A) \[15%\]
B) \[20.3%\]
C) \[66.7%\]
D) \[33.33%\]
Correct Answer: D
Solution :
Binding energy of satellite in the first case is \[=\frac{GMm}{2\,\,r}\] 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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