How this instrument works
An orbital period is the time a body takes to complete one full loop around another. Kepler's third law says that time is fixed entirely by the orbital radius and the mass of the body being orbited: T = 2π√(r³ ⁄ GM). Set the mass and the radius and the period falls out — no need to know the orbiting object's own mass, because a satellite's mass cancels out of the physics the same way a feather and a hammer fall at the same rate in a vacuum.
The formula comes straight from equating two expressions for the same force. Gravity supplies GMm/r² of pull; a circular path demands mv²/r of centripetal force to keep the orbiter turning instead of flying off straight. Set them equal, substitute v = 2πr/T for the speed of a circular lap, and solve for T — the r³ under the square root is what falls out of that algebra, not a coincidence. It is why period grows with radius to the 3/2 power rather than in step with it: double the radius and the period does not double, it multiplies by roughly 2.83.
The formula assumes a circular orbit around a spherically symmetric body, with the orbiting object's own mass small enough to ignore next to M — true for a satellite around a planet, false for two comparable stars orbiting their common center of mass. For a real ellipse, r becomes the semi-major axis rather than a constant radius. And the formula is silent on drag, oblateness, or the pull of a third body — real low-Earth satellites decay because thin atmosphere still reaches them; this instrument gives the clean two-body answer those effects perturb.
- Enter the mass of the body being orbited in "Central body mass", in kilograms — Earth's mass, 5.9722×10²⁴ kg, is the default.
- Enter "Orbital radius" as the distance from the orbited body's center, not altitude above its surface, in kilometres or metres.
- For a surface-launched satellite, add the central body's own radius to its altitude first — 400 km up around Earth means r = 6,371 + 400 = 6,771 km.
- Read "Orbital period" immediately; switch its unit menu between seconds, minutes, hours, and days to match the timescale you're thinking in.
Worked example — the ISS at 400 km altitude
The International Space Station flies roughly 400 km above Earth's surface, so its orbital radius is Earth's mean radius, 6,371 km, plus that altitude: r = 6,771 km, entered as 6,771,000 m. With Earth's mass M = 5.9722×10²⁴ kg and G = 6.6743×10⁻¹¹ m³ kg⁻¹ s⁻² held fixed, T = 2π√(r³ ⁄ GM) returns 5,544.84047021 seconds — 92.4140 minutes, or 1 hour 32 minutes and about 25 seconds for one full lap of the planet.
NASA's tracking data puts the station's real period at 5,553.6 seconds, 92.6 minutes — within nine seconds of this idealized circular-orbit figure, the small gap coming from the ISS's orbit being very slightly elliptical and gradually decaying rather than a perfect, fixed circle. Stretch the same radius out to 42,164 km, geostationary altitude, and the period lengthens to roughly 23 hours 56 minutes, because T scales with r to the 3/2 power, not in a straight line with it.
Questions
Why does the period grow with the cube of the radius instead of scaling directly with it?
Because the algebra behind Kepler's third law puts r³ under a square root, not r itself. Equating gravity (∝ 1/r²) to the centripetal force needed for a circular lap (∝ 1/r once orbital speed is substituted) and solving for time leaves T ∝ r^1.5. Practically, doubling the radius multiplies the period by about 2.83, not 2 — higher orbits are disproportionately slower.
Does "orbital radius" mean altitude above the surface?
No — it is the distance from the orbited body's center, not its surface. A satellite 400 km above Earth needs r equal to Earth's mean radius (6,371 km) plus that altitude, or 6,771 km. Entering bare altitude instead of center-to-center distance is the single most common mistake with this formula, and it noticeably shortens the computed period.
Why is geostationary orbit fixed at 42,164 km and nowhere else?
Because that is the one radius where Kepler's third law returns a period of 23 hours 56 minutes — one sidereal day, the time Earth itself takes to spin once. A satellite there completes exactly one lap per Earth rotation, so it hangs above the same point on the ground; any other radius drifts relative to the surface below it.
Can this formula handle an elliptical orbit, not just a circular one?
Yes, with one substitution: replace r with the orbit's semi-major axis, a, the average of its closest and farthest distances from the central body. Kepler's third law was originally stated for ellipses this way; the circular case used here is simply the special case where the near and far distances are equal.
Why does the satellite's own mass never appear in the formula?
Because it cancels. Gravity's pull on the orbiting body is proportional to its mass, and so is the centripetal force needed to keep it turning — set them equal and that mass divides out on both sides, the same reason a heavy and a light object fall together in a vacuum. Only the central body's mass, M, survives in the period.
How precisely is G known, and does that limit the answer?
The Newtonian constant of gravitation, G, is the least precisely measured fundamental constant, known to only about 1 part in 10,000 by CODATA's current recommendation. For planetary work this rarely matters because G and a planet's mass are usually reported together as the product GM, measured far more precisely by tracking real spacecraft than G or M could be measured apart.