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A particle is...

Question

A particle is released at a height $r_{E}$ (radius of earth ) above the earth 's surface. Determine its velocity when it hits the earth. Ignore air resistance.

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Solution

Potential energy when it is at a distance 2r from the centre= $- \frac{G M m}{2 r}$

Potential energy when it is on the ground= $- \frac{G M m}{r}$

This change in potential should be equal to the kinetic energy,

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

we get, $v^{2} = \frac{G M}{r} = g r$

$v = \sqrt{g r} = \sqrt{9.8 \times 6400000} = 7.9 \times 10^{3} m / s$

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