Question

Assume that a tunnel is dug across the earth (radius= $R$ ) passing through its Centre. Find the time a particle takes to cover the length of the tunnel if:

(a) it is projected into the tunnel with a speed of $\sqrt{g R}$ .

(b) it is released from a height $R$ above the tunnel.

(c) it is thrown vertically upward along the length of tunnel with a speed of $\sqrt{g R}$ .

Accepted Answer

(a) Net force on particle inside ( $r$ is distance from center of earth)
$m g_{i n} = m a$
$\Rightarrow$ $- m [\frac{G M r}{R^{3}}] = m a$
$\Rightarrow$ $a = - \frac{G m}{R^{3}} r$
We know acceleration on the surface of the earth $g = \frac{G M}{R^{2}}$
$∴ a = - \frac{g}{R} r$ This is equation of SHM. It means particle perform SHM inside earth.
Compare above equation with differential equation of SHM $a = - ω^{2} r$
$∴ ω= \sqrt{\frac{g}{R}}$
$T = 2 π \sqrt{\frac{R}{g}}$

Particle is projected inside tunnel with speed $v = \sqrt{g R}$ at $r = R$

Velocity of particle in SHM with displacement from mean position $v = ω \sqrt{A^{2} - r^{2}}$

$\Rightarrow \sqrt{g R} = \sqrt{\frac{g}{R}} \sqrt{A^{2} - R^{2}}$
$\Rightarrow R = \sqrt{A^{2} - R^{2}}$

$\Rightarrow A^{2} = 2 R^{2}$

$A = \sqrt{2} R$

Let equation of SHM of particle is $x = A \sin ω t$ Considering SHM start from mean position.
$R = \sqrt{2} R \sin ω t$
$\frac{1}{\sqrt{2}} = \sin ω t$
$∴ ω t = \frac{π}{4}$
$t = \frac{π}{4 ω} = \frac{T}{4}$ It is the time taken by particle from mean position (Center of earth) to surface of earth $(r = R)$
So total time taken by particle in tunnel $= 4 t = T$ $= 2 π \sqrt{\frac{R}{g}}$

(b) In this case particle released from $h = R$ , $u = 0$

Let speed of particle is $v$ when it entered in tunnel.
Energy conservation from $h = R$ and $h = 0$
$- \frac{GMm}{R + R} + 0 = - \frac{GMm}{R} + \frac{1}{2} {mv}^{2}$
$\frac{G M m}{R} - \frac{G M m}{2 R} = \frac{1}{2} m v^{2}$
$\frac{1}{2} m v^{2} = \frac{G M m}{2 R}$
$v = \sqrt{\frac{G M}{R}} = \sqrt{g R}$ Same as previous case
So time remains same as in previous case.
$t = 2 π \sqrt{\frac{R}{g}}$

(c) Velocity at the surface is same as in previous cases

Thus, the time remains in this case is also the same as previous case

$t = 2 π \sqrt{\frac{R}{g}}$

H C Verma Solutions for Chapter: Simple Harmonic Motion, Exercise 4: EXERCISES

H C Verma Physics Solutions for Exercise - H C Verma Solutions for Chapter: Simple Harmonic Motion, Exercise 4: EXERCISES

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