Circular orbit
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia article "Circular_orbit".

Two bodies with a slight difference in mass orbiting around a common barycenter with circular orbits.
Two bodies with a slight difference in mass orbiting around a common barycenter with circular orbits.
For other meanings of the term "orbit", see orbit (disambiguation)

In astrodynamics or celestial mechanics a circular orbit is an elliptic orbit with the eccentricity equal to 0. It is an example of a rotation around a fixed axis: this axis is the line through the center of mass perpendicular to the plane of motion.

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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:

Velocity

Under standard assumptions the orbital velocity of a body traveling along circular orbit, v_c\, can be computed as:

v_c=\sqrt{\mu\over{r}}

where:

Conclusion:

  • Velocity is constant along the path.

Orbital period

Under standard assumptions the orbital period (T\,\!) of a body traveling along circular orbit can be computed as:

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

where:

Energy

Under standard assumptions, specific orbital energy (\epsilon\,) is negative for a closed orbit and the orbital energy conservation equation (the Vis-viva equation) can take the form:

{v^2\over{2}}-{\mu\over{r}}=\epsilon< 0

where:

The boundary case is \epsilon\,=0 which corresponds to escape from the primary (parabolic orbit), with v=\sqrt{2}v_c=\sqrt{2\mu\over{r}} .

The virial theorem applies even without taking a time-average:

  • the potential energy of the system is equal to twice the total energy
  • the kinetic energy of the system is equal to minus the total energy

Thus 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.

Equation of motion

Under standard assumptions, the orbital equation becomes:

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

where:

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
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