A) \[\frac{E}{4}\]
B) \[\frac{E}{16}\]
C) \[\frac{E}{32}\]
D) \[\frac{E}{64}\]
Correct Answer: B
Solution :
minimum energy required (E) = - (Potential energy of object at surface of earth) \[=-\left( -\frac{GMm}{R} \right)=\frac{GMm}{R}\] Now \[{{M}_{earth}}=64{{M}_{moon}}\] \[\rho .\frac{4}{3}\pi R_{e}^{3}=64.\frac{4}{3}\pi R_{m}^{3}\Rightarrow {{R}_{e}}=4{{R}_{m}}\] Now\[\frac{{{E}_{moon}}}{{{E}_{earth}}}=\frac{{{M}_{moon}}}{{{M}_{earth}}}.\frac{{{R}_{earth}}}{{{R}_{moon}}}=\frac{1}{64}\times \frac{4}{1}\] \[\Rightarrow {{E}_{moon}}=\frac{E}{16}\]You need to login to perform this action.
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