A satellite is revolving round the earth with orbital speed $v_{0}$ . If it stops suddenly, find the speed with which it will strike surface of earth ( $v_{e}$ = escape velocity of a particle on earth's surface).

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Solution

Let $r =$ radius of satellite.

$v_{0} =$ the orbital speed of a satellite.

$v_{e} =$ escape velocity of a particle on earth's surface

$v =$ velocity of satellite at which it strike the surface of the earth

$M =$ mass of earth

$m =$ mass of satellite

The orbital speed of a satellite is given by

$v_{0}^{2} = \frac{G M}{r}$ . . . . .(1)

Escape velocity of a particle on the earth surface

$v_{e}^{2} = \frac{2 G M}{R}$ . . . . .(2)

Applying law of conservation of energy at point A and B,

$K . E =$ change in potential energy

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

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

$v^{2} = v_{e}^{2} - 2 v_{0}^{2}$ (from equation 1 and 2)

$v = \sqrt{v_{e}^{2} - 2 v_{0}^{2}}$

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A satellite is revolving round the earth with orbital speed v0. If it stops suddenly, find the speed with which it will strike surface of earth (ve= escape velocity of a particle on earth's surface).

A satellite is revolving round the earth with orbital speed $v_{0}$ . If it stops suddenly, find the speed with which it will strike surface of earth ( $v_{e}$ = escape velocity of a particle on earth's surface).