GravitationMedium
Question
A body of mass $m$ is taken from the surface of earth to a height equal to twice the radius of earth $\left( R_{e} \right)$. The increase in potential energy will be $\_\_\_\_$ . ( $g$ is acceleration due to gravity at the surface of earth)
Options
A.$\frac{1}{2}mgR_{e}$
B.$\frac{3}{4}mgR_{e}$
C.$\frac{1}{4}mgR_{e}$
D.$\frac{2}{3}mgR_{e}$
Solution
Sol. [Key Concept: $\Delta\ U = U(2R) - U(surface) = - \frac{GMm}{3R} + \frac{GMm}{R} = \frac{2}{3}\ mgR$.]
$${U_{1} = \frac{- GMm}{R_{e}} }{U_{2} = \frac{- GMm}{R_{e} + 2R_{e}} = - \frac{GMm}{3R_{e}} }{\Delta U = U_{2} - U_{1} = \frac{- GMm}{3R_{e}} + \frac{GMm}{R_{e}} }{\Delta U = \frac{2}{3}\frac{GMm}{R_{e}} = \frac{2}{3}\left\lbrack \frac{GM}{R_{e}^{2}} \right\rbrack\left( {mR}_{e} \right) }{\Delta U = \frac{2}{3}{mgR}_{e} }$$
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