Question

If g is the acceleration due to gravity on the surface of earth, find the gain in potential energy of an object of mass m raised from the surface of earth to a height equal to the radius R of the earth.

• A. $\frac{1}{2}mgR$.
• B. $\frac{1}{4}mgR$.
• C. $\frac{1}{3}mgR$.
• D. $\frac{2}{1}mgR$.

Answer 1

The correct option is A $\frac{1}{2}mgR$.

Let M, R be the mass and radius of the earth. Then
$g=GM/R^2$
or $GM=g{R}^2$ (i)
Potential energy of the object on the surface of earth
$U_1=\frac{-GMm}{R}$
The potential energy of the object at a height equal to radius of the earth is
$U_2=\frac{-GMm}{2R}$
Gain in PE is
$U_2-U_1=-\frac{GMm}{2R}+\frac{GMm}{R}=\frac{GMm}{2R}$
$=\frac{\left(g{R}^2\right)m}{2R}=\frac{1}{2}mgR$.