Orbit as continuous free fall

Observation

Drop a stone and it falls straight down. Throw it horizontally and it still falls, but it travels forward as it does, landing some distance away; throw it harder and it lands farther still. Newton imagined a cannon on a mountain so high it cleared the air, firing horizontally faster and faster. Each shot falls toward the Earth, but each also flies farther before it lands, because the ground curves away beneath it. At some speed the ground curves away exactly as fast as the ball falls toward it, and the ball never lands. It has entered orbit. The Moon, Newton saw, is doing nothing more exotic than this: it is perpetually falling toward the Earth and perpetually missing.

Variables

The gravitational parameter GM combines the gravitational constant with the mass of the body being orbited; for a satellite around the Earth it is a fixed number. The orbital radius r is measured from the center of the Earth, not the surface, so it is the Earth's radius plus the altitude. The orbital speed v is what the entry solves for. The period T is the time for one full revolution. This entry consumes the inverse-square law of gravitation and the unification of the falling apple with the orbiting Moon from the entry on universal gravitation; its own fresh burden is the general relationship between speed, altitude, and period, and the meaning of weightlessness.

Model

A body in a circular orbit is accelerating, even at constant speed, because its direction is always changing. Circular motion requires an inward acceleration of v squared over r. The only force available to supply it is gravity, which pulls with an acceleration of GM over r squared. Setting the supply equal to the demand is the entire model:

GM / r squared = v squared / r

Everything about circular orbits follows from balancing those two expressions.

Derivation

Solve the balance for speed. Multiplying both sides by r gives v squared equal to GM over r, so:

v = sqrt(GM / r)

The orbital speed is fixed entirely by the radius. A higher orbit is a slower orbit, and this is derived, not asserted: r appears in the denominator, so larger r means smaller v. The period follows from the circumference divided by the speed, T equal to 2 pi r over v, which after substituting for v gives T squared proportional to r cubed. That proportionality is Kepler's third law, recovered here for the circular case from Newtonian mechanics rather than taken as an empirical rule.

Weightlessness now has a clean explanation. An astronaut in orbit is not beyond gravity; gravity at a few hundred kilometers up is nearly as strong as at the surface. The astronaut floats because the station and everyone in it are all falling together at the same rate, so there is no contact force between them. Weightlessness is shared free fall, not the absence of gravity.

Worked example: low orbit against high orbit

A satellite in low Earth orbit sits a few hundred kilometers up, at a radius of roughly 6,800 kilometers from the center. Its speed works out to about 7.7 kilometers per second and its period to about 92 minutes. A geostationary satellite, which must complete one orbit per day to hover over a fixed point, sits far higher, at a radius near 42,000 kilometers. Its speed is only about 3.1 kilometers per second and its period is a full day. The higher orbit is slower and longer, exactly as v equal to sqrt(GM over r) demands. The numbers are not memorized; they are the equation evaluated at two radii.

Limits and boundary conditions

This treatment covers circular orbits only. Real orbits are generally ellipses, and there the speed varies along the path: the body moves fastest at closest approach and slowest when farthest away. The reason connects to the live prerequisite on angular momentum: gravity always points at the focus, so it exerts no torque about that point, so angular momentum is conserved and the body trades speed for distance as it moves. That is the fact the angular-momentum entry points here to complete. Escape velocity is the boundary case where the speed is large enough that the fall never repeats and the body leaves for good, equal to sqrt(2) times the circular speed at that radius. The account also idealizes a two-body system and ignores atmospheric drag, which in low orbit slowly removes energy and eventually deorbits a satellite.

Common mistakes

Three misconceptions are worth naming. First, that there is no gravity in space: gravity is what holds the orbit: without it the satellite would fly off in a straight line. Second, that a satellite needs its engines running to stay up: once in orbit it coasts, engines off, falling freely. Third, that going faster always means going higher: adding speed in low orbit does eventually raise the orbit, but the settled speed at the higher altitude is lower than the speed at the lower one, which catches many people out.

Test yourself

Take GM for Earth as 3.99 times ten to the fourteenth in SI units and find the altitude and speed for a satellite with a 90-minute period. Work from T equal to 2 pi r over v together with v equal to sqrt(GM over r), which combine to let you solve for r from T alone, then back out the altitude by subtracting the Earth's radius of about 6,370 kilometers and the speed from the orbital-speed equation. Then answer the harder question: if that satellite fires its engine briefly forward to climb into a higher orbit, is it moving faster or slower once it settles there? Success is the correct altitude and speed, near 280 kilometers and 7.7 kilometers per second, plus the counterintuitive answer that the higher orbit is the slower one.