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


1 Answer 1



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. 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.

The time it will take for the objects to collide can be calculated: $$t=\frac{5}{256}\frac{c^5}{G^3}\frac{r^4}{(m_1m_2)(m_1+m_2)}$$ 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 could offset some of the effects.

  • $\begingroup$ 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? $\endgroup$ Jul 25, 2015 at 23:05
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    $\begingroup$ 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. $\endgroup$ Jul 25, 2015 at 23:11
  • $\begingroup$ 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? $\endgroup$
    – user541686
    Jul 26, 2015 at 0:53
  • $\begingroup$ @WillihamTotland That number is, I would think, accurate. Like I wrote, the effect is non-negligible on most scales. $\endgroup$
    – HDE 226868
    Jul 26, 2015 at 1:37
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    $\begingroup$ @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. $\endgroup$
    – Stan Liou
    Jul 26, 2015 at 1:47

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