A) \[\frac{GMm}{R}\]
B) \[-\frac{13GMm}{14R}\]
C) \[\frac{GMm}{7R}\]
D) \[-\frac{GMm}{14R}\]
Correct Answer: B
Solution :
The energy of artificial satellite at the surface of the earth \[{{E}_{1}}=-\frac{GMm}{R}\] The energy of artificial satellite when\[R=7R\] \[{{E}_{2}}=-\frac{1}{2}\frac{GMm}{7R}\] Energy required for launching of satellite \[E={{E}_{1}}-{{E}_{2}}\] \[=-\frac{GMm}{R}+\frac{1}{2}\left( \frac{GMm}{7R} \right)=-\frac{13\,GMm}{14\,R}\]
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