SOLUTION TO EXERCISE 2


Let's break it down into
the disk and bulge components.  The disk ("frisbee") has a diameter
of about 30 kpc and a thickness of about 0.3 kpc; the bulge ("flattened
tennis ball") has a diameter of about 3 kpc.  So if we approximate the
shape of the disk as a very flat cylinder, the volume of the disk is

   V = 4 * pi * r^2 * h = 12.56 * (15 kpc)^2 * (0.3 kpc) = 848 kpc^3

We can also approximate the shape of the bulge as a sphere; if so,
the volume of the bulge is

   V = (4/3) * pi * r^3 = 4.18 * (1.5 kpc)^3 = 14.1 kpc^3

We also know that the closest star to the Sun is about 1.3 pc away,
and that we are in an "ordinary" part of the galactic disk.  So let's
make a reasonable estimate that the average distance between stars
throughout the galaxy is 1.3 pc.  This means that, on average, each
star occupies a volume equal to a sphere 1.3 pc (or 0.0013 kpc) in
diameter.  Thus each star occupies

   V = (4/3) * pi * r^3 = 4.18 * (0.0013 kpc)^3 = 9.18 * 10^(-9) kpc^3

So in the disk, there are a total of

   (848 kpc^3) / (9.18 * 10^(-9) kpc^3) = 9.2 * 10^10 stars.

The bulge is much smaller - 100 times less volume than the disk;
so it would contain about 1.5 * 10^9 stars if the number density of
stars in the bulge is the same as the disk.  You may have concluded
that the bulge is denser than the disk, however, if you noticed that
it's at the center of the disk and thus at the center of the Milky
Way's gravitational field; in that case, it'd have to have at least
ten times the density to affect our estimate by 10% - and we've made
so many reasonable approximations already that we're not justified
in claiming that level of precision in our calculation.  So, we are
all right in saying that we estimate there are 9.2 * 10^10 stars in
the Milky Way Galaxy, which is 92 billion; since we've made a number
of ballpark estimates, I'll use the nearest round number and say that
there are about 100 billion stars in the Milky Way galaxy.