A satellite in a circular orbit is in free fall: gravity provides exactly the centripetal force needed to maintain circular motion. Setting GMm/r² = mv²/r gives orbital speed v = √(GM/r) and period T = 2π r^(3/2) / √(GM). The total mechanical energy of a circular orbit is E = −GMm/(2r): negative, indicating a bound orbit. Elliptical orbits are more general and governed by the same gravitational law with energy E = −GMm/(2a), where a is the semi-major axis.
Derive orbital speed and period for circular orbits, then extend to geostationary orbit calculations (find the orbital radius where T = 24 hours). Compare the energy of a circular orbit at different radii to understand why lower orbits move faster.
You know from Newton's law of gravitation that every mass attracts every other, and from circular motion dynamics that something moving in a circle requires a centripetal force directed inward. Orbital mechanics combines these ideas: a satellite in a circular orbit is in a state of continuous free fall where gravity provides exactly the centripetal force needed to curve the trajectory into a circle. Setting gravitational force equal to centripetal force: GMm/r² = mv²/r. Solving for v gives the orbital speed v = √(GM/r).
Two features of this result are crucial. First, the satellite's mass m cancels — orbital speed is independent of the object's mass. A feather and a boulder at the same altitude orbit at the same speed. Second, v ∝ 1/√r: satellites in lower orbits move *faster*. This seems backwards — you might expect adding energy to speed things up — but adding energy raises the orbital radius, and the speed formula says radius and speed trade off as 1/√r. The orbital period follows as T = 2πr/v = 2πr^(3/2)/√(GM), which is Kepler's third law: period grows as r^(3/2), so higher orbits have longer periods. A geostationary satellite (T = 24 hours) must orbit at a specific radius — approximately 42,000 km from Earth's center — uniquely determined by this formula.
The total mechanical energy of a circular orbit is E = KE + PE = ½mv² − GMm/r. Substituting v² = GM/r gives E = −GMm/(2r). The negative sign is significant: it means the satellite is gravitationally bound and lacks the energy to escape. The more negative E is (smaller r), the more tightly bound the orbit. To raise a satellite to a higher orbit you must add energy — making E less negative. Elliptical orbits generalize this: E = −GMm/(2a), where a is the semi-major axis. A circle is the special case where a = r; the same energy formula holds throughout.
The misconception about weightlessness deserves careful attention. The ISS orbits at about 400 km altitude, where Earth's gravity is still roughly 90% of its surface value. Astronauts are not outside gravity's reach — they are in continuous free fall toward Earth. Because the station and every object inside it fall at the same rate, there is no relative acceleration between them. The floor of the station does not push up on your feet because it is falling at the same rate you are. Weightlessness is the subjective experience of free fall — a distinction that becomes vivid once you understand that gravity is providing the centripetal acceleration keeping the station in orbit rather than sending it crashing to Earth.