A) \[mgR\]
B) \[\frac{3}{2}mgR\]
C) \[\frac{mgR}{2}\]
D) \[\frac{mgR}{4}\]
Correct Answer: C
Solution :
The gravitational potential energy of mass m in the gravitational field of mass M at a distance R from centre of earth is \[{{U}_{i}}=-\frac{GMm}{R}\] If mass m is raised to a height R from earths surface. Then final potential energy of mass m is \[{{U}_{f}}=-\frac{GMm}{R+R}=-\frac{GMm}{2R}\] Hence, change in potential energy \[\Delta U={{U}_{f}}-{{U}_{i}}\] \[=-\frac{GMm}{2R}-\left( -\frac{GMm}{R} \right)\] \[=-\frac{GMm}{2R}+\frac{GMm}{R}\] \[=\frac{GMm}{2R}\] Also, we know that \[g=\frac{GM}{{{R}^{2}}}\] \[\Rightarrow \] \[GM=g{{R}^{2}}\] Hence, \[\Delta {{U}^{2}}=\frac{g{{R}^{2}}m}{2R}=\frac{mgR}{2}\]You need to login to perform this action.
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