Mass of the Milky Way: Blog 7, Worksheet 3.1, Problem 6

Now consider a spherically symmetric galaxy with a density profile: $$ρ(r) = \frac{C}{ 4πr^2(r + a )^2 }$$ where a and C are constants. 

(a) What is M(< r ) for this galaxy? You can leave your answer in terms of C and a. 
$$ρ(r) = \frac{C}{ 4πr^2(r + a )^2 }$$ We know from the previous problem that $$\frac{dM}{dr} = 4 \pi r^2 \rho (r)$$ Combining this information gives: $$\frac{dM}{dr} = 4 \pi r^2 \left( \frac{C}{ 4πr^2(r + a )^2 } \right)$$  And now we can solve for M $$\frac{dM}{dr} = \frac{C}{(r + a )^2}$$   $$dM = \frac{C}{(r + a )^2}dr$$   $$\int dM = \int_{0}^{r} \frac{C}{(r + a )^2}dr$$ $$M(r) = - \frac{C}{(r + a ) } |_{0}^{r} = \frac{ - C}{r + a } + \frac{C}{r + a}$$ $$M = \frac{Cr}{a^2 + ar}$$
(b) The total mass of the system is the the limit of M(< r ) as r --> $\infty$; calculate this total mass. 
$$lim_{x\to\infty} \frac{Cr}{a^2 + ar} = \frac{\infty}{\infty}$$ That is not very helpful for the sake of this problem. In order to solve this, we can think back to high school calculus and invoke L'Hopital's rule, which tells us that in the case that the limit of the top and bottom equal infinity, the limit derivative of the top and bottom of the fraction will equal the limit of the fraction: $$lim_{x\to\infty} \frac{C}{a}$$ $$M_{tot} = \frac{C}{a}$$

(c) If we write the total mass as Mtot, rewrite M(< r ) and ρ(r) to eliminate C in favor of Mtot. 

This problem is just a matter of combining two equations we already know: $$1) \: M_{tot} = \frac{C}{a}$$ can also be written as $$C = a \times M_{tot}$$ and $$2) \:  ρ(r) = \frac{C}{ 4πr^2(r + a )^2 }$$ Substituting for C gives: $$ρ(r) = \frac{a  M_{tot}}{ 4πr^2(r + a )^2 }$$

(d) What is the rotation curve v(r) for this galaxy? You should find that the circular velocity goes to a constant as r --> 0, i.e., at radii r << a, this galaxy appears to have a flat rotation curve. If we write this constant velocity as Vo, write M(< r ) and ρ(r) in terms of Vo, a, and r. What is Mtot in terms of Vo and a? 

We can use the formulas we have for density to solve for velocity: $$\rho (r)  = \frac{ V_c^2}{4 \pi r^2 G} =  \frac{a  M_{tot}}{ 4πr^2(r + a )^2 }$$ $$V_c^2 = \frac{Ga  M_{tot}}{(r+a)^2}$$   $$V_c = \left( \frac{Ga  M_{tot}}{(r+a)^2} \right)^{\frac{1}{2}}$$ Now we can find the limit of $V_c$ to get $V_0$ when r<<a. $$lim_{r\ to 0}V_c = \left( \frac{Ga  M_{tot}}{a^2} \right)^{\frac{1}{2}}$$ This means: $$V_0^2 = \frac{GM_{tot}}{a^2}$$ Now we want to rewrite M(< r) and ρ(r) in terms of Vo, a, and r. For M(< r), we only really need to get rid of C   $$M = \frac{Cr}{a^2 + ar}$$ We can use the equation derived above for Vo to find $M_{tot}$ $$\boxed{M_{tot} = \frac{V_0^2 a}{G}}$$ setting this equal to our equation for $M_{tot}$ in part b gives: $$\frac{C}{a} = \frac{V_0^2 a}{G}$$ $$C = \frac{V_0^2 a^2}{G}$$ $$M(r) = \frac{Cr}{a^2 + ar} = \frac{V_0^2 a^2 r}{G (a^2 + ar)}$$ $$\boxed{M(r) =\frac{V_0^2 ar}{Ga + Gr}}$$ $$\boxed{ ρ(r) = \frac{V_0^2 a^2}{ G4πr^2(r + a )^2 } }$$

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

For this last problem, we can just plug and chug. In cgs, $V_0 = 2.4 \times 10^7$ cm/s and $M_{\odot} = 2 \times 10^{33} g$.

Solving the $M_{tot}$ formula for a gives: $$a = \frac{GM_{tot}}{V_0^2} = \frac{(6.67 \times 10^{-8})(10^{12} \times 10^{33} ) }{(2.4 \times 10^7)^2 }= 1.2 \times 10^{23} \: cm \approx 40 \: kpc$$