Problem 11.42

11.42. a)

We know that the period is related to the velocity by the formula
$T = \frac{2\pi r}{v}$
so, plugging in what we know for the solar system, we find
$T = 7.13 \times 10^{15}s ~\simeq~ 2 \times 10^{8} years$

b)
Now, from Kepler's laws, we can relate the mass to the period by the relation
$M = \frac{4\pi^{2}d^{3}}{GT^{2}}$
where we need to convert d from the 30,000 light years into meters.
We make use of the fact that
1 light year = $9.46 \times 10^{15} m$
so, plugging in our value for T found in part a) and the distance d in meters, we find
$M = 2.66 \times 10^{41}kg$
or $M = 1.34 \times 10^{11}$ solar masses.
In other words, there is the mass equivalent of approximately 10¹¹ stars the mass of our sun in the Milky Way.