The time period of a satellite, revolving above earth's surface at...

The velocity of the satellite revolving round the Earth at a height $r$ from the centre of the Earth is given by

$v = \sqrt{\left(\dfrac{GM}{r}\right)} . . . \left(1\right)$

The formula to calculate the time period in the circular orbit is given by

$T = \dfrac{2\pir}{v} . . . \left(2\right)$

In this case: $r = R + R = 2R$

Substitute the expression for the velocity from equation (1) into equation (2) and simplify to obtain the required time period.

$\begin{cases}T & = & \dfrac{2\pir}{\sqrt{\dfrac{GM}{r}}} \\ & = & \dfrac{2\pi \times 2R}{\sqrt{\dfrac{gR}{2}}} \\ & = & \dfrac{4\piR\sqrt{2}}{\pi\sqrt{R}} \\ & = & 4\sqrt{2R} \\ & = & \sqrt{32R}\end{cases}$