Question

A satellite is revolving round the earth with orbital speed v_o. If it stops suddenly, the speed with which it will strike the surface of earth would be (v_e = escape velocity of a particle on earth's surface)

Difficult

Solution

$U_{2} = 2 (\frac{1}{2} m v_{o}^{2})$
Now, $\frac{1}{2} m v^{2} = - U_{2} + U_{1} = U_{1} - U_{2} = \frac{m v_{e}^{2}}{2} - m v_{o}^{2}$
$∴ v = \sqrt{v_{e}^{2} - 2 v_{o}^{2}}$

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