question_answer

The escape velocity of a sphere of mass m is given by (\[G=\]universal gravitational constant, \[{{M}_{e}}=\]mass of earth and \[{{R}_{e}}=\] radius of earth):        [AIPMT 1999]

A) \[\sqrt{\frac{G{{M}_{e}}}{{{R}_{e}}}}\]

B) \[\sqrt{\frac{2\,G{{M}_{e}}}{{{R}_{e}}}}\]

C) \[\sqrt{\frac{2\,Gm}{{{R}_{e}}}}\]

D) \[\frac{G{{M}_{e}}}{R_{e}^{2}}\]

Correct Answer: B

Solution :

Key Idea: If an energy equal to the binding energy of sphere on earth's surface is given to it in form of kinetic energy, it escapes the gravitational field of earth.

The binding energy of sphere of mass m (say) on the surface of earth kept at rest is . To escape it from earth's surface, this much energy in the form of kinetic energy is supplied to it.

So,

or          escape velocity

Here,      radius of earth,  mass of earth.