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Is it possible for an artificial satellite (in this case, an unmanned communications platform) to maintain a stable orbit directly around the barycenter of a Binary Star system, as in well inside the orbital paths of the two stars themselves?

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  • $\begingroup$ Quick clarification - how long for and how much/little mass do you want to expend "station keeping"? With a sufficiently handwavy "unlimited" drive you could easily have an artificial satellite that stays on a less stable orbit for longer than one that has no drives that's on a more stable orbit. See here:en.wikipedia.org/wiki/Orbital_station-keeping $\endgroup$ – Miller86 Jan 3 at 9:20
  • $\begingroup$ Well I was admittedly looking for an orbit that could remain stable on it's own, but I'm looking to use a large fleet of communication satellites all stationed around the barycenters of various binary systems. Given that each binary system would have its own properties, likely wildly different from one another, I could see station keeping being implemented for some of the more extreme cases. Thanks for the suggestion. $\endgroup$ – White76Knight Jan 3 at 10:33
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The barycenter is the common point around which both stars appear to orbit.

But it is only an artefact of their gravitational relationship, not a real thing. Any object located at the barycenter would not be affected in any special way, and there is certainly no mechanism by which it could naturally orbit about that point.

If both stars have equal masses, then the gravitational forces from each star are equal, and so in theory an object could remain stationary at that point. In practice though, it would need to be continually adjusting its position as the balance point is unstable, and even the slightest perturbation in the system would produce positive feedback and subject it to an ever increasing gravitational attraction toward one star.

If the stars have different masses, the barycenter is even more meaningless for the proposed situation. Consider our solar system, with everything but the Sun and Jupiter removed from it. The barycenter would be much closer to the Sun than to Jupiter (in fact only about 50,000 km from its surface) and so the gravitational force of the sun would far outweigh that of Jupiter. Any object placed there would immediately fall into the Sun. That the starting point happened to be the barycenter would be irrelevant.

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I would take a look at Lagrangian points. Areas of relative stability between 2 large bodies and a smaller 3rd one. There is a point (not the barycenter) where the gravity forces generally cancel out known as the L1 point. The L1 point is between the 2 suns, and is going to be somewhat closer to the smaller of the 2 suns. If they are relatively the same size, the L1 point will be very close to the barycenter, if the mass difference is large, the L1 point will be rather close to the smaller star. For example, the earth/sun L1 point is .01 AU (or 1/100th of the average distance between the sun and earth) away from the earth.

The L1 point is technically unstable, but even current tech considers station-keeping necessary to be fairly minor. There are several other locations that will give you a general "stable orbit" known as the L4 and L5 points, but they tend to be roughly the same orbit as the larger stars.

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  • $\begingroup$ I hadn't considered the Lagrangian Points, but they just might work. How would one calculate the distance of the L1 point between the two suns? $\endgroup$ – White76Knight Jan 3 at 21:03
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    $\begingroup$ The wiki I linked has the math... but it is quite beyond me sadly. A rough aproximation can be found by finding the point where the gravity strength is the same. Gravity goes up directly according to mass (so a star twice as heavy will have twice as much gravity) and goes down based on distance squared. (so if you get twice as far the gravity is only 1/4 as strong) if Star A is 4 times the mass of star B, the L1 point is going to be twice as far from Star A as Star B. If they are the same size the L1 point would be in the middle. Non-circular orbits are super super messy. $\endgroup$ – Corbin Matheson Jan 4 at 15:12

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