Homework 5 Solutions
Problem 1
A)
The mass deficit is 24 − 23.98504 = .01496 amu. This corresponds to an energy
release of 13.95 MeV. The reaction is exothermic.
B)
The change in mass is 24 − 15.99491 − 2 × 4.0026 = −.00011 amu. This is an
energy difference of −.103 MeV, so the reaction is endothermic.
C)
The mass difference is 18.99840+1.008−15.99491−4.0026 = .00889 amu, which
means the energy is 8.3 MeV, the reaction is exothermic.
Problem 2
We’ll start with the small stars. The luminosity is
L = 10−4.3 Ls = 1.92 × 1022 W
If all of the hydrogen is burned then the total energy is given by
E = .007 × .072Ms c2 = 9 × 1043 J
So the hydrogen burning lifetime is
E
= 5 × 1021 s = 1.5 × 1014 yr
L
For the big star we have
L = 106.006 Ls = 3.9 × 1032 W
Only 10% of the hydrogen is burned so we have
E = (.007)(.1)85Ms c2 = 1.1 × 1046 J
The lifetime is then
E
= 9 × 105 yr
L
1
Problem 3
R=
L
4πσT 4
1/2
Plugging in the information we get R.072 = .082 solar radii and R85 = 13.1 solar
radii. Then, the ratio is .0063.
Problem 4
The magnetic pressure is given by equation 11.10
B2
2µ0
Pm =
So, plugging in, Pm = 1.59 × 104 Newtons per square meter. Approximately
4/5 of the gas pressure.
Problem 5
The Jeans length is given by equation 12.16
RJ =
15kT
4πGµmH ρ0
1/2
We just have to plug in the information from the example to get
RJ = 4 × 1011 m
Problem 6
We can use equation 12.18 to write
MJ =
5
G
3/2
vT3
3
4πρ0
1/2
Now, we can use the ideal gas law to get the equation in term of the pressure.
We have
ρ0 kT
µmH
= ρ0 vT2
P0 =
Substituting for ρ0 and combining all of the numerical factors gives the answer.
2
Problem 7
A)
The mass ratio is equal to the ratio of distances ab /ac = .37. The sum of the
masses can be found from kepler’s third law. Then, the individual masses can
be solved for.
To use the third law we need the semi major axis, which is given by a = αd.
d is give by
1
= 5 pc
d=
.201
Then, we get for the semimajor axis(after appropriate unit conversions)
a = 5.12 × 101 2 meters
Now we can find the sum of the masses
mB + mC =
4π 2 3
a = .65 solar masses
GP 2
Solving for the individual masses, in solar masses, yields
mB = .478
mC = .177
B)
The luminosity, in terms of the solar luminosity, is
LB = 100(Ms −MB )/5 = .0116
C)
We use the Stefan-Boltzmann equation
s
LB
RB =
= 8.75 × 106 m
4πσTB4
This is about 1% the size of the sun, 1.4 times the size of the Earth, and 1.6
times the radius of Sirius B.
D)
The average density is calculated from the mass and radius to be 3.37 × 108
kilograms per cubic meter. This is approximately an order of magnitude less
than the average density of Sirius B. The density is determined by how much
electron degeneracy pressure is necessary to stop the collapse of the star. Sirius
B has a greater mass, therefore the degeneracy pressure, and hence the density,
must be higher.
3
E)
For 40 Eri B we have mV = 2.69 × 1051 kgm3 and for Sirius B mV = 1.45 ×
1051 kgm3 .
The departure from the mass volume relation is caused by the relativistic
considerations explained on page 570.
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