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Ast 4001
Answers for final exam, 16 December 2013
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1. (25 points) Each of these small problems should take only a minute or two to solve.
There's no need for explanations, but the quantitative answers must be approximately
correct. Quote relevant short formulae to show that you're not just guessing – but
points may be subtracted for formulae that are clearly irrelevant!
(a) In this course we noted some idealized power-law relations between mass, radius,
and central temperature for main-sequence stars in the mass range 0.8 -- 5 M sun .
Use them to deduce how a star's escape velocity depends on mass. In other words,
estimate the exponent a in V esc  M a.
Answer: V esc  M 1 / 8.
R  M 0.75 was used several times in class for the mid-range main sequence.
This was deduced from the H-R diagram combined with the mass-luminosity
relation – two observed relations. (The central temperature relation T c  M 0.25
was also quoted several times and this can also be used to find R  M 0.75.)
Then we use V esc2 = G M / R  M / R  M 0.25 so V esc  M 1 / 8.
Of course these proportionalities are empirical rules of thumb, not exact.
(b) Imagine an eclipsing binary star system. Both stars A and B have the same apparent
visual magnitude m A = m B = 7.75, so their combined brightness is like a star with
m = 7.0. Star B is hotter than A, and its surface area is only half that of A.
Estimate the combined apparent magnitude during eclipse, at a time when star B
is exactly between us and star A.
Answer:
m = 7.31 .
Since star B has half the surface area of star A, it also has half the cross-section
area; so during eclipse it blocks half of the light from A. Therefore, during eclipse
we see (normal brightness of B) plus (half of A’s normal brightness) = 75% of
the normal brightness. A factor of 0.75 corresponds to 0.31 magnitude fainter
than the normal combined brightness: magnitude 7.0 + 0.31.
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Problem 1 continued ---
(c) Some experts say that a habitable planet requires nearly a billion years to develop
recognizable life forms. Obviously, then, the planet's star must have a main-sequence
lifetime at least that long. This timescale depends on the approximate mass-luminosity
law and also on the criterion for leaving the main sequence.
Based on comparisons with the Sun, estimate the mass M / M sun of a star whose
main-sequence lifetime is 10 9 years.
Answer: M = roughly 2.5 M sun ; values between 2.2 and 3.2 are acceptable.
A billion years is about one-tenth the lifetime of the Sun. Since a star leaves
the main sequence when it has burned a nearly-constant fraction of its hydrogen,
its main-sequence lifetime is proportional to (fuel) / (rate of burning )  M / L .
Since the mass-luminosity relation is L  M 3.5 for stars in the relevant mass

range (or, anyway, between M 3 and M 4 ), we have (lifetime)  M  .
Thus M  (lifetime), and the lifetime is about one-tenth that of the Sun;
hence M = about 10 M sun .
(d) Two normal white dwarf stars 1 and 2 have the same chemical composition.
Star 1 has M = 0.6 M sun , R = 7500 km , and central density  c = 4 × 10 g cm.
Star 2 is denser, with  c = 9 × 10 g cm.
Estimate the mass and radius of star 2 , expressed in M sun and km.
Answers: 0.9 M sun and about 6550 km.
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Since R  M  for normal white dwarfs and they all have basically the
same structure, we know that  c  M / R 3  M 2 . Star 2 is stated to be denser
than star 1 by a factor of 2.25, so its mass must be larger by a factor of 1.5.
Thus its radius is smaller than star 1 by a factor of 1.5 = about 1.145.
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Problem 1 continued ---
(e) At the center of star X, the density is 400 g cmand the temperature is 10  K.
Photons provide about 1 percent of the total pressure there while the hot gas
accounts for 99 percent. In another star Y, the central temperature is 2 × 10K 
and half of the pressure is due to photons.
Estimate the approximate density at the center of star Y. Accuracy of ± 3% is
adequate.
Answer: About 32 g cm.
The ratio (radiation pressure) / (gas pressure) is proportional to T 3 /  .
In star X this ratio is about 0.01, and in star Y it’s 1.
Therefore ( T 3 /  ) Y  100 ( T 3 /  ) X . Rearranging factors,
we find  Y  0.01 ( T Y / T X ) 3  X = 0.08  X = 32 g cm.
