Circular orbit

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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.

1 Circular acceleration
2 Velocity
3 Orbital period
4 Energy
5 Equation of motion
6 Delta-v to reach a circular orbit
7 See also

[edit] 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 frequency, measured in radian per second.

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[edit] Velocity

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

$v=\sqrt{\mu\over{r}}$

where:

$r\,$ is radius of orbit equal to radial distance of orbiting body from central body,
$\mu\,$ is standard gravitational parameter.

Conclusion:

Velocity is constant along the path.

[edit] Orbital period

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

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

where:

$r\,$ is orbit radius equal to radial distance of orbiting body from central body,
$\mu\,$ is standard gravitational parameter.

Conclusions:

The orbital period is the same as that for an elliptic orbit with the semi-major axis ( $a\,\!$ ) equal to orbit radius.

[edit] Energy

Under standard assumptions, specific orbital energy ( $\epsilon\,$ ) is negative and the orbital energy conservation equation for this orbit takes the form:

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

where:

$v\,$ is orbital velocity of orbiting body,
$r\,$ is radius of orbit equal to radial distance of orbiting body from central body,
$\mu\,$ is standard gravitational parameter.

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!

[edit] Equation of motion

Under standard assumptions, the orbital equation becomes:

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

where:

$r\,$ is radial distance of orbiting body from central body,
$h\,$ is specific angular momentum of the orbiting body,
$\mu\,$ is standard gravitational parameter.

[edit] 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.

Categories: Astrodynamics | Orbits

Circular orbit

From Wikipedia, the free encyclopedia

Contents

[edit] Circular acceleration

[edit] Velocity

[edit] Orbital period

[edit] Energy

[edit] Equation of motion

[edit] Delta-v to reach a circular orbit

[edit] See also

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