The time period of a satellite, revolving above earth's surface at a height equal to R will be (Given g…

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

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

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

$T=\frac{2\pi r}{v} ...\left(2\right)$

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.

$\begin{array}{rcl}\mathrm{T}& =& \frac{2\pi r}{\sqrt{\frac{\mathrm{GM}}{r}}}\\ & =& \frac{2\pi \times 2R}{\sqrt{\frac{gR}{2}}}\\ & =& \frac{4\pi R\sqrt{2}}{\mathrm{\pi }\sqrt{R}}\\ & =& 4\sqrt{2\mathrm{R}}\\ & =& \sqrt{32\mathrm{R}}\end{array}$