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A tunnel is dug along the diameter of earth. There is a particle of mass m at centre of tunnel. Find minimum velocity given to the particle so that it just reaches the surface of earth (R=radius of earth, M= mass of earth)

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$\sqrt{\frac{G M}{2 R}}$

$\sqrt{\frac{3 G M}{R}}$

$\sqrt{\frac{2 G M}{R}}$

$\sqrt{\frac{G M}{R}}$

answer is A.

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Detailed Solution

At center of earth, total energy
$E_{1} = \frac{1}{2} m V^{2} - \frac{3 G M m}{2 R}$
If particle reach surface of earth V=0, total energy
$E_{2} = 0 - \frac{G M m}{R}$
Apply LCME
$V = \sqrt{\frac{G M}{R}}$ .

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