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 earth's field, body should fee given as much energy which is equal to binding energy of body at earth's surface. The binding energy of particle at earth's 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 earth's 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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