Question

An artificial satellite of mass m of a planet of 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 the satellite depends on velocity as $F=a{v}^2$ where a is constant, calculate the time the satellite will stay in orbit before it falls onto the planet's surface.

• A. $\frac{m\sqrt{R}(\sqrt{n}-1)}{3a\sqrt{\left(GM\right)}}$.
• B. $\frac{m\sqrt{R}(\sqrt{n}-1)}{2a\sqrt{\left(GM\right)}}$.
• C. $\frac{m\sqrt{R}(\sqrt{n}-1)}{a\sqrt{\left(GM\right)}}$.
• D. $\frac{m\sqrt{R}(\sqrt{n}-1)}{4a\sqrt{\left(GM\right)}}$.

Answer 1

The correct option is C $\frac{m\sqrt{R}(\sqrt{n}-1)}{a\sqrt{\left(GM\right)}}$.

$v_0=\sqrt{\frac{GM}{r}}$

$KE=\frac{1}{2}m{v}_0^2=\frac{GMm}{2r}$

$P.E=-\frac{GMm}{r}$

Total energy $E=KE+PE$

$E=-\frac{GMm}{2r}$

Now $\frac{dE}{dt}=F.v$

$\Rightarrow \frac{GMm}{2r^2}\left(\frac{dr}{dt}\right)=-a{v}^3$

$\Rightarrow -a{\left(\sqrt{\frac{GM}{r}}\right)}^3$

$\Rightarrow \frac{GMm}{2r^2}\left(\frac{dr}{dt}\right)=-a\frac{GM}{r}\sqrt{\frac{GM}{r}}$

$\Rightarrow \frac{a}{m}{\int }_0^{t}dt=-\frac{1}{2}\sqrt{\frac{1}{GM}}{\int }_{r_i}^{r_f}{r}^{-1/2}dr$

$\Rightarrow \frac{a}{m}{\int }_0^{t}dt=-\frac{1}{2}\sqrt{\frac{1}{GM}}{\int }_{nR}^R{r}^{-1/2}dr$

$t=\frac{m}{a}\sqrt{\frac{R}{GM}}(\sqrt{n}-1)$

Hence c is the correct answer