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Rotational Power

If a body is ...

Question

If a body is released from a great distance from the centre of the earth, find its velocity when it strikes the surface of the earth. Take $R = 6400 k m$

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Solution

At great distance, since the body is initially at rest, the total energy is 0.

Hence, from mechanical energy conservation,

$\frac{1}{2} m v^{2} - \frac{G M m}{R} = 0$

$v = \sqrt{\frac{2 G M}{R}} = \sqrt{2 g R} = \sqrt{2 (9.8) (6400000)} = 11.2 k m / s$

Answer is $11.2 k m / s$

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