A) One
B) Two
C) Half
D) Infinity
Correct Answer: B
Solution :
K.E. required for satellite to escape from earth's gravitational field \[\frac{1}{2}mv_{e}^{2}=\frac{1}{2}m{{\left( \sqrt{\frac{2GM}{R}} \right)}^{2}}=\frac{GMm}{R}\] K.E. required for satellite to move in circular orbit \[\frac{1}{2}mv_{0}^{2}=\frac{1}{2}m{{\left( \sqrt{\frac{GM}{R}} \right)}^{2}}=\frac{GMm}{2R}\] The ratio between these two energies = 2
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