In a circular orbit, the orbiting body is the same distance from the central body at all times. But since the body is moving, will doppler shift occur from the perspective of the central body?

  • $\begingroup$ This probably answers your question: en.wikipedia.org/wiki/… $\endgroup$ Mar 24 at 17:47
  • $\begingroup$ The orbiting body is the same distance from the the center of the central body. $\endgroup$
    – John Doty
    Mar 24 at 21:36

2 Answers 2


There are two effects to be taken account of, both of which are easy to handle for the case of a circular orbit around a spherically symmetric static body where the orbiting mass is much smaller than the central mass.

For a circular orbit with orbital speed $v$, the relative time dilation between an observer at radius $r_1$ (assumed stationary) and the orbiting object at $r_2$ is $$\frac{d\tau_2}{d\tau_1} = \left[ \frac{1 - r_s/r_2 - v^2/c^2}{1 - r_s/r_1}\right]^{1/2}\ , $$ where $r_s$ is the Schwarzschild radius and $\tau_1$ and $\tau_2$ are times measured on the clocks of the observer and the orbiting object. This expression easily falls out of the Schwarzschild spacetime interval for a static, spherically symmetric mass.

You can think of this as a combination of time dilation in Special Relativity (sometimes referred to as the transverse Doppler effect) and a gravitational time dilation caused by the difference in gravitational potential between the two observers. The latter is absolutely necessary in a problem describing an orbiting body (as opposed to something being whirled in a circle).

Since $r_s = 2GM/c^2$ and in a circular orbit, $v^2 = GM/r_2$ $$\frac{d\tau_2}{d\tau_1} = \left[ \frac{1 - 3GM/r_2 c^2}{1 - 2GM/r_1c^2}\right]^{1/2}\ . $$ Notice that the "Special Relativistic" ($v^2/c^2$) and General Relativistic ($r_s/r$) terms are similar in size, so the latter cannot be obviously neglected.

You did not specify exactly where the observer on the orbited body would be. You cannot place them at the origin in a gravitational problem (or if you were to assume a point central mass, then there would be an infinity in the denominator). Neither is it necessary or obvious that $r_2 \gg r_1$. Thus I would leave the expression for the relative time dilation as-is. However, if you can assume that $r_1, r_2 \gg r_s$ (much more likely to be true), then the terms can be binomially expanded and approximated as $$\frac{d\tau_2}{d\tau_1} = \frac{f_1}{f_2} \simeq 1 + \frac{GM}{r_1 c^2} -\frac{3GM}{2r_2 c^2}\ , $$ where $f_1$ and $f_2$ are the frequencies according to the stationary observer and the orbiting object.

You can see from this expression that depending on the exact values of $r_1$ and $r_2$ there may be a blueshift or a redshift, or even, when $r_2 = 1.5r_1$, no shift at all.

As a concrete example, for an observer on the Earth's surface (where their velocity is comparatively very small compared with orbiting bodies), a signal arriving from the ISS when it is overhead will be redshifted ($r_2 < 1.5r_1$), whereas signals from a much higher GPS satellite ($r_2 > 1.5r_1$) are blueshifted.

Since both of these terms are of order $v^2/c^2$ they will usually be dwarfed by any line of sight component of the velocity between the orbiting object and the observer, which will produce the usual longitudinal Doppler shift of order $v/c$.


In a circular orbit there is obviously no classical radial Doppler shift observed from either of the bodies as the distance is always constant. However Relativity predicts an additional frequency shift due to the time dilation effect in moving reference frames. This means we should see the clocks in the moving system running slower which causes a redshift. This is apparent regardless of the direction of motion. However, the effect is very small, only of the order $(v/c)^2$ where $c$ is the speed of light. The exact formula is

$$\nu=\nu_0 \sqrt{1-\frac{v^2}{c^2}} $$

See this link for some additional explanation https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Tatum)/15%3A_Special_Relativity/15.19%3A_The_Transverse_and_Oblique_Doppler_Effects

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    $\begingroup$ But note that if the orbit is gravitational, then the gravitational blueshift contributes at the same order in $v/c$ (but depends on the size of the central body) $\endgroup$
    – Sten
    Mar 24 at 19:25
  • $\begingroup$ @Sten How would this have anything to do with the orbital velocities? $\endgroup$
    – Thomas
    Mar 24 at 19:31
  • $\begingroup$ $v^2$ is proportional to the gravitational potential depth. $\endgroup$
    – Sten
    Mar 24 at 20:15
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    $\begingroup$ Indeed, there is an orbital radius at which the net shift is zero. $\endgroup$
    – ProfRob
    Mar 24 at 20:22
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    $\begingroup$ Question was about a circular orbit, so no need to discuss eccentricity or the possibility of the velocity being zero, and it mentioned a central body and an orbiting body, so no need to think about equal masses. The idea is that the question was about orbits, not plain circular motion, which suggests the asker is thinking about gravity, in which case it is useful to point out in a comment that the frequency shift is not given by transverse Doppler alone. $\endgroup$
    – Sten
    Mar 25 at 20:01

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