An artifical 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 resistance force on satellite to depend on velocity, as F = av² where 'a' is a constant, how long the satellite will stay in the orbit before it falls onto the planet's surface.

$(\sqrt{n}-1) \frac{m}{a \sqrt{g R}}$

Air resistance F = -av² where v = $\sqrt{\frac{GM}{r}}$ ; r = the distance of the satellite from earth’s center.

$\Rightarrow F=-\frac{G M a}{r}$

The work done by the resistance force

$=dW = F . dx=[Fvdt] $

$=\left[\frac{GM a}{r} \sqrt{\frac{GM}{r}} dt\right]$

$=\left[\frac{(G M)^{3 / 2} a}{r^{3 / 2}}d t\right]$            .......(1)

⇒ The loss of energy of the satellite = dE

$∴ \frac{d E}{d r}=\frac{d}{d r}\left[-\frac{G M ~ m}{2 r}\right]=\frac{G M m}{2 r^2}$

$\Rightarrow d E=\frac{G M m}{2 r^2} d r$           .......(2)

Since dE = dW (work energy theorem)

$\frac{G M m}{2 r^2} d r=\frac{(G M)^{3 / 2} a}{r^{3 / 2}} d t$

$\Rightarrow t=\frac{m}{2 a \sqrt{G M}} \int\limits_{n R}^R \frac{d r}{\sqrt{r}}$

$\Rightarrow t=\frac{m \sqrt{R}(\sqrt{n}-1)}{a \sqrt{G M}}=(\sqrt{n}-1) \frac{m}{a \sqrt{g R}}$