PH 510
HOMEWORK ASSIGNMENT: CHAPTER 15
Problem 15.1
“Estimate the Eddington limit for eta Car and compare your answer with the luminosity of
that star. Is your answer consistent with its behaviour? Why or why not?”
Solution:
LEdd =
4Gc
M
where the mean opacity in eta Car is expected to be primarilyue to scattering by free
electrons, i.e.
= 0.02(1 + X ) 0.02(1 + 0.7) = 0.034
in SI units.
This gives
4 (6.67 10 11 )(3.00 10 8 )
(120 1.99 10 30 )
0.034
33
= 1.77 10 W = 4.6 10 6 Lsolar
LEdd =
(using Carroll & Ostlie’s estimate of 120 Msolar for the mass of eta Car).
Carroll & Ostlie quote eta Car’s present luminosity as about 5 x 106 Lsolar . So:
Yes, the above calculation is consistent with eta Car’s behaviour, since:
(i):
(ii):
eta Car exists, i.e. it has not violently exceeded its Eddington limit , but
it is very unstable, therefore is probably very close to its Eddington limit.
Problem 15.3
“The angular extent of one of the lobes of eta Car is approximately 8.5 seconds of arc.
Assuming a constant expansion rate of the lobes of 650 km/s, estimate how long it has been
since the Great Eruption that produced the lobes. Is this likely to be an overestimate or an
underestimate? Justify your answer.”
Solution:
Using the relation between angular size () and physical size (s) at a distance r:
s = r
s = 2300 pc we get:
8.5 1 degree 2 radians
= 0.095 pc
1
3600 360 degrees
Expanding from zero size to this extent, at a constant speed of 650 km/s, would take
s 0.095 3.09 1013 seconds (since 1 pc = 3.09 x 1013 km)
t = = 650
v = 4.51 x 109 seconds (143 years).
One would expect the lobes to slow down as they hit the clouds of gas and dust that
surround eta Car. Then, the actual extent of the lobes is smaller than they would have been
without encountering these clouds, and therefore the Great Eruption probably occurred a
bit earlier than 143 years ago, i.e. the calculation is an underestimate.
[note to grader: Alternative arguments (either for an overestimate or an underestimate) that
are well reasoned can also earn full credit].
Problem 15.5
“Assume that the 1 solar mass core of a 10 solar mass star collapses to form a Type II
supernova. Assume further that 100% of the energy released by the collapsing core is
converted to neutrinos and that 1% of these neutrinos are absorbed by the overlying
envelope to power the ejection of the supernova remnant. Estimate the final radius of the
stellar remnant if sufficient energy is to be liberated to just barely eject the remaining 9
solar masses of material to infinity.Be sure to state clearly any additional assumptions you
make in determining your estimate of the final radius of the remnant.”
Solution:
This question needs to be read carefully to determine what needs to be calculated. There is
mention of a stellar remnant and of a supernova remnant. The former is the remainder after
the 1 solar mass core collapses, and the latter is the 9 solar mass envelope that needs to be
driven to infinity (i.e. become totally unbound gravitationally).
Because of the density gradient in a pre-supernova star, most of the 9 solar masses would
lie closer to the iron core than to the surface. The canonical size of the pre-collapse iron
core is about the same as the Earth, i.e. about 5 000 km, while the radius of the whole star
will be many millions of km. Let’s assume a radius of 100 000 km for the 9 solar mass
envelope. (This is very subjective; you could choose a substantially different radius, based
on more sophisticated arguments. The simplest choice of all is simply to choose the same
initial radius for the pre-collapse core and the envelope respectively. This makes the
solution of the problem very simple, but gives a quite inaccurate answer).
