Circular orbit

Velocity

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

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

where:

• ${\displaystyle G}$, is the gravitational constant
• ${\displaystyle M}$, is the mass 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 the standard gravitational parameter.
• ${\displaystyle r}$ is the distance from the center of mass.

Equation of motion

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

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

where:

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

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

Angular speed and orbital period

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

Hence the orbital period (${\displaystyle T}$) can be computed as:

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

Compare two proportional quantities, the free-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 a radial 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 from dimensional analysis.

Energy

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

${\displaystyle ϵ=-\frac{v^2}{2}}$

${\displaystyle ϵ=-\frac{\mu }{2r}}$

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 √2 times the speed in a circular orbit at that distance: the kinetic energy is twice as much, hence the total energy is zero.

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.

Orbital velocity in general relativity

In Schwarzschild 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.

Derivation

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

The four-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 the four-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, their four-velocity is proportional to the vector ${\displaystyle {\mathrm{\partial }}_t}$. The normalization condition implies that it is equal to:

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

The dot product of the four-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 the velocity:

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

Or, in SI units:

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