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?

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