A) \[\frac{E}{4}\]
B) \[\frac{E}{16}\]
C) \[\frac{E}{32}\]
D) \[\frac{E}{64}\]
Correct Answer: B
Solution :
[b] |
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.
You will be redirected in
3 sec