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Orbital period

Time an astronomical object takes to complete one orbit around another object

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For celestial objects in general, the orbital period is determined by a 360° revolution of one body around its primary, e.g. Earth around the Sun.

Periods in astronomy are expressed in units of time, usually hours, days, or years. Its reciprocal is the orbital frequency, a kind of revolution frequency, in units of hertz.

Small body orbiting a central body

According to Kepler's Third Law, the orbital period T of two point masses orbiting each other in a circular or elliptic orbit is:

T=2πa3GM

where:

For all ellipses with a given semi-major axis the orbital period is the same, regardless of eccentricity.

Inversely, for calculating the distance where a body has to orbit in order to have a given orbital period T:

a=GMT24π23

For instance, for completing an orbit every 24 hours around a mass of 100 kg, a small body has to orbit at a distance of 1.08 meters from the central body's center of mass.

In the special case of perfectly circular orbits, the semimajor axis a is equal to the radius of the orbit, and the orbital velocity is constant and equal to

vo=GMr

where:

  • r is the circular orbit's radius in meters,

This corresponds to 1⁄√2 times (≈ 0.707 times) the escape velocity.

Effect of central body's density

For a perfect sphere of uniform density, it is possible to rewrite the first equation without measuring the mass as:

T=a3r33πGρ

where:

  • r is the sphere's radius
  • a is the orbit's semi-major axis,
  • G is the gravitational constant,
  • ρ is the density of the sphere.

For instance, a small body in circular orbit 10.5 cm above the surface of a sphere of tungsten half a metre in radius would travel at slightly more than 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 for sustaining 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 ρ (in kg/m3), the above equation simplifies to

T=3πGρ

(since r now nearly equals a). Thus the orbital period in low orbit depends only on the density of the central body, regardless of its size.

So, for the Earth as the central body (or any other spherically symmetric body with the same mean density, about 5,515 kg/m3, e.g. Mercury with 5,427 kg/m3 and Venus with 5,243 kg/m3) we get:

T = 1.41 hours

and for a body made of water (ρ ≈ 1,000 kg/m3), or bodies with a similar density, e.g. Saturn's moons Iapetus with 1,088 kg/m3 and Tethys with 984 kg/m3 we get:

T = 3.30 hours

Thus, as an alternative for using a very small number like G, the strength of universal gravity can be described using some reference material, such as 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 be used as a kind of "universal" unit of time if we have a unit of density.

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