Question

The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the earth is:

• A. $\frac{n}{n+1}\left(mgR\right)$
  
• B. $\frac{n}{n-1}\left(mgR\right)$
  
• C. $nmgR$
  
• D. $\frac{\left(mgR\right)}{n}$

Answer 1

The correct option is A $\frac{n}{n+1}\left(mgR\right)$


General expression for Gravitation potential energy (GPE)
$=-\frac{GM}{x}m$
When x = R (distance/displacement from earth’s centre at surface) $GP{E}_1=\frac{-GMm}{R}$
At a height = nR
$GP{E}_2=\frac{-GMm}{R+nR}\Delta GPE=GP{E}_2-GP{E}_1=\frac{-GMm}{nR+R}-[-\frac{GMm}{R}]=GMm[\frac{1}{R}-\frac{1}{(n+1)R}]\Rightarrow \frac{GMm}{R}[\frac{1}{1}-\frac{1}{n+1}]=\frac{GMm}{R}\left(\frac{n}{n+1}\right)\Rightarrow \frac{GMm.R}{R^2}=mgR\Rightarrow g=\frac{GM}{R^2}=mgR\left(\frac{n}{n+1}\right)$