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Homework Solutions: Chapter 10, The Deaths of Stars.
[ RQ: 6, 8, 12
Problems: 3 , 5 , 9 ]
RQ 6: How can star clusters confirm our theories of stellar evolution?
Answer: Star clusters contain stars of a wide variety of masses that all formed at about the
same time. Since our models tell us that massive stars should evolve faster than low mass
stars, we expect to see differences in the star clusters based on the mass of the stars. One
clear test that our models are at least on the right track, is that we find no star clusters
with O and B main sequence stars, and a large number of giants and supergiants. All of
the star clusters show us that when giants and supergiants are present in the clusters, there
are few if any massive stars (O and B stars) on the main sequence. We can take models of
many different masses of stars and determine what a cluster should look like after a given
time. We can then compare the HR diagram from our model with that of the observations
of the actual cluster to see how well they agree. For most part, the agreement is quite
good.
RQ 8: What causes an aging giant star to produce a planetary nebula?
Answer: A planetary nebula is produced as the carbon-oxygen core of a star contracts and
its outer envelope expands. The intense radiation from the hotter inner layers ionizes the
innermost gas of the envelope and drives it outward from the star. As this gas overtakes
mass lost earlier in the stars life as a giant, material piles up to form a dense shell of
ionized gas. The gas is continually ionized by the radiation from the hot contracting core.
RQ 12: How can we understand the Algol paradox?
Answer: Algol is a binary star composed of a five solar mass main sequence star and a
one solar mass giant. This presents a paradox because massive stars evolve faster than
low mass stars, so the five solar mass stars should leave the main sequence before the one
solar mass star. The solution to this paradox is mass transfer. Suppose that the system
was originally composed of a five solar mass star and a one solar mass star. After about
200 million years the five solar mass star would run out of hydrogen and begin to expand
to become a red giant. If as it expands, the star fills its Roche lobe, material will be drawn
off of the five solar mass star and accreted onto the one solar mass star that is still on the
main sequence. As the giant continues to expand, more and more mass will be passed
from the giant to the star that is still on the main sequence. If four solar masses of
material can be drawn off of the giant and deposited on the main sequence star, the main
sequence star will end up with a mass of five solar masses and the giant will have a mass
of only one solar mass. Thus through mass transfer as the massive star expands to
become a giant, the star still on the main sequence can become more massive than the
giant.
PR 3: About how far apart are the stars in an open cluster? in a globular cluster? (Hint:
What share of the cluster’s volume belongs to a single star?)
Answer: An open cluster is a collection of 10 to 1000 stars in a region about 25 pc in
diameter.
i) If there are 10 stars then each star has a volume equal to,
4 3 12.53  V  817.7 pc 3
V 
10
Hence the maximum distance rmax between two stars in an open cluster is,
4 3
3
817.7  rmax
 rmax
 195.3  rmax  5.8 pc
3
ii) If there are 1000 stars then each star has a volume equal to,
3

4 3 12.5
V 
 V  8.18 pc 3
1000
Hence the minimum distance rmin between two stars in an open cluster is,
4 3
3
8.18  rmin
 rmin
 1.95  rmin  1.25 pc
3
Finally, 1.25 pc  r  5.8 pc
A globular cluster is a collection of 10 5 to 10 6 stars in a region about 20 pc in diameter.
i) If there are 10 6 stars in a region of 10 pc diameter then each star has a volume equal to,
4 3 103  V  0.0042 pc 3
V 
10 6
Hence the minimum distance rmin between two stars in a globular cluster is,
4 3
3
0.0042  rmin
 rmin
 0.001  rmin  0.1 pc
3
ii) If there are 10 5 stars then each star has a volume equal to,
4 3 103  V  0.042 pc 3
V 
10 5
Hence the maximum distance rmax between two stars in a globular cluster is,
4 3
3
0.042  rmax
 rmax
 0.01  rmax  0.21 pc
3
Finally, 0.1 pc  r  0.21 pc
PR 5: If the Ring Nebula is expanding at a velocity of 15 km/sec, typical of planetary
nebulae, how old is it?
Answer: The Ring Nebula has a diameter of about D=1 light-year. Its radius therefore is
R=0.5 ly which is also the distance of the gas shell from its center. The gas must have had
enough time to travel the distance of 0.5 ly at a speed of v=15 km/sec. Therefore, the age
of the nebula must be at least,
R 0.5  300000Km sec   1year
150000
T 
T 
year  T  10000 years
v
15Km sec 
15
The Ring Nebula must be at least 10000 years old.
PR 9: The Crab Nebula is now 1.35pc in radius and is expanding at 1400 Km sec . About
when did the supernova occur? (Hint: 1pc equals 3  1013 Km)
Answer: The speed of the expansion is constant and so we can write,
dis tan ce 1.35  3  1013 Km
T

 T  2.89  1010 sec
velocity
1400 Km sec
But,
 30days   24hours   60 min utes   60 sec onds 
  
1year  12months  
  


 month   day   hour   min ute 
1year  3.11  10 7 sec
Hence,
2.89  1010 sec
T
 T  929 years
3.11  10 7 sec year 