# Circulation patterns on a tidally-locked binary planet

A targeted spin-off from this question:

In a tidally-locked planet that rotates about a barycentric point that’s located outside of its own sphere, what would the coriolis forces be like? What would the circulation patterns look like, how similar or different might they be from those of Earth’s?

This needs to consider not only the direction of the centrifugal force, but those of the moving fluids (air, water weather patterns and mantle convection) as they move away from the sub-barycenter point to the half way great circle, and then converge again as they approach the antipode.

NOTE:

The coordinate systems refers to the sub-barycenter point as being one pole, and its antipode is the point opposite the sub-barycenter; the planet doesn’t have a rotation axis.

Something like winds and weather patterns driven by coriolis forces will expand over the surface of the planet until they reach the half-way mark where they will have expanded to their maximum size. This is also the half-way mark of a great circle (the shortest distance between two points on the surface of a sphere). Thereafter they will converge once more on their approach to the other, opposite pole, the antipode.

• In editing your question, I was stumped by this phrase " see the available area spread out". It didn't seem to fit into the flow of your reasoning. This may be more my problem than yours. So not knowing what changes were needed, I decided to let you as its author to make the necessary changes. Oh yes, and, it's a good question. I will look forward to the answers with interest. Oct 11, 2016 at 3:10
• @a4android the poles are points and the equator has maximum size. Something driven from one pole will have room to spread out as it moves away; past the half way point it is crowded back together as it approqches the other pole. (Here my coordinate systems refers to the sub-barycenter point as one pole; the planet doesn’t have a rotation axis.) Oct 11, 2016 at 12:48
• Any help improving the explaination would be apprciated. Oct 11, 2016 at 12:48
• Ask that question here: physics.stackexchange.com Oct 24, 2016 at 16:41
• This question appears less about world building and more about astronomy. Furthermore, it is plainly impossible to accurately predict the circulation patterns on a planet, considering we are unable to do that with precision for our own home planet. Distance to the parent star, type of parent star, density of atmosphere, gravity of the planet, composition of atmosphere, gravity of the planet, number and size of moons ... the variables are nearly endless and we don't even know with absolute confidence what effect and deeper implication each of them has alone and in conjunction with others. Oct 28, 2016 at 8:51

This is not a complete answer, but is some more accurate descriptions of the rotation-inducing forces.

To a first approximation, the angular momentum of a unit of air (or water) near the surface of the planet can be visualized by a croquet ball carved out of a piece of straight-grained wood: the central axis of the log is aligned with the rotation axis of the system.

If you carve a sphere out of the center of the log and then stain it, the grain will show contour lines representing points on the surface that are equal distances from the line of the rotation axis. You can carefully count rings starting from the center to label them, too.

In this (normal planet) case, the lines will follow the planet’s lattitude. Due to symmetry, every circle around a position of constant lattitude will have the same angular momentum. This causes air (or water) that changes lattitude to be deflected.

Now consider carving a ball out of one half of a log, representing the case where the rotation axis is outside of the sphere. Looking at a cross section through the equator you also see the sphere from above if you imagine the lines on where you know the surface is.

## todo illustration here

With a little imagination you can picture the resulting ball from another angle: the contour lines are very different.

First, they go in the opposite direction. You can see concentric loops around the antipode point and around the epibarycenter (baryepicenter?) point.

What this doesn’t show is that the angular momentum is not always parallel to the sphere’s surface, which is a considerably different effect.

Consider a contour line near the half-way point, and follow it around. At the north, the angular momentum is pointing parallel to the ground like we are used to. Likewise it is the same at the south. But at the spinward edge the spin vector is pointing straight up out of the ground, and at the antispinward edge it is pointing into the ground.

So, air (or water) transported to a different location will not simply be deflected east or west as on Earth, but up or down! In general, the deflection will be a complex vector combining a direction parallel to the contour line along the ground and a rising/falling component.

• We would love to see the "todo" illustration! Jun 15, 2019 at 13:17
• How's that illustration coming along?
– rek
Mar 21, 2022 at 2:45

I would think that the only thing that really matters is the orbital separation between the two components of the binary. That will set the orbital period of the binary-planet system. Since the planets are each tidally locked, their spin periods are the same as their orbital period. My understanding is that the Coriolis forcing on a planet's atmosphere is governed mainly by the spin rate. Fast spinning planets like Jupiter have a lot of convective (Hadley) cells between the equator and poles and so have a banded structure. Slow-spinning planets have fewer Hadley cells. Very slow-spinning planets would only have one. Having fewer Hadley cells should (I think) homogenize the temperature across the planet because the poles and equator would be in closer thermal contact.

I would expect the following correlation: Distant binary orbit -> long orbital period -> slow-spinning planets -> more uniform latitudinal temperature distribution across planet

Another thing to consider is that the binary planet is (I presume) orbiting a star, which means that the length of each planet's day is also the same as its spin period. And the energy deposited by the star is surely much much larger than the energy deposited by the other planet (by tides I would expect). So, I think the biggest factor of the other planet is simply in determining the spin rate. Tides must also play a role to some degree but that is likely to be on longer timescales, e.g. by making the planets move apart from each other. And if the planets themselves orbit relatively close to the star then star-planet-planet tidal effects would make this even more complicated.

• This would be a perfectly good answer had the OP not tagged his question hard-science. That tag is merciless and I'm not a fan of it. However, he tagged it that way and even gave readers fair warning. Consequently, this is a low-quality response due to its lack of mathematics and authority to backup the answer. Sorry.
– JBH
Dec 9, 2017 at 2:58

Forgive me for this kinda poor answer, but it's a start...

Assuming our Earth system, but with a mini-Earth (here called Lua) for a moon...

The Coriolis force on weather would be greatly weakened. Like mentioned in this question, The tidal lock would skew the Earth and Lua into a more egg like shape. This relative 'mountain' bulge would help dampen the winds, making the weather a bit more stable.

Of course heat distribution would also be a key driving force. So do we speed up Lua or slow Earth or both when this tidal lock is formed? I'm going to assume speed up Lua and compensate as needed for simplicity. Although, since this Lua is bigger, it's affect on the weather will be amplified. Yet always occurring on the same half of the planet means that the side locked to the other will experience colder temperatures and less wind. This area should become its own arctic at the equator.

This cold spot will actually mix with the air that regularly gets it's full days worth of sunlight, and create weather patterns similar to a planet tidally locked with a star (but much calmer and cooler). (wind patterns)

Weather however is complicated and hard to predict in normal cases. This is more of a guideline of points to consider, and an estimated guess on the result.

• This is not a hard-science answer. Maybe you can post these thoughts on the linked question or another “normal” question on the topic. Nov 2, 2016 at 23:16
• @JDługosz I'll make it a community wiki. I figured since it's been almost a month with no answers, maybe a little nudge would help? Nov 3, 2016 at 1:41