AST 341
Final Exam and Solutions
12 December 2012
1. Given that the Earth’s orbital velocity is 30 km/s, the solar constant is 1.4×106 erg/cm2 /s,
and that the mass deficit (the fraction of mass converted to energy in the pp reaction)
is 0.007, estimate the lifetime of the Sun. Use only these numbers and the constants on
page 1. If you need the solar mass, radius, or luminosity, calculate them from the given
information. Show all your work; state all assumptions you feel you need to make. (30
points)
You need to know the mass of the Sun. First, use the Earth’s velocity v to determine
the length of the AU (assume a circular orbit). The orbital period is 1 year (π×107
sec). I did give you the AU on page 1, so you could have skipped this step, or solved
for the length of the year.
v×1 year = 2π AU, or the AU = 30×π×107 / 2π ∼ 1.5×108 km (the more accurate
value is 1.496×108 km - estimates can be pretty good).
Then, get the mass of the Sun from Newtonian gravity. For example,
12+13
2
∼ 2 × 1033 gm.
gives M⊙ = v GAU = 9×1.5×10
6.67×10−8
GM⊙ M⊕
AU2
=
M⊕ v 2
AU
The solar luminosity L⊙ = 4π AU2 times the solar constant, or 4π×2.25×1026 ×1.4×106 ,
or about 4 × 1033 erg/s.
The energy comes from fusion of 4 protons into one helium nucleus. The number of
protons in the Sun is M⊙ /mH , or 2 × 1033 /1.67 × 10−24 ∼ 1057 . Each nuclear reaction
consumes 4 protons, and produces 4 × 0.007mH c2 erg, or about 4 × 10−5 ergs.
33
4×10
Assume the Sun is 100% hydrogen. To produce the solar luminosity requires 4×10
−5 ,
38
or 10 reactions/sec. With each reaction consuming 4 protons, we have 2.5 × 1056
possible reactions. For a constant L⊙ (appropriate for the main sequence), the time
56
to use up all the hydrogen is 2.5×10
, or 2.5 × 1018 seconds. If 10% the solar mass is
1038
consumed, the lifetime becomes about 2.5 × 1017 seconds, or about 9×109 years.
Alternatively, you can greatly simplify this by realizing that the total energy E produced over the entire main sequence lifetime of the Sun is 0.1 × 0.007 × M⊙ c2 , and the
lifetime τ is E/L. Therefore, τ = 0.1×0.007×2×1033 ×9×1020 /4×1033 ∼ 14×1017 sec,
or about 5×109 years.
2. Why is there a main sequence? Why is there no main sequence for brown dwarfs? (10
points)
Stars are stable, and are in hydrostatic equilibrium while they undergo core Hydrogen
burning. This phase lasts about 90% of a star’s life, therefore most stars will be found
in this phase. Hydrostatic equilibrium gives a single-valued relation between stellar
mass and core pressure, or temperature, and hence luminosity. This unique locus for
most of the stars in the H-R diagram is the main sequence.
Brown dwarfs never burn hydrogen stably, so they are only in quasi-equilibrium. They
continually move across the H-R diagram (towards lower L and cooler T).
3. An empirical fit to the upper main sequence shows that the luminosity increases as the
fourth power of the mass. How does the main sequence lifetime τ of a star depend on
its mass? (10 points)
The lifetime is given by the ratio of the amount of fuel (the mass), and is inversely
proportional to the rate at which the fuel is burned (the luminosity). Since L∝M4 , the
M
−3
lifetime τ ∼ M
4 , or τ ∼ M
4. (a) Why does the solar luminosity increase by about a factor of 2 while on the MS? (b)
Assuming that the radius does not change, what happens to Tef f ? (10 points)
(a) As you convert 4 protons into 1 Helium nucleus, you remove particles from the
core. P=nkT. P must offset the stellar mass to maintain hydrostatic equilibrium, and
cannot change. Since n decreases, T must increase to compensate. Nuclear reaction
rates scale as Tν , where ν > 0, so the energy generation, and luminosity, must increase.
(b) Approximate the Sun as a blackbody. L∼T4 , so the temperature will rise by about
1
2 4 , or about 20%.
5. High mass stars produce elements up to Fe56 by α burning. (20 points)
(a) Why do they stop there?
Fe56 has the highest binding energy of any nucleus - reactions that build up heavier
elements are endothermic in nature, and do not release energy.
(b) The binding energy of Fe56 is about 9 MeV per nucleon; that of He4 is about 7 MeV
per nucleon. If the main sequence lifetime of a 50 M⊙ star is about 5×106 years, what
is its average luminosity while on the main sequence?
Proceed as in problem 1. 10% of the stellar mass is in the core. This is 50 ×2 × 1033 ×
0.1/1.67 × 10−24 nucleons, or about 5×1057 nucleons. The total binding energy, which
is that produced in nuclear fusion (and available for the supernova), is 5×1057 nucleons
× 9 MeV/nucleon. 1 MeV is 1.6×10−6 erg. Multiply this out to get a total energy of
about 5×1052 erg. 5×1052 erg radiated in 5×106 years is an average luminosity of 1046
erg/year, or 3×1038 erg/s (about 105 L⊙ ).
(c) Such massive stars do not ascend the giant branch, rather they evolve horizontally
towards lower surface temperatures at roughly constant luminosity. Estimate the lifetime of this star from when it leaves the main sequence until it develops an iron core.
