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 $\sqrt{g\mathrm{R}}$ (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$\sqrt{g\mathrm{R}}$.

Answer 1

Given:
Radius of the earth is R.
Let M be the total mass of the earth and $\rho$ be the density.
Let mass of the part of earth having radius x be M'.
$$\begin{gathered} \therefore \frac{M\text{'}}{M}=\frac{\rho \times {\displaystyle \frac{4}{3}}\mathrm{\pi }{x}^3}{\rho \times {\displaystyle \frac{4}{3}}\mathrm{\pi }{R}^3}=\frac{x^3}{R^3} \\ \Rightarrow M\text{'}=\frac{M{x}^3}{R^3} \end{gathered}$$

Force on the particle is calculated as,
$$\begin{gathered} {\mathrm{F}}_x=\frac{GM\mathit{\text{'}}m}{x^2} \\ =\frac{GMm}{R^3}x\dots \left(1\right) \end{gathered}$$

Now, acceleration $\left(a_x\right)$ of mass M' at that position is given by,
$$\begin{gathered} a_x=\frac{GM}{R^3}x \\ \Rightarrow \frac{a_x}{x}={\omega }^2=\frac{GM}{R^3}=\frac{g}{R}\left(\because g=\frac{GM}{R^2}\right) \\ \mathrm{So},\mathrm{Time}\mathrm{period}\mathrm{of}\mathrm{oscillation},T=2\mathrm{\pi }\sqrt{\left(\frac{R}{g}\right)} \end{gathered}$$

(a) Velocity-displacement equation in S.H.M is written as,
$$\begin{gathered} V=\omega \sqrt{\left(A^2-y^2\right)} \\ \mathrm{where},A\mathrm{is}\mathrm{the}\mathrm{amplitude};\mathrm{and} \\ y\mathrm{is}\mathrm{the}\mathrm{displacement}. \end{gathered}$$

When the particle is at y = R,
The velocity of the particle is $\sqrt{gR}$ and $\omega =\sqrt{\frac{g}{R}}$.
On substituting these values in the velocity-displacement equation, we get:
$$\begin{gathered} \sqrt{gR}=\sqrt{\frac{g}{R}}\sqrt{A^2-R^2} \\ \Rightarrow R^2=A^2-R^2 \\ \Rightarrow A=\sqrt{2R} \end{gathered}$$

Let t₁ and t₂ be the time taken by the particle to reach the positions X and Y.
Now, phase of the particle at point X will be greater than $\frac{\mathrm{\pi }}{2}$ but less than $\mathrm{\pi }$.
Also, the phase of the particle on reaching Y will be greater than $\mathrm{\pi }$ but less than $\frac{3\mathrm{\pi }}{2}$.

Displacement-time relation is given by,
y = A sin ωt

Substituting y = R and A =$\sqrt{2R}$ , in the above relation, we get:
$R=\sqrt{2}R\sin \omega t_1$
$\Rightarrow \mathrm{\omega }{t}_1=\frac{3\mathrm{\pi }}{4}$

Also, $R=\sqrt{2}R\sin \omega t_2$

$$\begin{gathered} \Rightarrow \mathrm{\omega }{t}_2=\frac{5\mathrm{\pi }}{4} \\ \mathrm{So},\mathrm{\omega }\left(t_2-t_1\right)=\frac{\mathrm{\pi }}{2} \\ \Rightarrow t_2-t_1=\frac{\mathrm{\pi }}{2\omega }=\frac{{\displaystyle \mathrm{\pi }}}{2\left({\displaystyle \sqrt{\frac{g}{R}}}\right)} \end{gathered}$$

Time taken by the particle to travel from X to Y:
$t_2-t_1=\frac{\mathrm{\pi }}{2\omega }=\frac{\mathrm{\pi }}{2}\sqrt{\frac{R}{g}}$ s

(b) When the body is dropped from a height R

Using the principle of conservation of energy, we get:
Change in P.E. = Gain in K.E.
$$\begin{gathered} \Rightarrow \frac{\mathrm{GM}m}{\mathrm{R}}-\frac{\mathrm{GM}m}{2\mathrm{R}}=\frac{1}{2}m{v}^2 \\ \Rightarrow v=\sqrt{\left(g\mathrm{R}\right)} \end{gathered}$$

As the velocity is same as that at X, the body will take the same time to travel XY.


(c) The body is projected vertically upwards from the point X with a velocity $\sqrt{g\mathrm{R}}$. Its velocity becomes zero as it reaches the highest point.
The velocity of the body as it reaches X again will be,
$v=\sqrt{\left(g\mathrm{R}\right)}$
Hence, the body will take same time i.e. $\frac{\mathrm{\pi }}{2}\sqrt{\left(\frac{\mathrm{R}}{g}\right)}$ s to travel XY.