In binary star systems, could there be a planet the stars revolve around, with eternal day on all sides?

1st scenario: Imagine a binary system consisting of two Sunlike G5V stars of 1 solar mass each, orbiting each other, and at the barycenter between them is a planet (which would make the planet itself revolve at the same horizontal speed around its axis, making each sun appear always above the same location on the planet). If the stars revolved around the planet both sides would be illuminated similarly. I don't see why it should be impossible. A planet that once was the outermost one in orbit around one of the stars got ejected from its orbit by the other star's gravity, migrating into the barycenter between them.

2nd scenario: Imagine there's a planet at the Lagrangian point between Alpha Centauri A and B. If the stars revolved around the planet both sides would be illuminated similarly. Would that be possible?

Do situations like those above occur or was such even observed?

  • $\begingroup$ @antispinwards I expanded my question. $\endgroup$ – Giovanni Aug 15 '20 at 11:45

Could there be a planet at the barycenter between two or more stars revolving around each other?


The best case two star scenario is two stars of equal mass. In that case, the barycenter is midway between the two stars and coincides with the L1 Lagrange point. The L1 Lagrange point is metastable. Another name for metastable is unstable. Think of it as a very sharp pencil standing straight up. In theory, a pencil can be stood straight up. In practice, it falls over in a very short time.

If one of the two stars is more massive than the other the barycenter isn't even metastable. The barycenter is closer to the more massive star than it is to the less massive star, which in turn means the gravitational acceleration toward the more massive star is greater than it is to the less massive star. The object at the barycenter will orbit the more massive star at a rate greater then the two stars orbit one another. The less massive star will be a mere perturbation.

The same applies to more than two stars. While there are theoretical balanced on the tip of a standing up pencil that are metastable, these points are a space of measure zero. In other words, there is zero chance of this happening.

  • $\begingroup$ The James Webb telescope is to be placed at the L-point between the Earth and Sun. Why would that be stabler? $\endgroup$ – Giovanni Aug 16 '20 at 5:13
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    $\begingroup$ @Giovanni The JWST will be in a halo orbit around the L2 point, which puts the Earth between it and the Sun, and it will need to spend a few m/s of delta-V per year to stay near that point. Once the station keeping fuel runs out, it will remain in a solar orbit, but it won't stay near L2. $\endgroup$ – notovny Aug 16 '20 at 6:18
  • $\begingroup$ @notovny Thank you. Though if I understand you correctly, the JWST will be on the other side, where the Earth is between them, not the telescope between the Earth and Sun? $\endgroup$ – Giovanni Aug 16 '20 at 7:23
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    $\begingroup$ @Giovanni - The vicinity of the Sun-Earth L1 point is a good location for a spacecraft that observes the Sun, but not so good for observing the universe. The vicinity of the Sun-Earth L2 point is a good location for a spacecraft that observes the universe. Note that I wrote "vicinity of". The linear Lagrange points (L1, L2, and L3) are metastable. An immense amount of fuel would be needed to have a spacecraft stay at these points. It turns out that there are pseudo orbits about these points that are much less unstable in the sense that much less fuel is needed for stationkeeping. (continued) $\endgroup$ – David Hammen Aug 16 '20 at 11:19
  • $\begingroup$ The JWST will be placed in a halo orbit (one of three different kinds of pseudo orbits) about the Sun-Earth L2 point. The way to visualize these pseudo orbits is to imagine a frame of reference that rotates with the Earth's orbit about the Sun and that stretches and contracts as the Earth moves further from and closer to the Sun. The position of the Sun and Earth are fixed from this perspective. From this perspective, objects in one of these pseudo orbits (halo orbits, Lissajous orbits, and Lyapunov orbits) appear to orbit one of the Lagrange points. $\endgroup$ – David Hammen Aug 16 '20 at 11:25

No. Such an arrangement is at best "metastable". That is, although there are periodic solutions to the three body problem (stable orbits) an infintesimal perturbation (eg the proverbial butterfly flapping its wings) will push the system off the stable orbit and into chaos. Getting a planet to remain at the barycentre is like trying to balance a pencil on its sharpened point.

With two bodies, each orbits around the barycentre. But with three bodies, the bodies don't orbit around the three-way barycentre. And a planet placed near the barycentre of two stars will not tend to remain in orbit around that point.

The Lagrangian point L1 is also at best metastable. Satellites that orbit the sun at the Earth-Sun Lagrangian point need to fire their engines and do regular "station keeping" to keep them from drifting away.

The L4 and L5 points can be stable. Bodies at the L4 and 5 points are called "Trojans". However no Trojan exoplanets are known. A Trojan planet would see the two stars separated by (a variable amount averaging at) 60 degrees

  • $\begingroup$ But what if the planet was at the very barycenter between two stars of equal mass? I mean like the barycenter of the system being in the center of the planet's core. Would this really be unstable? $\endgroup$ – Giovanni Aug 15 '20 at 12:13
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    $\begingroup$ One way to see why it’s unstable is to think about what happens if the planet is pushed a tiny bit one way or the other. If the gravitational pulls of the two stars are exactly equal, then a perturbation that pushes the planet even a tiny bit one way will cause the pull of one star to now be stronger than the other, and the planet will keep going in that direction. You might say “but I will keep it at the center - I won’t allow it to be perturbed!” But that’s impossible unless you are in an empty universe; there are always tiny perturbations from other stars. $\endgroup$ – Eric Jensen Aug 15 '20 at 12:56
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    $\begingroup$ It's worth noting that the $\rm L_4$ and $\rm L_5$ points are only stable for mass ratios $m_1 / m_2 \gtrsim 25$, which is a fairly extreme mass ratio for binary stars. $\endgroup$ – user24157 Aug 15 '20 at 22:46

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