Blog #7: Milky Way Rotation

4. We actually observe a flat rotation curve in our own Milky Way. (You will show this with a radio telescope in your second lab!) This means that $v(r)$ is nearly constant for a larger range of distances. 

(a) Let's call this constant rotational velocity $V_c$. If the mass distribution of the Milky Way is spherically symmetric, what must be the $M(<r)$ as a function of $r$ in this case, in terms of $V_c$, $r$, and $G$?

Since the rotation curve for the galaxy is actually flat, we can use the equation of Kepler's Rotation Curve derived in Blog#6, part(c) to solve for a the mass distribution. Recall that Kepler's Rotation curve is as follows:

\begin{align}
v(r) &= \left(\frac{GM_{enc}}{r}\right)^{\frac{1}{2}}
\end{align}

However, in this problem, since the enclosed mass, $M_{enc}$ is a function of radius, $r$, and the velocity is constant, $V_c$, we can rewrite the equation as follows:

\begin{align}
V_c &= \left(\frac{GM(<r)}{r}\right)^{\frac{1}{2}}
\end{align}

Since we are looking for $M(<r)$, we can rearrange the equation to solve for $M(<r)$:

\begin{align}
V_c &= \left(\frac{GM(<r)}{r}\right)^{\frac{1}{2}}\\
V_c^2 &= \frac{GM(<r)}{r}\\
M(<r) &= \frac{V_c^2 r}{G}
\end{align}

(b) How does this compare with the picture of the galaxy you drew last week with most of the mass appearing to be in bulge? 

Looking at mass as function of radius, $M(<r) = \frac{V_c^2 r}{G}$, you can see that this equation implies that mass increases linearly the further away you go from the center of the galaxy. However, this shows that mass and radius have a linear relationship, meaning that the volume has to increase linearly. However, volume does not increase linearly in a spherical configuration. Therefore, looking at the galaxy as a cylinder, we can think of the height of the cylinder as the radius, which would increase volume linearly with a constant of $\pi r^2$. Therefore, this shows that the mass of the galaxy cannot be concentrated in a bulge, but rather in a more cylindrical shape.


(c) If the Milky Way rotation curve is observed to be flat ($V_c \approx 240$ km/s) out of 100 kpc, what is the total mass enclosed within 100 kpc? How does this compare with the mass in stars? 

We can solve for total mass enclosed within 100 kpc of the galaxy by plugging in values into the equation derived in part (a) as follows:

\begin{align}
M(<r) &= \frac{V_c^2 r}{G}\\
M(<r) &= \frac{(240 \frac{km}{s})^2 (100 \text{ kpc})}{4.3 \times 10^{-3} \frac{\text{ pc} \cdot \text{km}^2}{M_{\odot}\text{s}^2}}\\
M(<r) &= \frac{(240 \frac{km}{s})^2 (1 \times 10^5 \text{ pc})}{4.3 \times 10^{-3} \frac{\text{ pc} \cdot \text{km}^2}{M_{\odot}\text{s}^2}}\\
M(<r) &= 1.3 \times 10^{12} M_{\odot}
\end{align}

The mass of the Milky Way comes out to be $1.3 \times 10^{12} M_{\odot}$, which is huge! What is more interesting is that we know that the stellar mass of the Milky Way is about $10^{10} M_{\odot}$. If we subtract the stellar mass of the Milky Way with the mass of the entire Milky Way galaxy, $1.3 \times 10^{12} M_{\odot} - 10^{10} M_{\odot} = 1.29 \times 10^{12}$.

Therefore, stars make up only 1% of the mass of the Milky Way!