A) \[2E\]
B) \[\sqrt{E}\]
C) \[E/2\]
D) \[\sqrt{E/2}\]
E) \[E\]
Correct Answer: E
Solution :
Kinetic energy of satellite in its orbit \[E=\frac{1}{2}mv_{o}^{2}\] Or \[E=\frac{1}{2}m\left( \frac{GM}{r} \right)=\frac{GMm}{2r}\] Kinetic energy at escape velocity \[E=\frac{1}{2}mv_{e}^{2}\] \[=\frac{1}{2}m\left( \frac{2GM}{r} \right)=\frac{GMm}{r}\] \[=2E\] Therefore, additional kinetic energy required \[=2E-E=E\]
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