ASTR 1040 Spring, 2005
1. In this problem, you were given data on the magnitudes versus time for a Cepheid variable star, and you were given a period-luminosity relationship that you could use to find the average absolute visual magnitude of the star, once its period was established. The questions are: how far away (in light years) is the Cepheid; is it within our galaxy or in another galaxy; and what comments would you make on the usefulness of Cepheids for measuring the size scale of the universe?
First, construct a light curve for the star (see handout from class)
The period derived from this plot should be about 45 days, and the mean apparent magnitude is aproximately (27 + 28.5)/2 = 27.75. From the period-luminosity graph, we find that the mean absolute visual magnitude for a star having period P = 45 days (logP = 1.65) is about -5.4. Using the distance modulus relation m - M = 5log(d/10) then yields a distance of d = 4.3 X 10^7 pc = 43 Mpc = 1.4 X 10^8 light years. This is far beyond our own galaxy, but is only a very small fraction of the size of the universe, which is of order 10 billion light years. Thus Cepheids cannot directly provide information on the size scale of the universe, since even with the HST we cannot expect to detect Cepheids much fainter than this one.
2. Suppose the rotation curve for a spiral galaxy looks like this:
(a) What is the mass of the galaxy?
(See the original assignment or the answer sheet, handed out in class.)
Find the mass the same way we did for the Milky Way, by applying Kepler's third law to a star far enough from the center so that its orbit is approximately Keplerian (i.e., the star acts like a point mass orbiting a common center of mass with another point mass, the other object being the galaxy). It is therefore best to pick a point farther out than the peak in the curve above, because the peak represents the location where the galaxy rotation stops being rigid and where individual stars begin to act like free particles orbiting a single center of mass; i.e., the point where stellar orbits start to be approximately Keplerian.
I chose 6 kpc as the distance from the center; at that distance, the graph shows that the orbital velocity is about 80 km/sec. To use Kepler's third law, the next step is to calculate the orbital period for a star at 6 kpc from the center (i.e., 1.9 X 10^17 km), which you can find from P = 2πa/v, or P = 1.5 X 10^16 sec = 4.7 X 10^8 yr. The semimajor axis a, converted to AU, is 1.2 X 10^9 AU, So, using Kepler's third law, we find for the mass M = a^3/P^2 = (1.2 X 10^9)^3/(4.7 X 10^8)^2 = 7.8 X 10^9 solar masses. You should have found a similar answer for any point you chose that is at least 5 kpc from the center (i.e., at or beyond the point where the orbits become Keplerian).
(b) What assumptions were adopted in the technique you used?
The first assumption was that the stellar orbits, at least in the outer galaxy, are adequately represented by Kepler's laws, in particular Kepler's third law. This means that an individual star acts like a point mass orbiting another point mass (i.e., the galaxy itself), and that the stars do not interact with each other. This is approximately true only in the outer galaxy, beyond the point (about 5 kpc in this example) where the stars are locked into each other and the galaxy rotates like a rigid body.
The second assumption made was that all, or at least most, of the galaxy's mass is interior to the orbit of the star you chose to use in calculating the mass using Kepler's third law. Given the idealized rotation curve that was presented to you in this problem, that looks like a good assumption. But real spiral galaxy rotation curves don't look like this example; instead they generally show that the orbital speed does not drop off with distance, which indicates that there is a lot of mass (probably the majority of all the mass in the galaxy!) located much farther out.
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