If potential energy of a body of mass m on th | vedclass

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Question

(c)

The concept involved here is that,

Gravitational potential energy difference between any two points in a gravitational field is independent

Vedclass · Accepted Answer

$\frac{G M m h}{R(R+h)}$

Vedclass · Answer

(c)

The concept involved here is that,

Gravitational potential energy difference between any two points in a gravitational field is independent of the choice of reference.

When potential at infinity is assigned zero value,

Potential energy of a body of mass $m$ on the surface of earth $=\frac{-G M m}{R}=U_s$

Potential energy at height, $h=\frac{-G M m}{R+h}=U_n$

$U_s-U_h=+G M m\left(-\frac{1}{R}+\frac{1}{R+h}\right)$

$=G M m\left(\frac{-R-h+R}{R(R+h)}\right)$

$=\frac{-G M m h}{R(R+h)}$

Now, when potential at the surface is taken zero, Let $U_s^{\prime}, U_h^{\prime}$ be the new values of potential energy at the surface and height $h$ respectively,

And, $U_s-U_h=U_s^{\prime}-U_h^{\prime}$

$\Rightarrow \frac{-G M m h}{R(R+h)}=0-U_h^{\prime}$

$\Rightarrow \quad U_h^{\prime}=\frac{G M m h}{R(R+h)}$

If potential energy of a body of mass $m$ on the surface of earth is taken as zero then its potential energy at height $h$ above the surface of earth is [ $R$ is radius of earth and $M$ is mass of earth]

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