Circular orbit

For other meanings of the term "orbit", see orbit (disambiguation)

A circular orbit is the orbit at a fixed distance around any point by an object rotating around a fixed axis.

Below we consider 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 plane of motion.

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.

Circular acceleration

Transverse acceleration (perpendicular to velocity) causes change in direction. If it is constant in magnitude and changing in direction with the velocity, we get a circular motion. For this centripetal acceleration we have

$\mathbf{a} = - \frac{v^2}{r} \frac{\mathbf{r}}{r} = - \omega^2 \mathbf{r}$

where:

• $v\,$ is orbital velocity of orbiting body,
• $r\,$ is radius of the circle
• $\omega \$ is angular speed, measured in radians per second.

Velocity

The relative velocity is constant:

$v = \sqrt{ G(M\!+\!m) \over{r}} = \sqrt{\mu\over{r}}$

where:

• G is the gravitational constant
• M and m are the masses of the orbiting bodies.
• $\scriptstyle \mu = G(M\!+\!m)\,$ is the standard gravitational parameter.

Equation of motion

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

$r={{h^2}\over{\mu}}$

where:

• h = rv is specific angular momentum of the orbiting body.

This is just another way of writing μ = rv² again.

Angular speed and orbital period

ω²r³ = μ

Hence the orbital period ($T\,\!$) can be computed as:

$T=2\pi\sqrt{r^3\over{\mu}}$

Compare two proportional quantities, the free-fall time (time to fall to a point mass from rest)

$T_{ff}=\frac{\pi}{2\sqrt{2}}\sqrt{r^3\over{\mu}}$ (17.7 % of the orbital period in a circular orbit)

and the time to fall to a point mass in a radial parabolic orbit

$T_{par}=\frac{\sqrt{2}}{3}\sqrt{r^3\over{\mu}}$ (7.5 % of the orbital period in a circular orbit)

The fact that the formulas only differ by a constant factor is a priori clear from dimensional analysis.

Energy

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

${v^2\over{2}}=-\epsilon$

$-{\mu\over{r}}=2\epsilon$

Thus the virial theorem applies even without taking a time-average:

• 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 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

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.

See also

• Elliptic orbit
• List of orbits
• Two-body problem