Gravitational collapse and free fall time (spherical, pressure-free)

For a spherically symmetric distribution of mass, the acceleration felt by a test particle at radius $r$ is $-G M /r^2$ (negative because pointing in toward the center), regardless of the radial distribution of mass. This is a key part of the question, make sure you are comfortable with it. It is a concept that is related to Gauss' law of electromagnetism, if you've encountered that.

The total mass of the collapsing cloud is given by the initial uniform density times the volume, or $M = (4 \pi /3) r_0^3 \rho_0$

From Newton's second law, the equation of motion for a test particle at the edge of the cloud is then

$$ \frac{d^2r}{dt^2} = - \frac{4 \pi G r_0^3 \rho_0}{3r^2}$$

Now for some chain rule trickery (this is a nice trick, so it's good to remember it for similar differential equations):

$$ \frac{d}{dt} = \frac{dr}{dt} \frac{d}{dr}$$

Keeping in mind that $v \equiv \frac{dr}{dt}$, and using the chain rule subustitution just mentioned, the equation of motion is now

$$ v \frac{dv}{dr} = - \frac{4 \pi G r_0^3 \rho_0}{3r^2}$$

The point of doing all this is that the differential equaiton is now more clearly separable. You can solve it by integrating as follows

$$ \int v \, dv = - \frac{4 \pi G r_0^3 \rho_0}{3}\int\frac{dr}{r^2}$$

$$\frac{1}{2} v^2 = \frac{4 \pi G r_0^3 \rho_0}{3r} + C$$

(You also could have gotten to this point by relating gravitational potential energy to kinetic energy, and being careful about where you set the zero of the gravitational potential).

When $r = r_0$, $v = 0$, so $C = - \frac{4 \pi G r_0^2 \rho_0}{3}$ and

$$\frac{1}{2} v^2 = \frac{4 \pi G r_0^2 \rho_0}{3}\left(\frac{r_0}{r} - 1 \right)$$

$$ |v| = \sqrt{\frac{8 \pi G r_0^2 \rho_0}{3}\left(\frac{r_0}{r} - 1 \right)}$$

The total time can be found by integrating

$$t_{\rm collapse} = \int dt = \int \frac{dr}{|v|} = \sqrt{\frac{3}{8 \pi G r_0^2 \rho_0}}\int_0^{r_0}{\frac{dr}{\sqrt{\left(\frac{r_0}{r} - 1 \right)}}}$$

This is going to be a tricky integral, so let's non-dimensionalize it. Make a change of variable $u \equiv r/r_0$. Then we have

$$ t_{\rm collapse} = \sqrt{\frac{3}{8 \pi G \rho_0}}\int_0^{1}{\frac{du}{\sqrt{\frac{1}{u} - 1}}}$$

If you are really adept at trigonometric substitutions in integrals, here's your chance to shine. Otherwise, just use Wolfram Alpha or something similar to tell you that the integral evaulates to $\pi/2$. That gives, finally,

$$ t_{\rm collapse} = \sqrt{\frac{3 \pi}{32 G \rho_0}}$$

A partial answer only.

Given a spherical collapse, and ignoring relativistic effects, the time is the same as the time taken for a particle at the edge of the cloud to fall to the centre.

As all the mass is inside the edge, we can determine the mass pulling that edge in as the volume of the sphere of radius $$r_0$$ times the density.

I.e., the key idea is that the cloud is not important, just the total mass.

So the falling particle behaves as it if is falling to a point mass.

But this is just a special kind of orbit, very ellipsoidal. The period of the orbit:

http://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body

depends on the cube of the radius.

$$t = 2\pi\sqrt{\frac{a^3}{GM}}$$

As Wikipedia shows, this actually means that the period is independent of the radius:

$$t = \sqrt{\frac{3\pi}{G\rho_0}}$$

http://en.wikipedia.org/wiki/Orbital_period#Orbital_period_as_a_function_of_central_body.27s_density

The in-fall time is only the same as a quarter of an orbit. So the total time should be:

$$t = \sqrt{\frac{3\pi}{16G\rho_0}}$$

So I'm missing a factor of $$\sqrt{\frac{1}{2}}$$ here - hence the 'partial answer'.

What have I missed?