A) \[\left( \frac{n}{n+1} \right)mgR\]
B) \[\left( \frac{n}{n-1} \right)mgR\]
C) nmgR
D) \[\frac{mgR}{n}\]
Correct Answer: A
Solution :
[a] : Gravitational potential energy of mass m at any point at a distance r from the centre of earth is \[U=-\frac{GMm}{r}\] At the surface of earth r = R, \[\therefore \]\[{{U}_{s}}=-\frac{GMm}{R}=-mgR\left( \because g=\frac{GM}{{{R}^{2}}} \right)\] At the height h = nR from the surface of earth \[\therefore \]\[{{U}_{h}}=-\frac{GMm}{R(1+n)}=-\frac{mgR}{(1+n)}\] Change in gravitational potential energy is \[\Delta U={{U}_{h}}-{{U}_{s}}\] \[=-\frac{mgR}{(1+n)}-(-mgR)\] \[=-\frac{mgR}{1+n}+mgR=mgR\left( 1-\frac{1}{1+n} \right)\] \[=mgR\left( \frac{n}{1+n} \right)\]
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