Question

If $g$ is the acceleration due to gravity on the 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

The correct option is C

$\frac{1}{2}mgR$

Step 1: Given Data,

$h=R$

$h$= height from the surface of the earth.
$R$= Radius of the earth,

Step 2: Formula used,

The potential energy of an object at the surface of the earth,
$U_1=-\frac{GMm}{R}$

$G$= Gravitational constant,
$R$= Radius of the earth,

$m$= mass

$GM=g{R}^2$ , $g$ is the acceleration due to gravity on the earth's surface.

The potential energy of the object at a height $h=R$ from the surface of the earth,
$U_2=-\frac{GMm}{R+h}$
$=-\frac{GMm}{R+R}$

$h$= height from the surface of the earth.

Step 3: Calculating the gain in potential energy,

Hence, the gain in potential energy of the object
$\mathrm{\Delta }U=U_2-U_1$
$\mathrm{\Delta }U=-\frac{GMm}{R+R}+\frac{GMm}{R}$
$\mathrm{\Delta }U=-\frac{GMm}{2R}+\frac{GMm}{R}$
$\mathrm{\Delta }U=\frac{1}{2}\frac{GMm}{R}$
But we know that $GM=g{R}^2$
Hence, $\mathrm{\Delta }U=\frac{1}{2}\frac{g{R}^{2}m}{R}$
or,$\mathrm{\Delta }U=\frac{1}{2}gRm$
or,$\Delta U=\frac{1}{2}mgR$

Hence, the correct option is C.