What is the orbital period?

How it works

Kepler's third law, in Newton's form with both masses:

• Period: T = 2π × √(a³ ÷ μ).
• Semi-major axis: a = ∛(μ × T² ÷ 4π²).

Here a is the semi-major axis from the centre of the body orbited (the radius of a circular orbit) and μ = GM, the body's gravitational parameter. With an orbiting mass m, μ = GM + G × m; leave it empty when the orbiting body is tiny next to the central one. G = 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻².

Size of orbit as a semi-major axis or a height h above the mean surface, with a = R + h for a circular orbit and R the body's mean radius (typed for your own body). For a period, you type that size. For a semi-major axis found from a period, the answer is a, or the height h = a − R of a circular orbit of that radius; if a is less than R, the orbit would be inside the body and there is no answer.

The page also gives the period in s, min, h, days (86,400 s) and years (Julian years of 365.25 days), a in km and astronomical units, the speed of a circular orbit of radius a, √(μ ÷ a), in km/s, and, for a height found from a period, h in km.

Bodies (GM in km³/s², mean radius in km): Earth 398,600.435507 and 6,371.0084; Sun 132,712,440,041.279419 and 695,700; Moon 4,902.800118 and 1,737.4; Mercury 22,031.868551 and 2,439.4; Venus 324,858.592 and 6,051.8; Mars 42,828.37362 and 3,389.5; Jupiter 126,686,531.9 and 69,911; Saturn 37,931,206.23 and 58,232; Uranus 5,793,951.3 and 25,362; Neptune 6,835,099.97 and 24,622; Pluto 869.3 and 1,188.3. These GM values are taken as exact; 1 km³/s² = 10⁹ m³/s². For your own body, GM = G × M.

Units (exact): 1 au = 149,597,870,700 m; 1 km = 1,000 m; 1 mi = 1,609.344 m; 1 ft = 0.3048 m; 1 min = 60 s; 1 h = 3,600 s; 1 day = 86,400 s; 1 year = 365.25 days = 31,557,600 s; 1 t = 1,000 kg; 1 lb = 0.45359237 kg.

Rules

• Semi-major axis and period are numbers from 10⁻⁶ to 10¹² in the unit chosen beside them; height from 0 to 10¹² m; your body's mass from 10⁻⁶ kg to 10⁴⁰ kg and radius from 0.001 m to 10¹⁵ m; the orbiting mass from 0 to 10⁴⁰ kg. Limits include their ends; a height, radius or mass typed in another unit is converted first.
• A height found from a period that is below 0 (a less than R) has no answer.
• A result that is exactly 0 shows as 0. Any other result too large or too small to show as a number gives no answer; within these limits it does not happen.

Output format

The answer line reads T = x u for a period, with u the largest of years, days, h, min and s in which T, rounded to 6 significant figures, is at least 1 (s below 1 s). For a semi-major axis it reads a = x u, or h = x u for a height above the surface, with u the axis unit chosen (km, m, mi or au); h = a − R is worked out as 64-bit floats. Every number shows to 6 significant figures, rounded half up, with trailing zeros dropped and thousands commas, and below 10⁻⁶ or from 10¹⁵ up as a × 10ⁿ.

Typed values are read back as the exact decimals you typed, and every unit size and GM value is exact, so a³ ÷ μ is an exact fraction: its square root is rounded once, then multiplied by 2π. A semi-major axis uses the cube root of the exact μT² (rounded to a 64-bit float first) divided by ∛(4π²). A typed period gives the other period units as exact fractions rounded once.

Worked examples by hand

The ISS. 400 km above the Earth: a = 6,371.0084 + 400 = 6,771.0084 km, so T = 2π √(6,771,008.4³ ÷ 3.98600435507 × 10¹⁴) = 5,544.87 s = 92.4144 min, shown as T = 1.54024 h. A circular orbit there moves at √(398,600.435507 ÷ 6,771.0084) = 7.67259 km/s (OpenStax Example 13.9: 5.55 × 10³ s and 7.67 × 10³ m/s with rounded values).

The Moon. a = 384,400 km around the Earth, with the Moon's 7.346 × 10²² kg added: μ = 3.98600435507 × 10¹⁴ + 6.6743 × 10⁻¹¹ × 7.346 × 10²² = 4.03503376287 × 10¹⁴ m³/s², and T = 2π √((3.844 × 10⁸)³ ÷ μ) = 27.2846 days.

The Earth's year. a = 1 au around the Sun: T = 2π √(149,597,870,700³ ÷ 1.32712440041279419 × 10²⁰) = 31,558,196.0 s = 365.257 days, shown as T = 1.00002 years; the sidereal year is 365.25636 days, a little less because the Earth's own mass adds to the Sun's.

A geostationary orbit. One turn per sidereal day, T = 86,164.09054 s: a = ∛(3.98600435507 × 10¹⁴ × 86,164.09054² ÷ 4π²) = 42,164.2 km, and its height above the mean surface is 42,164.169 − 6,371.0084 = 35,793.2 km.

Halley's comet. OpenStax Example 13.13: T = 75.3 years = 2,376,287,280 s around the Sun: a = ∛(1.32712440041279419 × 10²⁰ × T² ÷ 4π²) = 2.66757 × 10¹² m = 17.8316 au (OpenStax: 17.8 AU).

OpenStax's Earth and Moon. M = 5.96 × 10²⁴ kg and a = 384,000 km: T = 2π √((3.84 × 10⁸)³ ÷ (6.6743 × 10⁻¹¹ × 5.96 × 10²⁴)) = 2,370,559 s, against the 2.36 × 10⁶ s of Example 13.10.