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First, I'd like to take the definition of a "double-planet" as two bodies orbiting each other where the center of gravity is not inside the larger body. Also, the system would have to fill other planet requirements (like emptying it's own orbital area around the star).

Now, let's say that the "binary system" is in an orbit close enough to a star to be tidally locked to the star. How would the tidal effects of the star's gravity effect the orbit of the planets around each other??

Let's take an example of a "double-planet" (of $0.075$ and $0.030\:\mathrm{M_{Earth}}$ with a separation of $1.17 \times 10^6\:\mathrm{km}$ orbiting their CM in about $450\:\mathrm{days}$. They are $0.085\:\mathrm{AU}$ from a star of mass $0.35\:\mathrm{M_{Sun}}$ and thus orbit the star in $15.3\:\mathrm{days}$. The planetary rotation/co-orbital period is much longer than the orbital period around the star, so I would think that the planets would have characteristics of a lone TL planet.

I am wondering how such a system would evolve over time. Would someone on the surface of one of the planets experience experience night and day in 15.3 day cycles? Would the difference in tidal effects of the star on the different sized planets cause the orbit to "precess" or have some other effect?

Imagine designing a long term calendar for this system!!

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  • $\begingroup$ This is a three body problem. There is no non-numerical solution. $\endgroup$ – adrianmcmenamin Sep 12 '17 at 8:31
  • $\begingroup$ A long term calendar would be pretty easy to design. It'd be pretty short and the last day on the calendar would say "Planet Destruction Day!". $\endgroup$ – zephyr Sep 12 '17 at 14:30
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Your scenario isn't stable. A simple way to explain this is to imagine that the planet's orbit each other at the same rate they orbit the star (your scenario has them orbiting even slower).

At the same rate of orbit, the synodic period essentially approaches infinity. (see diagram of the Moon's synodic orbit).

enter image description here

When this happens, the inner planet is at L1 to the outer planet and the outer planet is at L2 to the inner planet. That's ofcourse, not quite right as the different masses would have different Hill Spheres, and the mutual gravitation of two objects massive to have a barrycenter outside the more massive planet would combine into a slightly faster orbit, but it's close enough to demonstrate the instability.

The L1 and L2 points are on the border of the Hill Sphere and well outside the true region of stability.

No system can be stable where the Moon (or binary planet system) orbits around each other slower than they orbit around the central star. They need to be at least 50% to 67% closer in than that would allow.

Using the Orbital Period or "T" squared = distance ("a" semi-major axis) cubed, per Kepler's laws, the upper limit for a stable orbital period of a moon around a planet or a binary planet would be about 19%-35% the period of the planet's year. Any longer the Moon or binary system would risk instability.

You might be able to set up a synchronous system where the planets orbited each other 4 times for every orbit around the sun, where they'd end up in the same (basically eclipse) position every perihelion with the planet experiencing greater tides (smaller and/or more fluid planet) closer to the sun at perihelion. I suspect a 4-1 orbital period ratio is about as close as you're likely to get for any length of time for a stable binary planet-star, tidal locked system and that would be odd, but I see no reason why it wouldn't be stable. It would only be possible around a small star where the planets were fairly close so that all 3 could have significant tidal effects on each other. Without strong tides you don't get tidal locking.

You also can't have tidal locking with both objects when they have different orbital periods. Tidal locking means that the rotation period is equal to the (sidereal) orbital period.

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  • $\begingroup$ Of course, I should have known that past L1 the star has a stronger attraction than the planet. I didn't do the math. This makes me think of a follow up question, however. I will ask it as a separate question. $\endgroup$ – Jack R. Woods Sep 25 '17 at 21:16

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