Astronomy 102 Problem Set #7 Solutions

Problem #1: Suppose a supernova goes off in Galaxy M245 and ejects a spherical shell of material into the surrounding interstellar medium.

a) Spectra of the ejected gas shell indicate that Hydrogen-alpha emission (rest wavelength = 656nm) is doppler-shifted to a wavelength of 652.4 nm. Calculate the expansion speed of the ejected shell.

b) Ten years after the supernova appears, radio astronomers measure the size of the ejected shell to be 78 micro-arcseconds in diameter (1 micro-arcsecond = 10-6 arcseconds). Assuming that the shell has been expanding at the same rate for ten years, calculate the distance to the supernova (and therefore, the galaxy it's in).

Solution -- Part a): This is a straightforward doppler shift problem. Though many of you tried to do it relativistically, it's not necessary to do so. Since the wavelength has only been shifted by 3 nm or so, delta_lambda/lambda = 3/656 or about 0.005, so the speeds involved aren't near the speed of light at all. That means you can use our old non-relativistic formula for relating the wavelength shift to the speed

The change in wavelength is the difference between the rest wavelength, 656 nm, and the observed wavelength, 652.4 nm, or

We can now solve for the speed of the emitter, putting in values for the change in wavelength, the rest wavelength, and the speed of the wave

so,

Solution -- Part b): Now that we've figured out how fast the supernova shell is expanding, we can figure out how far it travels in 10 years:

That's the distance the shell of gas has expanded from the supernova, so it's the radius of the shell. The radio observation says that the diameter subtends 78 micro-arcseconds, so we need the diameter, which is twice the radius, or 1.04 x 1015 m. Now we have the linear size of the structure and the angle it subtends, so we can calculate the distance using the Observer's Triangle. However, to use the Observer's Triangle in the form you're used to, we need to convert the 78 micro-arcseconds to degrees:

Now we can use the Observer's Triangle:

Solving for distance, we get

Problem #2:a)If astronomers measure the redshift of a galaxy which is receding at a speed equal to 90% of the speed of light, what redshift will they measure? (Remember to use the relativistic redshift formula!)

b) The Lyman-alpha spectral line from atomic hydrogen (i.e., the transition from the first excited state to the ground state) is normally found in the ultraviolet region of the spectrum, and has a rest wavelength of 121.6 nm. At what wavelength would you expect to measure emission from this transition for atoms in the galaxy receding at 90% of the speed of light?

Solution -- Part a): The expression relating velocity and redshift you were given in class is 

                                                                 1/2
                                                      | 1 + v/c |
redshift = (change in wavelength)/(rest wavelength) = | ------- |   - 1
                                                      | 1 - v/c |
where (x)1/2 means "the square root of x."

We're told that the speed of the galaxy is 90% of the speed of light, or v = 0.9 x c. That means v/c = 0.9, so we can put this quantity directly into the redshift formula 

                      1/2
           | 1 + 0.9 |
redshift = | ------- |    - 1
           | 1 - 0.9 |
so

Solution -- Part b): This part of the question is "do you know what to do with a redshift once you've calculated it?" The redshift is the change in wavelength over the rest wavelength, so if the rest wavelength of the Lyman alpha line is 121.6 nm and the redshift of a galaxy moving at 90% of the speed of light is 3.36, then

Is this the wavelength of the observed emission? No it is not. You've just calculated the change in the wavelength. The observed wavelength then is Note that I knew to add the change in wavelength to the rest wavelength because the galaxy is moving away from me, as are all galaxies at cosmological distances.

Problem #3:a) What is the speed of a galaxy which has a measured redshift of 5.0?

b) How far away is this galaxy if we live in an open universe and Ho = 75 (km/s)/Mpc?

Solution -- Part a): This is another application of the speed-redshift relationship: 

                                                                 1/2
                                                      | 1 + v/c |
redshift = (change in wavelength)/(rest wavelength) = | ------- |   - 1
                                                      | 1 - v/c |

Here we're given the redshift and asked for the speed, so we'll have to do some manipulation. Start by adding one to both sides: 

                                         1/2
                              | 1 + v/c |
             5.0 + 1 = 6 =  = | ------- | 
                              | 1 - v/c |

Now square both sides 


                    | 1 + v/c |
            36 =  = | ------- | 
                    | 1 - v/c |
and multiply both sides by (1-v/c): Let's now add 36(v/c) to both sides Now subtract one and divide by 37 and we've discovered that the galaxy must be moving away from us at 35/37ths of the speed of light, or 0.946c, or 2.84 x 108 m/s.

Solution -- Part b): Now we know how fast it's moving, so we can use the Hubble Law to figure out how far away it is:

However, check your units! You'll need speed in km/s if you want to get an answer out in Mpc. I quoted my answer in m/s above, so I'll need to convert: Now,