<p>You check the periodic frequency shift with the Doppler effect. The light from the two stars is not very suitable because each light source moves. Much better: The binary system (total mass <span class="math-container">$m_A\approx 2m_{\odot}$</span>) generates a gravitational wave with high power that appears to come from the <em>stationary</em> center of gravity of both stars. There is no source of error when evaluating the GW.</p>
<p>Let's assume that this GW can be measured and that the two suns take 2.315 days to orbit. Then the pair emits a GW with <em>constant</em> frequency <span class="math-container">$f_{GW}=10~\mu$</span>Hz. If a planet <span class="math-container">$B$</span> orbits the pair at a distance of <span class="math-container">$r_B$</span>, the binary system orbits the common center of gravity of the entire system with radius <span class="math-container">$r_A$</span>. Because of <a href="https://en.wikipedia.org/wiki/Center_of_mass" rel="nofollow noreferrer">center of gravity theorem</a> <span class="math-container">$m_A\cdot r_A=m_B\cdot r_B$</span> applies. Because the planet forces the center of mass of the binary system to move, the instantaneous value of <span class="math-container">$f_{GW}$</span> is sometimes larger and sometimes smaller than the average value. The maximum frequency deviation is</p>
<p><span class="math-container">$\Delta f = f_{GW} {\left(\sqrt{\frac{v_{GW}+v_{orbit}}{v_{GW}-v_{orbit}}}-1 \right)}$</span></p>
<p>With <span class="math-container">$\Delta f$</span> you can calculate: a) at what speed does the binary system rotate around the common center of gravity of the triple.
b) at what speed does the planet rotate around the common center of gravity of the triple.</p>
<p>With these intermediate results you can calculate the orbital period and the mass of the planet.</p>

CC BY-SA 4.0
2023-11-19T19:54:21.513Z
Gravitational waves from planets are not detectable with the current generation of GW detectors. The current detector can only detect gravitational waves from compact bodies (black holes, neutron stars) and only when they are in the final stages of an inspiral. While it might be possible *in theory* to detect GW from regular binaries, it is far far beyond current technology.
James K
CC BY-SA 4.0
2023-11-20T08:29:53.623Z
This would be state of the art for a long time if all configurations (e.g. the ground pressure of the atmosphere or gravimeters) were systematically tested to see how strongly they react to GW and at which frequencies. Instead of experimenting, astronomers philosophize.
9herbert9
CC BY-SA 4.0
2023-11-19T18:40:25.733Z
2023-11-19T18:40:25.733Z
9herbert9
<p>The light from the two stars would be Doppler shifted in a sinusoidal pattern (for a circular orbit). The signals from the two stars would be in anti-phase and would oscillate at the orbital period of the binary.</p>
<p>The planet would cause a small pull on the binary system which would superimpose a further small sinusoidal signal (again, assuming a circular orbit), but at the longer orbital period of the planet around the binary system.</p>
<p>The process is illustrated below by adding two (red and blue) sine waves together, where the blue sine wave has a much smaller amplitude and lower frequency than the red sine wave. These sine waves encode the Doppler shift that would be measured from the absorption lines of one of the binary components. Thus the red wave would indicate the Doppler shift due to the motion of the binary system and the blue sine wave the additional modulation due to the smaller motion induced by the tug of the planet as it orbits. The black line then gives the superposition of these (they can basically be added if the velocities are non-relativistic) and you can see that it is subtly different from a monochromatic sine wave of fixed amplitude.</p>
<p><a href="https://i.sstatic.net/RoTRt.png" rel="noreferrer"><img src="https://i.sstatic.net/RoTRt.png" alt="enter image description here" /></a></p>
<p>For non-circular orbits, the pattern of Doppler shifts would be non-sinusoidal but still periodic in a similar way.</p>

CC BY-SA 4.0
2022-11-17T13:36:44.567Z
Your diagram illustrates perfectly why stiff glassware on a stiff surface is susceptible to reproducing powerful AM transmissions - I never understood that phenomenon before.
dotancohen
CC BY-SA 4.0
2022-11-17T13:58:34.950Z
@dotancohen actually my comparison with amplitude modulation was erroneous. The signals are added rather than being multiplied together.
ProfRob
CC BY-SA 4.0
2022-11-17T19:20:37.960Z
see also Fourier Transforms applied to Doppler shifts, position, intensity of the received light...
Sarah Messer
CC BY-SA 4.0
2022-11-16T20:59:38.990Z
2022-11-17T13:57:50.237Z
2022-11-17T13:57:50.237Z
ProfRob
<p>I read in Neil deGrasse Tyson's book <a href="https://en.wikipedia.org/wiki/Astrophysics_for_People_in_a_Hurry" rel="noreferrer">Astrophysics for People in a Hurry</a> that scientists can tell if a star has a planet orbiting it because the light appears to shake.</p>
<p>So if in the case of a binary star which is quite obviously shaking because its orbiting another star how would I know a planet is orbiting the 2 stars?</p>

<observational-astronomy><exoplanet><spectroscopy><binary-star>
How would a person know if a planet is orbiting a binary star?
Tardy
# How would a person know if a planet is orbiting a binary star?

