Astronomy 102 Problem Set #4 Solutions

Problem #1: When a massive star explodes as a supernova, it shines with a luminosity ten billion times greater than that of the Sun for a short period of time. If such a supernova explosion were to appear in our sky as bright as the Sun, how far away from us must it be located?

Solution: This is another problem where there's an easy way and a hard way to do it. The hard way is to calculate the flux of the Sun, and then use that value as the flux of the supernova to calculate its distance. This method will work fine, but it involves a lot of unnecessary multiplication.

First, make sure you understand what you're told in the problem. The first sentance tells you that the luminosity of the supernova is equal to 10 billion times the luminosity of the Sun; that is,

Secondly, we're told that the supernova and the Sun appear equally bright in the sky. What's that mean? Well, objects that are equaly bright have the same flux, since you eye is a flux-measuring machine. So, Now, we have flux and luminosity (albeit in ratio from), so we can use the relationship between flux, distance, and luminosity that we're now used to: However, instead of calculating the actual flux of the Sun as a numerical value, let's just call it fSun. Then, where LSun and DSun mean the luminosity and distance to the Sun, repsectively. We can write the same relationship for the supernova, Now we're told in the problem that the flux of the supernova equals the flux of the Sun, or fSN = fSun. That also means We can immediately gid rid of the 4 x pi's on either side of the equation, and then solving we DSN we get where as always, (x)1/2 means "the square root of x." At this point we can plug in the values from the problem and use our calculator. We're told that the luminosity of the supernova is 1 x 1011 L0, and since the luminosity of the Sun is 1 L0, the fraction LSN/LSun = 1 x 1011. The distance to the Sun is 1 Astronomical Unit, or 1 A.U. So, This are perfectly reasonable units for a distance, but if you'd rather, we can convert to more common units. or Personally, I find the last set of units most meaningful for this problem, since I have a feel for the distance to stars in parsecs. The star nearest to the Sun, Proxima Centauri, is about 1 parsec away. It's not a massive star, and there aren't any massive stars that close to us, so it appears that we won't be seeing supernovae as bright as the Sun anytime soon. That's a good thing, since a supernova that nearby would be really unhealthy.

Problem #2: Radio wavelength observations of a gas cloud near the Galactic Center show spectral line emission at a
frequency of 1421.65 MHz (1 MHz = 106 Hz). Astronomers have determined that this emission line is generated by the
spin-flip transition of atomic hydrogen, which has a rest frequency of 1420.406 MHz (Electrons can come in two "flavors", spin
up and spin down. Transitions between such state is much less energy intensive than, say, a transition from the ground state to
the first excited state in hydrogen). Based on these observations, calculate the speed of the gas cloud toward or away from the
Earth.

Solution: This ia a pretty straightforward Doppler application. The photons you detect have a frequency which is different from the rest frequency you expect for atomic hydrogen emission. Why? Because the source of these photons (a cloud of gas near the galactic center) is moving toward or away from you.

Well which way is it? Toward you or away from you? The photons you detect have been shifted to a higher frequency, so the waves muct have been squished together, and so the cloud must be moving toward you. To figure out how fast the cloud is moving, we need to measure the change in frequency due to the Doppler effect. The change in frequency is the difference between the frequency you detect and the rest frequency, or

Now let's insert that quantity into the Doppler folmula, where you no doubt recall that radio waves, like any electromagnetic waves, travel at the speed of light. Note that I'm using units of MHz on the left hand side of this equation. Since the left side is a ratio between the frequency shift and the frequency, I'm free to use any units provided I use the same units top and bottom. If I instead used units of Hz, both the top and bottoms numbers would be a factor of 1 million greater, but the ratio would turn out the same.

We can solve for the speed of the emitter as follows

Problem #3: a) If the interstellar medium has an average density of 4.5 x 10-18 kg/m3, what volume of interstellar medium contains enough mass to make a 1 Mo star?

b) If this volume were spherical in shape, what radius would it have? (Hint: the volume of a sphere is 4/3 x pi x radius3.)

Solution: The density of anything is simply the amount of mass that is contained in a unit of volume of the material. In this case, the density of the interstellar medium is such that there's 4.5 x 10-18 kg of stuff in a cubic meter. How much mass is in 2 cubic meters? Why 9.0 x 10-18 kg, of course. Clearly, if we're going to collect one solar mass of material, which is 2.0 x 1030 kg, we're gonna need a lot of cubic meters.

Density is just mass over volume, and for this probelm we know the mass we want to get and the density of the medium, so

and, solving for volume, and the units come out as cubic meters, which makes sense for a volume.

Now we're asked to calculate the radius of a sphere which has the above volume. We're given the equation for the volume of a sphere

so, and solving for radius where (x)1/3 means "the cube root of x." This answer is perfectly correct, ande the units make sense, but I like it more in units of parsecs, because I understand stellar distances in parsecs better. It might not surprise you to learn that this size is rather typical for the dense star-forming cores in molecular clouds.