The Milky Way is now a sphere (trust me)

We are bounded in a nutshell of Infinite space: Blog Post #6, Worksheet # 3.1, Problem #3: The Milky Way is now a sphere (wait for it)

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 3^rd can be expressed in terms of the orbital frequency \(\Omega \equiv  2\pi / P \) (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 (Gravitational Shell Theorem), 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) First off, let us establish what, exactly, is Kepler’s third law. As described by: \[P^2 = \frac{4 \pi^2 a^3}{ GM_{tot}} , \] the equivalency states how, depending on the semi major axis of the ellipse of the rotation of the celestial body (a) and the total mass of the object, the period of rotation changes. This same description of orbital period (P) can be expressed in terms of the orbital frequency, the amount of rotations in a given time. Using the equation \[\Omega \equiv  2\pi / P , \] for orbital frequency \(\Omega\), \(P\) can be solved for and thus get: \[ P \equiv  2\pi / \Omega , \] which can now be placed directly into Kepler’s Third Law Equation: \[(2\pi / \Omega)^2 = \frac{4 \pi^2 a^3}{ GM_{tot}} , \] and now solved for when the orbit approximates a circle and thus the semi major axis a is also the r of the circle,

\[\frac {4\pi^2}{\Omega^2} = \frac{4 \pi^2 r^3}{ GM_{tot}} , \]

\(4 \pi ^2\) is canceled on both sides and figures are cross multiplied,

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

Which is identical to the original equation the problem asked us to prove.

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

(b) What this problem refers to is the Shell Theorem, a description of astrophysical phenomenon, specifically gravity when great amounts of matter exist in an area, from a particular perspective. The shell theorem explains how, when an object is found in concentric circles (spheres in 3 dimensions) which all have matter, and thus gravity, then the particular object will feel only the gravity of the masses (shells) that are more inward than his present position. Thus the gravity of outside shells is negligible if existent, and all the gravitational force is felt towards the center of the gravitational shells.

If this idea is extrapolated for a spherical galaxy (as the problem describes), then the gravity any observer feels is only towards the center of the spherical galaxy, not towards that which is farther away from the center than the observer. This becomes an alteration for Kepler’s Third law equation we just saw, changing the \(M_{tot}\) to \(M_{tot}  - M_{ext}\), and producing: \[r^3 \Omega^2 = G (M_{tot}  - M_{ext}), \], and now describing \(M_{tot}  - M_{ext}\) as the mass enclosed to the inner spherical shells \(M_{enc}\) we now have the equation:  \[r^3 \Omega^2 = GM_{enc}, \] relating the Shell Theorem to the rotation of an object from the center of the galaxy.

(c) Now, to obtain \(v(r)\), the equation must be reworked with the knowledge of orbital velocities as well. Orbital velocity \(\omega\), which is the same as \(v(r)\), is defined by: \[\omega = \frac {2\pi}{P} r.\] In this iteration of the equation, the \(r\) is specifically separated to illustrate how  the value of orbital frequency \(\Omega\) is clearly present in orbital velocity. The equation can now be re written as \[\omega = \Omega r,\] and can now be solved as \[\Omega = \frac{\omega}{ r}.\] Placed into the original Keplerian equation as:\[r^3 \Omega^2 = GM_{enc}, \] \[r^3 \left(\frac{\omega}{ r}\right)^2 = GM_{enc}, \] \[r \omega^2 = GM_{enc}, \] \[\omega^2 = \frac{GM_{enc}}{r}, \] \[ v(r)  = \left(\frac{GM_{enc}}{r}\right)^\frac{1}{2} . \] This is the formula for predicting rotation curves and speeds of a galaxy.