A) \[\frac{U}{{{v}_{e}}}=m\sqrt{\frac{GM}{2R}}\]
B) \[\frac{U}{{{v}_{e}}}=m\sqrt{\frac{GM}{2R}}\]
C) \[\frac{U}{{{v}_{e}}}=m\sqrt{\frac{2GM}{R}}\]
D) \[\frac{U}{{{v}_{e}}}=m\frac{GM}{R}\]
Correct Answer: B
Solution :
The work obtained in bringing a body from infinity to a point in a gravitational field is called the gravitational potential energy of the body at that point. \[U=-\frac{G{{M}_{m}}}{R}\] ...(1) Where G is gravitational constant, M is mass of earth R its radius and m is mass of body. Also for a body projected upwards at a certain velocity of projection the body will go out of the gravitational field of the earth and will never return to the earth, this initial velocity is called escape velocity. \[{{v}_{e}}=\sqrt{\frac{2GM}{R}}\] ?(2) Since, work is required to take a body from earth's surface to infinity, we have \[U=+\frac{G{{M}_{m}}}{R}\] ...(3) Dividing Eq. (3) by (2), we get \[\frac{U}{{{v}_{e}}}=m\sqrt{\frac{GM}{2R}}\]
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