<p>Binary stars are usually in stable orbits due to energy conservation: since they do not lose energy, they stay in the same orbit. What can happen over their lifespan is that one of them becomes a red giant and the extended atmosphere transfers mass to the other, changing their orbits, or (more importantly for this question) that tidal forces and gas drag in the envelope causes them to spiral in. Over very long timescales they may also lose energy due to gravitational radiation. However, these are slow processes. There is <em>a lot</em> of energy and angular momentum to somehow get rid of.</p>
<p>For a semi-major axis <span class="math-container">$a$</span> the energy is <span class="math-container">$$E=-\frac{GM_1M_2}{2a}.$$</span> Interlopers with energy comparable to this could disrupt the system, but the question is instead about causing a merger.</p>
<p>To get a merger the closest distance has to be within <span class="math-container">$r_{min}=R_1+R_2$</span> where <span class="math-container">$R$</span> is the stellar radius. <span class="math-container">$r_{min}=(1-e)a$</span>, where <span class="math-container">$e$</span> is the eccentricity, which needs to be driven up near 1 so <span class="math-container">$e&gt; 1 - (R_1+R_2)/a$</span>. For <span class="math-container">$a=$</span> 1 AU and <span class="math-container">$R_1=R_2=$</span> 1 solar radius, <span class="math-container">$e&gt;0.9907$</span> is required.</p>
<p>The eccentricity is <span class="math-container">$e=\sqrt{1+2\epsilon h^2/\mu^2}$</span> where <span class="math-container">$\epsilon$</span> is the total energy divided by the reduced mass <span class="math-container">$M_1M_2/(M_1+M_2)$</span>, <span class="math-container">$h$</span> the angular momentum divided by the reduced mass, and <span class="math-container">$\mu=G(M_1+M_2)$</span>. If we want to boost it to a high eccentricity we need to change things so <span class="math-container">$1+2(\epsilon+\Delta \epsilon)(h+\Delta h)^2/\mu^2 = e^2$</span>.</p>
<p>If we start with <span class="math-container">$e=0$</span> we have <span class="math-container">$1 +2\epsilon h^2/\mu^2=0$</span>, or <span class="math-container">$\mu^2/2=-\epsilon h^2$</span> (remember that the energy is negative for bound orbits). So to first order, <span class="math-container">$2\Delta \epsilon (h^2 + 2h \Delta h) \approx e^2 \mu^2$</span> or for <span class="math-container">$e\approx 1$</span> <span class="math-container">$\Delta \epsilon \approx \mu^2/h^2$</span> if we ignore the change in angular momentum (we set <span class="math-container">$\Delta h=0$</span>). This is essentially a change on the order of <span class="math-container">$E$</span>.</p>
<p>So if you want a rogue planet to hit and cause a merger, it needs to have about as much kinetic energy as one of the stars - not an easy task, since even Jupiter is 1/1000 of a solar mass, so it would need to move at about 200 km/s to equal the kinetic energy of a solar mass star in a 1 AU orbit.</p>
<p>Were it to happen the time until the merger would be <a href="https://en.wikipedia.org/wiki/Free-fall_time" rel="nofollow noreferrer">the free-fall timescale</a> <span class="math-container">$$\tau = \frac{\pi a^{3/2}}{\sqrt{2G(M_1+M_2)}},$$</span> which in this case is 0.25 year, or about 91 days.</p>

CC BY-SA 4.0
2021-10-01T11:19:58.187Z
This is very illuminating~! Thanks for showing me the Math for this! 200km/s seems like a reasonable speed for a rogue planet I suppose. But that said Jupiter is still quite large. More food for thought I suppose. Thanks again!
Glitch
CC BY-SA 4.0
2021-10-03T16:22:52.940Z
In an inelastic collision between a star and planet, momentum ($mv$) is conserved, kinetic energy ($1/2mv^2$) doesn't get canceled out.  So I think your velocity estimate for a rogue planet to stop a star is too small.  If the resultant energy release exceeds the star's gravitational binding energy, it won't hold together.
Connor Garcia
CC BY-SA 4.0
2021-09-30T22:28:14.220Z
2021-09-30T22:28:14.220Z
Anders Sandberg
<p>The third 'invading' celestial body passing by or crashing into one of the stars could be a possible reason for the orbit collapse. Something that throws the stars* off-course speeding up their death dance. Is an orphaned/rogue planet capable of this? If so how long would it take? If I were to consider extreme events, how 'fast' can this process occur?**</p>
<p>*Main Sequence stars</p>
<p>**Say from the moment the third body starts to affect the system significantly till the first collision of the main bodies of the stars.</p>
<p>I hope this question is not off-topic. This is my first time asking on this site.</p>

CC BY-SA 4.0
2021-09-30T13:03:33.437Z
It depends on the mass and distance from the binary. Also, rogue planets will almost always be hyperbolic and won't be much of an influence.
WarpPrime
CC BY-SA 4.0
2021-09-30T13:43:26.250Z
*Interesting question!* My guess is that you could construct a scenario with a binary pair of low mass stars orbiting their center of mass at a very large distance (i.e. moving relatively slowly) and a third, much more massive start passing close by one of them sending it on an intercept course with the other. In other words, this is possible and the minimum time could be less than one orbital period. The more closely they orbit, the harder it would be for a passing third body to affect them *differently enough* that it sends them into each other.
uhoh
CC BY-SA 4.0
2021-09-30T13:43:42.987Z
So I would think the answer could be very roughly of order 1 year or even less, but that's just a guess, not an answer.
uhoh
CC BY-SA 4.0
2021-09-30T17:44:15.067Z
@uhoh The question is about planets, though, not more massive stars.
Peter Erwin
CC BY-SA 4.0
2021-09-30T17:46:40.883Z
@PeterErwin I saw "a third body" and "third 'invading' celestial body" and missed the planet specifier later. So it's a lot less possible then.
uhoh
CC BY-SA 4.0
2021-10-01T11:24:11.413Z
Apologies for my jargon. I'm not well versed in these terms. As for the question itself, yes it's a near impossible scenario. But I still wanted to see if it was possible, even if improbable. I'm quite lacking in orbital mathematics so I couldn't figure out whether this could be possible on my own.
Glitch
<binary-star><impact><rogue-planet><orbital-migration>
What can be the minimum time for a somewhat stable twin star orbit to collapse on being affected by a third body?
Glitch
# What can be the minimum time for a somewhat stable twin star orbit to collapse on being affected by a third body?

