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Picture the classic Earth-like tidally locked planet, with a frozen dark side, a boiling stormy light side, and a thin strip of habitable land along the terminator. The planet is similar in size and composition to Earth.

The year of the planet lasts around 10 Earth days, and it experiences a certain amount of libration (just as our Moon does), resulting in somewhat of a “day-night cycle” (the sun just goes up and down a few degrees over the horizon).

If this planet had a moon, I assume it would be also tidally locked to the star, which means no lunar tides. So what about solar tides? Well, I guess the “bulge” would always be facing to the sun, so the habitable coasts of the planet would be always in low tide, without fluctuation.

So, how can I have high and low tides on my planet?

Edit: the question isn’t about whether a tidally locked planet could or could not have a moon, but how —by what means— could the planet experience tides. By having a moon orbiting it is a valid answer, I guess, but I’m sure there are many other forms to have tides.

(And by tides I really mean periodical and predictable rising and lowering of the waters of the planet. Tidal forces seem the more plausible reason to me, but it could be periodical thaw of the glaciers, drastic evaporation, something to do with the Coriolis effect maybe?... I don’t know.)

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    $\begingroup$ If the planet had a moon, it would orbit the planet normally. If it was similar to Earth's moon, it would be tidally locked to the planet, not to the sun. $\endgroup$
    – Cadence
    Sep 23 at 22:04
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    $\begingroup$ And a world tide-locked to its sun won't keep a moon for an astronomically-long length of time. Tidal interactions would crank it in past the Roche limit, and tear it apart. $\endgroup$
    – notovny
    Sep 23 at 22:20
  • $\begingroup$ Tidal interactions might well disrupt a moon, but it depends on the relative sizes. In the case of our Moon it could approach to roughly 12,000 miles before being disrupted by Earth's gravity. $\endgroup$
    – Slarty
    Sep 23 at 23:00
  • $\begingroup$ I don't believe this quesiton is ready for prime time. If the moon orbits the planet, there will be tides (and that would be pretty cool, causing ice breakups along the equator and currents between the hemispheres that wouldn't exist without the moon). If the moon doesn't orbit the planet (realistic or not) then there aren't any tides. What's the actual question here? Is the moon orbiting or not? This question would have benefitted from our Sandbox. $\endgroup$
    – JBH
    Sep 24 at 1:54

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Elliptical orbit around the sun.

Your planet is Earth sized. And fairly well whipping around its sun if a year is 10 earth days. From OP:

The planet is similar in size and composition to Earth.

The year of the planet lasts around 10 Earth days,

I take it that the planet is close to the star. That would go with it being tidally locked as well.

Let us say your planet is not always the same distance from its sun. It has an elliptical orbit.

https://www.nasa.gov/audience/forstudents/5-8/features/nasa-knows/what-is-orbit-58.html

elliptical orbit

The tide generating force varies by the cube of distance.

https://oceanservice.noaa.gov/education/tutorial_tides/tides02_cause.html

Tidal forces are based on the gravitational attractive force. With regard to tidal forces on the Earth, the distance between two objects usually is more critical than their masses. Tidal generating forces vary inversely as the cube of the distance from the tide generating object. Gravitational attractive forces only vary inversely to the square of the distance between the objects

Our sun is 27 million times larger than our moon. Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth. If tidal forces were based solely on comparative masses, the sun should have a tide-generating force that is 27 million times greater than that of the moon. However, the sun is 390 times further from the Earth than is the moon. Thus, its tide-generating force is reduced by 3903, or about 59 million times less than the moon. Because of these conditions, the sun’s tide-generating force is about half that of the moon

The orbit does not need to be dramatically elliptical because the effect of distance is cubed and your planet is already very close as planets go. In a 10 day year there will thus be a high tide on the near approach and low tide when the planet is at maximum distance.

How exactly tides manifest when one side of the world is frozen is a matter to take up in your story. It will be interesting!

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  • $\begingroup$ The eccentricity of the orbit determines the amount of libration. I don't know the math, but I suspect that, with a sufficiently elliptical orbit, you could be tidally locked and still see daylight over most of the planet. $\endgroup$
    – David G.
    Sep 25 at 17:02

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