# Orbital speed

Template:Distinguish {{ safesubst:#invoke:Unsubst||$N=Refimprove |date=__DATE__ |$B= {{#invoke:Message box|ambox}} }}

The orbital speed of a body, generally a planet, a natural satellite, an artificial satellite, or a multiple star, is the speed at which it orbits around the barycenter of a system, usually around a more massive body. It can be used to refer to either the mean orbital speed, i.e. the average speed as it completes an orbit, or the speed at a particular point in its orbit.Template:Citation needed (lead)

The orbital speed at any position in the orbit can be computed from the distance to the central body at that position, and the specific orbital energy, which is independent of position: the kinetic energy is the total energy minus the potential energy.

In the case of radial motion:{{ safesubst:#invoke:Unsubst||date=__DATE__ |$B= {{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }} ## Transverse orbital speed The transverse orbital speed is inversely proportional to the distance to the central body because of the law of conservation of angular momentum, or equivalently, Kepler's second law. This states that as a body moves around its orbit during a fixed amount of time, the line from the barycenter to the body sweeps a constant area of the orbital plane, regardless of which part of its orbit the body traces during that period of time.  This law implies that the body moves faster near its periapsis than near its apoapsis, because at the smaller distance it needs to trace a greater arc to cover the same area. ## Mean orbital speed For orbits with small eccentricity, the length of the orbit is close to that of a circular one, and the mean orbital speed can be approximated either from observations of the orbital period and the semimajor axis of its orbit, or from knowledge of the masses of the two bodies and the semimajor axis.{{ safesubst:#invoke:Unsubst||date=__DATE__ |$B= {{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }}

$v_{o}\approx {2\pi a \over T}$ $v_{o}\approx {\sqrt {\mu \over a}}$ where v is the orbital velocity, a is the length of the semimajor axis, T is the orbital period, and μ=GM is the standard gravitational parameter. Note that this is only an approximation that holds true when the orbiting body is of considerably lesser mass than the central one, and eccentricity is close to zero.

Taking into account the mass of the orbiting body,

$v_{o}\approx {\sqrt {G(m_{1}+m_{2}) \over r}}$ where m1 is now the mass of the body under consideration, m2 is the mass of the body being orbited, r is specifically the distance between the two bodies (which is the sum of the distances from each to the center of mass), and G is the gravitational constant. This is still a simplified version; it doesn't allow for elliptical orbits, but it does at least allow for bodies of similar masses.

When one of the masses is almost negligible compared to the other mass as the case for Earth and Sun, one can approximate the previous formula to get:

$v_{o}\approx {\sqrt {\frac {GM}{r}}}$ or assuming r equal to the body's radius

$v_{o}\approx {\frac {v_{e}}{\sqrt {2}}}$ Where M is the (greater) mass around which this negligible mass or body is orbiting, and ve is the escape velocity.

For an object in an eccentric orbit orbiting a much larger body, the length of the orbit decreases with orbital eccentricity e, and is an ellipse. This can be used to obtain a more accurate estimate of the average orbital speed:

$v_{o}={\frac {2\pi a}{T}}\left[1-{\frac {1}{4}}e^{2}-{\frac {3}{64}}e^{4}-{\frac {5}{256}}e^{6}-{\frac {175}{16384}}e^{8}-\dots \right]$ The mean orbital speed decreases with eccentricity.

## Precise orbital speed

For the precise orbital speed of a body at any given point in its trajectory, both the mean distance and the precise distance are taken into account:

$v={\sqrt {\mu \left({2 \over r}-{1 \over a}\right)}}$ where μ is the standard gravitational parameter, r is the distance at which the speed is to be calculated, and a is the length of the semi-major axis of the elliptical orbit. For the Earth at perihelion,

$v={\sqrt {1.327\times 10^{20}~m^{3}s^{-2}\cdot \left({2 \over 1.471\times 10^{11}~m}-{1 \over 1.496\times 10^{11}~m}\right)}}\approx 30,300~m/s$ which is slightly faster than Earth's average orbital speed of 29,800 m/s, as expected from Kepler's 2nd Law.