<p><strong>Yes.</strong></p>

<p>Two bodies in orbit around each other will inevitably collide. The reason for this is that the system will give off energy in the form of <a href="http://en.wikipedia.org/wiki/Gravitational_wave" rel="nofollow">gravitational waves</a>. This effect is commonly cited in binary neutron star systems, where the two stars are isolated and close together. One of the most famous of these systems is the <a href="http://en.wikipedia.org/wiki/PSR_B1913%2B16" rel="nofollow">Hulse-Taylor binary</a>.</p>

<p>The time it will take for the objects to collide <a href="http://en.wikipedia.org/wiki/Gravitational_wave#Orbital_lifetime_limits_from_gravitational_radiation" rel="nofollow">can be calculated</a>:
$$t=\frac{5}{256}\frac{c^5}{G^3}\frac{r^4}{(m_1m_2)(m_1+m_2)}$$
where $r$ is the initial radius, $m_1$ and $m_2$ are the masses of the bodies, and $c$ and $G$ are the familiar constants, the speed of light in a vacuum and Newton's universal gravitational constant.</p>

<p><em>However</em>, <a href="http://en.wikipedia.org/wiki/Tidal_acceleration" rel="nofollow">tidal acceleration</a> could offset some of the effects.</p>

CC BY-SA 3.0
2015-07-25T23:05:43.473Z
Surely that's the absolute upper bound given no input of energy, not "the time"? I haven't done the math, but it seems to me that the formula provided is wont to spit out _hilariously_ huge numbers; to the point where stuff like passing stars and, more importantly, drag in the interplanetary medium, would have a noticeable effect?
Williham Totland
CC BY-SA 3.0
2015-07-25T23:11:02.473Z
Actually, I _did_ do the math for Sol / Terra; giving me, assuming I managed to plug everything in correctly, 10 trillion times the current age of the universe. So, you know, a _hilariously_ huge number.
Williham Totland
CC BY-SA 3.0
2015-07-26T00:53:16.740Z
Would this depend on whether the universe is closed or open? Like, if the universe is closed, then couldn't the gravitational waves come "back" to the same place? And in such a case, wouldn't the system potentially never lose energy?
user541686
CC BY-SA 3.0
2015-07-26T01:37:13.627Z
@WillihamTotland That number is, I would think, accurate. Like I wrote, the effect is non-negligible on most scales.
HDE 226868
CC BY-SA 3.0
2015-07-26T01:38:28.783Z
@Mehrdad I have no idea, but if the universe *was* closed - which all evidence indicates is not true - then it would take a very, very, very long time for the gravitational wave to somehow go back. The bodies might have already collided by then, and, in most cases, would have. If, that is, the Earth survived that long, which will not happen. In most systems, this would not be a problem. Perhaps in toy universes, but not in ours. None of the right conditions exist.
HDE 226868
CC BY-SA 3.0
2015-07-26T01:47:31.260Z
@Mehrdad their refocusing and absorption by the system is of nigh-infinitesimal probability. But to answer your question, the formula given is based on an a circular orbit in an otherwise empty and asymptotically flat spacetime. The contributions to the emitted radiation has "instantaneous" terms (really dependent on the retarded position) and "non-local" terms (dependent on prior history), which are smaller. Ignoring the latter and taking the leading-order post-Newtonian approximation of the should get us the result in the answer.
Stan Liou
CC BY-SA 3.0
2015-07-26T09:52:10.450Z
What assumption is made about the radiuses of the two bodies? This could be important in a close binary star system.
Peter Mortensen
CC BY-SA 3.0
2015-07-26T13:22:19.583Z
@PeterMortensen I don't think that the radii of the bodies themselves are considered, though mass loss/gain would be. The orbits, however, must be circular (or approximately so).
HDE 226868
CC BY-SA 3.0
2015-07-28T16:02:04.260Z
@barrycarter I think no such orbit exists.
HDE 226868
CC BY-SA 3.0
2015-07-25T22:37:22.380Z
2015-07-28T17:47:47.107Z
2015-07-28T17:47:47.107Z
HDE 226868
<p>If two celestial bodies are in orbit, will they always eventually collide if not acted upon by outside forces?</p>

<orbit>
Do all orbiting bodies eventually collide?
Douglas
# Do all orbiting bodies eventually collide?

(Asked by: username_0 on 2015-07-25)

If two celestial bodies are in orbit, will they always eventually collide if not acted upon by outside forces?

**Tags:** orbit

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## 1 answers

### Answer by username_1 on 2015-07-25. Score: 13 | Accepted answer

**Yes.**

Two bodies in orbit around each other will inevitably collide. The reason for this is that the system will give off energy in the form of [gravitational waves](http://en.wikipedia.org/wiki/Gravitational_wave). This effect is commonly cited in binary neutron star systems, where the two stars are isolated and close together. One of the most famous of these systems is the [Hulse-Taylor binary](http://en.wikipedia.org/wiki/PSR_B1913%2B16).

The time it will take for the objects to collide [can be calculated](http://en.wikipedia.org/wiki/Gravitational_wave#Orbital_lifetime_limits_from_gravitational_radiation):
$$t=\frac{5}{256}\frac{c^5}{G^3}\frac{r^4}{(m\_1m\_2)(m\_1+m\_2)}$$
where $r$ is the initial radius, $m\_1$ and $m\_2$ are the masses of the bodies, and $c$ and $G$ are the familiar constants, the speed of light in a vacuum and Newton's universal gravitational constant.

*However*, [tidal acceleration](http://en.wikipedia.org/wiki/Tidal_acceleration) could offset some of the effects.

> **Comments:**
> **username_2**: Surely that's the absolute upper bound given no input of energy, not "the time"? I haven't done the math, but it seems to me that the formula provided is wont to spit out _hilariously_ huge numbers; to the point where stuff like passing stars and, more importantly, drag in the interplanetary medium, would have a noticeable effect?  
> **username_2**: Actually, I _did_ do the math for Sol / Terra; giving me, assuming I managed to plug everything in correctly, 10 trillion times the current age of the universe. So, you know, a _hilariously_ huge number.  
> **username_3**: Would this depend on whether the universe is closed or open? Like, if the universe is closed, then couldn't the gravitational waves come "back" to the same place? And in such a case, wouldn't the system potentially never lose energy?  
> **username_1**: @WillihamTotland That number is, I would think, accurate. Like I wrote, the effect is non-negligible on most scales.  
> **username_1**: @Mehrdad I have no idea, but if the universe *was* closed - which all evidence indicates is not true - then it would take a very, very, very long time for the gravitational wave to somehow go back. The bodies might have already collided by then, and, in most cases, would have. If, that is, the Earth survived that long, which will not happen. In most systems, this would not be a problem. Perhaps in toy universes, but not in ours. None of the right conditions exist.  
> **username_4**: @Mehrdad their refocusing and absorption by the system is of nigh-infinitesimal probability. But to answer your question, the formula given is based on an a circular orbit in an otherwise empty and asymptotically flat spacetime. The contributions to the emitted radiation has "instantaneous" terms (really dependent on the retarded position) and "non-local" terms (dependent on prior history), which are smaller. Ignoring the latter and taking the leading-order post-Newtonian approximation of the should get us the result in the answer.  
> **username_5**: What assumption is made about the radiuses of the two bodies? This could be important in a close binary star system.  
> **username_1**: @PeterMortensen I don't think that the radii of the bodies themselves are considered, though mass loss/gain would be. The orbits, however, must be circular (or approximately so).  
> **username_1**: @barrycarter I think no such orbit exists.  

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