An artificial satellite of a planet revolves in a circular orbit whose radius exceeds the radius of the planet 7 times. In the process of motion, the satellite experiences slight resistance due to cosmic dust. Assuming the resistive force depends on the speed of the satellite as $F = {kv}^{2}$ (where k is a constant), The time taken by the satellite will stay in orbit until it falls on to the planet’s surface. is $\frac{m}{k \sqrt{gR}} (\sqrt{x} - 1)$ (Given $m =$ mass of the satellite, $g =$ acceleration due to gravity at the surface of planet and $R =$ radius of the planet). Find the value of x.

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Detailed Solution

Let r be the orbital radius

$\frac{GMm}{r^{2}} = \frac{{mV}^{2}}{r} \Rightarrow V = \sqrt{\frac{GM}{r}}$

Total energy of the satellite

$E = \frac{- GMm}{r} + \frac{1}{2} {mV}^{2} \Rightarrow E = \frac{- GMm}{2 r}$

$\frac{dE}{dr} = \frac{GMm}{2 r^{2}}$ ......(1)

$\frac{dE}{dt} = p = \vec{F} \cdot \vec{V}$

$\Rightarrow \frac{dE}{dt} = - F \cdot V = - k V^{2} V = - k V^{3}$

$\frac{dE}{dt} = - k {(\frac{GM}{r})}^{3 / 2}$ ........(2)

From (1) and (2) $\frac{GMm}{2 r^{2}} dr = - k {(\frac{GM}{r})}^{3 / 2} . dt$

$∴ t = \int dt = \frac{m}{2 k \sqrt{GM}} \int_{7 R}^{R} \frac{d r}{\sqrt{r}}; t = \frac{m}{k \sqrt{gR}} [\sqrt{7} - 1]$

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