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A hole is drilled from the surface of earth to its centre . A particle is dropped from rest at the surface of earth . The speed of the particle when it reaches the centre of the earth in terms of its escape velocity on the surface of earth V_e is

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a

$\frac{V_{e}}{2}$

b

V_e

c

$\sqrt{2} V_{e}$

d

$\frac{V_{e}}{\sqrt{2}}$

answer is D.

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Detailed Solution

Kinetic energy at centre=potential energy at surface -potential energy at centre.
Or $\frac{1}{2} {mv}^{2} = - \frac{GMm}{R} - (\frac{- 3}{2} \frac{GMm}{R})$
Or $\frac{1}{2} {mv}^{2} = \frac{1}{2} \frac{GMm}{R}$
Or $V = \sqrt{\frac{GM}{R}} = \sqrt{gR} = \frac{V_{e}}{\sqrt{2}}$

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