Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simpl

Question

Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is $100 g$ . The time period of the motion of the particle will be (approximately)

(take $g = 10 {ms}^{- 2}$ , radius of earth $= 6400 km$ )

TestHub · Accepted Answer

The gravitational force acting on a particle of mass $m$ at a distance $r (< R)$ from the centre of Earth is

$F = - G \frac{M^{'} m}{r^{2}}$ ....(1)

where $M^{'}$ is the mass of portion of Earth enclosed by a sphere of radius $r$ concentric with Earth. Here $-$ sign indicates radially inward direction of force.

Taking the density of Earth to be uniform

$\frac{M^{'}}{4 {πr}^{3}} = \frac{M}{4 {πR}^{3}}$

$\Rightarrow M^{'} = \frac{M r^{3}}{R^{3}}$ ....(2)

Substituting $M^{'}$ from equation 2 into equation 1:

$F = - (\frac{G M m}{R^{3}}) r$

Which is similar to the equation of motion $F = - k x$ with $k = G \frac{M m}{R^{3}} = \frac{m g}{R} (∵ g = G \frac{M}{R^{2}})$

For such motion, the time period is given by:

$T = 2 π \sqrt{\frac{m}{k}}$

Substituting k from above,

$T = 2 π \sqrt{\frac{R}{g}}$

Putting the values, we get

$T = 2 \times 3.14 \sqrt{\frac{6400 \times 10^{3}}{10}} = 2 \times 3.14 \times 800 \approx 1 hour 24 minutes$

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