Circular orbit

 ( Orbits ) 

A circular orbit is an orbit with a fixed distance around the barycenter; that is, in the shape of a circle. In this case, not only the distance, but also the speed, angular speed, Potential energy and kinetic energy are constant. There is no periapsis or apoapsis. This orbit has no Radial orbit.

Listed below is 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 orbital plane.

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Circular acceleration Transverse acceleration (perpendicular to velocity) causes a change in direction. If it is constant in magnitude and changing in direction with the velocity, circular motion ensues. Taking two derivatives of the particle's coordinates concerning time gives the centripetal acceleration

$a\backslash ,\; =\; \backslash frac\; \{v^2\}\; \{r\}\; \backslash ,\; =\; \{\backslash omega^2\}\; \{r\}$

where:

• $v\backslash ,$ is Kinetic energy of the orbiting body,
• $r\backslash ,$ is radius of the circle
• $\backslash omega\; \backslash$ is angular speed, measured in per unit time.

The formula is dimensionless, describing a ratio true for all units of measure applied uniformly across the formula. If the numerical value $\backslash mathbf\{a\}$ is measured in meters per second squared, then the numerical values $v\backslash ,$ will be in meters per second, $r\backslash ,$ in meters, and $\backslash omega\; \backslash$ in radians per second.

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Velocity The speed (or the magnitude of velocity) relative to the centre of mass is constant:

(2025). 9781108411981, Cambridge University Press. ISBN 9781108411981

$v\; =\; \backslash sqrt\{\; GM\backslash !\; \backslash over\{r\}\}\; =\; \backslash sqrt\{\backslash mu\backslash over\{r\}\}$

where:

• $G$, is the gravitational constant
• $M$, is the mass of both orbiting bodies $(M\_1+M\_2)$, although in common practice, if the greater mass is significantly larger, the lesser mass is often neglected, with minimal change in the result.
• $\backslash mu\; =\; GM$, is the standard gravitational parameter.
• $r$ is the distance from the center of mass.

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

$r=$

where:

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

This is because $\backslash mu=rv^2$

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Angular speed and orbital period

$\backslash omega^2\; r^3=\backslash mu$

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

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

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

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

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

$T\_\backslash text\{par\}=\backslash frac\{\backslash sqrt\{2\}\}\{3\}\backslash sqrt\{r^3\backslash over\{\backslash mu\}\}$ (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.

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

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

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

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 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 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 In Schwarzschild metric, the orbital velocity for a circular orbit with radius $r$ is given by the following formula:

$v\; =\; \backslash sqrt\{\backslash frac\{GM\}\{r-r\_S\}\}$

where $\backslash scriptstyle\; r\_S\; =\; \backslash frac\{2GM\}\{c^2\}$ is the Schwarzschild radius of the central body.

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Derivation For the sake of convenience, the derivation will be written in units in which $\backslash scriptstyle\; c=G=1$.

The four-velocity of a body on a circular orbit is given by:

$u^\backslash mu\; =\; (\backslash dot\{t\},\; 0,\; 0,\; \backslash dot\{\backslash phi\})$

($\backslash scriptstyle\; r$ is constant on a circular orbit, and the coordinates can be chosen so that $\backslash scriptstyle\; \backslash theta=\backslash frac\{\backslash pi\}\{2\}$). The dot above a variable denotes derivation with respect to proper time $\backslash scriptstyle\; \backslash tau$.

For a massive particle, the components of the four-velocity satisfy the following equation:

$\backslash left(1-\backslash frac\{2M\}\{r\}\backslash right)\; \backslash dot\{t\}^2\; -\; r^2\; \backslash dot\{\backslash phi\}^2\; =\; 1$

We use the geodesic equation:

$\backslash ddot\{x\}^\backslash mu\; +\; \backslash Gamma^\backslash mu\_\{\backslash nu\backslash sigma\}\backslash dot\{x\}^\backslash nu\backslash dot\{x\}^\backslash sigma\; =\; 0$

The only nontrivial equation is the one for $\backslash scriptstyle\; \backslash mu\; =\; r$. It gives:

$\backslash frac\{M\}\{r^2\}\backslash left(1-\backslash frac\{2M\}\{r\}\backslash right)\backslash dot\{t\}^2\; -\; r\backslash left(1-\backslash frac\{2M\}\{r\}\backslash right)\backslash dot\{\backslash phi\}^2\; =\; 0$

From this, we get:

$\backslash dot\{\backslash phi\}^2\; =\; \backslash frac\{M\}\{r^3\}\backslash dot\{t\}^2$

Substituting this into the equation for a massive particle gives:

$\backslash left(1-\backslash frac\{2M\}\{r\}\backslash right)\; \backslash dot\{t\}^2\; -\; \backslash frac\{M\}\{r\}\; \backslash dot\{t\}^2\; =\; 1$

Hence:

$\backslash dot\{t\}^2\; =\; \backslash frac\{r\}\{r-3M\}$

Assume we have an observer at radius $\backslash scriptstyle\; r$, who is not moving with respect to the central body, that is, their four-velocity is proportional to the vector $\backslash scriptstyle\; \backslash partial\_t$. The normalization condition implies that it is equal to:

$v^\backslash mu\; =\; \backslash left(\backslash sqrt\{\backslash frac\{r\}\{r-2M\}\},0,0,0\backslash right)$

The dot product of the Four-velocity of the observer and the orbiting body equals the gamma factor for the orbiting body relative to the observer, hence:

$\backslash gamma\; =\; g\_\{\backslash mu\backslash nu\}u^\backslash mu\; v^\backslash nu\; =\; \backslash left(1-\backslash frac\{2M\}\{r\}\backslash right)\; \backslash sqrt\{\backslash frac\{r\}\{r-3M\}\}\; \backslash sqrt\{\backslash frac\{r\}\{r-2M\}\}\; =\; \backslash sqrt\{\backslash frac\{r-2M\}\{r-3M\}\}$

This gives the Kinetic energy:

$v\; =\; \backslash sqrt\{\backslash frac\{M\}\{r-2M\}\}$

Or, in SI units:

$v\; =\; \backslash sqrt\{\backslash frac\{GM\}\{r-r\_S\}\}$

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See also

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

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