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| {{Unreferenced|date=December 2009}}
| | 44 year old Optical Mechanic Edgar from Port McNicoll, loves to spend some time bicycling, diet and consuming out. Just had a family visit to Kernave Archaeological Site (Cultural Reserve of Kernave). |
| [[File:Gravity Wells Potential Plus Kinetic Energy - Circle-Ellipse-Parabola-Hyperbola.png|thumb|250px|A circular orbit is depicted in the top-left quadrant of this diagram, where the [[gravity well|gravitational potential well]] of the central mass shows potential energy, and the kinetic energy of the orbital speed is shown in red. The height of the kinetic energy remains constant throughout the constant speed circular orbit.]]
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| :''For other meanings of the term "orbit", see [[orbit (disambiguation)]]''
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| A '''circular orbit''' is the [[orbit]] at a fixed distance around any point by an object [[rotation around a fixed axis|rotating around a fixed axis]].
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| 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.
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| 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.
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| ==Circular acceleration==
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| [[:wikt:transverse|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
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| :<math> \mathbf{a} = - \frac{v^2}{r} \frac{\mathbf{r}}{r} = - \omega^2 \mathbf{r}</math>
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| where:
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| *<math>v\,</math> is [[Kinetic energy|orbital velocity]] of orbiting body,
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| *<math>r\,</math> is [[radius]] of the circle
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| *<math> \omega \ </math> is [[angular speed]], measured in radians per second.
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| ==Velocity==
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| The relative velocity is constant:
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| :<math> v = \sqrt{ G(M\!+\!m) \over{r}} = \sqrt{\mu\over{r}} </math>
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| where:
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| * ''G'' is the [[gravitational constant]]
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| * ''M'' and ''m'' are the masses of the orbiting bodies.
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| *<math> \scriptstyle \mu = G(M\!+\!m)\,</math> is the [[standard gravitational parameter]].
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| ==Equation of motion==
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| The [[orbit equation]] in polar coordinates, which in general gives ''r'' in terms of ''θ'', reduces to:
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| :<math>r={{h^2}\over{\mu}}</math>
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| where:
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| *<math>h=rv</math> is [[specific angular momentum]] of the orbiting body.
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| This is just another way of writing <math>\mu=rv^2</math> again.
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| ==Angular speed and orbital period==
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| :<math>\omega^2 r^3=\mu</math>
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| Hence the [[orbital period]] (<math>T\,\!</math>) can be computed as:
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| :<math>T=2\pi\sqrt{r^3\over{\mu}}</math>
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| Compare two proportional quantities, the [[free-fall time]] (time to fall to a point mass from rest)
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| :<math>T_{ff}=\frac{\pi}{2\sqrt{2}}\sqrt{r^3\over{\mu}}</math> (17.7 % of the orbital period in a circular orbit)
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| and the time to fall to a point mass in a [[radial parabolic orbit]] | |
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| :<math>T_{par}=\frac{\sqrt{2}}{3}\sqrt{r^3\over{\mu}}</math> (7.5 % of the orbital period in a circular orbit)
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| The fact that the formulas only differ by a constant factor is a priori clear from [[dimensional analysis]].
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| ==Energy==
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| The [[specific orbital energy]] (<math>\epsilon\,</math>) is negative, and
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| :<math>{v^2\over{2}}=-\epsilon</math>
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| :<math>-{\mu\over{r}}=2\epsilon</math>
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| Thus the [[virial theorem]] applies even without taking a time-average:
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| *the kinetic energy of the system is equal to the absolute value of the total energy
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| *the potential energy of the system is equal to twice the total energy
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| 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.
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| ==Delta-v to reach a circular orbit==
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| 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]].
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| == Orbital velocity in general relativity ==
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| In [[Schwarzschild metric]], the orbital velocity for a circular orbit with radius <math>R</math> is given by the following formula:
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| :<math>v = \sqrt{\frac{GM}{r-r_S}}</math>
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| where <math>\scriptstyle r_S = \frac{2GM}{c^2}</math> is the Schwarzschild radius of the central body.
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| === Derivation ===
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| For the sake of convenience, the derivation will be written in units in which <math>\scriptstyle c=G=1</math>.
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| The [[four-velocity]] of a body on a circular orbit is given by:
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| :<math>u^\mu = (\dot{t}, 0, 0, \dot{\phi})</math>
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| (<math>\scriptstyle r</math> is constant on a circular orbit, and the coordinates can be chosen so that <math>\scriptstyle \theta=\frac{\pi}{2}</math>). The dot above a variable denotes derivation with respect to proper time <math>\scriptstyle \tau</math>. | |
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| For a massive particle, the components of the [[four-velocity]] satisfy the following equation:
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| :<math>\left(1-\frac{2M}{r}\right) \dot{t}^2 - r^2 \dot{\phi}^2 = 1</math>
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| We use the geodesic equation:
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| :<math>\ddot{x}^\mu + \Gamma^\mu_{\nu\sigma}\dot{x}^\nu\dot{x}^\sigma = 0</math>
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| The only nontrivial equation is the one for <math>\scriptstyle \mu = r</math>. It gives:
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| :<math>\frac{M}{r^2}\left(1-\frac{2M}{r}\right)\dot{t}^2 - r\left(1-\frac{2M}{r}\right)\dot{\phi}^2 = 0</math>
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| From this, we get:
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| :<math>\dot{\phi}^2 = \frac{M}{r^3}\dot{t}^2</math>
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| Substituting this into the equation for a massive particle gives:
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| :<math>\left(1-\frac{2M}{r}\right) \dot{t}^2 - \frac{M}{r} \dot{t}^2 = 1</math>
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| Hence:
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| :<math>\dot{t}^2 = \frac{r}{r-3M}</math>
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| Assume we have an observer at radius <math>\scriptstyle r</math>, who is not moving with respect to the central body, that is, his [[four-velocity]] is proportional to the vector <math>\scriptstyle \partial_t</math>. The normalization condition implies that it is equal to:
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| :<math>v^\mu = \left(\sqrt{\frac{r}{r-2M}},0,0,0\right)</math>
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| The dot product of the [[Four-velocity|four-velocities]] of the observer and the orbiting body equals the gamma factor for the orbiting body relative to the observer, hence:
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| :<math>\gamma = g_{\mu\nu}u^\mu v^\nu = \left(1-\frac{2M}{r}\right) \sqrt{\frac{r}{r-3M}} \sqrt{\frac{r}{r-2M}} = \sqrt{\frac{r-2M}{r-3M}}</math>
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| This gives the [[Kinetic energy|velocity]]:
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| :<math>v = \sqrt{\frac{M}{r-2M}}</math>
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| Or, in SI units:
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| :<math>v = \sqrt{\frac{GM}{r-r_S}}</math>
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| ==See also==
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| *[[Elliptic orbit]]
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| *[[List of orbits]]
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| *[[Two-body problem]]
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| {{Orbits}}
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| {{DEFAULTSORT:Circular Orbit}}
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| [[Category:Orbits]]
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44 year old Optical Mechanic Edgar from Port McNicoll, loves to spend some time bicycling, diet and consuming out. Just had a family visit to Kernave Archaeological Site (Cultural Reserve of Kernave).