Rankers Physics

Gravitational Potential Energy: Practice Problem & Solution

A body of mass '$m$' taken from the earth's surface to the height equal to twice the radius ($R$) of the earth. The change in potential energy of body will be: (2013)
$\frac{1}{3}mgR$
$2 mgR$
$\frac{2}{3}mgR$
$3 mgR$

Solution Explained:

To solve this problem, we apply the core principles of Gravitational Potential Energy. Understanding the underlying formula is key to arriving at the correct answer below:

Change in potential energy $\Delta U = \frac{mgh}{1+h/R}$. For $h = 2R$, $\Delta U = \frac{mg(2R)}{1+2} = \frac{2}{3}mgR$.

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