Homework 1 Solutions
Problem 1
A)
For an elliptical orbit we just plug into equation 2.30
p
Ltotal = µ GM a(1 − e2 ) = 1.927 × 1043 kg m2 /s
B)
We will use L = mrv. We need the radius and the orbital velocity. The distance
between the sun and the center of mass is given by
as = µa/Ms = 7.426 × 108 m
and the orbital velocity is
v = 2πas /P = 12.46 m/s
So we get
Ls = Ms as vs = 1.840 × 1040 kg m2 /s
C)
We use the same formulas as for the previous part, but with the values for
Jupiter instead of those of the sun. We get
L = 1.929 × 1043 kg m2 /s
This is very close to the number found in part a, which makes sense because
the angular momentum of the sun is so small compared to the total and the
circular orbit approximation is good.
D)
We use
L = Iω =
1
2 2
mr ω
5
Plugging in the numbers given for the sun and Jupiter we get
LSrotational = 1.078 × 1042 kg m2 /s
LJrotational = 6.312 × 1038 kg m2 /s
E)
Jupiter’s orbital angular momentum makes the largest contribution.
Problem 2
A)
If we use units of AUs and years Kepler’s third law becomes an equality. This
gives us the semi-major axis from the period.
a = P 2/3 = 17.9AU
B)
We will use Newton’s version of Kepler’s third law given by equation 2.37. We
can neglect the mass of the comet because it so much smaller than the solar
mass.
4π 2 3
a = 1.98 × 1030 kg
MS =
GP 2
C)
The answers are immediately obtained upon plugging into equations 2.5 and 2.6
rp = a(1 − e) = .585 AU
ra = a(1 + e) = 35.2 AU
D)
the velocities are given by equations 2.33, 2.34, and 2.36 in the text.
s
vp =
GM (1 + e)
= 55 km/s
rp
s
GM (1 − e)
= .91 km/s
ra
r
GMS
vsemi-minor =
= 7.0 km/s
a
va =
2
E)
The ratio of kinetic energies is given by the square of the ratio of velocities.
vp2
= 3650
va2
Problem 3
Thermal equilibrium requires that energy is emitted at the same rate it is absorbed. First, we will find the energy input from the sun. The solar constant
is 1,361 W/m2 but most of the Earth’s surface is not directly facing the sun.
Instead the energy absorbed is the same as a disk with the same radius as the
Earth that is directly facing the sun. Then we must also take the albedo into
account.
So, the power per square meter due to the sun is
P = 1, 361 ×
1
πr2
2
(1 − a) = 1, 361 × × .61 = 207.55 W/m
2
4πr
4
First of all, we can see that the internal heat flow is negligible, so we will
ignore it. Now, to find the equilibrium temperature we plug this into the StefanBoltzmann law.
207.55 = σT 4
T = 246 K
The difference between this and the actual mean temperature is primarily
due to the greenhouse effect.
Problem 4
We need to find the maximum of the Planck function, so we take the derivative
and set it equal to zero.
8πhc
1
λ5 ehc/λkT − 1
∂B
hcehc/λkT
5
= 8πhc
−
=0
∂λ
kT λ7 (ehc/λkT − 1)2
λ6 (ehc/λkT − 1)
B=
We set x ≡ hc/λkT and simplify to get
x
ex
−5=0
ex − 1
3
We can solve this equation numerically to get x ≈ 4.97. Then,
hc
= 4.97
λkT
λT = 2.9 × 10−3 m K
Problem 5
A)
Equation 3.17 gives
L = 4πR2 σT 4 = 1.17 × 1031 W = 30, 500LS
B)
Use equation 3.7.
M = MS − 2.5 log(L/LS ) = −6.47
C)
The apparent magnitude is given by equation 3.6.
m = M + 5 log(d/10 pc) = −1.02
D)
The distance modulus is m − M = 5.45.
E)
Equation 3.18 gives
F = σT 4 = 3.48 × 101 0 W/m
2
F)
Use the inverse square law
f=
L
2
= 6.5 × 10−8 W/m
4πr2
This divided by the solar constant is 4.7 × 10−11 .
G)
Using Wien’s law
λmax =
2.9 × 10−3
= 103.6 nm
28000
4
Problem 6
A)
The Planck function is
2hc2
1
λ5 ehc/λkT − 1
In the classical limit we can Taylor expand the exponential and drop higher
order terms.
ex ≈ 1 + x
B=
Then we get
B=
2hc2
1
λ5 hc/λkT
=
2ckT
λ4
B)
4 ´ 1013
3 ´ 1013
2 ´ 1013
1 ´ 1013
1000
2000
3000
4000
Problem 7
A)
We need to do the integral
∞
Z
L=
0
8π 2 R2 hc2
dλ
− 1)
λ5 (ehc/λkT
We will perform a change of variables
hc
λkT
hc
du = − 2
dλ
λ kT
u=
5
5000
Substituting, and remembering to take into account the integration limits
to cancel the minus sign, we get
Z
8π 2 R2 k 4 T 4 ∞ u3 du
L=
c2 h3
eu − 1
0
8π 6 R2 k 4 T 4
=
15c2 h3
B)
Comparing with the Stefan-Boltzmann law gives
8π 6 R2 k 4 T 4
= 4πR2 σT 4
15c2 h3
2π 5 k 4
σ=
15c2 h3
C)
Plugging in yields σ = 5.67 × 10−8 Wm−2 K−4
Problem 8
A)
We use
U − B = −2.5 log
B365 ∆λU
B440 ∆λB
+ CU −B = −1.08
Doing the same for B − V yields −.23.
B)
First we find the distance
d=
1
= 216 pc
p
Now, we can use equation 3.6 and 3.30
MV = M − m + V = V − 5 log
6
d
10
= −5.05