Blog #6: WS3.1, #3

3. We will approximate the shape of our Galaxy as 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 approximately circular.
a) Show that Kepler's 3rd can be expressed in terms of the orbital frequency $\Omega \equiv 2\pi /P$ (i.e. orbits/time) and the distance from the center $GM_{tot} = r^3\Omega ^2$.

Kepler's 3rd states $P^2 = \frac{4\pi ^2a^3}{GM_{tot}}$ where P is the orbital period, a is the semimajor axis, $M_{tot}$ is the sum of the masses in the system, and G is the gravitational constant.
$a \equiv r$ (distance from the object to the galactic center)
$P = \frac{2\pi}{\Omega}$
$(\frac{2\pi}{\Omega})^2 = \frac{4\pi ^2r^3}{GM_{tot}}$
$GM_{tot} = r^3\Omega ^2$

b) Assume that the Milky Way has a spherical mass distribution -- this is a good approximation when talking about the total mass distribution. Rewrite the above for an object orbiting a radius r from the center of the galaxy. 

Since the mass inside the orbit of the object can be approximated as a point mass at the center of the galaxy, the equation remains essentially the same, but with the mass and radius adjusted for the orbiting object.
$r^3\Omega ^2 = GM_{enc}$ where $M_{enc}$ is the mass enclosed within the object's orbit.

c) Next, let's call the velocity of this object at a 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. It describes the motion of the planets in the solar system, since the Sun has nearly all of the mass. 

$\Omega = \frac{v(r)}{r}$ (orbital velocity is the ratio between linear velocity and radius)
$r^3(\frac{v(r)}{r})^2 = GM_{enc}$
$rv^2(r) = GM_{enc}$
$v^2(r) = \frac{GM_{enc}}{r}$
$v(r) = (\frac{GM_{enc}}{r})^{1/2}$