Time Period Of Satellite

The period of a circular orbit follows from $T=\frac{2\pi r}{v_o}$ with orbital speed $v_o=\sqrt{\frac{GM}{r}}$, which combines into $T=2\pi \sqrt{\frac{r^3}{GM}}$. For a near-surface orbit $r\approx R$ and $GM=g{R}^2$, so $T=2\pi \sqrt{\frac{R}{g}}$. Substituting, $T=2\pi \sqrt{\frac{6.4\times 10^6}{9.8}}=2\pi \sqrt{6.53\times 10^5}=2\pi \times 808\approx 5080$ s, which is about 85 minutes. The option of 24 hours is the geostationary period at a far larger radius, not a low orbit. The option of 60 minutes is too short and would require an unrealistically small radius. The option of 120 minutes overestimates the period for such a low altitude. This matches the well-known NCERT result that low-Earth-orbit satellites circle the planet in roughly an hour and a half. Interestingly, the period of a grazing orbit depends only on the planet's mean density and not on its size, because substituting $M=\frac{4}{3}\pi R^3\rho$ into the period expression eliminates the radius entirely, leaving $T=\sqrt{3\pi /(G\rho )}$. The general form $T^2\propto r^3$ is simply Kepler's third law applied to artificial satellites, with the same proportionality governing the Moon and all natural orbiting bodies. A magnitude check confirms the period depends only on $R$ and $g$ for a grazing orbit, giving a value near 85 minutes.