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 $\sqrt{gR}$ (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 $\sqrt{gR}$

Answer 1

Let M be the total mass of the earth.

At any position x.

$\therefore \frac{M_1}{M}=\frac{p\times \frac{4}{3}\pi x^3}{p\times \frac{4}{3}\pi r^3}\times \frac{x^3}{R^3}$

$\Rightarrow M_1=\frac{M{x}^3}{R^3}$

So, force on the particle is given by

$\therefore F_x=\frac{GMm}{x^2}$

$=\frac{GMm}{R^3}x...\left(1\right)$

So, acceleration of the mass 'M' at that position is given by,

$a_x=\frac{GM}{R^3}$

$\Rightarrow \frac{ax}{x}={\omega }^2$

$=\frac{GM}{R^3}=\frac{g}{R}(g=\frac{GM}{R^2})$

So, $T=2\pi \sqrt{\left(\frac{R}{g}\right)}$ = Time period of oscillation

(a) Now using velocity = displacement equation

$v=\omega \sqrt{(A^2-Y^2)}$ [where, A = amplitude]

Given when,

$y=R,v=\sqrt{gR,}w=\sqrt{\frac{g}{R}}$

$\Rightarrow \sqrt{gR}=\sqrt{\frac{\left(\frac{g}{R}\right)}{(A^2-R^2)}}$

[because $\omega =kg/r$]

$\Rightarrow R^2=A^2-R^2$

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

[Now, the phase of the particle at the point P is greater than $\frac{\pi }{2}$ has less than $\pi$ and at Q is greater than $\pi$ but less than $\frac{3\pi }{2}$. Let the times taken by the particle to reach the positions P and Q be $t_1$ and $t_2$ respectively, then using displacement equation].

$y=rsin\omega t$

We have,

$R=\sqrt{2}Rsin\omega t$

$\Rightarrow \omega t_1=\frac{3\pi }{4}$

We have,

$R=\sqrt{2}Rsin\omega t_1$

$\Rightarrow \omega t_2=\frac{5\pi }{4}$

So, $\omega (t_2-t_1)=\frac{\pi }{2}$

$\Rightarrow t_2-t_1=\frac{\pi }{2\omega }\frac{\frac{\pi }{2}}{\left(\frac{R}{g}\right)}$

Time taken by the particle to travel from P to Q is

$t_2-t_1=\frac{\pi }{2}\left(\frac{R}{g}\right)sec.$

(b) When the body is dropped from a height r, then applying conservation of energy, change in P.E. = gain in K.E.

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

$\Rightarrow v=\sqrt{\left(gR\right)}$

Since, the velocity is same at P, as in part (a) the body will take same time to travel PQ.

(c) When the body is projected vertically upward from P with a velocity.

$\sqrt{\left(gR\right)}$ its velocity will be zero at the highest point.

The velocity of the body, when reaches p, again will be

$v=\sqrt{\left(gR\right)}$

Hence the body will take same time

$\frac{\pi }{2}\sqrt{\left(\frac{R}{g}\right)}$ to travel P.Q.