This articlerelies largely or entirely on asingle source. Relevant discussion may be found on thetalk page. Please helpimprove this article byintroducing citations to additional sources. Find sources: "Circular orbit" – news ·newspapers ·books ·scholar ·JSTOR(April 2020)

Acircular orbit is anorbit with a fixed distance around thebarycenter; that is, in the shape of acircle.In this case, not only the distance, but also the speed,angular speed,potential andkinetic energy are constant. There is noperiapsis or apoapsis. This orbit has noradial version.

Listed below is a circular orbit inastrodynamics orcelestial mechanics under standard assumptions. Here thecentripetal force is thegravitational force, and the axis mentioned above is the line through thecenter of the central massperpendicular to theorbital plane.

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 thecentripetal acceleration

${\displaystyle a=\frac{v^2}{r}={\omega }^{2}r}$

where:

• ${\displaystyle v}$ isthe orbital velocity of the orbiting body,
• ${\displaystyle r}$ isradius of the circle
• ${\displaystyle \omega }$ isangular speed, measured inradians per unit time.

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

The speed (or the magnitude of velocity) relative to the centre of mass is constant:^[1]: 30

${\displaystyle v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{\mu }{r}}}$

where:

• ${\displaystyle G}$, is thegravitational constant
• ${\displaystyle M}$, is themass of both orbiting bodies${\displaystyle (M_1+M_2)}$, although in common practice, if the greater mass is significantly larger, the lesser mass is often neglected, with minimal change in the result.
• ${\displaystyle \mu =GM}$, is thestandard gravitational parameter.
• ${\displaystyle r}$ is the distance from the center of mass.

Theorbit equation in polar coordinates, which in general givesr in terms ofθ, reduces to:^{[clarification needed][citation needed]}

${\displaystyle r=\frac{h^2}{\mu }}$

where:

• ${\displaystyle h=rv}$ isspecific angular momentum of the orbiting body.

This is because${\displaystyle \mu =r{v}^2}$

${\displaystyle {\omega }^2{r}^3=\mu }$

Hence theorbital period (${\displaystyle T}$) can be computed as:^[1]: 28

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

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

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

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

${\displaystyle T_{\text{par}}=\frac{\sqrt{2}}{3}\sqrt{\frac{r^3}{\mu }}}$ (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 fromdimensional analysis.^{[citation needed]}

Thespecific orbital energy (${\displaystyle ϵ}$) is negative, and

${\displaystyle ϵ=-\frac{v^2}{2}}$

${\displaystyle ϵ=-\frac{\mu }{2r}}$

Thus thevirial theorem^[1]: 72 applies even without taking a time-average:^{[citation needed]}

• 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

Theescape 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.^{[citation needed]}

Maneuvering into a large circular orbit, e.g. ageostationary orbit, requires a largerdelta-v than anescape orbit, although the latter implies getting arbitrarily far away and having more energy than needed for theorbital speed of the circular orbit. It is also a matter of maneuvering into the orbit. See alsoHohmann transfer orbit.

InSchwarzschild metric, the orbital velocity for a circular orbit with radius${\displaystyle r}$ is given by the following formula:

${\displaystyle v=\sqrt{\frac{GM}{r-r_S}}}$

where${\displaystyle r_S=\frac{2GM}{c^2}}$ is the Schwarzschild radius of the central body.

For the sake of convenience, the derivation will be written in units in which${\displaystyle c=G=1}$.

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

${\displaystyle u^{\mu }=(\dot{t},0,0,\dot{\varphi })}$

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

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

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

We use the geodesic equation:

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

The only nontrivial equation is the one for${\displaystyle \mu =r}$. It gives:

${\displaystyle \frac{M}{r^2}\left(1-\frac{2M}{r}\right){\dot{t}}^2-r\left(1-\frac{2M}{r}\right){\dot{\varphi }}^2=0}$

From this, we get:

${\displaystyle {\dot{\varphi }}^2=\frac{M}{r^3}{\dot{t}}^2}$

Substituting this into the equation for a massive particle gives:

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

Hence:

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

Assume we have an observer at radius${\displaystyle r}$, who is not moving with respect to the central body, that is, theirfour-velocity is proportional to the vector${\displaystyle {\mathrm{\partial }}_t}$. The normalization condition implies that it is equal to:

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

Thedot product of thefour-velocities of the observer and the orbiting body equals the gamma factor for the orbiting body relative to the observer, hence:

${\displaystyle \gamma =g_{\mu \nu }{u}^{\mu }{v}^{\nu }=\left(1-\frac{2M}{r}\right)\sqrt{\frac{r}{r-3M}}\sqrt{\frac{r}{r-2M}}=\sqrt{\frac{r-2M}{r-3M}}}$

This gives thevelocity:

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

Or, in SI units:

${\displaystyle v=\sqrt{\frac{GM}{r-r_S}}}$

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

• Orbits

• Articles with short description
• Short description matches Wikidata
• Articles needing additional references from April 2020
• All articles needing additional references
• Wikipedia articles needing clarification from January 2016
• All articles with unsourced statements
• Articles with unsourced statements from August 2019