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We've discovered stars that are contact binaries. This means that they orbit so close together that their photosphere reaches through their Roche lobes and links the two stars together.

Contact binary pair

This started me thinking about the possibilities with respect to gas giants though, could in theory gas giants orbit each other also so close together that their atmospheres merge? Would they be able to hold together in that state or would they rip each other apart?

ArtOfCode
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Tim B
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3 Answers3

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Probably not.

A direct quote from here would be nice, but I'm not sure if there's any relevant copyright on the text, so I'll just summarize. Here are the steps to forming a contact binary star system:

  1. A binary star system forms
  2. After billions - perhaps only millions, in larger stars - of years, one of the stars becomes a giant.
  3. The giant grows and grows until its outer layers reach its Roche Lobe.
  4. One of two things can happen: If the other star, too, is large, the stars may quickly form a contact binary. If the other star is small, matter transfer from the larger star to the smaller star may take place and then a contact binary may form.

Here's why I don't think this could happen in planetary systems:

  • Planets don't grow much once they become full-fledged planets. They can shrink and release energy via the Kelvin-Helmholtz mechanism, but they generally can't grow. They can grow when they're in the protoplanet stage, or when they've just reached planethood - whenever that happens - and are still accreting excess material
  • Gas giants generally collect all the gas in their immediate area. This means that its more likely that one will form first and take all the gas from another . . . unless matter transfer happens, as detailed above.
  • There are a lot of collisions among protoplanets. It's unlikely that two massive planets will form near one another without colliding. They'll each have sustained collisions with other protoplanets and planetesimals; I would think these collisions should make them be at a higher risk for collision with each other.
  • Gas giants often migrate from the outer reaches of the stellar system, so in their early days, their orbits may be unstable.

Maybe a temporary common envelope could form at some point, though the system would be unstable.

We can calculate the approximate radius of the Roche lobe. Let's assume that the planets are both the same mass, so $M_1=M_2$ and $\frac{M_1}{M_2}=1$. Given the distance between the planets to be $A$, the radius of the Roche Lobe $r_R$ is $$r_R=\left(0.38+0.2 \log \frac{M_1}{M_2} \right) A$$ $$r_R=(0.38+0)A$$ $$r_R=0.38A$$ Let's say that the gas giants are just large enough and close enough that they reach the outer edges of their Roche lobes. We'll also say that they are about the same mass and radius as Jupiter. $$r_R=r_J=0.38A$$ $$A=2.61R_J$$ That's problematic, because we can also calculate the Roche limit of the planets - the point at which one will be torn apart by the other. In this case, the more massive of the two has a mass $M_p$, and the less massive has a mass $M_2$. $$d=1.26 R_J \left(\frac{M_p}{M_s} \right) ^{\frac{1}{3}}$$ $$d \approx. 1.26R_J$$ So they're way to close! The more massive one will tear the less massive one apart - in essence, they'll collide.

Regarding what TimB wrote in the bounty - I don't have the expertise necessary to do an analysis like that, so I'll leave that to someone else. I wouldn't be able to do it as well as it deserves.

HDE 226868
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  • Wouldn't they have to orbit each other as well? – bowlturner Jan 17 '15 at 15:10
  • @bowlturner The gas giants? Yes. Here, we'd be looking at a very short period. – HDE 226868 Jan 17 '15 at 15:12
  • Does the Roche limit behaviour change because they're gas giants though? There isn't really anything to tear apart since the cores are much further apart - they just end up sharing a common atmosphere. – Tim B Jan 18 '15 at 12:40
  • @TimB Hm. I don't know what would happen in that scenario. If the cores were within each other's Roche lobes, then one would be torn apart. I'm not sure what will happen here. – HDE 226868 Jan 18 '15 at 18:08
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    You can quote from anything you want. That would not be a copyright concern as it was definitely fall under "fair use". – Chris Harrison Mar 15 '15 at 06:55
  • @ChrisHarrison Ah, thanks for reminding about that. When I wrote this I had seen some copyright issues on other SEs, and as I typically check copyright before quoting, I was unsure about this one. – HDE 226868 Mar 15 '15 at 14:29
  • @HDE226868 Really? You had copyright issues? I'm quite surprised by that. You should be able to quote whatever you like (unless of course you were copying and pasting the whole page without using including any commentary of your own). I've found that people shout "copyright infringement!" all too easily. – Chris Harrison Mar 16 '15 at 05:11
  • @ChrisHarrison Not me - I always source my stuff - but others. I'll add in a section once I figure out what to add. – HDE 226868 Mar 16 '15 at 21:46
  • Is there any reason you used the rigid body equation for the Roche Limit calculation instead of the fluid body one? @HDE226868 – Austin A Jan 17 '18 at 21:24
  • @HDE226868 your conclusion doesn't make any sense. Writing the constraints as inequalities makes things more clear. R>0.38A <=> A<2.61R so the planets exceed their Roche lobes. A>1.26R separation is greater than Roche limit. So 1.26R<A<2.61R there is a huge range over which this works. It is much smaller if using the uniform density fluid approximation instead (2.44R<A<2.61R), but even still, non empty. – Lex Jan 17 '18 at 21:54
  • @AustinA The reason is that I was careless; I apologize for the error. I'll redo the calculations tomorrow for a fluid body treatment; thank you for catching this. – HDE 226868 Jan 17 '18 at 22:27
  • @HDE226868, Seems like the problems come from the fact that one is bigger than the other. But what if they were so close in size that the difference was negligible? Or IF you had two Jupiter duplicates (hand wave the odds of that) whose cores were NOT in each others Roche lobes, would the atmospheres share then? Would that create greater stability? – Len Jan 26 '18 at 19:04
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I am really not 100% sure on my math here, but I think it checks out:

Using HDE's numbers:

A=2.61RJ

d≈.1.26RJ

It seems logical to me that the cores will be outside their partner's Roche limit if the radius of each core is <= 2.61 - 1.26 * 2 = .09RJ * 2 = .18RJ.

Estimates of Jupiter's Core put it at 4-14% of Jupiter's Mass. Let's go with the 4% number because we're just seeing if this is feasible. If we assume the core is average with the mass of Jupiter, then it would need to be about .26RJ - so way big.

From here we should be able to calculate the minimum mass needed of the core (more mass = smaller in radius) for the cores to be outside the Roche radius. Unfortunately I need to head to dinner so I can't finish it up right now, I'll try and get some pen and paper later and work through the formulas. I don't think it's looking good though, the core would need to be really massive to get under that .18RJ.

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Dan Smolinske
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    This answer seems to be heading in the right direction to get the bounty but as you say is incomplete right now. I don't suppose you are able to finish it before the bounty expires? – Tim B Mar 10 '15 at 08:12
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While HDE 226868 has a bunch of reasons against them I don't think they add up to a total showstopper.

As he says, they can't form near each other--but that doesn't preclude a binary pair from a capture event.

Now suppose the capture ended up with a retrograde--they'll spiral in due to tidal forces. At some point you'll end up with a contact binary. Soon after that they go splat.

Loren Pechtel
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