An artificial satellite (mass $m$ ) of a planet (mass $M$ ) revolves in a circular orbit whose radius is $n$ times the radius $R$ of the planet. In the process of motion, the satellite experiences a slight resistance due to cosmic dust. Assuming the force of resistance on satellite to depend on velocity, as $F = a v^{2}$ where $^{'} a^{'}$ is a constant, calculate how long the satellite will stay in the orbit before it falls onto the planet's surface.

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Solution

Air resistance $F = - a v^{2}$ , where orbital velocity $v = \sqrt{\frac{G M}{R}}; r =$ the distance of the satellite from planet's center.
$\Rightarrow F = - \frac{G M a}{r}$
The work done by the resistance force,
$= d W = F . d x = [F v . d t] = [\frac{G M a}{r} \sqrt{\frac{G M}{r}} d r]$
$= [\frac{(G M)^{3 / 2} a}{r^{3 / 2}} d t] . . . . . (1)$
$\Rightarrow$ The loss of energy of the satellite $= d E$
$∴ \frac{d E}{d r} = \frac{d}{d r} [- \frac{G M m}{2 r}] = \frac{G M m}{2 r^{2}}$
$\Rightarrow d E = \frac{G M m}{2 r^{2}} d r . . . . . (2)$
Since $d E = - d W$ (work energy theorem)
$- \frac{G M m}{2 r^{2}} d r = \frac{(G M)^{3 / 2} a}{r^{3 / 2}} d t$
$\Rightarrow t = \frac{m}{2 a \sqrt{G M}} \int_{n R}^{R} \frac{d r}{\sqrt{r}}$
$\Rightarrow t = \frac{m \sqrt{R} (\sqrt{n} - 1)}{a \sqrt{G M}}$
$= (\sqrt{n} - 1) = \frac{m}{a \sqrt{g R}}$ .

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