Answer

Verified

447.3k+ views

**Hint:**The constant force of gravity experienced by the massive body results in the development of an effect called gravitational compression. We will use the expression for the gravitational force experienced by a massive body and determine the pressure at the center of the earth by using the pressure-force relationship.

__Complete step by step answer:__

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}$

**Note:**In the core or center of a star such as the Sun, the gravitational pressure can be balanced by the outward thermal pressure from the fusion reactions, temporarily halting the effect of gravitational compression. At the center of a star or planet, gravitational compression can produce heat as well.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

The polyarch xylem is found in case of a Monocot leaf class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Casparian strips are present in of the root A Epiblema class 12 biology CBSE

How do you graph the function fx 4x class 9 maths CBSE