Blog #8: Milky Way Rotation

5. $M(<r)$ is related to the mass density $\rho(r)$ by the integral:

\begin{align}
M(<r) = \int_{0}^{r} 4\pi r'^2 \rho (r')dr'
\end{align}

 (Recall that the $4 \pi r'^2$ comes from the surface area of each spherical shell, and the $dr'$ is the thickness of each thin shell) 

The fundamental theorem of calculus then implies that $4 \pi r^2 \rho(r) = \frac{dM(<r)}{dr}$. For the case in question 4, what is $\rho(r)$? Is the density finite as $r \rightarrow 0$ in the case of a flat rotation curve?

The case in question question 4 assumes that we have a flat rotation curve in the Milky Way galaxy, where the mass can be measured as a function of the radius, $r$, as $M(<r)$ with the following equation:

\begin{align}
M(<r) &= \frac{V_c^2 r}{G}
\end{align}
where, $V_c$ is the constant velocity, $r$ is the radius of the enclosed mass, $G$ is the universal gravitational constant.

Knowing $M(<r)$, we can use the equation derived from the fundamental theorem of calculus and solve for the density, $\rho(r)$ as follows:

\begin{align}
4 \pi r^2 \rho(r) &= \frac{dM(<r)}{dr}\\
4 \pi r^2 \rho(r) &= \frac{d\left(\frac{V_c^2 r}{G}\right)}{dr}\\
4 \pi r^2 \rho(r) &= \frac{dr\left(\frac{V_c^2}{G}\right)}{dr}\\
4 \pi r^2 \rho(r) &= \frac{V_c^2}{G}\\
\rho(r) &= \frac{V_c^2}{4 G \pi r^2 }
\end{align}

Now that we have solved for $\rho(r)$, we have to consider what happens when you get to the center of the galaxy, where $r \rightarrow 0$. As $r \rightarrow 0, \rho(r) \rightarrow \infty$, which is the point of singularity. 

This shows that as you approach center of the Milky Way galaxy, the density is infinite, which is interesting, because right at the center of the Milky Way galaxy exists a supermassive black hole!