Circular orbit
From Wikipedia, the free encyclopedia
- 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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[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
- <math> \mathbf{a} = - \frac{v^2}{r} \frac{\mathbf{r}}{r} = - \omega^2 \mathbf{r}</math>
where:
- <math>v\,</math> is orbital velocity of orbiting body,
- <math>r\,</math> is radius of the circle
- <math> \omega \ </math> is angular frequency, measured in radian per second.
[edit] Velocity
Under standard assumptions the orbital velocity (<math>v\,</math>) of a body traveling along circular orbit can be computed as:
- <math>v=\sqrt{\mu\over{r}}</math>
where:
- <math>r\,</math> is radius of orbit equal to radial distance of orbiting body from central body,
- <math>\mu\,</math> is standard gravitational parameter.
Conclusion:
- Velocity is constant along the path.
[edit] Orbital period
Under standard assumptions the orbital period (<math>T\,\!</math>) of a body traveling along circular orbit can be computed as:
- <math>T={2\pi\over{\sqrt{\mu}}}r^{3\over{2}}</math>
where:
- <math>r\,</math> is orbit radius equal to radial distance of orbiting body from central body,
- <math>\mu\,</math> is standard gravitational parameter.
Conclusions:
- The orbital period is the same as that for an elliptic orbit with the semi-major axis (<math>a\,\!</math>) equal to orbit radius.
[edit] Energy
Under standard assumptions, specific orbital energy (<math>\epsilon\,</math>) is negative and the orbital energy conservation equation for this orbit takes the form:
- <math>{v^2\over{2}}-{\mu\over{r}}=-{\mu\over{2r}}=\epsilon< 0\,\!</math>
where:
- <math>v\,</math> is orbital velocity of orbiting body,
- <math>r\,</math> is radius of orbit equal to radial distance of orbiting body from central body,
- <math>\mu\,</math> 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:
- <math>r={{h^2}\over{\mu}}</math>
where:
- <math>r\,</math> is radial distance of orbiting body from central body,
- <math>h\,</math> is specific angular momentum of the orbiting body,
- <math>\mu\,</math> 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.
