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PH 510
HOMEWORK ASSIGNMENT: CHAPTER 13
Problem 13.2
“Estimate the Kelvin-Helmholtz timescale for a 5 solar mass star on the subgiant branch
and compare your result with the amount of time the star spends between points 4 and 5 in
Figure 13.1”
Solution:
The KH timescale is estimated by the formula
GM 2
KH = 0.3
÷L
R
Figure 13.1 suggests the following estimates for the average luminosity of a 5 solar mass
star on the subgiant branch:
L = 10 3.1 Lsolar
Since we are after the total contraction energy on the subgiant branch, we use the final
temperature on this branch to estimate the final radius:
T final = 10 3.7 K,
L R final = 4óT 4 KH = 0.3
1/ 2
,
GM 2
4T 4
3/ 2
L
(
)
1/ 2
= 1.25 1011 s (4 000 years)
On the other hand, Table 13.1 indicates that a 5 solar mass star spends about 350 000 years
on the subgiant branch (between points 4 and 5). Why the discrepancy?
(4 marks)
[A discussion along the following lines earns a few bonus marks: The collapse that occurs
is not of the entire star, but only of the helium core. A fair estimate of the mass of this core
is about 0.3 x the stellar mass (the Schoenberg-Chandrasekhar limit), and a fair estimate of
the size of the core after collapse is 6 x 10-4 of the stellar radius (see Carroll & Ostlie,
figure 13.7, and accommodate a drop in radius by a factor of 100 as the temperature rises
from about 107 K to about 108 K between hydrogen exhaustion and helium core ignition,
using a blackbody approximation for constant core luminosity). If these values are
substituted into the above calculation, the KH timescale comes out as 600 000 years. Given
the many rough approximations made, this is remarkably close to the value of 350 000
years produced by much more refined models. A mild downscale of the core mass from 0.3
to 0.23 x the stellar mass delivers 350 000 years].
Problem 13.5
“Use equation (10.27) to show that the ignition of the triple alpha process at the tip of the
red giant branch ought to occur at more than 108 K.”
Solution:
Eqn (10.27):
Tq =
Z12 Z 22 e 4 μ m
12 2 02 h 2 k
For 2 helium-4 nuclei forming a Be-8 nucleus, the Z values are both equal to 2, and the
reduced mass μm = 2 x mproton (more accurately, 2 atomic mass units). Using the correct
values for the physical constants, this delivers a reaction temperature of 6.2 x 108 K.
(3 marks)
[Strictly speaking, the triple alpha reaction requires the Be-8 to interact with another He-4
nucleus, to form carbon-12. Applying the same formula to this reaction step delivers a
temperature of 3.3 x 109 K. Including this earns the student a bonus mark]
Problem 13.6
“[The Reimers approximation for the mass loss rate of an AGB star] is given by
L
M& = 4 10 13
M solar year 1
gR
if L, g and R are all expressed in solar units (i.e. relative to the values of L, g and R for the
sun; gsolar = 274 m s-2 ). [The value of μ can be taken as unity in this problem].
(a) Explain qualitatively why L, g and R enter this equation in the way they do.
(b) Estimate the mass loss rate of a 1 solar mass AGB star that has a luminosity of 7 000
Lsolar and a temperature of 3 000 K.”
Solution:
(a) The larger L is, the greater the radiation pressure. Specifically,
P=
1
3 aT
4
and L = 4R 2T 4 ;
which shows that the pressure is directly proportional to the luminosity at constant radius.
Since
M& Fapplied P,
M& L
Furthermore:
gR =
GM
GM
R=
=
2
R
R
1
2
2 v esc ,
where vesc
is the escape velocity. Essentially, gR embodies the work done per unit mass to remove
material from the stellar surface. The mass loss rate should be inversely proportional to the
magnitude of this quantity.
(6 marks)
(b) Using the Reimers formula for M = 1 solar mass, L = 7 000 Lsolar, and T = 3 000 K
delivers a mass loss rate of 8.7 x 10-7 solar masses per year.
[Note: en route to this answer the star’s values of g and of R must be calculated (using the
given values of stellar mass, luminosity and temperature); the former from the definition of
g:
GM
g= 2
R
and the latter from the blackbody relation: L = 4R 2T 4
(4 marks)
Problem 13.7
(a) “Show that the Reimers [formula] can be written in the form
LR
M& = 4 10 13
M solar per year
M
where L, R and M are all expressed in solar units”.
(b) ”Assuming (incorrectly) that L, R and do not change with time, derive an expression
for the mass of the star as a function of time. Let M =M0 when the mass loss phase
begins.
(c) “Using L = 7 000 Lsolar, R = 310 Rsolar, M0 = 1 Msolar, and = 1, make a graph of the
star’s mass as a function of time.”
(d) “How long would it take for a star with an initial mass of 1 Msola, to be reduced to a
mass of 0.6 Msolar ?”
Solution:
(a)
g
R
1
= solar solar
gR g star
Rstar
and
g star =
=
GM star
2
Rstar
,
1
gR
M solar
R
star
M star M solar
QED
(2 marks)
(b)
M
t
MdM = (4 10
M0
13
LR )dt
0
1
M 2 M 02 = (4 10 13LR )t ,
2
(
)
M (t ) = M 02 (8 10 13LRt
)
(4 marks)
(c)
The vertical axis shows M(t) in solar units; the horizontal axis shows time in years
(4 marks)
(d)
We require M(t) = 0.6 Msolar , and simply calculate t:
(0.6) 2 = 1 8 10 13LRt ,
0.64 = 8 10 13LRt ,
t =
0.64
8 1011
=
= 370 000 years
8 10 13LR 7000 310
(3 marks)
Problem 13.8
“The Helix nebula is a planetary nebula with an angular diameter of 16 minutes of arc that
lies approximately 213 pc from Earth.
(a) Calculate the diameter of the nebula;
(b) Assuming that the nebula is expanding away from the central star at a constant
velocity of 20 km/s, estimate its age.
Solution:
Using the relation between angular size () and physical size (s) at a distance r:
s = r
we get:
s = 213 pc 16 1 degree 2 radians
= 0.99 pc
1
60
360 degrees
Expanding from zero size to this extent, at a constant speed of 20 km/s, would take
s 0.99 3.09 1013 seconds (since 1 pc = 3.09 x 1013 km)
t = = 20
v = 1.53 x 1012 seconds (48 500 years).
(4 marks)
[I will grade the bonus question 13.13 myself]