Question

A ball of mass $m$ is dropped from a height $h$ equal to the radius of the earth above the tunnel dug through the earth as shown in the figure. Choose the correct options. (Mass of earth $=M$)

• A. Particle will execute simple harmonic motion.
• B. Particle passes the centre of earth with a speed of $\sqrt{\frac{2GM}{R}}$
• C. Motion of the particle is periodic
• D. Particle will oscillate through the earth to a height $h$ on both sides

Answer 1

Answer: (B), (C), (D)

Particle will oscillate through the earth to a height $h$ on both sides

The ball performs $\text{SHM}$ inside the tunnel but not outside the tunnel surface because of the variation in the gravitational fields inside and outside the earth.

But from the energy conservation we can say that the ball will execute the periodic motion about the centre of the earth.

So, the motion is periodic with amplitude $(R+h)$.

Let the speed of the ball be $v$ while passing the centre.

Applying energy conservation principle between centre and the initial point,

$K.E_i+P.E_i=K.E_{centre}+P.E_{centre}$

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

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

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

$\therefore v=\sqrt{\frac{2GM}{R}}$

Hence, options (a), (c) and (d) are correct answers.

Why this question: This question checks the knowledge of the difference between periodic motion and S.H.M, along with the application of energy conservation principle.