Question

A tunnel is dug along the diameter of Earth.There is a particle of mass $m$ at the centre of tunnel.The maximum velocity given to the particle, so that it just reaches the surface of earth is: $(R$ is radius of Earth and $M$ is the mass of the earth $)$

• A. $\sqrt{\frac{GM}{R}}$
• B. $\sqrt{\frac{GM}{2R}}$
• C. $\sqrt{\frac{2GM}{R}}$
• D. Zero

Answer 1

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


Initially, the particle is at centre $\text{O}$ of the earth. Let the velocity imparted be $v$. So, the total energy of the particle at this point will be

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

Now, for the particle to just reach the surface $.i.e$, $v_s=0$. The total energy at this point will be

$E_f=K.E_f+P.E_f=0+\frac{-GMm}{R}$

Applying energy conservation principle,

$E_i=E_f$

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

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

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

Hence, option (a) is the correct answer.

Why this question? To give the idea of the concept of conservation of energy principle along with the expression of gravitational potential due to a solid sphere at inside points.