A) 11.8 km/s
B) 16.5 km/s
C) 11.2 km/s
D) 14.5 km/s.
Correct Answer: C
Solution :
Key Idea: To escape from earths field, body should be given as much energy which is equal to binding energy of body at earths surface. The binding energy of particle at earths surface kept at rest is\[\frac{GMm}{R}.\]If this much energy in the form of kinetic energy is supplied to the particle, it leaves the earths gravitational field. So, if\[{{v}_{e}}\]is the escape velocity of the particle, then \[\frac{1}{2}mv_{e}^{2}=\frac{GMm}{R}\] or \[{{v}_{e}}=\sqrt{\frac{2GM}{R}}\] or \[{{v}_{e}}=\sqrt{2g\,R}\] \[\left( \because \,g=\frac{GM}{{{R}^{2}}} \right)\] Since, escape velocity is independent of angle of projection, so it will be as before i.e., 11.2 km/s.
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