Download Problem Set 6 for Astro 320 Read sections 11.2

Problem Set 6 for Astro 320
Read sections 11.2-11.3, skimming pp. 375-377, and read chapter 13,
skimming pp. 453-456. Some questions require numerical information in
this text not mentioned in class.
Due Nov. 18, by 5 pm, to the Astro 320 drop box.
Problem 1: a) C & O, problem 11.2a.
The Sun’s luminosity is 3.8 × 1026 W, or J/s. That translates, via E = mc2 , to m = E/c2 =
3.8 × 1026 /(3 × 108 )2 = 4 × 109 kg/s. Per year, that’s 3.16 × 107 ∗ 4 × 109 = 1.26 × 1017
kg/year, or 6.3 × 10−14 M /year.
b) C & O, problem 11.2b.
The solar wind mass loss rate is dM/dt = 4πr2 nmH v, so for v = 500 km/s, n = 7 × 106 m−3 ,
and r =1 AU, this gives 3 × 10−14 M /year. So the solar wind takes half as much mass from
the Sun as its own fusion does.
c) C & O, problem 11.2c.
Over the Sun’s main-sequence lifetime (∼ 1010 years; e.g. Fig. 13.1), the loss from nuclear
fusion is 6.3 × 10−14 ∗ 1010 = 6.3 × 10−4 M .
From solar wind, it’s 3 × 10−4 M . So, no substantial change.
Problem 2: a) C & O, problem 11.12.
The magnetic pressure in the center of a sunspot is P = B 2 /(2µ0 ) = 0.22 /(2 ∗ 1.26 × 10−6 ) =
1.6 × 104 N m−2 . This is actually reasonably close to the gas pressure elsewhere in the
photosphere (2×104 ); the magnetic pressure in the sunspot replaces much of the gas pressure,
allowing the material to cool without condensing and sinking immediately.
b) C & O, problem 11.14.
If CMEs ejects 1013 kg of material per day, that’s ∼ 3.7 × 1015 kg/year, or 1.9 × 10−15
M /year; only 6% of the loss of mass by the solar wind.
Problem 3: C & O, problem 13.2.
We estimate the Kelvin-Helmholtz timescale for a 5 M star on the subgiant branch. We
can find the luminosity by using Fig. 13.1, using position 3 or 4 (or perhaps 5, depending
on how we interpret “subgiant”).
For point 4, we estimate L=103.2 L , T=104.1 K, and can
p
use L = 4πR2 σT 4 to find R = L/(4πσT 4 ) = 5.8 × 109 m. Alternatively, one can look at
Figure 13.7, check that the L and T agree with point 3, and just use R=3.94 R .
Now, the KH timescale is just ∆E/L, where ∆E = P E/2(virial theorem)= (3/10)GM 2 /R =
0.3 ∗ 6.7 × 10−11 ∗ (5 ∗ 2 × 1030 )2 /5.8 × 109 = 3.5 × 1041 J. So tKH = 3.5 × 1041 /6.0 × 1029 =
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5.8 × 1011 s, or 1.8 × 104 years.
The time between points 4 and 5 on Fig. 13.1 is 94.92-94.57 million years, or 3.5 × 105 years.
These numbers aren’t really close (due to the large number of inappropriate assumptions–
e.g. the KH timescale is calculated for the star, assuming constant density, rather than its
core), but they are within a factor of 20.
Problem 4: C& O, problem 13.11.
a) The main sequence lifetime of a 0.8 M star is, from Table 13.1, about 25 billion years
(depending on whether point 2 or 3 is chosen). As the universe is only 13.6 billion years old,
no stars of 0.8 M have left the main sequence yet.
b) A globular cluster’s main-sequence turnoff indicates the age of the cluster, as a star at the
turnoff has an age equal to its main-sequence lifetime. Thus, if you know the main-sequence
lifetime of stars of that mass, you can assess the age of the cluster. A globular cluster
with turnoff stars of 0.8 M (or less) would have to be about 25 billion years old. Since the
universe isn’t that old, such clusters don’t yet exist.
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