<p>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.</p>
<p>For a circular orbit with orbital speed <span class="math-container">$v$</span>, the relative time dilation between an observer at radius <span class="math-container">$r_1$</span> (assumed stationary) and the orbiting object at <span class="math-container">$r_2$</span> is
<span class="math-container">$$\frac{d\tau_2}{d\tau_1} = \left[ \frac{1 - r_s/r_2 - v^2/c^2}{1 - r_s/r_1}\right]^{1/2}\ , $$</span>
where <span class="math-container">$r_s$</span> is the Schwarzschild radius and <span class="math-container">$\tau_1$</span> and <span class="math-container">$\tau_2$</span> 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.</p>
<p>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).</p>
<p>Since <span class="math-container">$r_s = 2GM/c^2$</span> and in a circular orbit,  <span class="math-container">$v^2 = GM/r_2$</span>
<span class="math-container">$$\frac{d\tau_2}{d\tau_1} = \left[ \frac{1 - 3GM/r_2 c^2}{1 - 2GM/r_1c^2}\right]^{1/2}\ . $$</span>
Notice that the &quot;Special Relativistic&quot; (<span class="math-container">$v^2/c^2$</span>) and General Relativistic (<span class="math-container">$r_s/r$</span>) terms are similar in size, so the latter cannot be obviously neglected.</p>
<p>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 <span class="math-container">$r_2 \gg r_1$</span>. Thus I would leave the expression for the relative time dilation as-is. However, if you can assume that <span class="math-container">$r_1, r_2 \gg r_s$</span> (much more likely to be true), then the terms can be binomially expanded and approximated as
<span class="math-container">$$\frac{d\tau_2}{d\tau_1} = \frac{f_1}{f_2} \simeq 1 + \frac{GM}{r_1 c^2} -\frac{3GM}{2r_2 c^2}\ , $$</span>
where <span class="math-container">$f_1$</span> and <span class="math-container">$f_2$</span> are the frequencies according to the stationary observer and the orbiting object.</p>
<p>You can see from this expression that depending on the exact values of <span class="math-container">$r_1$</span> and <span class="math-container">$r_2$</span> there may be a blueshift or a redshift, or even, when <span class="math-container">$r_2 = 1.5r_1$</span>, no shift at all.</p>
<p>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 (<span class="math-container">$r_2 &lt; 1.5r_1$</span>), whereas signals from a much higher GPS satellite (<span class="math-container">$r_2 &gt; 1.5r_1$</span>) are blueshifted.</p>
<p>Since both of these terms are of order <span class="math-container">$v^2/c^2$</span> 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 <span class="math-container">$v/c$</span>.</p>

CC BY-SA 4.0
2024-03-25T21:30:53.397Z
2024-03-25T21:40:38.333Z
2024-03-25T21:40:38.333Z
ProfRob
<p>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 <span class="math-container">$(v/c)^2$</span> where <span class="math-container">$c$</span> is the speed of light. The exact formula is</p>
<p><span class="math-container">$$\nu=\nu_0 \sqrt{1-\frac{v^2}{c^2}} $$</span></p>
<p>See this link for some additional explanation  <a href="https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Tatum)/15%3A_Special_Relativity/15.19%3A_The_Transverse_and_Oblique_Doppler_Effects" rel="nofollow noreferrer">https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Tatum)/15%3A_Special_Relativity/15.19%3A_The_Transverse_and_Oblique_Doppler_Effects</a></p>

