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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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- 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) The (too simple) traditional
theory for the origin of spiral galaxies like our Milky Way was that it
formed
a.from a large cloud of
material that broke offf a larger galaxy. b. as a large rotating spherical
cloud of gas which collapsed and flattened. c.from collisions of
many smaller galxies d.from stars which formed independently in a
large cluster of galaxies.
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- Question 5 (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.
- Question 6 (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.
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- Question 7 (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 8 (1 point)
Which of the following would you not expect to find in the Population II
halo?
a.B star b.red
dwarf c.neutron star d.white dwarf
- Because B stars must be young stars, and there is no new star formation
in the halo.
- Question 9 (1 point) The black hole in the center of our
Milky Way galaxy seems to be about 2.6 million times the mass of the
sun. What is the radius of its event horizon? Think about how big this
is compared to other astronomical objects (e.g., Earth, the sun, the
solar system, the distances between stars).
a.About a billion km.
b.About 4 thousand km. c.About 4 million km.
d.About 8 thousand km. e. About 8 million km.
- Easy way. 1 solar mass
Schwarzschild radius is 3km, so 3x2.6 million is 8 million km.
- Question 10 (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.
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