A) 15%
B) 20.3%
C) 66.7%
D) 33.33%
Correct Answer: D
Solution :
Binding energy of satellite in the first case is \[=\frac{GMm}{2r}\] where r is the radius of orbit. In second case \[BE=\frac{GMm}{2\times \frac{3r}{r}}\] \[\therefore \] \[\Delta E=\frac{GMm}{r}\left( \frac{1}{2}-\frac{1}{3} \right)=\frac{GMm}{6r}\] % increase in energy of a satellite \[=\frac{\frac{GMm}{6r}}{\frac{GMm}{2r}}\times 100\] \[=\frac{2}{6}\times 100=33.33%\]You need to login to perform this action.
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