Then, the gravitational binding energy that needs to be eradicated by the iron core is
3 GM 2
(6.67 10 11 (9 1.99 10 30 ) 2
= 0.6
Nm = 1.28 10 44 J
8
5 R
1 10
(assuming a constant density envelope, of course). The collapse of the core (from an
initial radius of, say, 5000 km) is required to produce 100 times as much energy, according
to the given assumptions. This requires a collapse to a final radius Rf , such that
3 G (1M solar ) 2 3 G (1M solar ) 2
= 1.28 10 46 ,
6
5
Rf
5 5 10
1
1
1.28 10 46 5 / 3
=
,
R f 5 10 6 6.67 10 11 (1.99 10 30 ) 2
R f = 12.5 km
(again assuming a constant density core). Clearly, the choice of initial radius for the core
doesn’t make a substantial difference to this answer, while the choice of average radius for
the 9 solar mass envelope does. If this choice is larger than my estimate of 100 000 km, the
final radius of the collapsed core will be larger than 12.5 km (and vice versa). The
relationship is actually exactly linear, as is clear from the equations used.
[A clever student might have anticipated the energy required to dissociate all the iron
nuclei into unbound protons and neutrons during collapse. Following example 15.3.1 in
Carroll & Ostlie, this is easily calculated as 0.13 of the gravitational energy released by
collapsing to a radius of 12.5 km. In order to accommodate this energy drain as well as
manage the expansion of the 9 solar mass envelope to infinity, the core would need to
collapse to about 11 km instead of 12.5 km].
Problem 15.6
(a) “The angular size of the Crab SNR is (4 minutes of arc) x (2 minutes of arc) and its
distance from Earth is approximately 2 000 pc (see figure 15.4). Estimate the linear
dimensions of the nebula.”
(b) “Using the measured expansion rate of the Crab and ignoring any accelerations since
the time of the supernova explosion, estimate the age of the nebula.”
Solution:
(a) Following the procedure in 15.3 above, we find linear dimensions of 2.33 pc x 1.16 pc
(b) Using an expansion rate of 1400 km s-1 (I needed to look this up on the Internet), we
find an age (as in problem 15.3) of 1630 years. Since we know that the nebula is only 954
years old, this implies that it must have expanded more quickly in the past.
Problem 15.11
“The energy released during the decay of one cobalt-56 atom is 3.72 MeV. If 0.075 solar
masses of cobalt was produced by the decay of nickel-56 in SN 1987A, estimate the amount
of energy released per second through the radioactive decay of cobalt,
(a) just after formation of the cobalt;
(b) one year after the explosion;
(c) compare your answers with the light curve of SN 1987A given in figure 15.12.”
Solution:
The energy released per second will depend on the number of cobalt nuclei available,
following the standard radioactive rate formula
dN
ln 2
= N = N
dt
1/ 2
(a) At the start, the number of cobalt nuclei is
0.075 1.99 10 30
= 1.60 10 54
27
56 1.67 10
Hence, given the half-life of 77.7 days, i.e. 6.71 x 106 seconds, the equation
delivers a rate of 1.65 x 1047 decays per second,
i.e. 1.65 x 1047 x 3.72 x 106 x 1.60 x 10-19 J s-1 = 9.84 x 1034 Watt.
This matches the observations shown in figure 15.12 just about perfectly!
(b) After a year, the number of cobalt nuclei is
N (1 year) = N 0 e t = 1.60 10 54 e 1.0710
7
3.16107
= 5.47 10 52
This corresponds to 5.65 x 1045 decays per second,
i.e. 1.65 x 1047 x 3.72 x 106 x 1.60 x 10-19 J s-1 = 3.36 x 1033 Watt.
Again, this matches the observations shown in figure 15.12 just about perfectly!
Problem 15.12
“The neutrino flux from SN 1987A was estimated to be 1.3 x 1014 per square meter at the
location of Earth. If the average energy per neutrino was approximately 4.2 MeV, estimate
the amount of energy released via neutrinos during the supernova explosion”
Solution:
The flux ratio of neutrinos drops as the square of distance. Using the given distance of 50
kpc (Carroll & Ostlie page 524) to SN 1987A, and assuming a radius of about 1 000 km for
the neutrino-emitting core (see Carroll & Ostlie pages 531-2), we extrapolate to a flux of
2
50000 3.09 1013 = 3.10 10 44 per square meter.
1.3 10 1000
14
For this flux, the total number of neutrinos emitted over the surface of a sphere with radius
of 1 000 km is 3.90 x 1057.
If each neutrino carries 4.2 MeV of energy, the total energy carried by neutrinos is 2.62 x
1045 J.