Orbits & Kepler's Laws
A satellite in orbit isn't defying gravity — it's in permanent free fall, forever missing the ground because it's moving sideways fast enough to match the curve of the planet beneath it. That single idea, that gravity supplies exactly the centripetal force needed, is enough to derive how orbital period depends on orbital radius, and precisely how high a satellite must sit to stay parked over one spot on Earth forever.
Gravity as the centripetal force
A satellite or planet moving in a stable circular orbit is continuously accelerating toward the centre of its orbit — not speeding up or slowing down, but constantly changing direction, itself a form of acceleration. This is centripetal acceleration, and for an orbiting body it's provided entirely by the gravitational attraction between it and whatever it's orbiting. There's no separate "orbital force": gravity is the centripetal force here.
This gives a single equation describing every stable circular orbit: the gravitational force between the two bodies equals the centripetal force required to keep the orbiting body moving in its circle.
Kepler's third law
Orbital speed can also be written in terms of the orbital period $T$ — the time to complete one full orbit — since speed is simply the circumference of the orbit divided by the time taken to travel it: $v = \dfrac{2\pi r}{T}$. Combining this with the expression for orbital speed derived above leads directly to Kepler's third law, one of the most exam-useful results in this entire topic.
Physically, orbits further out don't just take longer to complete — they take proportionally much longer, since $T$ grows with $r^{3/2}$, faster than $r$ itself. Double the orbital radius and the period doesn't just double; it increases by a factor of $2^{3/2}\approx2.83$.
The constant of proportionality, $\dfrac{4\pi^2}{GM}$, depends only on $G$ (a universal constant) and $M$, the mass of whatever is being orbited — not on the mass or speed of the orbiting body. Every satellite orbiting Earth, from a tiny CubeSat to the ISS, obeys the exact same $T^2 \propto r^3$ relationship with the exact same constant — and every planet orbiting the Sun obeys the same relationship with a different constant, fixed by the Sun's mass instead of Earth's.
Geostationary orbits
A geostationary orbit is one where a satellite's orbital period exactly matches Earth's rotation period — 24 hours, or $T = 86\,400\,\text{s}$. Because the satellite takes exactly as long to complete one orbit as Earth takes to complete one rotation, it stays above the same fixed point on Earth's surface at all times, appearing motionless in the sky as seen from the ground.
Worked example: the altitude of a geostationary orbit
Rearranging Kepler's third law for $r$: $r^3 = \dfrac{GMT^2}{4\pi^2}$, so $r = \left(\dfrac{GMT^2}{4\pi^2}\right)^{1/3}$. Using Earth's mass $M = 5.97\times10^{24}\,\text{kg}$ and $T = 86\,400\,\text{s}$:
That's the orbital radius, measured from the centre of the Earth. Subtracting Earth's radius (about 6,370 km) gives the altitude above the surface:
So a geostationary satellite must sit at an altitude of roughly 35,800 km — a single value fixed entirely by Earth's mass and its 24-hour rotation period, with no freedom to choose a different altitude and remain geostationary.
A satellite in a stable orbit must have Earth's centre at the centre of its circular path, since gravity always pulls directly toward that centre. If such an orbit were tilted relative to the equator, its circular path would cross the equatorial plane twice per orbit, so the satellite would drift alternately north and south, never staying above one fixed point. Only an orbit lying exactly in the plane of Earth's equator — centred on Earth's centre, tilted at zero degrees — keeps the satellite above a single fixed line of longitude at every instant, matching Earth's rotation beneath it. This is why every true geostationary satellite sits directly above the equator, never above London or New York.
Because they appear fixed in the sky from any one location, geostationary satellites are heavily used for satellite television, telecommunications relays and weather monitoring, where a ground-based dish needs to point at a single unmoving spot rather than track a moving satellite.
Worked example: comparing two orbits with T² ∝ r³
Kepler's third law also gives a fast way to compare two orbits around the same central body without recalculating $G$ and $M$, since $\dfrac{4\pi^2}{GM}$ is identical for both:
Suppose a low Earth orbit satellite orbits at $r_1 = 6.77\times10^6\,\text{m}$ (altitude ~400 km) with period $T_1$, and the Moon orbits Earth at $r_2 = 3.84\times10^8\,\text{m}$ with a known period $T_2 = 27.3$ days. Finding the low Earth orbit's period directly from $T^2 = 4\pi^2 r^3/GM$:
Checking this against the Moon confirms the same constant applies to both: $\dfrac{T_1^2}{r_1^3} = \dfrac{(5540)^2}{(6.77\times10^6)^3} \approx 9.93\times10^{-14}\,\text{s}^2\text{m}^{-3}$, and $\dfrac{T_2^2}{r_2^3}$ (with $T_2$ converted to seconds) gives the same value, exactly as Kepler's third law predicts. This ratio technique — setting $T_1^2/r_1^3 = T_2^2/r_2^3$ directly, without calculating $M$ — is a common, efficient exam technique whenever two orbits share the same central body.
Key takeaways
- For a stable circular orbit, gravitational force supplies the centripetal force: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$, giving orbital speed $v = \sqrt{\dfrac{GM}{r}}$, independent of the orbiting body's own mass.
- Kepler's third law: $T^2 = \dfrac{4\pi^2 r^3}{GM}$, i.e. $T^2 \propto r^3$, derived by substituting $v = 2\pi r/T$ into the orbital speed equation.
- The constant $\dfrac{4\pi^2}{GM}$ depends only on the central mass $M$, so it's identical for every body orbiting that same mass, regardless of the orbiting body's own mass.
- A geostationary orbit has $T = 24\,\text{h} = 86\,400\,\text{s}$, giving an orbital radius of about 42,000 km from Earth's centre — an altitude of roughly 35,800 km above the surface.
- Geostationary orbits must be equatorial: only an orbit centred on Earth's centre and lying in the equatorial plane keeps a satellite above one fixed point on the ground at all times.
- For two orbits around the same central body, $\dfrac{T_1^2}{r_1^3} = \dfrac{T_2^2}{r_2^3}$ lets you compare periods and radii directly without calculating $M$.