Blog #6: Milky Way Rotation

3) Last time we approximated the shape of our Galaxy as a cylinder. This time it will be a sphere. If there are no other large-scale forces other than gravity (a good approximation in most galaxies), then an object's orbit around the galactic center will be the approximate center. 

(a) Show that Kepler's 3rd can be expressed in terms of the orbital frequency $\Omega \equiv \frac{2 \pi}{P}$ (i. e. orbits/time) and the distance from the center

\begin{align}
r^3 \Omega^2 \equiv G M_{tot}
\end{align}

Knowing that the orbital frequency $\Omega \equiv \frac{2 \pi}{P}$, we can rearrange it to solve for the period, $P = \frac{2\pi}{\Omega}$. Now that we have period, $P$, in terms of orbital frequency $\Omega$, we can substitute it into the equation for Kepler's Third Law to get Kepler's Third Law in terms of orbital period as follows:

\begin{align}
P^2 &= \frac{4\pi^2 r^3}{GM_{tot}}\\
\left( \frac{2\pi}{\Omega}\right)^2 &= \frac{4\pi^2 r^3}{GM_{tot}}\\
\frac{4 \pi^2}{\Omega^2} &=  \frac{4\pi^2 r^3}{GM_{tot}}\\
\frac{1}{\Omega^2} &=  \frac{ r^3}{GM_{tot}}\\
\Omega^2 r^3 &= GM_{tot}
\end{align}

(b) Now, assume that the Milky Way has a spherical mass distribution - this is a good approximation when talking about the total mass distribution. Using what you learned from Problem 2, rewrite the above for an object orbiting a radius $r$ from the center of the galaxy. 

Assuming that the Milky Way has a spherical mass distribution implies that the total mass that is being considered is a function of how far from the center of the sphere you are at.

Basically, at any given radius $r$ from the center of the sphere, the total mass that is enclosed in that mass, $M_{enc}$ includes the gravitational effects of the total mass enclosed within that radius, and is not affected by any mass beyond that radius.

Therefore, you can rewrite the equation is part (a) to the following:

\begin{align}
\Omega^2 r^3 &= GM_{enc}
\end{align}

(c) Next, let's call the velocity of this object at distance $r$ away from the center, $v(r)$. Use Kepler's Third Law as expressed above to derive $v(r)$ for a mass $m$ if the central mass is concentrated in a single point at the center (with mass $M_{enc}$), in terms of $M_{enc}$, $G$, and $r$. This is known as the Keplerian rotation curve. As you saw earlier, it describes the motion of the planets in the solar system, since the Sun has nearly all of the mass. 

So far, the equations we have about orbital period and Kepler's Third Law are:

\begin{align}
(1)  & \Omega^2 r^3 = GM_{enc}\\
(2)  & \Omega = \frac{2 \pi}{P}
\end{align}

In order to relate these two by velocity as a function of radius, $v(r)$, we need to think about rotational velocity. Since velocity is defined as $v = \frac{\text{distance}}{\text{time}}$, we can think of the circumference of an orbit, $2\pi r$ as the distance, and the period of the orbit, $P$, as the time, which gives us:

\begin{align}
v(r) = \frac{2 \pi r}{P}
\end{align}

Knowing these equations, we can solve for orbital period, $\Omega$ in terms of $v(r)$ as follows:

\begin{align}
\Omega = \frac{2 \pi}{P} &\text{   and   } v(r) = \frac{2 \pi r}{P}\\
v(r) &= \Omega r\\
\Omega &= \frac{v(r)}{r}
\end{align}

Finally, we can substitute this equation into Kepler's Third Law in terms of orbital velocity to find Kepler's Rotation Curve as follows:

\begin{align}
\Omega^2 r^3 &= GM_{enc}\\
(\frac{v(r)}{r})^2 r^3 &= GM_{enc}\\
v(r)^2 r &= GM_{enc}\\
v(r) &= \left(\frac{GM_{enc}}{r}\right)^{\frac{1}{2}}
\end{align}