<p>No. Gravitational waves are generated by (according to wikipedia):</p>
<blockquote>
<p>the motion or acceleration of gravitating masses</p>
</blockquote>
<p>Not all gravitational motions result in merging. For instance, the Moon's orbit around the Earth generates gravitational waves, and the Moon is moving away from Earth at about an inch per year. There are plenty of orbits like this in the universe that won't result in coalescence, and these all emit gravitational waves. Since you specifically say that a situation without ejection, this question is essentially asking whether a perfect orbit is possible, which is answered at this question:
<a href="https://physics.stackexchange.com/questions/821872/can-perfectly-stable-orbits-exist-in-gr">https://physics.stackexchange.com/questions/821872/can-perfectly-stable-orbits-exist-in-gr</a></p>

CC BY-SA 4.0
2024-08-18T22:46:45.550Z
what about the orbits I put in the OP? @Astrovis
vengaq
CC BY-SA 4.0
2024-08-19T23:18:47.997Z
@vengaq, these orbits also cause the motion or acceleration of gravitating masses, so they would not be infinitely stable.
Astrovis
CC BY-SA 4.0
2024-08-20T06:51:27.300Z
Ignoring what happens when the sun becomes a red giant, or possible interactions with other bodies, tidal drag will stop raising the moon's altitude once the earth's rotation slows enough to be tidally locked to the moon.  After that point, the moon would slowly lose altitude due to GW radiation.
Scott McPeak
CC BY-SA 4.0
2024-08-18T18:33:45.813Z
2024-08-18T18:33:45.813Z
Astrovis
<p>Another factor is things have only a finite lifetime. In the case of Earth orbiting the Sun (reported by <a href="https://en.wikipedia.org/wiki/Gravitational_wave" rel="nofollow noreferrer">Wikipedia</a>):</p>
<blockquote>
<p>[T]he total energy of the Earth orbiting the Sun (kinetic energy + gravitational potential energy) is about <span class="math-container">$1.14×10^{36}$</span> joules of which only <span class="math-container">$200$</span> watts (joules per second) is lost through gravitational radiation, leading to a decay in the orbit by about <span class="math-container">$1×10^{−15}$</span> meters per day or roughly the diameter of a proton.</p>
</blockquote>
<p>So, ignoring other effects such as the Sun losing mass which would make the orbit spiral <em>outward</em>, it would take almost 30 billion years for the orbit to spiral inward by ten millimeters through gravitational wave emission. Long before that time the Sun will have burned up its hydrogen fuel, passed through its red giant phase, and consumed the Earth through tidal drag during said red giant phase. Trust me, it's gonna get ugly!</p>
<p>The reason for this low inspiral rate is that essentially the relative loss of radius due to gravitational wave emission is proportional <span class="math-container">$v^8/c^8$</span>, where <span class="math-container">$v$</span> is the orbital velocity (we have <span class="math-container">$dr/(r\space dt)\sim1/r^4$</span> and <span class="math-container">$r\sim1/v^2$</span> at low velocities). This may be compared with <span class="math-container">$v^2/c^2$</span> for &quot;first-order&quot; relativistic effects such as time dilation. Therefore, gravitational-wave emissions have very little impact unless the orbital velocity is a quite large fraction of the speed of light. Such orbital velocities typically require a pair of tightly orbiting stars.</p>

