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A particle is dropped from a height equal to the radius of the earth above the tunnel dug through the earth as shown in the figure.
$R$ : Radius of earth.
$M$ : Mass of earth.

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Particle will oscillate through the earth to a height R on both sides.

Motion of the particle is periodic.

Particle will execute simple harmonic motion.

Particle passes the centre of earth with a speed $= \sqrt{\frac{2 GM}{R}} .$

answer is A, C, D.

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Detailed Solution

$\frac{- G M m}{2 R} = \frac{1}{2} m v^{2} - \frac{3}{2} \frac{G M m}{R}$ At an outside point

$F α \frac{1}{x^{2}} (not on SHM)$

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