Blog #7: WS3.1, #6

6. Now consider a spherically symmetric galaxy with a density profile: 
\[\rho (r) = \frac{C}{4\pi r^2(r+a)^2}\] where a and C are constants. 
a) What is M(<r) for this galaxy in terms of C and a? 

The mass would be the integral of the density function to the appropriate radius, so \(M(<r) = \int_{0}^{r} \rho (r')dV\)
\(dV = 4\pi r'^2 dr'\) because you're integrating over a spherical volume.
\(M(<r) = \int_{0}^{r} \frac{C}{4\pi r'^2(r'+a)^2}4\pi r'^2dr'\)
\(M(<r) = \int_{0}^{r} \frac{C}{(r'+a)^2}dr'\)
\(M(<r) = \int_{0}^{r} C(r'+a)^{-2}\)
\(M(<r) = \frac{-C}{r'+a}]_{0}^{r}\)
After plugging in the upper and lower limits for r', his simplifies to \(\frac{Cr}{a(r+a)}\).

b) The total mass of the system is the limit of M(<r) as \(\rightarrow \infty\); calculate this total mass. 

\(\lim_{x\rightarrow\infty}\frac{Cr}{a(r+a)}\)
As r approaches infinity, it will become much larger than a, so (r+a) can therefore be approximated as simply r; this allows simplification to the solution \(\frac{C}{a}\).

c) If we write the total mass as \(M_{tot}\), rewrite M(<r) and \(\rho (r)\) to eliminate C in favor of \(M_{tot}\). 

The total mass includes all mass regardless of distance from the center of the galaxy, so it should be the mass as r approaches infinity (the answer to part b): \(M_{tot} = \frac{C}{a}\).
\(C = aM_{tot}\)
\(\rho (r) = \frac{aM_{tot}}{4\pi r^2(r+a)^2}\)

d) What is the rotation curve for the galaxy? The circular velocity should go to a constant as \(r\rightarrow 0\), i.e. at radii r<<a, this galaxy appears to have a flat rotation curve. If we write this constant velocity as \(v_0\), write M(<r) and \(\rho (r)\) in terms of \(v_0\), a, and r. What is \(M_{tot}\) in terms of \(v_0\) and a? 

Rotation curve: \(v(r) = (\frac{GM(<r)}{r})^{1/2}\) (from problem 3c)
Constant velocity: \(v_0 = (\frac{GM_{tot}}{a})^{1/2}\)
\(M_{tot} = \frac{v_0^2a}{G}\)
Since mass and radius scale linearly, we can use the ratio \(\frac{M(<r)}{M_{tot}} = \frac{r}{r+a}\)
\(M(<r) = \frac{r}{r+a}M_{tot}\)
\(M(<r) = \frac{r}{r+a}\frac{v_0^2a}{G} = \frac{v_0^2ra}{G(r+a)}\)
\(\rho(r) = \frac{a^2v_0^2}{4\pi r^2G(r+a)^2}\) based on the equation from part c.

e) If we consider this as a model of the Milky Way, what is a in kpc if \(M_{tot} = 10^{12}M_\odot\) and \(v_0\) = 240km/s?

\(v_0 = 240km/s \rightarrow 240*10^3km/s\)
\(M_{tot} = 10^{12}*10^{33}g = 10^{45}g \rightarrow 10^{42}kg\)
\(G = 6.67*10^{-11}m^3/kgs^2\)
\(240*10^3m/s = (\frac{(6.67*10^{-11}m^3/kgs^2)(10^{42}kg)}{a})^{1/2}\)
\(a = 1.16*10^{21}m \rightarrow 37.7kpc\)