Time Period Of Satellite

Question

A small research satellite is set into a circular orbit very close to the Earth's surface so its orbital radius is essentially the Earth's radius. Taking g = 9.8 m/s^2 and R = 6.4 × 10^6 m, roughly what is its orbital period?

RankGuru · Accepted Answer

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 $GM = gR^2$, so $T = 2\pi\sqrt{\frac{R}{g}}$. Substituting, $T = 2\pi\sqrt{\frac{6.4	imes10^6}{9.8}} = 2\pi\sqrt{6.53	imes10^5} = 2\pi	imes808 \approx 5080$ 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 $M = \frac{4}{3}\pi R^3 ho$ into the period expression eliminates the radius entirely, leaving $T = \sqrt{3\pi/(Gho)}$. 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.

Time Period Of Satellite

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