A) \[\frac{mgh}{R}\]
B) \[\frac{2}{3}\,mgR\]
C) \[\frac{3}{4}mgR\]
D) \[\frac{mgR}{2}\]
Correct Answer: C
Solution :
Key Idea: The gravitational potential energy of mass m in the gravitational field of mass M at distance r from it is \[U=-\frac{GMm}{r}\]. From key idea, potential energy \[U=-\frac{GMm}{r}\] At earth's surface, r = R \[\therefore {{U}_{e}}=-\frac{GMm}{R}\] Now, if a body is taken to height h = 3R, then the potential energy is given by \[{{U}_{h}}=-\frac{GMm}{R+h}\,(\because \,r=h+R)\] \[=\,-\frac{GMm}{4R}\] Thus, change in gravitational potential energy, \[\Delta U={{U}_{h}}-{{U}_{e}}\] \[=-\frac{GMm}{4R}-\left( -\frac{GMm}{R} \right)\] \[=-\frac{GMm}{4R}+\frac{GMm}{R}\] \[=\frac{3}{4}\frac{GMm}{R}\] \[\therefore \Delta U=\frac{3}{4}\frac{g\,{{R}^{2}}\,m}{R}(\because \,\,GM=g{{R}^{2}})\] \[=\frac{3}{4}\,mgR\]
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