CC BY-SA 4.0
2024-03-24T19:25:45.220Z
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)
Sten
CC BY-SA 4.0
2024-03-24T19:31:12.493Z
@Sten How would this have anything to do with the orbital velocities?
Thomas
CC BY-SA 4.0
2024-03-24T20:15:14.090Z
$v^2$ is proportional to the gravitational potential depth.
Sten
CC BY-SA 4.0
2024-03-24T20:22:48.673Z
Indeed, there is an orbital radius at which the net shift is zero.
ProfRob
CC BY-SA 4.0
2024-03-25T18:29:16.353Z
@Sten The OP was asking about the effect of the orbital velocity on the Doppler shift. There is no direct correlation between the gravitational frequency shift and the orbital velocity. It only depends on the masses and does not vary for an eccentric orbit. Even for zero relative velocity it would be the same. What's more, it depends on *both* masses, with one causing a redshift, the other a blueshift. For equal masses there would be no gravitational shift at all. The time dilation redshift in contrast *does* depend directly on velocity, but should obviously not be called Doppler shift.
Thomas
CC BY-SA 4.0
2024-03-25T20:01:48.600Z
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.
Sten
CC BY-SA 4.0
2024-03-25T21:35:58.527Z
In real life, if you're communicating with an Earth orbiting satellite, there's a significant doppler shift unless it's in a geosynchronous orbit.
John Doty
CC BY-SA 4.0
2024-03-25T22:47:22.167Z
@Sten The question was about the orbital motion causing a Doppler shift *("But since the body is moving, will doppler shift occur from the perspective of the central body?")* . There is zero causal connection between the gravitational frequency shift and the orbital motion, so bringing it in here is at least potentially very misleading. The time dilation shift in contrast *is* causally connected to the orbital motion (although calling it transverse "Doppler" shift is potentially misleading as well).
Thomas
CC BY-SA 4.0
2024-03-24T18:14:52.630Z
2024-03-25T18:30:36.297Z
2024-03-25T18:30:36.297Z
Thomas
<p>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?</p>

CC BY-SA 4.0
2024-03-24T17:47:17.980Z
This probably answers your question: https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#One_object_in_circular_motion_around_the_other
Pierre Paquette
CC BY-SA 4.0
2024-03-24T21:36:26.790Z
The orbiting body is the same distance from the the *center* of the central body.
John Doty
<orbit><light><doppler-effect>
Will a circular orbit cause doppler shift?
Astrovis
# Will a circular orbit cause doppler shift?

(Asked by: username_0 on 2024-03-24)

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?

**Tags:** orbit, light, doppler-effect

> **Comments:**
> **username_1**: This probably answers your question: https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#One_object_in_circular_motion_around_the_other  
> **username_2**: The orbiting body is the same distance from the the *center* of the central body.  

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## 2 answers

### Answer by username_3 on 2024-03-25. Score: 2 | Accepted answer

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 $v/c$.


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### Answer by username_4 on 2024-03-24. Score: 1

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>

> **Comments:**
> **username_5**: 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)  
> **username_4**: @username_5 How would this have anything to do with the orbital velocities?  
> **username_5**: $v^2$ is proportional to the gravitational potential depth.  
> **username_3**: Indeed, there is an orbital radius at which the net shift is zero.  
> **username_4**: @username_5 The OP was asking about the effect of the orbital velocity on the Doppler shift. There is no direct correlation between the gravitational frequency shift and the orbital velocity. It only depends on the masses and does not vary for an eccentric orbit. Even for zero relative velocity it would be the same. What's more, it depends on *both* masses, with one causing a redshift, the other a blueshift. For equal masses there would be no gravitational shift at all. The time dilation redshift in contrast *does* depend directly on velocity, but should obviously not be called Doppler shift.  
> **username_5**: 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.  
> **username_2**: In real life, if you're communicating with an Earth orbiting satellite, there's a significant doppler shift unless it's in a geosynchronous orbit.  
> **username_4**: @username_5 The question was about the orbital motion causing a Doppler shift *("But since the body is moving, will doppler shift occur from the perspective of the central body?")* . There is zero causal connection between the gravitational frequency shift and the orbital motion, so bringing it in here is at least potentially very misleading. The time dilation shift in contrast *is* causally connected to the orbital motion (although calling it transverse "Doppler" shift is potentially misleading as well).  

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