Gravitational work done by a uniform cone on a particle I saw the problem 1.213 of Irodov's Problems in General physics and modified it.
So the modified problem is a particle of mass $m$ is on the tip of cone of mass $M$ and volume $V$. What work needs to be performed by external agent in the process to get the particle from the tip to infinity against the gravitational force of the cone?
I've tried hard. But unsuccessful, any hint?
 A: In order to know what work is needed, one way is to study the gravitationnal potential energy $E_p$ of $m$: in fact, if we consider that the $m$ is at rest in the beginning and in the end of the experience, then $W = \Delta E_p$. However, choosing the convention $E_p(\infty) = 0$, this relation becomes $W=-E_p$. So, first of all we will find a relation between the geometry of the cone and its volume; then, we will find the potential energy due to a infinitesimal slice of cone; finally, we will integrate this, which will gives us the total potential energy of $m$.


The volume between $h$ and $h+dh$ is, forgetting the terms in $d^2h$ is $$dV = \pi r^2 dh$$ However, Thalès's theorem states that
$$ \frac{r}{R} = \frac{h}{H}$$
So $$V = \int_0^H dV = \int_0^H \frac{\pi R^2}{H^2}h^2dh = \frac{\pi}{3}R^2H$$


Now we will find the potential energy $dE_p$ related to the attraction of the slice delimited between $h$ and $h+dh$. First of all, we must know the potential energy $d^2E_p$ relate dto the attraction of a tiny ring starting at $\rho$ and ending at $\rho+d\rho$. This slice has a mass
$$ \delta m = \frac{2\pi\rho\, d\rho\, dh}{V}M$$
So we have $$d^2E_p = -\frac{Gm\,\delta m}{l}= -\frac{2\pi Gm\rho\, d\rho\,dh M}{Vl}$$
However, $$l^2 = \rho^2 + (h-H)^2$$
So $$l\,dl = \rho\,d\rho$$
Finally, $$-d^2E_p = \frac{2\pi GMm}{V}dh\,dl$$
We have $l \in [H-h, \sqrt{(H-h)^2+\rho^2}]$, so the slice $[h,h+dh]$ causes a potential energy $$dE_p = \int d^2E_p = -\frac{2\pi GMm}{V}(\sqrt{(H-h)^2+r^2}-(H-h))dh$$
Thalès's theorem states that $$\frac{r}{R} = \frac{H-h}{H}$$
So, finally, $$dE_p = -\frac{2\pi GMm}{V}(\sqrt{1+\frac{R^2}{H^2}}-1)(H-h)dh$$

We can now find what the total potential energy of $m$ on the tip of the cone is. $h$ varies in $[0,H]$, so we have
$$ E_p = \int dE_p = -\frac{\pi GMm H^2}{V}(\sqrt{1+\frac{R^2}{H^2}}-1)$$
However, $V = \frac{\pi}{3}R^2H$, so
$$ E_p = -\frac{3GMm}{R^2}(\sqrt{H^2+R^2}-H)$$
The minimum work needed to bring the mass $m$ from the tip of the cone to the infinity is thus $$W = \frac{3GMm}{R^2}(\sqrt{H^2+R^2}-H)$$
