Orbital period

The orbital period, also called the revolution period, is the time an astronomical object needs to make one full orbit around another. In astronomy, this term typically describes planets or asteroids circling the Sun, moons orbiting planets, exoplanets around other stars, or binary stars. It can also refer to the time a satellite takes to complete one orbit around a planet or moon.

For celestial objects generally, the orbital period is defined by a 360° revolution of one body around its primary—for example, Earth around the Sun. Periods in astronomy are expressed in units of time, commonly hours, days, or years. The reciprocal of the orbital period is the orbital frequency, a type of revolution frequency measured in hertz.

For a small body orbiting a central body, Kepler’s Third Law gives the orbital period \( T \) for two point masses in a circular or elliptical orbit as \( T = 2\pi \sqrt{a^3 / (GM)} \), where \( a \) is the orbit’s semi-major axis, \( G \) is the gravitational constant, and \( M \) is the mass of the more massive body. For all ellipses with a given semi-major axis, the orbital period is the same, regardless of eccentricity. Conversely, to find the distance a body must orbit to achieve a specific period \( T \), the formula is \( a = \sqrt[3]{GMT^2 / (4\pi^2)} \). For example, to complete an orbit every 24 hours around a 100 kg mass, a small body must orbit at a distance of 1.08 meters from the central body’s center of mass.

In the special case of a perfectly circular orbit, the semi-major axis equals the orbit’s radius \( r \), and the orbital velocity is constant: \( v_o = \sqrt{GM / r} \). This velocity is about 0.707 times the escape velocity.

The central body’s density also affects the orbital period. For a perfect sphere of uniform density, the period can be expressed without measuring mass as \( T = \sqrt{(a^3 / r^3) \cdot (3\pi / (G\rho))} \), where \( r \) is the sphere’s radius and \( \rho \) is its density. For instance, a small body in a circular orbit 10.5 cm above the surface of a tungsten sphere half a meter in radius would travel at slightly over 1 mm/s, completing an orbit every hour. If the same sphere were made of lead, the small body would need to orbit just 6.7 mm above the surface to maintain the same orbital period.

When a very small body is in a circular orbit barely above the surface of a sphere of any radius and mean density \( \rho \) (in kg/m³), the equation simplifies to \( T = \sqrt{3\pi / (G\rho)} \) (since \( r \) is nearly equal to \( a \)). Thus, the orbital period in low orbit depends only on the central body’s density, not its size. For Earth (or any spherically symmetric body with the same mean density, about 5,515 kg/m³, such as Mercury at 5,427 kg/m³ or Venus at 5,243 kg/m³), this gives \( T = 1.41 \) hours. For a body made of water (density about 1,000 kg/m³), or bodies with similar density like Saturn’s moons Iapetus (1,088 kg/m³) and Tethys (984 kg/m³), the period is \( T = 3.30 \) hours. So, instead of using the very small number \( G \), the strength of universal gravity can be described using a reference material like water: the orbital period for an orbit just above the surface of a spherical body of water is 3 hours and 18 minutes. Conversely, this can serve as a kind of universal unit of time if a unit of density is defined.

When both orbiting bodies’ masses must be considered, the orbital period \( T \) is calculated as \( T = 2\pi \sqrt{a^3 / (G(M_1 + M_2))} \), where \( a \) is the sum of the semi-major axes of the ellipses in which the two bodies move, or equivalently the semi-major axis of the ellipse of one body relative to the other. For binary stars, this formula allows astronomers to determine the sum of the stars’ masses. For exoplanets, the orbital period can be used to find the semi-major axis of the planet’s orbit, given the star’s mass.