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)
Solution
Gravitational acceleration at a distance of $r$ from centre of earth is given by
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$g^{\prime}=\frac{g}{R} r$
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Where $R$ is the radius of earth
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So, $\frac{d^{2} r}{d t^{2}}=-\frac{g}{R} r$
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$\Rightarrow \quad T=2 \pi \sqrt{\frac{R}{g}}=2 \pi \sqrt{\frac{6400000}{10}}$
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$=2 \pi \times 800 \mathrm{sec}$
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$=5024 ~ \mathrm{sec}$
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= 1 hour 24 minutes (approx.)
About this question
Subject: Physics · Chapter: Gravitation · Topic: Newton's Law of Gravitation
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