CC BY-SA 4.0
2024-08-19T00:57:31.670Z
I'm aware that this would take extremely long time scales but I'm interested in the end result, rather than the time it would take to get there. Having said this, in other comments in this question eshaya mentioned that If one star in a binary system were disrupted into a gaseous ring, then the quadrupole momentum would be invariant and the pair would not emit gravity waves. I also linked to some 3-body orbits in OP. Would any of these avoid coalescence (assuming they are completely isolated so no external perturbations modify the system)? @OscarLanzi
vengaq
CC BY-SA 4.0
2024-08-19T19:48:50.630Z
@vengaq Anything in those time scales necessary for gravitational waves to make a difference is no longer a star. None of the current solutions to the 3-body problem have been observed in nature. So are you looking to know if gravitational waves will eventually destabilize an otherwise stable orbit? There are 3 end results total; merger, ejection, and stable. Stable is the only configuration that hasn't been observed. The linked examples require equal masses and zero disturbance to the orbit. Possibly a better question for the physics SE.
David S
CC BY-SA 4.0
2024-08-19T00:37:14.060Z
2024-08-19T10:52:00.460Z
2024-08-19T10:52:00.460Z
Oscar Lanzi
<p>All orbits emit gravitational waves since there is an asymmetric distribution which makes the quadrupole momentum not invariant</p>
<p>This means that over long time scales, orbiting bodies would emit gravitational waves and as a result, they would eventually coalesce. However, is this always the case? Or are there some orbital configurations where the members don't necessarily coalesce (without ejecting any member of the system)?</p>
<p>Are any of these orbits examples of this?:</p>
<p><a href="https://commons.wikimedia.org/wiki/File:Three-body_Problem_Animation_with_COM.gif#mw-jump-to-license" rel="nofollow noreferrer">https://commons.wikimedia.org/wiki/File:Three-body_Problem_Animation_with_COM.gif#mw-jump-to-license</a></p>
<p><a href="https://commons.wikimedia.org/wiki/File:Three_body_problem_figure-8_orbit_animation.gif#mw-jump-to-license" rel="nofollow noreferrer">https://commons.wikimedia.org/wiki/File:Three_body_problem_figure-8_orbit_animation.gif#mw-jump-to-license</a></p>
<p><a href="https://commons.wikimedia.org/wiki/File:5_4_800_36_downscaled.gif#mw-jump-to-license" rel="nofollow noreferrer">https://commons.wikimedia.org/wiki/File:5_4_800_36_downscaled.gif#mw-jump-to-license</a></p>

CC BY-SA 4.0
2024-08-18T19:35:01.070Z
Orbiting systems emit because the quadrupole is NOT invariant, it rotates.  If one star in a binary system were disrupted into a gaseous ring, then it would be invariant and the pair would not emit gravity waves.
eshaya
CC BY-SA 4.0
2024-08-18T21:10:23.057Z
whoops, that was a mistake @eshaya what would happen with that ring in that situation? would it fall into the other star? Also, what about the orbits in the linked I put in OP?
vengaq
CC BY-SA 4.0
2024-08-18T22:55:03.717Z
Not in any meaningful way.  The loss of energy to gravitational waves in systems with non-degenerate matter is so small that the time scale for the disruption of the system to multi-body instability is vastly shorter that the timescale to any perceptible loss of orbital energy to gravitational waves.(unless you assume the two bodies are the only things in the universe)
antlersoft
CC BY-SA 4.0
2024-08-18T23:00:36.793Z
@antlersoft exactly I was assuming these systems were isolated. For example in the other comment eshaya mentioned that If one star in a binary system were disrupted into a gaseous ring, then the quadrupole momentum would be invariant and the pair would not emit gravity waves. I also linked to some 3-body orbits in OP. Would any of these avoid coalescence (assuming they are completely isolated so no external perturbations modify the system)?
vengaq
CC BY-SA 4.0
2024-08-19T01:02:57.050Z
Why are you using those "#mw-jump-to-license" links?
PM 2Ring
CC BY-SA 4.0
2024-08-19T06:37:02.950Z
@PM2Ring I got them from wikipedia
vengaq
<orbit><orbital-mechanics><general-relativity><gravitational-waves><orbital-resonance>
Will all orbits emitting gravitational waves inevitably coalesce?
vengaq
# Will all orbits emitting gravitational waves inevitably coalesce?

(Asked by: username_0 on 2024-08-18)

All orbits emit gravitational waves since there is an asymmetric distribution which makes the quadrupole momentum not invariant

This means that over long time scales, orbiting bodies would emit gravitational waves and as a result, they would eventually coalesce. However, is this always the case? Or are there some orbital configurations where the members don't necessarily coalesce (without ejecting any member of the system)?

Are any of these orbits examples of this?:

<https://commons.wikimedia.org/wiki/File:Three-body_Problem_Animation_with_COM.gif#mw-jump-to-license>

<https://commons.wikimedia.org/wiki/File:Three_body_problem_figure-8_orbit_animation.gif#mw-jump-to-license>

<https://commons.wikimedia.org/wiki/File:5_4_800_36_downscaled.gif#mw-jump-to-license>

