# How would the ocean currents of a tidally locked super-Earth work? [closed]

Suppose for a moment that a large planet orbits an M0 red dwarf star (51% as wide, 60% as massive and only 7% as bright as our sun). It itself is 230% as wide and 700% as massive as Earth. There is no land on the surface. Like, at all. Atmosphere is 90 times thicker than Earth's (in other words, 100% as thick as Venus.)

Now, this SE is full of questions regarding the climate of tidally locked planets orbiting red dwarves, but what about ocean currents? How would they work on a tidally-locked super-Earth?

• How deep is the water? Is there land just a few feet down, or is the seafloor a mile beneath the surface?
– Tom
Feb 24, 2022 at 2:20
• This seems like it's highly dependent upon the geography the world in question. Can you edit this to be more specific about what you mean by "how the ocean currents work"? Feb 24, 2022 at 2:24
• Assuming this planet is in the habitable zone of its star? Frozen oceans make currents easy! Feb 24, 2022 at 2:55
• What does “work” mean exactly? This is a universe of answers. Focus on one problem that isn’t coming together in your story. Feb 24, 2022 at 3:00
• Your regular reminder that you've managed to accept answers on less than 10% of the last 60 questions you've asked, and that's just over the last few months. Consider perhaps going back and clarifying why the many answers you've been given are unacceptable, or actually accepting some of them. Maybe consider re-reading the relevant bits of the site documentation. Feb 24, 2022 at 8:20

## 2 Answers

Atmospheres are, for the most part, heated from below, with circulation driven by convection.

Oceans, on the other hand, absorb light more strongly, and are heated from the top--thus remaining stagnant unless other forces come into play.

Assuming that seafloor topography is negligible and heat transport is sufficient to prevent ice from dominating the dark side, your basic ocean currents will thus largely mirror the surface wind flows, to which they are coupled by friction. In very broad strokes, this means equatorial currents carrying water away from a center slightly east of the subsolar point, and polar currents carrying water back from the dark side to the dayside, slightly offset from a perfect longitudinal flow due to Coriolis effects and the greater inertia of water compared to the driving wind currents, with four major gyres spaced around the terminator.

They'd operate more predictably than they do on non-tidally-locked worlds

(The following to be spoken in your best Rod Serling voice...)

Imagine if you will1 a ball, held in place in front of a hair dryer with a block of ice behind it. You observe that water is on the surface of the ball and the heat from the hair dryer causes the warming water to expand, resulting in convective currents flowing to the back side of the ball, where the water is cooled by the ice. As the water pressure on the backside of the ball increases, water is forced back to the side with the hair dryer. You realize, to your horror, that the result looks suspiciously like currents.

You have a water world, so it's going to have gyers centered on the solar terminator. Heated water from the sun-side pushed toward the opposite-side. Cooled water pushes back.

Now, it's highly unlikely that your plant's non-water surface (you know, the ground beneath your keel) is smooth as a billiard ball. So, unless the water is always fairly deep, the undersea geography will cause more interesting currents. At a guess (and I'll admit, it's a bit of a guess on my part), shallower areas will tend to provide the channel for hot-to-cold currents while deeper areas will provide the cold-to-hot return.

So, while your oceanic currents would be much simpler than you'd find on a rotating planet with land, I don't think they'd be boring.

Just in case you were wondering, I'd expect that wind would basically do the same thing — but with complications. Evaporative moisture is part of the mix, meaning you'll have some honking massive storms at the terminator. I'm also ignoring things like ice at the poles, which will complicate both the currents and the wind. And I'm definitely avoiding clouds. But if you're to consider them, that's all to your good, right? Because they just add complexity to the equation.

1Rod Serling never actually said the phrase, "imagine if you will." But pretty much everyone who hears the phrase today hears it in Rod Serling's voice. You go Rod!