Question

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

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

Answer 1

Answer: (C)

$\frac{1}{2}\,mgR$

Answer 2

The potential energy of an object at the surface of the earth
$U_1=-\frac{GMm}{R}$ .....(i)
The potential energy of the object at a height
$h= R$ from the surface of the earth
$U_2=-\frac{GMm}{R+ h}$
$=-\frac{GMm}{R+R}$ .....(ii)
Hence, the gain in potential energy of the object
$\Delta U=U_2-U_1$
$\Delta U=-\frac{GMm}{R+R}+\frac{GMm}{R}$
$\Delta U=-\frac{GMm}{2R}+\frac{GMm}{R}$
$\Delta U=\frac{1}{2}\frac{GMm}{R}$
But we know that $GM =gR^{2}$
Hence, $\Delta U=\frac{1}{2}\frac{gR^2m}{R}$
or $\Delta U=\frac{1}{2} g R m$
or $\Delta U=\frac{1}{2}mgR$