Second Hour Exam |
1998 April 15 |
(Show your work!! I will be very generous with partial credit!!!)
Problem 1) By measuring the period of its variation in flux, you determine that the luminosity of a Cepheid variable star the M99 galaxy is 50,000 Lo (1 Lo = 3.8 x 1026 W) You measure its average flux to be 3.3 x 10-16 W/m2.
Part 1) Calculate the distance to the M99 galaxy.
Solution: From the front of your exam, you have the relationship between flux, luminosity, and distance:
Solving for distance, we get,
Plugging in numbers,
Part 2) The M99 galaxy has an angular size of 0.9 degrees. Calculate its linear size.
Solution: You solved for the distance to the galaxy in the Part 1 of this problem. Now that you're given its angular size, you can calculate the linear size of the galaxy with the Observer's Triangle:
This time, we're solving for w, the linear size of the galaxy, so
Problem 2) During the course of its Main Sequence lifetime, a typical O star converts 1 Mo (= 2.0 x 1030 kg) of hydrogen into helium in the following net reaction:
Part 1) Given that the mass of one helium atom (mHe = 6.63 x 10-24 kg) is 99.3% of the mass of four hydrogen atoms (mH =1.67 x 10-24 kg), calculate how much energy is generated during the O star's Main Sequence lifetime.
Solution: There's a lot of information here, only some of which you really need. The easiest way to answer this question is to consider the fact the 99.3% of the 1 solar mass of hydrogen is converted to helium. What happens to the other 0.7%? It gets converted into energy.
Therefore the total mass converted to energy by an O star during its Main Sequence lifetime is
The energy created when this mass is converted to energy can be calculated from Einstein's famous equation
Part 2) If the O star spends 10 million years (= 3.16 x 1014 seconds) on the Main Sequence, calculate its average luminosity during this time.
Solution: We've just calculate how much energy the O star produces during its Main Sequence lifetime. Now we're told how long it lives on the Main Sequence, and we're asked to calculate the average luminosity, or energy emitted per unit time.
Well, since luminosity is energy per time, we can take the total energy emitted and divide it by the total time in seconds, and we'll get the average luminosity in energy/second. So,
Problem 3) Radio wavelength observations of one of the clouds swirling around a black hole in the center of galaxy Lepus X-2 show that radiation from the hydrogen spin-flip transition (rest frequency = 1420.406 MHz) is detected at a frequency of 1421.229 MHz.
Part 1) Calculate the speed of this cloud and whether it's moving toward or away from us.
Solution: This is a pretty straightforward Doppler problem. You're given the rest frequency of the photons and their observed frequency. Since the observed frequency is higher than the rest frequency, the light waves must be "squished together" more and that means that the source is coming toward us.
Now, for the actual value of the velocity, we need to use the Doppler relation from the front of your exam:
speed of emitter change in frequency
---------------------- = ----------------------
speed of wave rest frequency
The change in frequency is the difference between the observed and rest
frequencies, or
Now we can plug numbers into the Doppler relation:
and, solving for the speed of the emitter, and noting that the units of MHz cancel out, we get
Part 2) If this gas cloud is located 0.1 pc from the black hole and it is orbiting it in a circle, calculate the mass of the black hole.
Solution: Now that we know the speed of the cloud, and we're told the distance at which this cloud orbits the black hole, we can calculate the mass responsible for the cloud's motion. Te dynamical mass of an object (in this case, the black hole) can be calculated from the orbital properties of anything orbiting it, as follows:
The only messy part of this problem is that we need to convert our distance units from parsecs to m, so that the units cancel out (look at the units for G), and we get an answer in kg. The unit conversion proceeds as follows:
Now we can use the dynamical mass formula directly,