Problem source

It is given that the gravitational force between a disc, of radius $a,$ thickness $\delta x$ and uniform density $\rho,$ and a particle of mass $m$ at a distance $b(\geqslant0)$ from the disc on its axis
is 
\[
2\pi mk\rho\delta x\left(1-\frac{b}{(a^{2}+b^{2})^{\frac{1}{2}}}\right),
\]
where $k$ is a constant. Show that the gravitational force on a particle of mass $m$ at the surface of a uniform sphere of mass $M$ and radius $r$ is $kmM/r^{2}.$ Deduce that in a spherical cloud of particles of uniform density, which all attract one another gravitationally, the radius $r$ and inward velocity $v=-\dfrac{\d r}{\d t}$
of a particle at the surface satisfy the equation 
\[
v\frac{\mathrm{d}v}{\mathrm{d}r}=-\frac{kM}{r^{2}},
\]
where $M$ is the mass of the cloud. 
At time $t=0$, the cloud is instantaneously at rest and has radius $R$. Show that $r=R\cos^{2}\alpha$ after a time 
\[
\left(\frac{R^{3}}{2kM}\right)^{\frac{1}{2}}(\alpha+\tfrac{1}{2}\sin2\alpha).
\]

Solution source

Suppose we divide a sphere of radius $r$ up into slices of thickness $\delta x$. Then the force acting on $P$ will be:

\begin{align*}
F &= \sum_{\text{slices}} 2\pi mk\rho\delta x\left(1-\frac{b}{(a^{2}+b^{2})^{\frac{1}{2}}}\right) \\
&=  \sum_{i=-r/\delta x}^{r/\delta x} 2\pi mk\frac{M}{\frac43 \pi r^3}\delta x\left(1-\frac{i \delta x}{((1-(i\delta x)^2)+(i \delta x)^{2})^{\frac{1}{2}}}\right) \\
&\to \int_{-r}^r \frac{1}{2} \frac{mkM}{r^3}(1-t) \d t \\
&=\frac{mkM}{r^2}
\end{align*}

We can see that the particle will have a force attracting it towards the centre, with magnitude $\frac{kmM}{r^2}$, therefore and since $\frac{\d v}{\d t} = \frac{\d v}{\d r} \frac{\d r}{\d t}$ we must have:

$v \frac{\d v}{\d r}m = - \frac{kmM}{r^2}$ and dividing by $m$ we get exactly the result we seek.

\begin{align*}
&& v \frac{\d v}{\d r} &= \frac{-kM}{r^2} \\
\Rightarrow && \frac{v^2}{2}+C &= \frac{kM}{r} \\
r = R, v =0: && C &= \frac{kM}{R} \\
\Rightarrow && v^2&= 2kM\left ( \frac1r - \frac1R\right ) \\
\Rightarrow && \frac{\d r}{\d t} &= -\sqrt{2kM\left ( \frac1r - \frac1R\right )} \\
\Rightarrow &&  -\sqrt{2kM}T  &= \int_{r=R}^{r=R\cos^2 \alpha} \frac{1}{\sqrt{\frac1r-\frac1R}} \d r \\
r = R\cos^2 \theta: && -\sqrt{2kM}T &= \int_{\theta = 0}^{\theta = \alpha} \frac{\sqrt{R}}{\sqrt{\sec^2 \theta - 1}} \cdot R \cdot 2 \cdot (-\cos \theta) \cdot \sin \theta \d \theta  \\
\Rightarrow && T &= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha \frac{2 \cos \theta \sin \theta}{\sqrt{\sec^2 \theta - 1}} \d \theta \\
&&&= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha \frac{2 \cos \theta \sin \theta}{\tan \theta} \d \theta \\
&&&=  \sqrt{\frac{R^3}{2kM}} \int_0^\alpha 2\cos^2 \theta \d \theta \\
&&&= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha 1 + \cos 2 \theta\d \theta \\
&&&= \sqrt{\frac{R^3}{2kM}} \left [1 + \frac12 \sin 2 \theta \right]_0^\alpha \\
&&&= \sqrt{\frac{R^3}{2kM}} \left (\alpha + \frac12 \sin 2 \alpha \right) \\
\end{align*}