A satellite revolves around a planet in circular orbit of radius R (much larger than the radius of the planet) with a time-period of revolution T. If the satellite is stopped and then released in its orbit : (Assume that the satellite experiences gravitational force due to the planet only).

(a) It will fall because mg is acting on it towards the centre of planet and initial velocity is zero. It will move in straight line.

(c) Time of fall can be found as follows : By energy conservation

$\frac{1}{2}\mathrm{m}{v}^2-\frac{\mathrm{GMm}}{\mathrm{r}}=0-\frac{\mathrm{GMn}}{\mathrm{r}}$.....(1)

Using this we get $V=f\left(r\right)$

Now use  $V=-\frac{dr}{dt}\Rightarrow f\left(r\right)=-\frac{dr}{dt}$

$\Rightarrow {\int }_{R\text{'}}^R\frac{dr}{f\left(r\right)}=-{\int }_0^{t}dt$

R’$=$radius of the planet.

In the final expression (or in the beginning itself)

${\mathrm{R}}^{\mathrm{\prime }}\to 0(\because \mathrm{R}>>{\mathrm{R}}^{\mathrm{\prime }})$

You will get

$t=\frac{T}{4\sqrt{2}}$

Here $\frac{\mathrm{GMm}}{{\mathrm{R}}^2}=\mathrm{m}{\left(\frac{2\pi }{\mathrm{T}}\right)}^2\mathrm{R}$