How galactic density falls using Oort constants

Question

I am given the observations that Oort constants $A$ and $B$ are, respectively: $14.5 \text{km s}^{-1} \text{kpc}^{-1}$ and $-12 \text{km s}^{-1} \text{kpc}^{-1}$. From these, I am supposed to conclude that the galaxy density falls off as $\sim R^{-2}$. (Source of problem: Paper 2 Question 7 of 2016, Cambridge Astrophysics Tripos - see page 17.)

I don't see how I can pull this off, would appreciate any help.

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In the earlier part of the question, I proved that at the solar location, it is true that

$$\frac{\partial^{2} \Phi}{\partial z^{2}}=4 \pi G \rho_{0}+2\left(A^{2}-B^{2}\right)$$

which might help somehow, but I don't see how.

Answer 1

From the definitions of the Oort constants we know that $$A + B = -\frac{dv}{dr} $$ at the solar radius.

But $A + B$ is small, implying that $dv/dr \simeq 0^{\dagger}$ and we are dealing with a flat(ish) rotation curve.

Assuming a spherically symmetric mass distribution with a density that is proportional to $r^\alpha$, we can write the centripetal acceleration at a radius $R$ $$\frac{v^2}{R} = \int^R_0 \frac{4\pi G r^2 \rho(r)}{R^2} \ dr\ . $$ Then, if $v$ is constant, we can just deal in proportionalities: $$ R^{-1} \propto R^{-2}\int^R_0 r^2 r^\alpha \ dr\ , $$ $$ R^{-1} \propto R^{-2} R^{3+\alpha}\ ,$$ and so $\alpha =-2$.

Another variant is to tell you that $A=0$ in some galaxy. This means $dv/dr = v_0/r_0$, which means $v \propto r$ (i.e. solid body rotation). This in turn means $\alpha =0$.

$\dagger$ by small, what we mean is that the velocity gradient is small compared with $v_0/R_0$ where $v_0$ is the rotation speed where the Oort constants are defined. Since $A-B = v_0/R_0$ this means that $$ |A+B| \ll |A-B| $$ which in this case is 2.5 vs 26.5 km/s per kpc.