(Asked by: username_0 on 2021-09-30)

The third 'invading' celestial body passing by or crashing into one of the stars could be a possible reason for the orbit collapse. Something that throws the stars\* off-course speeding up their death dance. Is an orphaned/rogue planet capable of this? If so how long would it take? If I were to consider extreme events, how 'fast' can this process occur?\*\*

\*Main Sequence stars

\*\*Say from the moment the third body starts to affect the system significantly till the first collision of the main bodies of the stars.

I hope this question is not off-topic. This is my first time asking on this site.

**Tags:** binary-star, impact, rogue-planet, orbital-migration

> **Comments:**
> **username_1**: It depends on the mass and distance from the binary. Also, rogue planets will almost always be hyperbolic and won't be much of an influence.  
> **username_2**: *Interesting question!* My guess is that you could construct a scenario with a binary pair of low mass stars orbiting their center of mass at a very large distance (i.e. moving relatively slowly) and a third, much more massive start passing close by one of them sending it on an intercept course with the other. In other words, this is possible and the minimum time could be less than one orbital period. The more closely they orbit, the harder it would be for a passing third body to affect them *differently enough* that it sends them into each other.  
> **username_2**: So I would think the answer could be very roughly of order 1 year or even less, but that's just a guess, not an answer.  
> **username_3**: @username_2 The question is about planets, though, not more massive stars.  
> **username_2**: @username_3 I saw "a third body" and "third 'invading' celestial body" and missed the planet specifier later. So it's a lot less possible then.  
> **username_0**: Apologies for my jargon. I'm not well versed in these terms. As for the question itself, yes it's a near impossible scenario. But I still wanted to see if it was possible, even if improbable. I'm quite lacking in orbital mathematics so I couldn't figure out whether this could be possible on my own.  

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## 1 answers

### Answer by username_4 on 2021-09-30. Score: 3 | Accepted answer

Binary stars are usually in stable orbits due to energy conservation: since they do not lose energy, they stay in the same orbit. What can happen over their lifespan is that one of them becomes a red giant and the extended atmosphere transfers mass to the other, changing their orbits, or (more importantly for this question) that tidal forces and gas drag in the envelope causes them to spiral in. Over very long timescales they may also lose energy due to gravitational radiation. However, these are slow processes. There is *a lot* of energy and angular momentum to somehow get rid of.

For a semi-major axis $a$ the energy is $$E=-\frac{GM\_1M\_2}{2a}.$$ Interlopers with energy comparable to this could disrupt the system, but the question is instead about causing a merger.

To get a merger the closest distance has to be within $r\_{min}=R\_1+R\_2$ where $R$ is the stellar radius. $r\_{min}=(1-e)a$, where $e$ is the eccentricity, which needs to be driven up near 1 so $e> 1 - (R\_1+R\_2)/a$. For $a=$ 1 AU and $R\_1=R\_2=$ 1 solar radius, $e>0.9907$ is required.

The eccentricity is $e=\sqrt{1+2\epsilon h^2/\mu^2}$ where $\epsilon$ is the total energy divided by the reduced mass $M\_1M\_2/(M\_1+M\_2)$, $h$ the angular momentum divided by the reduced mass, and $\mu=G(M\_1+M\_2)$. If we want to boost it to a high eccentricity we need to change things so $1+2(\epsilon+\Delta \epsilon)(h+\Delta h)^2/\mu^2 = e^2$.

If we start with $e=0$ we have $1 +2\epsilon h^2/\mu^2=0$, or $\mu^2/2=-\epsilon h^2$ (remember that the energy is negative for bound orbits). So to first order, $2\Delta \epsilon (h^2 + 2h \Delta h) \approx e^2 \mu^2$ or for $e\approx 1$ $\Delta \epsilon \approx \mu^2/h^2$ if we ignore the change in angular momentum (we set $\Delta h=0$). This is essentially a change on the order of $E$.

So if you want a rogue planet to hit and cause a merger, it needs to have about as much kinetic energy as one of the stars - not an easy task, since even Jupiter is 1/1000 of a solar mass, so it would need to move at about 200 km/s to equal the kinetic energy of a solar mass star in a 1 AU orbit.

Were it to happen the time until the merger would be [the free-fall timescale](https://en.wikipedia.org/wiki/Free-fall_time) $$\tau = \frac{\pi a^{3/2}}{\sqrt{2G(M\_1+M\_2)}},$$ which in this case is 0.25 year, or about 91 days.

> **Comments:**
> **username_0**: This is very illuminating~! Thanks for showing me the Math for this! 200km/s seems like a reasonable speed for a rogue planet I suppose. But that said Jupiter is still quite large. More food for thought I suppose. Thanks again!  
> **username_5**: In an inelastic collision between a star and planet, momentum ($mv$) is conserved, kinetic energy ($1/2mv^2$) doesn't get canceled out.  So I think your velocity estimate for a rogue planet to stop a star is too small.  If the resultant energy release exceeds the star's gravitational binding energy, it won't hold together.  

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