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hwk09
Ast 4001, 2015 November 24
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1. Suppose that a neutron star has M = 1.5 Msun and R = 10 km.
(a) Calculate av , its average mass density. Look up the mass and size of a typical
atomic nucleus, e.g. 16O , and compare its mass density to the neutron star's.
Answers: The neutron star's average mass density is about 7. 2  10 14 g cm .
An 16O nucleus, with R  3.0  10 cm and M  2.65  10 g , has
density  2.3  10 14 g cm . The point is that a neutron star is denser than
an atomic nucleus. (Why?)
(b) Calculate the neutron star's escape velocity (i.e., escape from its surface), compared
to the speed of light: v/c = ?
Strictly speaking there's a relativistic correction
2
of order (v/c) , but ignore that detail and use Newtonian physics.
Answer: In the Newtonian approximation, v esc = ( 2 G M / R ) 1 / 2  2.0  10 5 km/s
so v esc / c  0.7. At that speed, kinetic energy is roughly 0.3  rest mass energy.
In other words, a neutron star has a semi-relativistic escape speed -- because it's only
two or three times larger than a black hole of the same mass. ( R BH = 2 G M / c 2 . )
(c) Estimate the neutron star's total gravitational potential energy EG with two different
assumptions: (1) its density is uniform, and (2) its density distribution is a polytrope
with index n = 1.5. Compare each estimate with the star's rest-mass energy, i.e.,
calculate EG / M c 2. (Again, ignore relativity effects.)
Answers: Again using Newtonian physics, a famous freshman-physics factoid is the
potential energy of a uniform-density sphere:  0.6 G M 2 / R , which amounts to
3.6  10 53 ergs in this case. The expression for an n = 1.5 polytrope is approximately
 0.857 G M 2 / R  5.1  10 53 ergs. Meanwhile M c 2  2.7  10 54 ergs for the
neutron star. Thus EG / M c 2   0.13 and  0.19 for the two simplified models.
Since these are appreciably smaller than unity, the correct general-relativity value
won't be very much different.
(d) Neutron stars can be hot. Estimate the surface temperature necessary to produce the
same photon luminosity as the Sun. What average photon energy would this imply?
(Express it in eV. Is this a UV photon, or is it X-ray, or gamma-ray? )
Answers: According to the familiar L  R 2 T 4 rule, a neutron star with surface
temperature  1.5 million K would have the same luminosity as the Sun. Its
average photon energy would be approximately 2.7 k T  350 eV, in the "soft
X-ray" part of the EM spectrum.
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4001hwk09ans - p2
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2. Simple estimates for a typical supernova's energy-output budget. Imagine a core-collapse
SN whose progenitor star originally had mass M = 15 M sun. This explosion creates the
neutron star described in problem 1 above. Express each energy in ergs.
(a) Radiative energy: Suppose the peak luminosity has absolute bolometric magnitude
M bol   and the duration above half-peak luminosity is t  50 days.
Make an order-of-magnitude estimate of the total radiated energy.
Answer: Roughly 10 49 ergs. M bol   corresponds to L  5  10 8 L sun
 2  10 42 ergs s. Multiply by 50 days and we get 8.6  10 48 ergs.
(b) Kinetic energy: About 10 M sun of material is ejected at average speed  3000 km/s.
(Some mass was lost earlier in a stellar wind.) Evaluate the kinetic energy of the ejecta.
Answer: Roughly 10 51 ergs, 100 the radiated energy. ( M v 2 / 2  9  10 50 ergs . )
(c) Neutronization neutrinos: Assume that the SN core -- the material that became a neutron
star -- had equal numbers of protons and neutrons before the collapse. (Largely 56Ni.)
Calculate the number of neutrinos emitted in the neutronization process. Assuming
that the average neutrino has energy 1 MeV, calculate the total energy carried away
by these neutrinos, expressed in ergs.
Answers: Most of the protons are converted to neutrons, and each proton-to-neutron
conversion creates one neutrino. There are about 9  10 56 protons in the pre-SN
material that became a neutron star, so that is the number of neutronization-neutrinos.
Concerning energy, 9  10 56 MeV  1.4  10 51 ergs. This may be an underestimate
because the average neutrino energy may be larger than 1 MeV.
(In reality, plasmon neutrinos carry far more. According to problem 1c, the formation
of a neutron star releases more than 10 53 ergs and this has to be removed somehow.
"Somehow" is mostly neutrinos made by processes other than p  n.)
(d) For comparison, evaluate the rest-mass energy of the Sun, M sun c 2.
Answer: 1.8  10 54 ergs. The order of magnitude, 10 54 ergs, is worth remembering.
(Also remember that the number of baryons in the Sun is of order 10 57 .)
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----- problem 3 is on next page -----
4001hwk09ans - p3
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3. In an order-of-magnitude sense, the fundamental pulsation period of a star is fairly
close to the "dynamical timescale" t dyn = (G), where we'll assume in this
problem that  = average internal mass density. Evaluate the dynamical timescale
for each of the following objects. You may need to look up most of the necessary
masses and sizes, or even make intelligent guesses; in any case specify what numbers
you've assumed. Express each value in an appropriate unit of time for that object:
s, ms, seconds, minutes, hours, days, months, years, or whatever fits it best.
(a) The Sun.
( density  1.4 g cm)
(b) Planet Earth. Answer: A little less than half an hour.
( density  5.5 g cm)
(c) The red supergiant star Orionis = Betelgeuse.
Answer: Roughly 1.5 year. This star allegedly has M  7.7 M sun and
R  1180 R sun , implying average density  6.6  10  g cm .
(d) Sirius B.
Answer: About 2.5 seconds. ( density  2.3  10  g cm . )
(e) The neutron star described in problem 1. Answer: About 0.14 millisecond.
(f) An interstellar giant molecular cloud with density  104 H-atoms per cm3 .
Answer: A little less than 1 million years.
(g) (This question is a little different) If the observable universe has radius  13 billion
lightyears, estimate the mass density that would give it a dynamical timescale of
13 billion years. If this density were ordinary matter, express it in baryons per cm3 .
(In reality, as most people know, most of it is not ordinary matter.)
Calculate the total mass expressed in M sun, pretending that it’s a sphere with
familiar Euclidian geometry.
Answers: A mass density of 9  10  g cm would give (G ) 13 Gy.
If this were ordinary matter, it would require only 5  10  baryon per cm .
The mass within a "Euclidian" radius of 13 billion lightyears would be about
7  10  g = 3.5  10  M sun , enough to make a trillion respectable-sized
galaxies. In relativistic cosmology, the transition between open and closed
universes, i.e., a "flat space" universe, has numbers comparable to these.
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