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1
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- Question 1 (1 point) The use of Cepheid variable stars to
find distances relies on the fact that
a. their luminosities are
related to their ages. b. their luminosities are all the same.
c. their radial velocities are related to their periods. d.
their luminosities are related to their periods.
- Question 2 (1 point)
You observe two Cepheid variable stars, A and B, which have the same
average apparent magnitude. Star A brightens and dims with a period of 5
days, while star B has a period of 18 days. Which star is closer to
Earth?
a.Star A b.Star B
c.Star A and B are at the same distance d.You cannot tell
from this information
- Make the following argument to solve this one easily. The period-luminosity relationship is
precise, but precision is not needed here. The star with the longer period will
be more luminous, so that is star B.
But we want to know the distance.
They both appear to have the same average apparent
magnitude. If one is more
luminous than the other, the more luminous one must be farther way – if
it wasn’t, it would appear brighter.
Therefore the less luminous star is closer, and that is star A.
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2
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- Question 3 (1 point)
Which of the following is not a characteristic of the stars of the disk
component of our galaxy?
a. circular orbits b.
random highly inclined orbits c.higher metal abundance d.
young stars e. star formation regions
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Question 4 (1 point) If the sun is 5 billion years
old, how many times has it orbited the galaxy? Assume a circular orbit
for the sun around the center of the galaxy, and look up in the text its
orbital speed and distance from the galactic center.
a. 5 times b. 5 billion
times c. 200 times d. 20 times e. 5000 times
- Approach this problem like a time, rate, distance problem to figure out
an orbital period, then see how many go into 5 billion years. If the orbital speed is 220 km/s, and
the oribital radius is 8.5 kpc, we can get the period (time): time =
speed/distance and we must convert.
8.5 kpc = 8500 pc x 3.1 x 10 13 meters/pc = 2.6x1017
km. So time = 2πx2.6x1017
km/220 km/s = 7.5x1015 s.
What is that in years?
There are about 31 million seconds per year, so this period is
7.5x1015/3.1x107 = 2.4x108 years, or
about 240 million years. 5
billion divided by 240 million is 5x109/2.4x108 = 20.8.
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3
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- Question 5 (1 point) If all
the mass in our galaxy were centrally concentrated, we'd expect
velocities to fall with increasing distance according to Kepler's laws.
This is not seen in the disks of spiral galaxies. Galactic rotation
curves appear "flat" with increasing distance. This must be
due to
a. The gravitational
influence of massive globular clusters in the halo. b. The
difficulty in measuring velocities of stars in the galactic disk because
of all the gas and dust. c. The fact that Kepler's laws do not
apply over the 25 kpc size of the Milky Way galaxy because the effect of
gravity travels only at the speed of light. d. The Spiral density
wave traveling at a different speed than the stars in the disk. e.
The gravitaional influence of "dark matter" in the halo.
- Question 6 (1 point) The location of the sun relative to the
center of our galaxy was FIRST CORRECTLY determined by
a. the relative brightness
of the visible Milky Way in different directions. b. the positions of the globular clusters
and the properties of the variable stars in them. c. the position
of supernova. d. radio maps of the galaxy.
- Question 7 (1 pt) If you find a metal-rich star it is more likely to be
a member of which Population? A.
Population I (the disk)
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4
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- Question 8 (1 point) One fundamental requirement for an
object to be a good tracer (i.e. marker) of the galaxy's spiral arms is
that it be
a.moving with small radial
velocity (relative to earth). b.moving with large radial velocity
(relative to earth). c. a short lived and therefore young object.
d. a very long lived and therefore probably old object.
- Question 9 (1 point) What is the most common type of galaxy,
based on what we see in our own Local Group?
a. irregular galalaxies b. flocculant spiral galaxies
c. grand design spiral galaxies d. giant ellipticals
e. dwarf ellipticals
- This one you should know from the figure of the Local Group in the
slides/text. I’d make sure I knew
this for the exam!
- Question 10 (1 point) Observations of galaxies and clusters
of galaxies indicate that roughly ____ percent of the universe is dark
matter.
a.5%b.25%c.50%d.75%e. 95%
- You have to look this one up, but it is a number you should know.
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5
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- Question 11 (1 point) An elliptical galaxy could
- a. evolve into an irregular galaxy when it has used up all of its gas
and dust.
- b. be formed from the collision and merger of spiral galaxies.
- c. become a starburst galaxy if it were to move through the hot
intergalactic medium of a cluster.
- d. evolve from a single spiral galaxy when that spiral used up all of
its gas and dust.
- Elliptical galaxies have already used up their gas and dust, and it is
thought they can be formed from spiral mergers.
- Question 12 ( 1 point) If a galaxy has a radial velocity
(redshift) of 5000 km/s, how far away is it? Assume a Hubble Constant of
70 km/s/Mpc.
a.7 Mpc b.20 Mpc c.49 Mpc d.71 Mpc e.100 Mpc
- V = Hxd, so d = v/H = 5000 km/s/70 km/s/Mpc – the km/s cancel out to
give 5000/70 Mpc = 71 Mpc
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Notes