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

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

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

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

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i t w i l l r e a c h w i t h t h e h e l p o f a n e g l i g i b l e v e l o c i t y

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Solution

The correct option is D $i t w i l l r e a c h w i t h t h e h e l p o f a n e g l i g i b l e v e l o c i t y$
Initially, the particle is at centre of the earth. Let the velocity imparted be $v$ .

Thus $K . E_{i} + P . E_{i} = \frac{1}{2} m v^{2} + (- \frac{3 G M m}{2 R})$

Now for the particle to just reach the surface i-e, $v_{s} = 0$

$K . E_{f} + P . E_{f} = 0 + (- \frac{G M m}{R})$

Applying conservation of energy, $T . E_{i} = T . E_{f}$

$\frac{1}{2} m v^{2} + (- \frac{3 G M m}{2 R}) = 0 + \frac{- G M m}{R}$

$⟹ v = \sqrt{\frac{G M}{R}}$

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