Question

The radius of earth is R and acceleration due to gravity on its surface is g. The gravitational potential at the centre of earth is
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Answer 1

Answer:

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Explanation:

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Answer 2

What is Gravitational Potential?

The gravitational potential at a point is the work done to bring a unit mass from infinity to that point. It is also defined as the gravitational potential energy of a unit mass at that point.

Derivation for the Expression of Gravitational Potential at the Center of the Earth

By definition, it is the work done by the gravitational force of attraction of the earth to bring a point mass from infinity to the center of the earth. But we consider the work done to bring it from the surface of the earth to the center.

We know the gravitational force of attraction is,

$\displaystyle\longrightarrow\sf{F=\dfrac{GMm}{x^2}}$

where,

• $\sf{G=}$ gravitational constant

• $\sf{M=}$ mass of the earth

• $\sf{m=}$ mass of the body

• $\sf{x=}$ distance of the body from the center of the earth

For a unit mass $\sf{(m=1),}$

$\displaystyle\longrightarrow\sf{F=\dfrac{GM}{x^2}}$

Let the unit mass be at a distance $\sf{x}$ from the surface of the earth but inside the earth, so that its distance from the center is $\sf{R-x.}$

Thus the gravitational force of attraction will be,

$\displaystyle\longrightarrow\sf{F=\dfrac{GM'}{(R-x)^2}}$

where $\sf{R}$ is the radius of the earth and $\sf{M'}$ is the mass of the earth at the depth $\sf{x.}$

But since the density of the earth remains unchanged, we have,

$\displaystyle\longrightarrow\sf{M=\rho V}$

where $\sf{V}$ is the volume of the sphere. Since earth is spherical in shape,

$\displaystyle\longrightarrow\sf{M=\rho\cdot\dfrac{4}{3}\pi R^3}$

$\displaystyle\longrightarrow\sf{M=\dfrac{4\pi\rho\,R^3}{3}}$

At the depth $\sf{x,}$ radius becomes $\sf{R-x}$ and thus the mass will be,

$\displaystyle\longrightarrow\sf{M'=\dfrac{4\pi\rho(R-x)^3}{3}}$

Then the force will be,

$\displaystyle\longrightarrow\sf{F=\dfrac{4\pi\rho(R-x)^3G}{3(R-x)^2}}$

$\displaystyle\longrightarrow\sf{F=\dfrac{4\pi\rho(R-x)G}{3}}$

Work done to displace it through a small distance dx towards the center is,

$\displaystyle\longrightarrow\sf{dW=\dfrac{4\pi\rho(R-x)G}{3}\ dx}$

Then total work done to bring it from the surface of the earth $\sf{(d=0)}$ to the center $\sf{(d=-R)}$ is,

$\displaystyle\longrightarrow\sf{W=\int\limits_0^_-R}\dfrac{4\pi\rho(R-x)G}{3}\ dx}$

$\displaystyle\longrightarrow\sf{W=\dfrac{4\pi\rho\,G}{3}\int\limits_0^{-R}(R-x)\ dx}$

$\displaystyle\longrightarrow\sf{W=-\dfrac{4\pi\rho\,G}{3}\left[\dfrac{(R-x)^2}{2}\right]_0^{-R}}$

$\displaystyle\longrightarrow\sf{W=-\dfrac{4\pi\rho\,G}{3}\left[\dfrac{(R+R)^2}{2}-\dfrac{(R-0)^2}{2}\right]}$

$\displaystyle\longrightarrow\sf{W=2\pi\rho\,GR^2}$

$\displaystyle\longrightarrow\sf{W=2\pi\cdot\dfrac{M}{\dfrac{4}{3}\pi\,R^3}\cdot GR^2}$

$\displaystyle\longrightarrow\sf{W=2\cdot\dfrac{3M}{4R}\cdot G}$

$\displaystyle\longrightarrow\sf{W=\dfrac{3GM}{2R}}$

This work done is the gravitational potential at the center of the earth.

$\longrightarrow\sf{U=\dfrac{3GM}{2R}}$

But,

$\longrightarrow\sf{g=\dfrac{GM}{R^2}\quad\implies\quad gR=\dfrac{GM}{R}}$

Then,

$\longrightarrow\large\boxed{\sf{U=\dfrac{3}{2}\,gR}}$