Question 132: Gravitation

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 m s$$^{-2}$$ , radius of earth = 6400 km)

Accepted Answer

Gravitational acceleration at a distance of $r$ from centre of earth is given by
$$
g^{\prime}=\frac{g}{R} r
$$
Where $R$ is the radius of earth
So, $\frac{d^{2} r}{d t^{2}}=-\frac{g}{R} r$
$\Rightarrow \quad T=2 \pi \sqrt{\frac{R}{g}}=2 \pi \sqrt{\frac{6400000}{10}}$
$=2 \pi 	imes 800 \mathrm{sec}$
$=5024 ~ \mathrm{sec}$
= 1 hour 24 minutes (approx.)

Answer

12 hours

Answer

1 hour 24 minutes

Answer

24 hours

Answer

1 hour 40 minutes

Gravitation

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