A) \[\frac{1}{2}mgR\]
B) \[2\,mgR\]
C) \[mgR\]
D) \[\frac{1}{4}mgR\]
Correct Answer: A
Solution :
At the earth surface, potential energy of the earth mass system \[{{U}_{1}}=\frac{GMm}{R}\] The potential energy of the system. When mass m is at a height R from the earth surface \[{{U}_{2}}=\frac{GMm}{(R+R)}=\frac{GMm}{2R}\] Now change in potential energy = Work done by gravity \[\Rightarrow \]\[W=\frac{GMm}{R}-\frac{GMm}{2R}=\frac{GMm}{2R}\times \frac{R}{R}\] \[=\frac{1}{2}\times \frac{GMm}{{{R}^{2}}}\times R=\frac{1}{2}mgR\] \[(\therefore g=GM\text{/}{{R}^{2}})\]
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