Question

A tunnel is dug along the diameter of the earth. There is a particle of mass m at the centre of the tunnel. Find the minimum velocity given to the particle so that just reaches to the surface of the earth. (R=radius of earth)

• A. $\sqrt{\frac{GM}{R}}$
• B. $\sqrt{\frac{GM}{2R}}$
• C. $\sqrt{\frac{2GM}{R}}$
• D. None of the above

Answer 1

The correct option is A $\sqrt{\frac{GM}{R}}$

By conservation of energy,

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

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

A gravitational potential at centre $=-\frac{3}{2}\frac{GM}{R}$

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

Therefore,

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

$=\frac{1}{2}m{v}^2=\frac{GMm}{2R}$

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