A diametrical tunnel is dug across the earth. A ball is dropped into the tunnel from one side. The velocity of the ball when it reaches the centre of the earth is

\sqrt{0.5 g R}

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\sqrt{g R}

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\sqrt{2.5 g R}

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\sqrt{7.1 g R}

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Solution

The correct option is B $\sqrt{g R}$
$Δ K . E = Δ U g r a v i t a t i o n a l p o t e n t i a l a t t h e c e n t r e o f e a r t h = - \frac{3}{2} \frac{G M}{R} \Rightarrow \frac{1}{2} M v^{2} = G M_{e} M (\frac{3}{2 R} - \frac{1}{R}) ∴ v = \sqrt{\frac{G M_{e}}{R}} = \sqrt{g R}$

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= - \frac{3}{2} \frac{G M}{R}

A diametrical tunnel is dug across the earth. A ball is dropped into the tunnel from one side.

The velocity of the ball when it reaches the centre of the earth is

[Given: gravitational potential at the centre of earth = - \frac{3}{2} (\frac{G M}{R})

\frac{R}{2}

M

R

R / 2

O

C

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