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2. (24 points) These questions require clear, brief explanations but no calculations.
Answers must be legible as well as clear!
(a) How or why does helium burning in a star produce some oxygen?
Answer: C + He → O . Carbon is obviously present, because it’s being
produced by the triple-alpha reaction He + He + He. Therefore C nuclei
sometimes react with He; the potential barrier is much higher than for He + He,
but on the other hand C + He doesn’t need 3 particles. Of course the relative
efficiencies of He + He + He and C + He depend on temperature and density.
(b) Why do massive main-sequence stars have convective cores?
Answer: Massive stars burn hydrogen via the CNO cycle, which is extremely
sensitive to temperature because the potential barriers are higher than those found
in the pp process. Because of this strong temperature dependence, most of the
energy release occurs in a very small zone at the center of the star – it’s almost like
a point source. Therefore the energy flux F  L / r 2 is very large in the region
where r is smallest. This high flux is too large to be carried by radiation; it
satisfies the Schwarzschild criterion for convection to occur.
(The pp chain, by contrast, occurs in a more extended region, so L( r ) / r 2
does not become as large at small r .)
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(c) Consider this weak reaction: p+ + e → n +  e . Why does it almost never occur
in the core of a main sequence star like the Sun?
Answer: The rest mass energy of a neutron is larger than a proton plus electron.
Consequently this reaction is endothermic, which means energetically unfavorable.
(d) On the other hand, p+ + e → n +  e does occur in a neutron star or a sufficiently
massive white dwarf. Why? Give the fundamental physical reason, not just a
statement that the density is high.
Answer: At extremely high density, the Fermi energy becomes larger than the
rest-mass energy difference mentioned in part (c) -- i.e., larger than the mass
difference (neutron) minus (proton and electron). The Fermi energy refers
to the most energetic electrons. In that case, the total energy of the system
decreases if the quoted reaction makes some electrons disappear.
(e) Nuclear reactions never become significant in the lowest-mass “stars.” Explain why.
Answer: Low-mass stars have high densities. If the density is high enough,
then the electrons become degenerate and the star can support itself by
degeneracy pressure, independent of the temperature. In that case the object
doesn’t need to become hot enough for strong nuclear reactions.
(f) For hydrogen burning (either the pp chain or the CNO cycle), the nuclear energy
generation rate per unit mass is    f H ( T ) . Helium burning, however, has
a different density dependence:    2 f He ( T ) . Briefly explain why these differ.
Answer: Hydrogen burning involves only binary reactions, particle1 + particle2.
The rate per unit volume is proportional to n (particle 1) n (particle 2)    ,
so the rate per unit mass is proportional to . Helium burning, however,
requires a three-particle reaction He + He + He. Its rate per unit volume must
therefore be proportional to   so its rate per unit mass is proportional to   .
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3. (13 points) Most stellar hydrogen burning occurs via either the pp chain or the CNO
cycle. Each of these two processes has a big advantage and also a major impediment,
compared to the other.
(a) Briefly explain both the relative advantage and the disadvantage of each.
Explanation: The main advantage of the pp chain is its relatively low potential
barriers, e.g. between two protons. Its big disadvantage is the need for a
weak-interaction process, 1H + 1H → 2H + e+ +  e .
The CNO cycle is the converse: it has much higher potential barriers but
no need for a weak decay.
(b) Consequently the pp chain dominates in low-mass stars but the CNO cycle
dominates above about 2 M sun . Explain why.
Explanation: The disadvantage of high potential barriers becomes less serious
at high particle energies, i.e. at high temperatures. The disadvantage of a
weak decay, on the other hand, does not depend much on temperature.
Therefore, if the temperature rises high enough, then the CNO cycle’s
advantage outweighs its disadvantage. It will dominate at temperatures
higher than that. … And higher-mass stars tend to have higher central
temperatures, because they need to generate much higher luminosities
than low-mass stars. Therefore the CNO cycle dominates for high masses,
while the pp chain dominates for low-mass stars.
