Circular orbit

A circular orbit is an orbit with a fixed distance around the barycenter; that is, in the shape of a circle. In this case, not only the distance, but also the speed, angular speed, potential and kinetic energy are constant. There is no periapsis or apoapsis. This orbit has no radial version.

For other uses of "orbit", see Orbit (disambiguation).

Listed below is a circular orbit in astrodynamics or celestial mechanics under standard assumptions. Here the centripetal force is the gravitational force, and the axis mentioned above is the line through the center of the central mass perpendicular to the orbital plane.

$v\,$ is of the orbiting body,

the orbital velocity

$r\,$ is of the circle

radius

$\omega \$ is , measured in radians per unit time.

angular speed

Transverse acceleration (perpendicular to velocity) causes a change in direction. If it is constant in magnitude and changing in direction with the velocity, circular motion ensues. Taking two derivatives of the particle's coordinates concerning time gives the centripetal acceleration

where:

The formula is dimensionless, describing a ratio true for all units of measure applied uniformly across the formula. If the numerical value $\mathbf {a}$ is measured in meters per second squared, then the numerical values $v\,$ will be in meters per second, $r\,$ in meters, and $\omega \$ in radians per second.

$G$ , is the

gravitational constant

$M$ , is the of both orbiting bodies $(M_{1}+M_{2})$ , although in common practice, if the greater mass is significantly larger, the lesser mass is often neglected, with minimal change in the result.

mass

$\mu =GM$ , is the .

standard gravitational parameter

The speed (or the magnitude of velocity) relative to the central object is constant:^[1]^: 30

where:

$h=rv$ is of the orbiting body.

specific angular momentum

The orbit equation in polar coordinates, which in general gives r in terms of θ, reduces to:

where:

This is because $\mu =rv^{2}$

the kinetic energy of the system is equal to the absolute value of the total energy

the potential energy of the system is equal to twice the total energy

The specific orbital energy ( $\epsilon \,$ ) is negative, and

Thus the virial theorem^[1]^: 72 applies even without taking a time-average:

The escape velocity from any distance is √2 times the speed in a circular orbit at that distance: the kinetic energy is twice as much, hence the total energy is zero.

Delta-v to reach a circular orbit[edit]

Maneuvering into a large circular orbit, e.g. a geostationary orbit, requires a larger delta-v than an escape orbit, although the latter implies getting arbitrarily far away and having more energy than needed for the orbital speed of the circular orbit. It is also a matter of maneuvering into the orbit. See also Hohmann transfer orbit.