The time period of a satellite, revolving above earth's surface at a height equal to $\mathrm{R}$ will be

(Given $g=\pi^{2} \mathrm{~m} / \mathrm{s}^{2}, \mathrm{R}=$ radius of earth)

<p>For a satellite orbiting the Earth at a height equal to Earth&#39;s radius, its distance from the center of the Earth will be 2R, where R is the radius of the Earth.</p> <p>Using the formula for the gravitational force:</p> <p>$F = G\frac{Mm}{r^2}$</p> <p>where F is the gravitational force, G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the distance from the center of the Earth.</p> <p>The centripetal force acting on the satellite is given by:</p> <p>$F_c = \frac{mv^2}{r}$</p> <p>Equating the gravitational force and the centripetal force, we get:</p> <p>$G\frac{Mm}{(2R)^2} = \frac{mv^2}{2R}$</p> <p>Solving for the orbital speed v, we get:</p> <p>$v^2 = \frac{GM}{2R}$</p> <p>The circumference of the satellite&#39;s orbit is given by:</p> <p>$C = 2\pi(2R) = 4\pi R$</p> <p>The time period T of the satellite&#39;s orbit can be calculated as the ratio of the circumference to the orbital speed:</p> <p>$T = \frac{C}{v} = \frac{4\pi R}{\sqrt{\frac{GM}{2R}}}$</p> <p>Given that the acceleration due to gravity at the Earth&#39;s surface is $g = \pi^2 \,\mathrm{m/s^2}$, we can express the gravitational constant G in terms of the Earth&#39;s radius R and mass M:</p> <p>$g = \frac{GM}{R^2} \Rightarrow GM = gR^2 = \pi^2 R^2$</p> <p>Substituting the expression for GM into the equation for the time period T, we get:</p> <p>$$T = \frac{4\pi R}{\sqrt{\frac{\pi^2 R^2}{2R}}} = \frac{4\pi R}{\sqrt{\frac{\pi^2 R}{2}}}$$</p> <p>$$T = \frac{4\pi R}{\sqrt{\pi^2}\sqrt{\frac{R}{2}}} = \frac{4\pi R}{\pi\sqrt{\frac{R}{2}}} = \frac{4R}{\sqrt{\frac{R}{2}}}$$</p> <p>Multiplying the numerator and denominator by $\sqrt{2}$, we get:</p> <p>$T = \frac{4R\sqrt{2}}{\sqrt{R}} = 4\sqrt{2R} = \sqrt{32R}$</p>