Documentation / @interstellar-tools/equations / keplerPeriod

Function: keplerPeriod()

function keplerPeriod(a: number, mu: number): number;

Defined in: orbits/kepler-period.ts:59

Kepler's 3rd law - orbital period ($T$) from semi-major axis ($a$) and gravitational parameter ($\mu$).

Definition

$$T=2\pi \sqrt{\frac{a^3}{\mu }}$$

Units

• Inputs: ($a$) in meters (m), ($\mu$) in m³/s².
• Output: ($T$) in seconds (s).

INFO

• Valid for Keplerian (two-body) bound orbits with ($a>0$) (ellipses). Not defined for parabolic/hyperbolic trajectories.
• ($\mu$) is the standard gravitational parameter of the central body (e.g., Earth ($\approx 3.986004418\times {10}^{14}{\text{m}}^3/{\text{s}}^2$)).

In practiceGeostationary orbit

23.93 h

Orbital period 42 164 km from Earth's centre

That's one sidereal day, the time Earth takes to turn once relative to the stars, so a satellite there stays over the same spot on the equator.

keplerPeriod(42164e3, 3.986004418e14); // ≈ 86 164 s

Parameters

Parameter	Type	Description
a	number	Semi-major axis (m), must be finite and > 0.
mu	number	Gravitational parameter (m³/s²), must be finite and > 0.

Returns

number

Orbital period ($T$) (s).

Throws

Error If inputs are non-finite or out of domain.

Examples

// GEO: a ≈ 42,164 km → T ≈ 86,164 s (one sidereal day)
const a = 42164e3; // m
const muEarth = 3.986004418e14; // m³/s²
const T = keplerPeriod(a, muEarth); // ≈ 86164 s

// LEO-ish circular: a ≈ Earth radius + 400 km
const a = 6378e3 + 400e3; // m
const muEarth = 3.986004418e14; // m³/s²
const T = keplerPeriod(a, muEarth); // ≈ 5550 s (~92.5 min)