First slide

Gravitational potential and potential energy

Question

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. The minimum velocity given to the particle so that it just reaches to the surface of the earth is

Difficult

A

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

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

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

$\sqrt{\frac{G M}{4 R}}$

Solution

Given Mass of Earth =M
Radius of Earth =R
Let the minimum speed imparted to the particle of mass m so that it just reaches surface of earth is v.
Applying the principle of conservation of energy,
$\frac{1}{2} m v^{2} + (- \frac{3}{2} \frac{G M}{R} m) = - \frac{G M}{R} m + 0$
$\frac{1}{2} v^{2} = \frac{- G M}{R} + \frac{3}{2} \frac{G M}{R}$
$= \frac{G M}{R}$ , we get $v = \sqrt{\frac{G M}{R}}$ .
Therefore, the correct answer is (A).

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