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Orbital Period⚠ unverified

Aerospace / Orbital · Period of an orbit

Parameters

InputSymbolUnitDefaultDescription
aam6771000.0Semi-major axis
muμm^3/s^2398600000000000.0Gravitational parameter
OutputSymbolUnitDescription
TTsPeriod

The science & history

Understanding the Parameters

Derivation (Approaching a Proof)

Circular case first (transparent). For a circular orbit the body travels the circumference $2\pi r$ at the constant circular speed $v = \sqrt{\mu/r}$. Period is distance over speed:

$$T = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{\mu/r}} = 2\pi r \sqrt{\frac{r}{\mu}} = 2\pi\sqrt{\frac{r^3}{\mu}}.$$

With $r = a$ for a circle, this is already the result.

General ellipse (via Kepler's second law). For any orbit, Kepler's second law says the radius vector sweeps area at a constant rate $dA/dt = h/2$, where $h$ is the specific angular momentum. Integrating over one full period sweeps the entire area of the ellipse, $A = \pi a b$ (semi-axes $a$ and $b$):

$$T = \frac{A}{dA/dt} = \frac{\pi a b}{h/2} = \frac{2\pi a b}{h}.$$

For an ellipse the semi-minor axis is $b = a\sqrt{1-e^2}$, and the angular momentum is $h = \sqrt{\mu\,a(1-e^2)}$ (a standard orbit relation). Substitute both:

$$T = \frac{2\pi a\cdot a\sqrt{1-e^2}}{\sqrt{\mu\,a(1-e^2)}} = \frac{2\pi a^2\sqrt{1-e^2}}{\sqrt{\mu\,a}\,\sqrt{1-e^2}} = \frac{2\pi a^2}{\sqrt{\mu\,a}} = 2\pi\sqrt{\frac{a^3}{\mu}}. \qquad\blacksquare$$

The eccentricity cancels completely — the origin of the "$a$ only" rule. Squaring gives the familiar $T^2 = \dfrac{4\pi^2}{\mu}\,a^3$, Kepler's third law with Newton's constant.

Dimensional check. $$2\pi\sqrt{\frac{a^3}{\mu}} = \sqrt{\frac{\text{m}^3}{\text{m}^3/\text{s}^2}} = \sqrt{\text{s}^2} = \text{s}. \checkmark$$ ($2\pi$ is dimensionless.)

History and Development

Related Concepts: Circular Orbit Velocity, Orbital Velocity, Vis Viva Energy, Escape Velocity, Time Since Perigee, Gravitational Force, Angular Momentum

Notes: Kepler's third law (Newtonian form). Depends only on $a$, not eccentricity — same-$a$ orbits share a period. $T\propto a^{3/2}$ (LEO ~90 min, GPS ~12 h, GEO 1 sidereal day, Moon 27.3 d). $a$ = semi-major axis = mean of perigee/apogee radii. $T^2=4\pi^2a^3/\mu$ ⇒ orbits weigh the primary ($\mu=4\pi^2a^3/T^2$). Defaults ⇒ $T\approx92.4$ min.

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