A) \[mgR\]
B) \[\frac{mgR}{2}\]
C) \[2mgR\]
D) \[\frac{3}{2}mgR\]
Correct Answer: B
Solution :
Potential energy at earths surface, \[{{U}_{i}}=-\frac{GMm}{R}\] Potential energy at a point, distant h from the eathers surface, \[{{U}_{f}}=\frac{GMm}{R+h}\] Given, \[h=R\] So, \[{{U}_{f}}=-\frac{GMm}{2R}\] \[\therefore \] Change in potential energy \[\Delta U={{U}_{f}}-{{U}_{i}}\] or \[=-\frac{GMm}{2R}-\left( -\frac{GMm}{R} \right)\] or \[=\frac{GMm}{2R}\] or \[=\frac{mgR}{2}\] \[[\because \,\,GM=g{{R}^{2}}]\]
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