Geostationary and polar satellites

Three conditions for geostationary. A satellite only appears fixed over one spot on the ground if all three hold at once: it moves west to east, the same sense as Earth's spin; its orbital plane coincides with the equatorial plane; and its period matches Earth's rotation period, $T \approx 24\ \text{h}$ (strictly one sidereal day, $23\ \text{h}\ 56\ \text{min}$). Miss any one — say, tilt the plane — and the satellite still completes one orbit per day (it is geosynchronous) but traces a figure-of-eight in the sky instead of sitting still.

Fixing the radius. Because circular orbital motion obeys the same period–radius law derived earlier for any satellite, demanding $T = 24\ \text{h}$ pins down one unique radius, the parking orbit:

where: $T$ is the orbital period (s), $r$ is the orbital radius measured from Earth's centre (m), $G$ is the universal gravitational constant, and $M$ is Earth's mass, with $GM \approx 4.0\times10^{14}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}$ for Earth. Solving with $T = 24\ \text{h}$ gives $r \approx 4.2\times10^7\ \text{m} \approx 42{,}000\ \text{km}$, and subtracting Earth's radius $R \approx 6.4\times10^6\ \text{m}$ gives a height above the surface of $h = r - R \approx 3.6\times10^7\ \text{m} \approx 36{,}000\ \text{km}$. This single radius is why every geostationary satellite — regardless of who launched it — sits in the same narrow equatorial belt, just spaced apart in longitude.

Why geostationary orbits are useful. A ground antenna aimed at a geostationary satellite never has to move, because the satellite is always in the same place in the sky. That makes it ideal for continuous communication links: television broadcast, telephone relay, and weather satellites that stare at the same hemisphere around the clock (INSAT is a familiar example). The tradeoff is coverage — a satellite fixed over the Indian Ocean cannot see the poles or the far side of the Earth, and its great distance ($\approx 36{,}000\ \text{km}$ up) introduces a noticeable signal delay.

Polar satellites: the opposite tradeoff. A polar satellite orbits in a plane passing close to both poles, at a much lower altitude — typically a few hundred kilometres to about $1000\ \text{km}$. Because $r$ is small, $T^2 \propto r^3$ forces the period to be small too: roughly $90$ to $120$ minutes. Crucially, the orbital plane stays (nearly) fixed in space while the Earth spins beneath it once every 24 hours, so the ground track of each successive pass is shifted westward — the satellite scans a new strip of the globe every orbit. After enough passes (about 14 a day) it has covered essentially the entire planet, poles included, which a geostationary satellite fixed over the equator can never do. This makes polar orbits the natural choice for remote sensing, weather imaging, environmental monitoring, and reconnaissance, where the low altitude also gives sharper resolution than a distant geostationary platform ever could.

where: this is Kepler's third law for circular orbits (derived from the same $T=2\pi\sqrt{r^3/GM}$ relation above by squaring both sides); it is the reason the two orbit families sit at such different radii and periods — a factor of roughly $40$ in radius between polar and geostationary orbits produces a factor of over $300$ in period.

A ground-station engineer needs to confirm the parking-orbit radius before scheduling a launch. Take $GM = 4.0\times10^{14}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}$, $T = 24\ \text{h} = 86400\ \text{s}$, and Earth's radius $R = 6.4\times10^6\ \text{m}$. Find (a) the orbital radius $r$, and (b) the height $h$ above the surface.

Step 1. Square the period: $T^2 = (86400)^2 = 7.4666\times10^9\ \text{s}^2$.

Step 2. Compute $GMT^2 = (4.0\times10^{14})(7.4666\times10^9) = 2.9866\times10^{24}$, and divide by $4\pi^2 \approx 39.48$ to get $r^3 = \dfrac{2.9866\times10^{24}}{39.48} \approx 7.5648\times10^{22}\ \text{m}^3$.

Step 3. Take the cube root: $r \approx 4.23\times10^{7}\ \text{m} \approx 42{,}300\ \text{km}$, matching the known geostationary radius of about $42{,}200\ \text{km}$.

Step 4. Subtract Earth's radius for the height above the surface: $h = r - R = 4.23\times10^7 - 0.64\times10^7 \approx 3.59\times10^7\ \text{m} \approx 35{,}900\ \text{km}$. Answer: $r \approx 42{,}300\ \text{km}$ from Earth's centre, giving a height of about $35{,}900\ \text{km}$ above the surface — consistent with the standard textbook figures of $\approx 42{,}200\ \text{km}$ and $\approx 36{,}000\ \text{km}$.

Competitive stretch

If the engineer instead had a polar satellite with $T = 100\ \text{min} = 6000\ \text{s}$, the same formula gives $r^3 = \dfrac{(4.0\times10^{14})(6000)^2}{4\pi^2} \approx 3.65\times10^{20}\ \text{m}^3$, so $r \approx 7.15\times10^6\ \text{m} \approx 7150\ \text{km}$ — barely $700\ \text{km}$ above Earth's surface, over five hundred times closer than the geostationary radius, exactly the huge gap in $r$ that $T^2\propto r^3$ predicts from the huge gap in $T$.

• Assuming any satellite with $T = 24\ \text{h}$ is automatically geostationary — a satellite with the right period but an inclined or polar plane is geosynchronous only; it still drifts north-south in the sky instead of sitting still.
• Confusing orbital radius $r$ (measured from Earth's centre) with height $h$ above the surface — always subtract $R$: $h = r - R$, never use $r$ where the question asks for altitude.
• Thinking polar satellites are "slower" because they cover more ground over time — they are in fact much faster in orbit and have far shorter periods than geostationary satellites, because their much smaller $r$ demands a much smaller $T$ under $T^2\propto r^3$.
• Picturing a geostationary satellite as motionless in space — it is moving at roughly $3.1\ \text{km/s}$ around Earth; it only appears fixed relative to a point on the rotating ground.

Ask the tutor: Want to work through what parking-orbit radius a different period would require, or see why the same $T^2\propto r^3$ law also governs planetary orbits around the Sun — ask the AI tutor about geostationary and polar satellites.