Motion Of A Ball In Tunnel Through The Earth MCQ - Practice Questions with Answers

Case I: If the tunnel is along a diameter and the ball is released from the surface. If the ball at any time is at a distance y from the center of the earth as shown in the below figure,

So the restoring force will act on the ball due to gravitation between the ball and the earth.

Acceleration of the particle at the distance y from the center of the earth is given by

$\begin{aligned} & a={\frac{-G M y}{R^3}}_{\text {and }} g=\frac{G M}{R^2} \\ & \quad \text { So } a=\frac{-\left(g R^2\right) y}{R^3} \Rightarrow a=-\frac{g}{R} y \\ & \text { Comparing with } a=-\omega^2 y \\ & \quad \omega^2=\frac{g}{R} \quad \omega=\sqrt{\frac{g}{R}} \Rightarrow T=2 \pi \sqrt{\left(\frac{R}{g}\right)}=84.6 \mathrm{~min}\end{aligned}$

Case II: If the tunnel is along a chord and the ball is released from the surface. If the ball at any time is at a distance x from the centre of the tunnel, as shown in the below figure 

then the acceleration of the particle at the distance y from the center of the earth

$$
a=\frac{-G M y}{R^3}
$$

and using $g={\frac{G M}{R^2}}_{\text {we get }} a=\frac{-\left(g R^2\right) y}{R^3} \Rightarrow a=-\frac{g}{R} y$
This acceleration will be towards the center of the earth.
So the component of acceleration towards the center of the tunnel.

$$
a^{\prime}=a \sin \theta=\left(-\frac{g}{R} y\right)\left(\frac{x}{y}\right)=-\frac{g}{R} x
$$

Comparing with $a^{\prime}=-\omega^2 x$

$$
\omega^2=\frac{g}{R} \Rightarrow \omega=\sqrt{\frac{g}{R}} \Rightarrow T=2 \pi \sqrt{\frac{R}{g}}=84.6 \mathrm{~min}
$$

Note: The time period of oscillation is the same in both cases whether the tunnel is along a diameter or along the chord.