M(r < r) is related to the mass density ρ(r) by the integral: M(<r)=∫r04π′2ρ(r′)dr′
(Recall that the 4πr2 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)=V2crG
Taking the derivative gives us: dMdr=V2cG
which we can set equal to dMdr=4πr2ρ(r)
And solve for density. 4πr2ρ(r)=V2cG
ρ(r)=V2c4πr2G
This calculated density would be infinite in the case of a flat rotation curve.
Great job! 5/5
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