The time period of a satellite, revolving above earth's surface at a height equal to R will be (Given g = π 2 m s - 2 , R = radius of earth)

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The velocity of the satellite revolving round the Earth at a height $r$ from the centre of the Earth is given by
$v = \sqrt{(\frac{G M}{r})} . . . (1)$
The formula to calculate the time period in the circular orbit is given by
$T = \frac{2 π r}{v} . . . (2)$
In this case: $r = R + R = 2 R$
Substitute the expression for the velocity from equation (1) into equation (2) and simplify to obtain the required time period.
$\begin{array}{rcl} T & = & \frac{2 π r}{\sqrt{\frac{GM}{r}}} \\ = & \frac{2 π \times 2 R}{\sqrt{\frac{g R}{2}}} \\ = & \frac{4 π R \sqrt{2}}{π \sqrt{R}} \\ = & 4 \sqrt{2 R} \\ = & \sqrt{32 R} \end{array}$

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