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’ll move in straight line. 
(C) Time of fall can be found by two methods:
Method-I: By energy conservation 
$\frac{1}{2}m{v}^2-\frac{GMm}{r}=0-\frac{GMm}{R}-\left(1\right)$  Using this we get V = f(r). 
Now use 
$V=-\frac{dr}{dt}\Rightarrow f\left(r\right)=-\frac{dr}{dt}$ 
$\Rightarrow \underset{R}{\overset{R}{\int }}\frac{dr}{f\left(r\right)}=-\underset{0}{\overset{t}{\int }}dt;R\text{'}$ = radius of the planet.
In the final expression (or in the beginning itself) $R\text{'}\to 0$  $\left\{\because R>>R\text{'}\right\}$    
you will get  $t=\frac{T}{4\sqrt{2}}$
Here  $\frac{GMm}{R^2}=m{\left(\frac{2\pi }{T}\right)}^{2}R$
Method-II: 
Kepler’s Law:  $T^2\alpha r^3$.
Assume that the satellite moves in elliptical path with maximum and minimum distances from centre as R and R’.
$\because R>>R\text{'}$   $\therefore$Velocity at R is very small $\left(\simeq 0\right)$. When it reaches R’ then it touches the surface of the planet. This motion (from R to R’) is almost, same as given in the question.

$\begin{array}{l}\text{ Now }\frac{{\mathrm{T}}_1^2}{{\mathrm{T}}_2^2}=\frac{{\mathrm{r}}_1^3}{{\mathrm{r}}_2^3},{\mathrm{T}}_1=\mathrm{T},{\mathrm{r}}_1=\mathrm{R}\\ {\mathrm{r}}_2=\frac{\mathrm{R}+{\mathrm{R}}^{\mathrm{\prime }}}{2}\simeq \frac{\mathrm{R}}{2}\therefore {\mathrm{T}}_2=\frac{\mathrm{T}}{4\sqrt{2}}\end{array}$