Blog Post 6, Worksheet 3.1, Problem 3

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 approximately circular. 

(a) Show that Kepler's 3rd law can be expressed in terms of the orbital frequency \(\Omega \equiv 2π/P\) (i.e. orbits/time) and the distance from the center: \[r^3\Omega^3 = GM_{\text{ tot}}\]
As Kepler's Third Law expresses the relationship between period, P, and semimajor axis, a, we need to solve the given relationship between orbital frequency and period for P in order to be able to make the proper substitution in Kepler's Third Law: \[\Omega = 2π/P\] \[ P = \frac{2π}{\Omega}\] Now, we can plug this definition for P into Kepler's Third Law, as given in Problem 1: \[P^2 = \frac{4π^2a^3}{GM_{\text{tot}}}\] \[(\frac{2π}{\Omega})^2 = \frac{4π^2a^3}{GM_{\text{ tot}}}\] \[ \frac{4π^2}{\Omega^2} = \frac{4π^2a^3}{GM_{\text{tot}}}\] The \(4π^2\) terms cancel, and as we are referring to orbits as cylindrical (as opposed to elliptical), we can replace a, the semimajor axis of an ellipse, with r, the radius of a circle: \[ GM_{\text{tot}} = r^3 \Omega^2\] Thus, Kepler's Third law can be expressed in terms of the orbital frequency and the distance from the center.

(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.

From the Shell Theorem, we know that an object orbiting a galaxy at a distance r from the center will only experience the gravitational effects from the mass located between the center of the galaxy up to a distance r from the center of the galaxy. As a result, \(M_{\text{ enc}}\) can be written as M(<r), or the mass located in the galaxy up to a distance r from the center, and the equation above can be written as: \[ r^3 \Omega^2 = GM(<r)\]
(c) Next, let's call the velocity of the 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 \(M_{\text{enc}}\)), in terms of \(M_{\text{enc}}\), 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.

As we are trying to determine the velocity, v, for a mass m in terms of \(M_{\text{enc}}\), G, and r, we can use the standard relationship between linear velocity v and orbital frequency \(\Omega\) to introduce velocity into the equation for Kepler's Third Law: \[ v = \Omega r\] \[\Omega = \frac{v}{r}\] Now we can plug this definition for the orbital frequency into Kepler's Third Law as expressed above: \[ r^3 \Omega^2 = GM(<r)\] \[ r^3 (\frac{v}{r})^2 = GM(<r)\] \[ \frac{r^3}{r^2} v^2 = GM(<r)\] \[ v^2 r = GM(<r)\] \[v^2 = \frac{GM(<r)}{r}\] \[v = (\frac{GM(<r)}{r})^{1/2}\] Thus, the relationship between v and \(M_{\text{enc}}\), G, and r is given by: \[v = (\frac{GM_{\text{enc}}}{r})^{1/2}\]