**Tags:** orbit, orbital-mechanics, general-relativity, gravitational-waves, orbital-resonance
> **Comments:**
> **username_1**: Orbiting systems emit because the quadrupole is NOT invariant, it rotates.  If one star in a binary system were disrupted into a gaseous ring, then it would be invariant and the pair would not emit gravity waves.  
> **username_0**: whoops, that was a mistake @eshaya what would happen with that ring in that situation? would it fall into the other star? Also, what about the orbits in the linked I put in OP?  
> **username_2**: Not in any meaningful way.  The loss of energy to gravitational waves in systems with non-degenerate matter is so small that the time scale for the disruption of the system to multi-body instability is vastly shorter that the timescale to any perceptible loss of orbital energy to gravitational waves.(unless you assume the two bodies are the only things in the universe)  
> **username_0**: @antlersoft exactly I was assuming these systems were isolated. For example in the other comment eshaya mentioned that If one star in a binary system were disrupted into a gaseous ring, then the quadrupole momentum would be invariant and the pair would not emit gravity waves. I also linked to some 3-body orbits in OP. Would any of these avoid coalescence (assuming they are completely isolated so no external perturbations modify the system)?  
> **username_3**: Why are you using those "#mw-jump-to-license" links?  
> **username_0**: @PM2Ring I got them from wikipedia  

---

## 2 answers

### Answer by username_4 on 2024-08-18. Score: 5

No. Gravitational waves are generated by (according to wikipedia):

> the motion or acceleration of gravitating masses

Not all gravitational motions result in merging. For instance, the Moon's orbit around the Earth generates gravitational waves, and the Moon is moving away from Earth at about an inch per year. There are plenty of orbits like this in the universe that won't result in coalescence, and these all emit gravitational waves. Since you specifically say that a situation without ejection, this question is essentially asking whether a perfect orbit is possible, which is answered at this question:
<https://physics.stackexchange.com/questions/821872/can-perfectly-stable-orbits-exist-in-gr>

> **Comments:**
> **username_0**: what about the orbits I put in the OP? @Astrovis  
> **username_4**: @vengaq, these orbits also cause the motion or acceleration of gravitating masses, so they would not be infinitely stable.  
> **username_5**: Ignoring what happens when the sun becomes a red giant, or possible interactions with other bodies, tidal drag will stop raising the moon's altitude once the earth's rotation slows enough to be tidally locked to the moon.  After that point, the moon would slowly lose altitude due to GW radiation.  

---

### Answer by username_6 on 2024-08-19. Score: 4

Another factor is things have only a finite lifetime. In the case of Earth orbiting the Sun (reported by [Wikipedia](https://en.wikipedia.org/wiki/Gravitational_wave)):

> [T]he total energy of the Earth orbiting the Sun (kinetic energy + gravitational potential energy) is about $1.14×10^{36}$ joules of which only $200$ watts (joules per second) is lost through gravitational radiation, leading to a decay in the orbit by about $1×10^{−15}$ meters per day or roughly the diameter of a proton.

So, ignoring other effects such as the Sun losing mass which would make the orbit spiral *outward*, it would take almost 30 billion years for the orbit to spiral inward by ten millimeters through gravitational wave emission. Long before that time the Sun will have burned up its hydrogen fuel, passed through its red giant phase, and consumed the Earth through tidal drag during said red giant phase. Trust me, it's gonna get ugly!

The reason for this low inspiral rate is that essentially the relative loss of radius due to gravitational wave emission is proportional $v^8/c^8$, where $v$ is the orbital velocity (we have $dr/(r\space dt)\sim1/r^4$ and $r\sim1/v^2$ at low velocities). This may be compared with $v^2/c^2$ for "first-order" relativistic effects such as time dilation. Therefore, gravitational-wave emissions have very little impact unless the orbital velocity is a quite large fraction of the speed of light. Such orbital velocities typically require a pair of tightly orbiting stars.

> **Comments:**
> **username_0**: I'm aware that this would take extremely long time scales but I'm interested in the end result, rather than the time it would take to get there. Having said this, in other comments in this question eshaya mentioned that If one star in a binary system were disrupted into a gaseous ring, then the quadrupole momentum would be invariant and the pair would not emit gravity waves. I also linked to some 3-body orbits in OP. Would any of these avoid coalescence (assuming they are completely isolated so no external perturbations modify the system)? @OscarLanzi  
> **username_7**: @vengaq Anything in those time scales necessary for gravitational waves to make a difference is no longer a star. None of the current solutions to the 3-body problem have been observed in nature. So are you looking to know if gravitational waves will eventually destabilize an otherwise stable orbit? There are 3 end results total; merger, ejection, and stable. Stable is the only configuration that hasn't been observed. The linked examples require equal masses and zero disturbance to the orbit. Possibly a better question for the physics SE.  

---

