Question

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]

• A. $\frac{- GMm }{R+h}$
• B. $\frac{- GMm }{h}$
• C. $\frac{G M m h}{R(R+h)}$
• D. $\frac{ GMmh }{h+2 R}$

Answer 1

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} $
$\Rightarrow U_{h}^{\prime}=\frac{G M m h}{R(R+h)}$