A) 22.3%
B) 33.3%
C) 66.7%
D) 100%
Correct Answer: C
Solution :
Binding energy of satellite in 1st case \[{{U}_{1}}=-\left( -\frac{GMm}{r}+\frac{1}{2}\frac{GMm}{r} \right)\] \[=\frac{GMm}{2r}\] Binding energy of satellite in 2nd case \[{{U}_{2}}=-\left[ -\frac{GMm}{3r}+\frac{1}{2}\frac{GMm}{3r} \right]\] \[=\frac{GMm}{6r}\] Energy increased \[\Delta E={{U}_{1}}-{{U}_{2}}\] \[=\frac{GMm}{r}\left[ \frac{1}{2}-\frac{1}{6} \right]\] \[=\frac{GMm}{3r}\] % increase in energy \[=\frac{\Delta E}{{{U}_{1}}}\times 100\] \[=\frac{GMm/3r}{GMm/2r}\times 100\] \[=66.7%\]
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