A) K
B) 2K
C) mg R
D) mK
Correct Answer: B
Solution :
The escape velocity is given by \[{{V}_{e}}=\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}\] So, the kinetic energy of the satellite to escape from the earths gravitational field is \[KE=\frac{1}{2}mv_{e}^{2}=\frac{1}{2}\frac{m\times 2G{{M}_{e}}}{{{R}_{e}}}=\frac{G{{M}_{e}}m}{{{R}_{e}}}\] Because the kinetic energy of satellite while rotating \[K=\frac{GMem}{2{{R}_{e}}}\] So, the kinetic energy \[KE=2K\]
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