Because the rotation curve for the Milky Way Galaxy is approximately flat, a star in a circular orbit 13 kpc from the galactic center has the same orbital speed as the Sun, namely 220 km/sec. How much mass is interior to this star's orbit (in solar masses)?

Question

Quizplus · Accepted Answer

We can use the circular velocity equation, $v_c = \sqrt{\frac{GM(<r)}{r}}$, where $v_c$ is the circular velocity, $M(<r)$ is the mass interior to the star's orbit, and $r$ is the distance from the galactic center. Since the star is at the same distance as the Sun ($r=13$ kpc) and has the same circular velocity ($v_c = 220$ km/sec), we can solve for $M(<r)$:
\begin{align*}
v_c^2 &= \frac{GM(<r)}{r} \
M(<r) &= \frac{v_c^2r}{G} \
&= \frac{(220	ext{ km/s})^2 	imes 13	ext{ kpc}}{G} \
&\approx 1.5 	imes 10^{11}M_\odot
\end{align*}
where $G$ is the gravitational constant. The answer is therefore (B), 1.5 × 10¹¹.

Because the Rotation Curve for the Milky Way Galaxy Is