Questions & Answers

Question

Answers

Answer

Verified

129k+ views

Gravitational compression is described as a phenomenon in which gravitational force, acting on the mass of an object, compresses it and reduces its size which results in increasing the density of the object.

We are given a uniform sphere that has a mass $M$ and a radius of $R$. Let’s say its mass density is $\rho $.

We have partitioned the solid sphere into thin spherical layers and consider a layer of thickness $dr$ lying at a distance $r$ from the center of the ball. Each spherical layer possesses on the layers within it. The considered spherical layer is attracted to the part of the sphere lying within it. The outer part of the sphere will not act on it.

Therefore, for a spherical layer of thickness $dr$ at a distance $r$ from the centre of the ball,

$dF=dP\cdot 4\pi {{r}^{2}}$

Where,

$dF$ is the force experienced by the spherical layer

$dP$ is the amount of pressure acting on the spherical layer

$4\pi {{r}^{2}}$ is the area of the spherical layer

Or,

$dP\cdot 4\pi {{r}^{2}}=\dfrac{G\left( \dfrac{4}{3}\pi {{r}^{3}}\rho \right)\left( 4\pi {{r}^{2}}dr\rho \right)}{{{r}^{2}}}$

Where,

$\rho $ is the mean density of the sphere

$G$ is the gravitational constant

$dP=\dfrac{4}{3}\pi G{{\rho }^{2}}rdr$

Thus,

$P=\int\limits_{r}^{R}{dP}$

Put, $dP=\dfrac{4}{3}\pi G{{\rho }^{2}}rdr$

We get,

$P=\int\limits_{r}^{R}{\dfrac{2\pi }{3}G{{\rho }^{2}}\left( {{R}^{2}}-{{r}^{2}} \right)}$

The pressure will be vanished at$r=R$

Put, $P=\dfrac{M}{\dfrac{4}{3}\pi {{R}^{3}}}$

$P=\dfrac{3}{8}\left( 1-\dfrac{{{r}^{2}}}{{{R}^{2}}} \right)\dfrac{G{{M}^{2}}}{\pi {{R}^{4}}}$

Now,

By putting $r=0$, we will get the pressure at the sphere’s centre

For finding the pressure at the centre of earth,

$\begin{align}

& M=5.972\times {{10}^{24}}Kg \\

& \rho =5.5\times {{10}^{3}}kg{{m}^{-3}} \\

& R=6400Km=6.4\times {{10}^{6}}m \\

\end{align}$

$\begin{align}

& P=\dfrac{3}{8}\left( 1-\dfrac{{{r}^{2}}}{{{R}^{2}}} \right)\dfrac{G{{M}^{2}}}{\pi {{R}^{4}}} \\

& P=\dfrac{3}{8}\times \dfrac{6.67\times {{10}^{-11}}\times {{\left( 5.972\times {{10}^{24}} \right)}^{2}}}{\dfrac{22}{7}\times {{\left( 6.4\times {{10}^{6}} \right)}^{4}}} \\

\end{align}$

We get,

$\begin{align}

& P=1.73\times {{10}^{11}}Pa \\

& P=1.72\times {{10}^{6}}atms \\

\end{align}$