 × 10
12 m
Plugging into Kepler’s third law yields m a
+ m b
= 3 .
24 solar masses
Finally, we can use this along with m a a a
= m b a b to find the individual masses.
m a
= 2 .
21 solar masses m b
= 1 .
03 solar masses
1

B)
Use equation 3.7 with M
1
= 4 .
74 and L
1
= 1.
L = 100
(4 .
74 − M ) / 5
This yields for the luminosities, in terms of the solar luminosity
L
A
= 22 .
5
L
B
= .
024
C
We will use the Stefan-Boltzmann equation
L = 4 πR
2
σT
4
R = 5 .
85 × 10
6 m
= .
0084 R sol
= .
92 R
Earth
Problem 7.13
The fraction of light blocked is given by the ratio of the cross sectional area of
Jupiter and the sun, which is the ratio of the radii squared.
R 2
J
R 2 sol
= .
01
Problem 8.1
Use k = 8 .
62 × 10
− 5 eV/K and take 300K for room temperature
8 .
62 × 10
−
5 × 300 =
1
38 .
7 eV
For the other two cases T = 11600 K and T = 157773 K.
Problem 8.7
We use g n
= 2 n 2 and E n
= − 13 .
6 /n 2 to write
Z ≈ 2 + 5 .
8 × 10
− 5
+ 1 .
45 × 10
− 5 which is very close to 2.
2

Problem 8.9
First, we must write N
II
/N t in terms of N
II
/N
I
, then we can use equation 8.8.
N
II
N
I
+ N
II
N
II
/N
I
=
1 + ( N
II
/N
I
)
For convenience, define f ≡ N
II
/N t
. Then, we can use the substitutions given in the problem to rewrite the electron number density that appears in the Saha equation as f ρ n e
= m p
Then, we can write the Saha equation as
N
II
N
I
= m p f ρ
2 πm h e
2 kT
So our equation is f 1 + m p f ρ f + m p
ρ f
2
+ f m p
ρ
2 πm h e
2 kT
2 πm e kT h 2
2 πm e kT h 2
3 / 2 e
− χ
1
/kT
!
− m p f ρ
3 / 2 e
− χ
1
/kT − m p f ρ
3 / 2 e
− χ
1
/kT − m p
ρ
3 / 2 e
− χ
1
/kT
2 πm h
2 πm h e
2 e
2
2 πm e h 2 kT kT kT
3 / 2
3 / 2
3 / 2 e
− χ
1
/kT e
− χ
1
/kT e
− χ
1
/kT
= 0
= 0
= 0
Where the last line is obtained by multiplying both sides by f.
B)
First we will define b ( T ) ≡ m p
ρ
2 πm h e
2 kT
3 / 2 e
− χ
1
/kT
= 4 .
02 T
Now, we plug this into the quadratic formula to get
− b +
√ b 2 + 4 b f =
2
The plot of this function is shown in figure ??
3 / 2 e
− 157773 /T
Problem 8.12
We use the same method as the previous problem and write f =
N
II
/N
I
1 + N
II
/N
I
3

1.0
0.8
0.6
0.4
0.2
10 000 15 000 20 000
Figure 1: The fraction of ionized hydrogen versus temperature in Kelvins
25 000
Plugging all of the values from the problem into the Saha equation and then plugging that into the above equation yields f = .
71
The discrepancy is caused by the fact that the density at the center of the sun is so high that it effectively decreases the ionization energy, which results in more ionization occurring.
Problem 8.13
We will use the alternative form of the Saha equation given by equation 8.9
to calculate N
III
/N
II
. From the problem we know that χ
III
= 11 .
9 eV and
Z
III
= 1, the rest of the necessary information is given in the problem. Plugging in yields
N
III
= .
002
N
II
This means that almost all of the calcium atoms are CA II, and therefore available for forming the H and K lines.
4

Problem 9.7
We know that you see down to an optical depth of 2 / 3. So, the distance is given by the equation
2 s =
3 κρ
Plugging in κ = .
03 square meters per kilogram and ρ = 1 .
2 kilograms per cubic meter yields.
s = 18 .
52 m
5