μ = GM is far more precisely known than G or M separately, so every orbit formula here uses it directly.
T = 2π√(a³/μ): the period depends on the semi-major axis alone, the orbit’s size regardless of shape.
v² = μ(2/r − 1/a): speed at any point. Circular is v² = μ/r; escape is the limit as apoapsis goes to infinity, v² = 2μ/r.
Solved from Kepler’s equation M = E − e·sin E, which is what makes the point speed up at periapsis: equal areas in equal time.
They are falling, continuously, but also moving sideways fast enough that the curve of their fall matches the curve of the planet falling away beneath them. Orbit is free-fall with enough horizontal speed to keep missing the ground. Higher orbits move slower because gravity is weaker further out, so less speed is needed to keep circling.
Both come from vis-viva: circular v² = μ/r, escape v² = 2μ/r. The ratio is exactly √2, at any altitude, for any body.
The altitude falls out of the period: match Earth’s rotation and the orbit radius is about 42,164 km, 35,786 km up. The period to match is the sidereal day, one rotation relative to the stars, about four minutes shorter than the solar day; radius goes as the two-thirds power of period, so using 86,400 s puts the answer roughly 77 km wrong. It is a consequence, not a design choice.
Two burns on opposite sides of an ellipse is the minimum-energy route between circular orbits: LEO to geostationary costs about 3.9 km/s over five and a quarter hours. Past a radius ratio of roughly 11.94 a three-burn bi-elliptic route out to a distant apoapsis is cheaper, because a craft crawling at apoapsis needs very little push to change orbit. The price is time: years where Hohmann takes days.
Kepler’s first two laws: orbits are ellipses with the central body at one focus, and a line from the body to the orbiter sweeps equal areas in equal time, so a satellite is fastest at perigee and slowest at apogee. A circular orbit is the special case where the two coincide. Altitude is measured from the surface; the semi-major axis from the centre, and for an ellipse is the average of periapsis and apoapsis distance.
Gravity is weaker further out, so less centripetal force is available — and less force means less speed is needed to keep circling rather than falling in. It's the same reason outer planets crawl around the Sun far more slowly than Mercury does.
It's the one altitude where the orbital period exactly matches Earth's rotation — the satellite traces its orbit at the same rate the ground turns beneath it, so from the surface it appears to hang motionless in the sky.
Both come from the same vis-viva equation — circular orbit is v² = μ/r, escape is the limit as apoapsis goes to infinity, v² = 2μ/r. The ratio between them is exactly √2, at any altitude, for any body.
Yes — the orbiting point's position is solved from the true time-domain equations of orbital motion (via Kepler's equation), not just an ellipse traced at a fixed angular rate, so it genuinely speeds up approaching periapsis and slows down approaching apoapsis, sweeping out equal areas in equal time.
For a circular orbit, v = √(μ / r), where μ = GM (about 398,600 km³/s² for Earth) and r is the distance from the planet’s centre, its radius plus the altitude. For the ISS at about 400 km, r = 6,771 km and v = 7.67 km/s.
T = 2π √(r³ / μ), with r measured from the planet’s centre. At the ISS’s 6,771 km that is about 92.4 minutes. It is Kepler’s third law: the period squared grows with the cube of the orbital radius.
Rearrange the period formula: r = ³√(μT² / 4π²). For a period of one sidereal day (86,164 s) that gives 42,164 km from Earth’s centre, which is the geostationary altitude of 35,786 km above the surface.
Releases in which this page changed, newest last. Derived from the archived copy of every release, not from notes written afterwards — so it reflects what actually shipped. Site-wide passes are left out; they are in the full changelog.