(Asked by: username_0 on 2022-11-16)

I read in Neil deGrasse Tyson's book [Astrophysics for People in a Hurry](https://en.wikipedia.org/wiki/Astrophysics_for_People_in_a_Hurry) that scientists can tell if a star has a planet orbiting it because the light appears to shake.

So if in the case of a binary star which is quite obviously shaking because its orbiting another star how would I know a planet is orbiting the 2 stars?

**Tags:** observational-astronomy, exoplanet, spectroscopy, binary-star


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## 2 answers

### Answer by username_3 on 2022-11-16. Score: 18 | Accepted answer

The light from the two stars would be Doppler shifted in a sinusoidal pattern (for a circular orbit). The signals from the two stars would be in anti-phase and would oscillate at the orbital period of the binary.

The planet would cause a small pull on the binary system which would superimpose a further small sinusoidal signal (again, assuming a circular orbit), but at the longer orbital period of the planet around the binary system.

The process is illustrated below by adding two (red and blue) sine waves together, where the blue sine wave has a much smaller amplitude and lower frequency than the red sine wave. These sine waves encode the Doppler shift that would be measured from the absorption lines of one of the binary components. Thus the red wave would indicate the Doppler shift due to the motion of the binary system and the blue sine wave the additional modulation due to the smaller motion induced by the tug of the planet as it orbits. The black line then gives the superposition of these (they can basically be added if the velocities are non-relativistic) and you can see that it is subtly different from a monochromatic sine wave of fixed amplitude.

[![enter image description here](https://i.sstatic.net/RoTRt.png)](https://i.sstatic.net/RoTRt.png)

For non-circular orbits, the pattern of Doppler shifts would be non-sinusoidal but still periodic in a similar way.

> **Comments:**
> **username_4**: Your diagram illustrates perfectly why stiff glassware on a stiff surface is susceptible to reproducing powerful AM transmissions - I never understood that phenomenon before.  
> **username_3**: @username_4 actually my comparison with amplitude modulation was erroneous. The signals are added rather than being multiplied together.  
> **username_5**: see also Fourier Transforms applied to Doppler shifts, position, intensity of the received light...  

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### Answer by username_1 on 2023-11-19. Score: -3

You check the periodic frequency shift with the Doppler effect. The light from the two stars is not very suitable because each light source moves. Much better: The binary system (total mass $m\_A\approx 2m\_{\odot}$) generates a gravitational wave with high power that appears to come from the *stationary* center of gravity of both stars. There is no source of error when evaluating the GW.

Let's assume that this GW can be measured and that the two suns take 2.315 days to orbit. Then the pair emits a GW with *constant* frequency $f\_{GW}=10~\mu$Hz. If a planet $B$ orbits the pair at a distance of $r\_B$, the binary system orbits the common center of gravity of the entire system with radius $r\_A$. Because of [center of gravity theorem](https://en.wikipedia.org/wiki/Center_of_mass) $m\_A\cdot r\_A=m\_B\cdot r\_B$ applies. Because the planet forces the center of mass of the binary system to move, the instantaneous value of $f\_{GW}$ is sometimes larger and sometimes smaller than the average value. The maximum frequency deviation is

$\Delta f = f\_{GW} {\left(\sqrt{\frac{v\_{GW}+v\_{orbit}}{v\_{GW}-v\_{orbit}}}-1 \right)}$

With $\Delta f$ you can calculate: a) at what speed does the binary system rotate around the common center of gravity of the triple.
b) at what speed does the planet rotate around the common center of gravity of the triple.

With these intermediate results you can calculate the orbital period and the mass of the planet.

> **Comments:**
> **username_2**: Gravitational waves from planets are not detectable with the current generation of GW detectors. The current detector can only detect gravitational waves from compact bodies (black holes, neutron stars) and only when they are in the final stages of an inspiral. While it might be possible *in theory* to detect GW from regular binaries, it is far far beyond current technology.  
> **username_1**: This would be state of the art for a long time if all configurations (e.g. the ground pressure of the atmosphere or gravimeters) were systematically tested to see how strongly they react to GW and at which frequencies. Instead of experimenting, astronomers philosophize.  

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