Another way of saying almost the same thing is that high potential barriers
lead to a stronger temperature-dependence in a reaction. Therefore the
CNO cycle is more sensitive to temperature than the pp chain is. Since they
both increase with temperature, and the CNO cycle has a steeper slope,
the CNO cycle must win above some critical temperature, while the pp chain
wins at temperatures below that value.
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4. (13 points) As a massive star evolves, its first few nuclear burning stages are easy to
describe and have fairly clear consequences. Thus hydrogen burning makes helium;
helium burning creates carbon and oxygen; and carbon burning produces a mixture
of silicon, magnesium, and several other species. Each reaction has an obvious
direction.
“Silicon burning,” however, is different, and not so straightforward. Briefly describe
how it differs from the reactions noted above, and why it involves many simultaneous
reactions. State why silicon burning needs high temperatures for a reason other than
barrier penetration. Also state why it eventually comes to an end.
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5. (14 points) Consider an ionized, non-degenerate pure helium gas. It’s in thermodynamic
equilibrium, so hot that there are practically no He0 atoms. The gas is entirely He+, He++,
associated electrons, and Planckian photons.
(a) Let x be the fraction of doubly ionized ions: x = n ( He++ ) / n He where n He means
n ( He+ ) + n ( He++ ) . A fairly simple function of n He and x is constant if the
temperature is constant. Write an expression for this ionization-equilibrium function
F ( n He , x ) . Caution: One factor is different from the formula that we used
for hydrogen ionization.
Answer: { ( 1 + x ) x / ( 1
The Saha equation for
x ) } n He is constant if T is constant.
He+ + photon
 He++ + e
is
n ( He++ ) n ( e ) / n ( He+ ) = f ( T ) .
We can substitute n ( He++ ) = x n He and n ( He+ ) = ( 1
x ) n He . Moreover,

+
helium must provide all the electrons, therefore n ( e ) = n ( He ) + 2 n ( He++ )
= ( 1 – x ) n He + 2 x n He = ( 1 + x ) n He .
Putting all these terms into the Saha expression, we get
{ (1 + x) x / (1
x ) } n He = f ( T ) .
(b) At some particular temperature, x = 0.2 at density n He = n and x = 0.8 at
density n He = n . Calculate the ratio n  / n  .
Answer: n  / n  = 2.4 .
expression in part (a).
Just substitute the given values of x into the constant
(c) The macroscopic ideal gas law requires a “gas constant”  : P =  k T . In fully
ionized pure hydrogen, its value is  = 1.65 × 10 cm s K
Using the hydrogen case as a convenient reference standard, calculate  for fully
ionized helium, i.e. with x = 1.0 . For simplicity, pretend that the mass of a
helium atom is exactly four times that of a hydrogen atom.
Answer:
 ( fully ionized He ) =
0.375  ( fully ionized H ) = 6.19 × 10 cm s K
The macroscopic gas constant  is simply k / m av where k is Boltzmann’s
constant and m av is the average mass per particle. In fully ionized hydrogen
we have two particles per hydrogen-mass, so m av = 0.5 m H . Fully ionized
helium, however, has three particles per helium-mass, and m He = 4 m H .
Hence the average mass particle is larger than hydrogen by a factor of 8 / 3 .
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6. (16 points) Adiabatic index  relates pressure to density. Most parts of most stars
have  1.666 , but in some locations  can be less than 1.4.
(a) Why does this affect the stability of a star? Don’t explain in detail, but merely quote
the Famous Hugely Important Fact about  vs. stellar stability.
(b) Describe two different types of region where  can drop below 1.5 inside a star –
the reasons must be different for the two cases. State what general kinds of star
can have each type of region.
(c) Briefly explain why the adiabatic index decreases in each of the two cases in part (b).
In one of them, a qualitative thought experiment is the best way to do this. Clarity
is essential.
(d) Suppose that  is a known function of density  . If some gas initially has density
 and pressure P 1 , and then an adiabatic expansion or contraction changes the
density to  , the resulting pressure can be calculated from
P 2 / P 1 = a well-defined expression involving the function (  ) ,
and the values  and  .
Write a formula for this expression. Its final form should not contain any logarithmic
terms or factors, but it may include a definite integral.
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