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?

• A. $\sqrt{\frac{GM}{R}}$
• B. $\sqrt{\frac{GM}{2R}}$
• C. $\sqrt{\frac{2GM}{R}}$
• D. It will reach with the help of a negligible velocity

Answer 1

Answer: (A)

$\sqrt{\frac{GM}{R}}$

BY conservation of energy

Gain in K.E$=$ change in potential energy

$\frac{1}{2}m{v}^2=m[-\frac{GM}{R}-[-\frac{3GM}{2R}\left]\right]$

As Gravitational potential at centre$=\frac{-3}{2}\frac{GM}{R}$

Gravitational potential at surface$=-\frac{GM}{R}$

Therefore

$\frac{1}{2}m{v}^2=-\frac{GmM}{R}+\frac{3}{2}\frac{GMm}{R}=\frac{1}{2}m{v}^2=\frac{GMm}{2R}v=\sqrt{\frac{GM}{R}}$