Download ASTRONOMY 120

Astronomy 120
HOMEWORK - Chapter 20
Stellar Evolution
Use a calculator whenever necessary.
For full credit, always show your work and explain how you got your answer in full, complete
sentences on a separate sheet of paper.
Be careful about units!
Please CIRCLE or put a box around your final answer if it is numerical.
If you wish, you may discuss the questions with friends, but please turn in your own hand-written
solutions, with questions answered in your own way.
1. Chaisson Review and Discussion 20.2
How do astronomers test the theory of stellar evolution? (3 points)
Stars change so slowly over time, that we have no hope of observing the changes they go
through directly in a human lifetime or even in all of human history. However, we have a
galaxy full of many stars at different stages of development. By compiling statistics about the
properties and types of stars, we can use our knowledge of how gasses behave and how stars
fuse atoms to piece together these different stages into a coherent “life story” for a star. If
our theory of stellar evolution can reproduce the stellar populations we see in the Galaxy, it
is a pretty good theory.
2. Chaisson Review and Discussion 20.4
Why is the depletion of hydrogen in the core of a star such an important event? (3 points)
When the hydrogen is depleted in the core of a star, it can no longer fuse hydrogen into
helium. With no energy source to provide large amounts of outward pressure in the core,
gravity is then able to collapse the core, forcing major changes in the entire structure of the
star.
3. Chaisson Review and Discussion 20.5
What makes an ordinary star become a red giant? (3 points)
When a star runs out of hydrogen in its core, energy generation by fusion stops, and the core
collapses. As a result, the core’s temperature increases and additional energy is radiated
away. With a higher temperature, the fusion in the hydrogen shell around the core becomes
more efficient. So the core puts out even more energy than it did as a main sequence star.
The increased gas pressure pushes on the outer part of the star, expanding it. As the gas in
the outer layers expands and becomes less dense, it cools down, and the star’s color changes
to red.
4. Chaisson Review and Discussion 20.6
Roughly how big (in A.U.) will the Sun become when it enters the red-giant phase? (3 points)
A star like the Sun will evolve into a red giant with a size about 100 times its current size.
This is equivalent to about 70 million km, or almost half an AU.
5. Chaisson Review and Discussion 20.8
Do all stars eventually fuse helium in their cores? (3 points)
A star must have sufficient mass to collapse its core enough to reach the critical temperature
of about 100 million K to fuse helium into carbon. Only stars with a mass more than about
25% of the Sun’s mass can manage. In less massive stars, the core stabilizes before the
temperature can be reached.
6. Chaisson Review and Discussion 20.9
What is a helium flash? (4 points)
By the time the helium core of a red giant has collapsed for millions of years, the core has a
high density of electrons that produce a pressure unlike that of normal gas. This electron
pressure is not influenced by temperature. When the core temperature finally reaches about
100 million K, helium begins to fuse into carbon. Because of electron pressure is dominant,
this increase in temperature does not increase the pressure enough to expand and stabilize
the core. The fusion of helium raises the core temperature, producing more and more helium
fusion without raising the pressure, so that over a few hours a large quantity of helium fuses
into carbon. This rapid fusion of helium is known as the helium flash.
7. Chaisson Review and Discussion 20.11
How do the late evolutionary stages of high-mass stars differ from those of low-mass stars?
(4 points)
As a result of helium fusion, low mass stars eventually form a carbon core that collapses but
cannot collapse enough to attain a high enough temperature to allow the fusion of carbon.
The outer part of the star continues to expand and as the final shells of hydrogen and helium
fusion die out, this outer part of the star is ejected into space. This cloud of gas is known as a
planetary nebula. The core of the star remains, continues to cool, and is known as a white
dwarf. In contrast, more massive stars are able to reach the core temperatures required for
more complex nuclear fusion processes: the fusion of carbon, oxygen, and heavier elements.
As the fusion processes get more complex, they take less and less time. The most massive
stars can fuse elements to create iron cores, and shortly thereafter they explode.
8. Chaisson Review and Discussion 20.12
What is the internal structure of a star on the asymptotic-giant branch? (3 points)
The star has a carbon core in which a small amount fuses with helium to form oxygen.
Around the core, helium continues to fuse into carbon, and outside this region, hydrogen
fuses into helium. The temperature of the core is about 300 million K, too cool to fuse carbon
on a large scale. The heat of the interior is enough to expand the star to incredible size –
hundreds of times its previous radius.
9. Chaisson Review and Discussion 20.13
What is a planetary nebula? Why do many planetary nebulae appear as rings? (4 points)
A planetary nebula is the ejected outer layer of a giant star. Because the ejection was
uniform and fairly gradual, it is in the shape of a spherical shell and is composed of
relatively cool, thin gas. This shell often appears as a ring; the thickest parts are seen in
cross-section and look like a ring and the parts toward the core are thin and emit little light.
The farther away from the center (from our perspective) we look, the thicker the layer of gas
we are looking through.
10. Chaisson Review and Discussion 20.14
What are white dwarfs? What is their ultimate fate? (4 points)
As a dying star sheds its outer layers to make a planetary nebula, the dead core of the star is
exposed. In the case of a solar-mass star, this core will be made of carbon. The core
stabilizes at a radius roughly equal to that of the Earth. This means it has an incredibly high
density, a million times the density of the Sun, as it still represents about 10% of the star’s
original mass packed into such a small space. The small surface area also translates into a
low luminosity, despite its white-hot temperature. Since it is not actively generating energy,
it will gradually cool down and fade from white to red to black.
11. Chaisson Review and Discussion 20.18
How can astronomers measure the age of a star cluster? (3 points)
The age of a star cluster is determined by the type of star that is getting ready to leave its
main sequence. As a cluster evolves, its more massive main sequence stars become red giants
and die out, leaving only the fainter, low-mass main sequence stars. We note the temperature
and luminosity of stars that are about to leave the main sequence, and translate that into a
mass. The mass and luminosity help us calculate the main sequence lifetime of that star,
which is a good estimate of the age of the cluster.
12. Chaisson Review and Discussion 20.19
What are the Roche lobes of a binary system? (3 points)
In a binary star system, the region around each star in which its gravity dominates is known
as its Roche lobe. When stars are close to each other, the regions are teardrop-shaped and
meet at the Lagrangian point. The two joined lobes look like a three-dimensional figure-8.
13. Chaisson Problem 20.2
Use the radius-luminosity-temperature relation to calculate the radius of a red supergiant with
temperature 3000 K (half the solar value) and total luminosity 10,000 times that of the Sun.
How many planets of our solar system would this star engulf? (5 points)
We use the formula developed in the previous chapter:
2
 RRSG 
LRSG

 
L
 R 
4
 T   10, 000  5780 

 

  138, 000
 TRSG   1  3000 
4
Taking the square root, we find that the radius of the supergiant is 370 times that of the Sun.
This is about 1.7 AU, and would engulf all four planets of the inner solar system.
14. Chaisson Problem 20.3
What would be the luminosity of the Sun if its surface temperature were 3000 K and its
radius were…
a) 1 A.U. (4 points)
b) 5 A.U. (3 points)
4
2
T   R 
L
We can use the formula New   New   New  to find the new luminosity.
L
 T   R 
4
2
4
2
L
 3000   150, 000, 000 
a) New  
 
  3370 L (4 points)
L
 5780   686, 000 
L
 3000   750, 000, 000 
b) New  
 
  84,300 L (4 points)
L
 5780   686, 000 
This is 25 times greater than the luminosity in part a), because of the 5 times increase in
radius.