You have generated 7 MeV per nucleon on the main sequence, there are only 2 left to
go. If only the core is involved in the nuclear reactions, the lifetime is 2/7 the main
sequence lifetime. Since shell burning adds to the core mass, this is an underestimate,
but the livetiune is of order 2/7 × 5×106 years, or a few million years.
(d) What is the final evolutionary state of such a star?
A core-collapse (type II) supernova. The core will probably collapse into a black hole,
since the core mass of about 5 M⊙ exceeds the likely upper bound for neutron star
masses.
2
3 GM
6. The Kelvin-Helmholtz timescale τK−H = 10
, where L, M, and R are the luminosity,
LR
mass, and radius of the star. (15 points)
(a) Explain what τK−H measures.
τK−H measures the timescale to radiate away the thermal energy (half the potential
energy, by the virial theorem) in a star.
(b) When is it important in the evolution of a star?
This is the timescale for the collapse to the pre-main sequence, or for the expansion
up the giant branch.
(c) Verify by dimensional analysis that τK−H is a time.
cm3
2 gmcm2
The units are gmsec
2 × gm / sec2 sec /cm = seconds, QED.
7. Consider a cataclysmic variable with a 2 hour orbital period. The white dwarf is 1 M⊙ ;
the donor star is 0.1 M⊙ . (15 points)
(a) About how much of its Roche Lobe does the white dwarf fill?
Not much. To get the orbital radii, plug into Kepler’s third law: a3 =G/4π 2 (M+m)P2 .
M+m = 1.1 × 2 × 1033 gm; P=7200 sec, so P2 ∼ 5×107 sec2 . Solve for a. I get about
2.5×1010 cm. The radii of the Roche lobes are inversely proportional to the masses, so
the Roche radius of the WD is about 2.2×1010 cm. A white dwarf is about the radius of
the Earth (6×108 cm) or 1% of the solar radius, so it occupies about 3% of the radius,
or about 3×10−5 of the volume.
(b) What is the distance from the center of the donor star to the inner LaGrangian
point?
2.5×109 cm, since Roche lobe radii are inversely proportional to the stellar masses.
(c) If you assume that on the lower main sequence R∝M, then would a normal main
sequence star be filling its Roche Lobe?
The “normal” 0.1 M⊙ main sequence star would have a radius of about 7×109 cm, so
a “normal” main sequence star would indeed be filling (or overfilling) its Roche lobe,
and accreting onto the white dwarf. (You should be able to estimate the radius of the
Sun in a number of ways, either from the solar constant, Tef f m and the luminosity, or
the angular diameter and the distance, or the mass and mean density.)
8. The figure shows spectra of 5 main sequence stars. The spectra of each star go from
0 to 1, but are offset from one-another by 1 flux unit for clarity. Place these in a
temperature sequence, with the hottest star first. Explain your reasoning, and identify
the prominent spectral features. Estimate spectral types. (15 points)
The hottest star is b. It has the bluest continuum, and is a B0 star. The strongest lines
are the Balmer lines of hydrogen; the weak lines are mostly Helium. Star e (A0) has the
strongest Balmer lines. It is hotter than star c (F0); it has a bluer continuum and more
metallic lines between the Hydrogen lines. Star a (G0) has only weak Balmer lines.
The strongest lines are Ca II K&H The coolest is star d (M0), which has very little
blue flux. There are few strong absorption lines; broad features are due to molecules,
mostly TiO.
9. The solar corona is fully ionized. The mean electron density ne ∼ 1010 cm−3 . The
temperature is about 106 K. Assume that the coronal radius is approximately equal to
the solar radius.
(a) What is the approximate optical depth of the corona to electron scattering?
The optical depth τ = nlσ, where sigma is the scattering cross section. In this case
σ = σT . Plug in n = 1010 cm−3 and l = 7 × 1010 cm to get τ = 7 × 6.6 × 1010+10−25 ∼
4.5 × 10−5 . The corona is optically thin.
(b) What are the units of the optical depth?
optical depth is dimensionless.
q
10. Use the virial theorem to show that the circular orbital velocity is given by GM/R,
where M is the mass of the central object and R is the radius of the orbit. (10 points)
The virial theorem states that 2K = -U in a system in equilibrium, where K is the
kinetic energy and U is the potential energy. The kinetic energy of an object in orbit is
1
mv 2 . The potential energy is -GM/R for gravity, where M is the mass of the central
2
2
object
q and R is the distance to the object. Equating 2K = mv to -U = -GM/R gives
v = GM/R. QED
11. Write the equation of detailed balance. Define all the terms. (10 points)
What comes up must come down. Consider a two level atom with an upper state U
and a lower state L. There are nL and nU electrons in the lower and upper states,
respectively. Electrons in state L can be excited to state U by absorbing a photon; the
rate for absorption is Iν BLU , where Iν is the intensity of the radiation field. Electrons
in the upper state can spontaneously de-excite by emitting a photon (rate = AU L )
or can radiate by stimulated emission (rate=I BU L ). In equilibrium, the ratios of the
populations in the two states is constant. Hence the number of upward transitions,
nL Iν BLU must equal the number of downward transitions nU (AU L + Iν BU L ).
nU (AU L + Iν BU L ) = nL Iν BLU , or
nU
nL
=
AU L +Iν BU L
nL Iν BLU
is the detailed balance equation.