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Here are examples of how to calculate orbital velocity of a given radius

Orbital velocity in general relativity

The orbital velocity for a circular orbit with radius R is given by the following formula:

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

where \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 \scriptstyle c=G=1.

Time Derivation Notation:

A variety of notations are used to denote the time derivative.

\frac {dx} {dt}

A very common short-hand notation used, especially in mathematics and physics, is the ‘over-dot’. I.E.


(This is called Newton’s notation)

Higher time derivatives are also used: the second derivative with respect to time is written as

\frac {d^2x} {dt^2}

with the corresponding shorthand of \ddot{x}.

As a generalization, the time derivative of a vector, say:

 \vec V = \left[ v_1,\ v_2,\ v_3, \cdots \right] \ ,

is defined as the vector whose components are the derivatives of the components of the original vector. That is,

 \frac {d \vec V } {dt} = \left[ \frac{ d v_1 }{dt},\frac {d  v_2 }{dt},\frac {d  v_3 }{dt}, \cdots \right] \ .

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

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

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

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

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

We use the geodesic equation:

\ddot{x}^\mu + \Gamma^\mu_{\nu\sigma}\dot{x}^\nu\dot{x}^\sigma = 0

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

\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:

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

Substituting this into the equation for a massive particle gives:

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


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

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

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:

\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:

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

Or, in SI units:

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

Circular acceleration

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

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


  • v\, is orbital velocity of orbiting body,
  • r\, is radius of the circle
  •  \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 of  \mathbf{a} is measured in meters per second per second, then the numerical values for v\, will be in meters per second, r\, in meters, and  \omega \ in radians per second.


The relative velocity is constant:

 v = \sqrt{ G(M\!+\!m) \over{r}} = \sqrt{\mu\over{r}}


  • G is the gravitational constant
  • M and m are the masses of the orbiting bodies.
  •  \scriptstyle \mu = G(M\!+\!m)\, is the standard gravitational parameter.

Equation of motion

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



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

This is just another way of writing \mu=rv^2 again.

Angular speed and orbital period

\omega^2 r^3=\mu

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


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

T_{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

T_{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 it is clear from dimensional analysis.


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


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.

How to enter dot(s) or tilde notations above a time derivative – refer to Keyboards below this link.

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