How this instrument works
Orbital period, T, is the time one complete lap takes — perigee to perigee, or any reference point back to itself. This instrument reaches it from just two numbers: the orbit's semi-major axis, a, and the mass, M, of the body being circled. Because gravity supplies the only inward pull and a closed orbit has to keep re-tracing itself, the two are locked together by T = 2π√(a³ ⁄ GM); nothing about the orbiting body's own mass, its speed at any one instant, or the shape of the ellipse beyond a needs to be known first.
Mass and distance pull in opposite directions inside that square root, and swapping either one shows how much each matters. Hold the Moon's real distance, 384,400 km, fixed and swap Earth's mass for the Sun's 1.989×10³⁰ kg, and the period collapses from 27.45 days to about 68.5 minutes — proof that GM, not distance, does most of the work near a star (that point sits deep inside the Sun itself, so treat it as arithmetic, not a real orbit). Hold the mass fixed instead and double the distance to 768,800 km, and the period does not double — it stretches to roughly 2.83 times as long, about 77.6 days, because a enters as a cube under a square root, a three-halves power law rather than a straight line.
The formula assumes a clean two-body orbit: a point-mass or spherical central body, an orbiting mass small enough to ignore, and nothing else tugging. Feed it the Moon's textbook numbers and the readout lands a few hours later than the sidereal month astronomers actually clock. Most of that slack traces to the Sun's steady tug reshaping the Moon's path lap by lap — a wobble early skywatchers named evection and logged with the naked eye long before Newton could explain it — with a smaller share coming from treating the Moon as weightless when it is really pulling back on Earth too. A related mistake is entering tonight's Earth-Moon distance rather than the orbit's long-run average: the real separation swings between roughly 363,300 km at perigee and 405,500 km at apogee across each lap, and only the semi-major axis, the mean of the two, belongs in this formula. Mission designers plotting a lunar parking orbit, and anyone checking a textbook's Kepler claim against real astronomical data, both lean on exactly this arithmetic.
- Enter the orbit's semi-major axis in the Orbital semi-major axis field, in kilometres — the time-averaged distance from the central body's centre, not today's exact separation.
- Enter the mass of the object being orbited into the Central body mass, kg field; Earth's value, 5.972×10²⁴ kg, loads by default and can be replaced with the Sun's, another planet's, or any other body's mass.
- Leave the orbiting body's own mass out entirely — it cancels out of the physics and has no field to enter it into.
- Read the answer in the Orbital period field, then flip its unit toggle between days and hours to match the scale of orbit you are checking.
Worked example — the Moon's orbit, checked against reality
Set Orbital semi-major axis to the Moon's real value, 384,400 km (384,400,000 m), and Central body mass, kg to Earth's 5.972×10²⁴ kg. Cubing the axis gives a³ = 5.6800×10²⁵ m³; multiplying G by the mass gives GM = 3.9857×10¹⁴ m³ ⁄ s²; dividing the two leaves 1.4251×10¹¹ s². The square root of that is 377,503 s, and 2π times it lands on T = 2,371,877.064 seconds — the exact figure this instrument returns for those two inputs.
Divide by 86,400 seconds a day and the answer reads 27.452 days. Astronomers clock the Moon's actual sidereal month at 27.322 days — a gap of about three hours and eight minutes, small enough to confirm the formula is doing real physics, large enough to matter for anyone planning a precise rendezvous, and explained by the Sun's pull on the Moon's path plus the Moon's own mass, both left out of this clean two-body form.
Questions
Why does the semi-major axis matter more than today's Earth-Moon distance?
Because the real orbit is an ellipse, not a circle, so the actual separation changes throughout each lap — roughly 363,300 km at perigee out to about 405,500 km at apogee for the Moon. The semi-major axis, 384,400 km, is the long-run average of those two extremes, and that average is the single number Kepler's third law is built around. Plugging in whatever distance a given night happens to show scatters the answer depending on where the Moon sits in its orbit that day.
Why doesn't this match the Moon's real 27.3-day month exactly?
It comes within a few hours, because this is the idealized two-body answer, not the full picture. Two things are missing: the Sun's steady pull steadily reshapes the Moon's path, a wobble skywatchers were logging as far back as Ptolemy, and the calculation treats the Moon as weightless when it is really about one-eightieth of Earth's mass and tugs back just enough to matter. Both effects trim the real period down from what the bare formula predicts.
What happens if Earth's mass is swapped for the Sun's at the same distance?
The period collapses from 27.45 days to about 68.5 minutes (4,109.93 seconds), holding the semi-major axis at 384,400 km throughout. That distance sits deep inside the Sun, 696,000 km in radius, so it is not a real orbit — but the arithmetic still shows how strongly GM, not distance, controls the pace: the Sun outweighs Earth by a factor of about 333,000, and period falls with the inverse square root of that.
Does doubling the orbital distance double the period?
No. Moving from the Moon's actual 384,400 km out to 768,800 km, with Earth's mass unchanged, stretches the period from 27.45 days to about 77.6 days — roughly 2.83 times longer, not twice. The semi-major axis enters the formula cubed and then under a square root, so period scales with distance to the three-halves power, a curve that steepens the further out you go.
Can I use this for a spacecraft on its way to the Moon, not the Moon itself?
Only indirectly. This returns the period of a stable, closed orbit at a given semi-major axis — useful for a parking orbit or a ballistic capture around the Moon or Earth, not for a one-way translunar coast, which is a single elliptical arc interrupted partway by a lunar encounter rather than completed as a lap. Mission designers use this formula for the orbit segments and separate n-body software for the transfer itself.
Why enter the mass of the body being orbited, not the orbiting body?
Because in the regime this formula covers, one mass is negligible next to the other, and only the heavier one's gravity sets the pace. Enter Earth's mass to time the Moon's lap, the Sun's to time Earth's, or a planet's to time one of its own moons. Swapping in the lighter body's mass by mistake — a common slip since both are just called mass on the label — returns a period off by many orders of magnitude.