Blog Post 7, WS 3.1, Problem 3: Keplerian Rotation Curve

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 can be expressed in terms of the orbital frequency Ω =2π/P (i.e. orbits/time) and the distance from the center \(r^3\Omega^2=GM_{tot}\)

(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.

(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. As you saw earlier, it describes the motion of the planets in the solar system, since the Sun has nearly all of the mass.

(a) Show that Kepler’s 3rd can be expressed in terms of the orbital frequency Ω =2π/P (i.e. orbits/time) and the distance from the center \(r^3\Omega^2=GM_{tot}\) 

Let's start with Kepler's 3rd law.
\[P^2=\frac{4\pi^2r^3}{GM}\]

Next Let's solve our new equation for P.

\[P=\frac{2\pi}{\Omega}\]

Now we substitute the second equation into the first.

\[\frac{4 \pi^2}{\Omega^2}=\frac{4\pi^2r^3}{GM}\]

The \(4\pi^2\)'s cancel, leaving us with our answer:

\[\boxed{r^3\Omega^2=GM}\]

(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. 

In Problem 2 we determined that the mass inside a certain radius of a spherical mass distribution is defined as:

\[M_{enc}=\frac{4}{3}\pi \rho r^3\]

Where \(\rho\) is a mass density constant.

Substituting in for M, we get:

\[r^3\Omega^2=G\frac{4}{3}\pi \rho r^3\]

\[\boxed{\Omega=\left(\frac{4}{3}G\pi\rho\right)^{1/2}}\]

(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. As you saw earlier, it describes the motion of the planets in the solar system, since the Sun has nearly all of the mass.

Let's start with Kepler's Third as usual.

\[r^3\Omega^2=GM_{enc}\]

Earlier we defined \(\Omega=\frac{2\pi}{P}\).  With a simple modification of a factor of r, we can relate it to V. \(\Omega r=\frac{2\pi r}{P}=V\)

Substituting in, we get:

\[V^2(r)r=GM_{enc}\]

\[\boxed{V(r)=\left(\frac{GM_{enc}}{r}\right)^{1/2}}\]

I worked with B. Brzycki, G. Grell, and N. James on this problem.