Question

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 :

• A. $\frac{v_e}{2}$
• B. $v_e$
• C. $\sqrt{2}{v}_e$
• D. $\frac{v_e}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{v_e}{\sqrt{2}}$

Lets $m=$ mass of the particle

$M=$ mass of the earth

$R=$ radius of the earth

$v=$ speed of particle when it reaches the center of the earth

applying conservation of total energy,

$(K.E{)}_1+(P.E{)}_1=(K.E{)}_2+(P.E{)}_2$

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

$-\frac{GM}{R}+\frac{3GM}{2R}=\frac{1}{2}{v}^2$

$v^2=\frac{GM}{R}$

$v=\sqrt{\frac{GM}{R}}$. . . . (1)

We know that the escape velocity of the particle from the earth is given by,

$v_e=\sqrt{\frac{2GM}{R}}$. . . . .(2)

$v_e=\sqrt{2}\sqrt{\frac{GM}{R}}$

$v_e=\sqrt{2}v$ (from equation 1)

$v=\frac{v_e}{\sqrt{2}}$

The correct option is D.