Blog Post 8, Worksheet 3.1, Problem 5

5. M(<r) is related to the mass density \(\rho(r)\) by the integral: \[M(<r) = \int_0^r 4πr'^2\rho(r')dr'\] (Recall that the \(4πr'^2\) comes from the surface area of each spherical shell, and the dr' is the thickness of each thin shell). The fundamental theorem of calculus then implies that \(4πr^2\rho(r) = dM(<r)/dr\). For the case in question 4, what is \(\rho(r)\)? Is the density finite as r --> 0 in the case of a flat rotation curve?

This problem gives the relationship between mass, M(<r), and mass density \(\rho(r)\). In problem 4, we found that the mass, M(<r), is given by: \[ M(<r) = \frac{V_c^2r}{G},\] where \(V_c\) is the constant rotational velocity of an object in the galaxy, r is the distance from the center of the galaxy, and G is the gravitational constant. We now have two equations for M(<r) - the integral from the problem, and our equation from question 4. We can set these two definitions for M(<r) equal to each to solve for \(\rho(r)\). \[ \int_0^r 4πr'^2\rho(r')dr' = \frac{V_c^2r}{G}\] Taking the derivative of each side with respect to r, we get: \[4πr^2\rho(r) = \frac{dM(<r)}{dr} = \frac{V_c^2}{G}\] Rearranging this equality, we get: \[ \rho(r) = \frac{V_c^2}{4πr^2G}\] As a result, the density is not finite as r --> 0 as in the case of a flat rotation curve, because as r goes to zero, \(\rho(r)\) goes to infinity.