M(r < r) is related to the mass density ρ(r) by the integral: \[ M(< r) = \int_{0}^{r} 4π'^{2} ρ(r') dr'\](Recall that the \(4πr^{2} \) comes from the surface area of each spherical shell, and the dr' is the thickness of each thin shell; talk to a TF if this is not clear.) The fundamental theorem of calculus then implies that 4πr^2ρ(r) = dM(< r)/dr. For the case in question 4, what is ρ(r)? Is the density finite as r --> 0 in the case of a flat rotation curve?
The Milky Way's rotation curve
This calculation is simply a matter of using what we already know. We have from question 4: \[M(r) = \frac{ V_c^2 r }{G} \] Taking the derivative gives us: \[\frac{dM}{dr} = \frac{ V_c^2}{G} \] which we can set equal to \[\frac{dM}{dr} = 4 \pi r^2 \rho (r) \] And solve for density. \[4 \pi r^2 \rho (r) = \frac{ V_c^2}{G} \] \[ \rho (r) = \frac{ V_c^2}{4 \pi r^2 G} \]
This calculated density would be infinite in the case of a flat rotation curve.
Great job! 5/5
ReplyDelete