Question

A uniform sphere has a mass $M$ and radius $R$. Find the pressure $p$ inside the sphere, caused by gravitational compression, as a function of the distance $r$ from its centre. Evaluate $p$ at the centre of the Earth, assuming it to be a uniform sphere.

Answer 1

We partition the solid sphere into thin spherical layers and consider a layer of thickness $dr$ lying at a distance $r$ from the centre of the ball. Each sperical layer presses on the layers within it. The considered layer is attracted to the part of the sphere lying within it (the outer part does not act on the layer). Hence for the considered layer
$dp4\pi r^2=dF$
or, $dp4\pi r^2=\frac{\gamma \left(\frac{4}{3}\pi r^3\rho \right)\left(4\pi r^{2}dr\rho \right)}{r^2}$
(where $\rho$ is the mean density of sphere)
or, $dp=\frac{4}{3}\pi \gamma {\rho }^{2}rdr$
Thus $p={\int }_r^{R}dp=\frac{2\pi }{3}\gamma {\rho }^2(R^2-r^2)$
(The pressure must vanish at $r=R$.)
or, $p=\frac{3}{8}(1-\left(\frac{r^2}{R^2}\right))\frac{\gamma M^2}{\pi R^4}$, Putting $\rho =\frac{M}{\left(\frac{4}{3}\right)\pi R^3}$
Putting $r=0$, we have the pressure at sphere's centre, and treating it as the Earth where mean density is equal to $\rho =5.5\times {10}^{3}kg/m^3$ and $R=64\times {10}^{2}km$
we have, $p=1.73\times {10}^{11}Pa$ or $1.72\times {10}^{6}atms$.