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

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

where:

• ${\displaystyle v\,}$  is the orbital velocity of the orbiting body,
• ${\displaystyle r\,}$  is radius of the circle
• ${\displaystyle \omega \ }$  is angular speed, measured in radians 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 ${\displaystyle \mathbf {a} }$  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 central object is constant:^[1]: 30

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

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.

The orbit equation in polar coordinates, which in general gives r in terms of θ, reduces to:^{[clarification needed][citation needed]}

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

where:

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

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

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

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

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

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

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

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

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

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

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

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

Thus the virial 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

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.^{[citation needed]}

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.

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 ${\displaystyle \scriptstyle r_{S}={\frac {2GM}{c^{2}}}}$  is the Schwarzschild radius of the central body.

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

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

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

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

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

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

We use the geodesic equation:

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

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

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

From this, we get:

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

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 ${\displaystyle \scriptstyle r}$ , who is not moving with respect to the central body, that is, their four-velocity is proportional to the vector ${\displaystyle \scriptstyle \partial _{t}}$ . 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:

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

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