Question

Assume that a tunnel is dug across the earth(radius R) passing through its center. Find the time a particle takes to cover the length of the tunnel if
a) it is projected into the tunnel with a velocity (gR)^1/2
b)it is released from a height R above the tunnel
c)it is thrown vertically upward along the length of the tunnel with a speed (gR)^1/2

Answer 1

Answer on Question #51673, Physics, Mechanics | Kinematics | Dynamics

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

a) it is projected into the tunnel with a velocity $(\mathrm{gR})^{\wedge}1 / 2$

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

c) it is thrown vertically upward along the length of the tunnel with a speed $(\mathrm{gR})^{\wedge}1 / 2$

Solution:

(a) Let $T$ be the time period of the oscillatory motion of the particle, $x_{t}, v_{t}$ and $a_{t}$ the displacement from the center, velocity and acceleration at time $t$. Then

$$a_{t} = \frac{F}{m} = GM \frac{x_{t}^{3}}{R^{3} x_{t}^{2}} = GM x_{t} / R^{3}$$

The period of oscillation

$$T = 2 \pi \left(\frac{x_{t}}{a_{t}}\right)^{\frac{1}{2}} = 2 \pi \left(\frac{R^{3}}{GM}\right)^{\frac{1}{2}} = 2 \pi \left(\frac{R}{g}\right)^{\frac{1}{2}}$$

The angular velocity

$$\omega = \left(\frac{g}{R}\right)^{\frac{1}{2}}$$

If A be the amplitude of the motion then

$$v_{t}^{2} = \omega^{2} (A^{2} - R^{2}) = \frac{g (A^{2} - R^{2})}{R}$$

But at the surface of the Earth we have $v_{t} = (gR)^{\frac{1}{2}}$

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

Let $t = t_1$ and $t = t_2$ be two successive instants when value of $x_t$ changes from $+R$ to $-R$ and therefore $t_2 - t_1$ is the time taken to cover the length of the tunnel. But

$$R = \sqrt{2} R \sin \omega t_{1}$$ $$\sin \omega t_{1} = \frac{1}{\sqrt{2}}$$ $$\omega t_{1} = \frac{3\pi}{4}$$

and

$$- R = \sqrt{2} R \sin \omega t_{2}$$ $$\sin \omega t_{2} = -\frac{1}{\sqrt{2}}$$ $$\omega t_{2} = \frac{5\pi}{4}$$ $$\omega t_{2} - \omega t_{1} = \frac{\pi}{2}$$ $$t _ {2} - t _ {1} = \frac {\pi}{2 \omega} = \frac {\pi}{2} \left(\frac {R}{g}\right) ^ {\frac {1}{2}}$$

b) In this case the particle is released from height R above the tunnel being a case of free fall motion till the particle reaches surface of the earth. The velocity of the particle when it reaches the surface of the earth is

$$v = (2 g R) ^ {\frac {1}{2}}$$

Therefore

$$2 g R = \frac {g (A ^ {2} - R ^ {2})}{R}$$ $$2 R ^ {2} = A ^ {2} - R ^ {2}$$ $$A = \sqrt {3} R$$

Let $t = t_1$ and $t = t_2$ be two successive instants when value of $x_t$ changes from $+R$ to $-R$ and therefore $t_2 - t_1$ is the time taken to cover the length $o$ the tunnel. But

$$R = \sqrt {3} R \sin \omega t _ {1}$$ $$\sin \omega t _ {1} = \frac {1}{\sqrt {3}}$$ $$\omega t _ {1} = \sin^ {- 1} \frac {1}{\sqrt {3}} = 0.615 \mathrm{rad}$$

and

$$- R = \sqrt {3} R \sin \omega t _ {2}$$ $$\sin \omega t _ {2} = - \frac {1}{\sqrt {3}}$$ $$\omega t _ {2} = - 0.615 \mathrm{rad}$$ $$\omega t _ {1} - \omega t _ {2} = 0.615 * 2 = 1.23 \mathrm{rad}$$ $$t _ {2} - t _ {1} = \frac {1.23}{\omega} = 1.23 \left(\frac {R}{g}\right) ^ {\frac {1}{2}}$$

c) In this case the particle is thrown up with velocity $(gR)^{\frac{5}{2}}$ it comes back to the surface of the earth with the same speed and then covers the length of the tunnel in the same time as in case (a) above.