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I want to create a planet that has the longest possible length of time that the tide is low to allow land animals to forage in the exposed shoreline ecosystems.

What arrangement of star, planet and moon sizes and distances or other factors would allow a habitable earth-like planet to have the longest period of high and low tide?

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  • $\begingroup$ The timing and duration of tides and the tidal range depend on the topography of the shore and the general shape of the ocean. You may want to look into how those factors will make the most out of your astronomical setup. $\endgroup$
    – Cadence
    May 20, 2022 at 0:00
  • $\begingroup$ Give your moon a very, very slow orbit. $\endgroup$
    – Carlos Arturo Serrano
    May 20, 2022 at 6:04
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    $\begingroup$ Are you okay with having really long days to go along with your really long tides? The easiest way to do it would just be to make your planet rotate very slowly. $\endgroup$
    – N. Virgo
    May 20, 2022 at 6:06
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    $\begingroup$ @CarlosArturoSerrano: that's no right. You want the moon to have a slow orbit relative to the planet's rotation, so that the moon's orbit is almost geosynhcronous. $\endgroup$
    – TonyK
    May 20, 2022 at 14:44

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Gas giant moon, almost tidally locked.

Our moon travels around our planet and causes the tide as it goes pulling the ocean towards it. A smaller high tide occurs on the far side of the moon.

tide

https://scijinks.gov/tides/

Your planet is the moon of a gas giant. As it travels around its giant, the giant pulls the ocean towards it in a serious way. Your planet is almost tidally locked, rotating to face the giant in sync with its orbit around it. The same section of planet faces the giant for long periods of time.

Eventually your planet will be tidally locked and the high and low tide areas will be more or less fixed, changing only because of the more distant effects of the star. But it is not quite locked. The planet still rotates very slowly and so the high tide and low tide very gradually move across the latitudes.

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    $\begingroup$ A couple quick points: (1) While it's true that the high tide on the far side of the earth is smaller, it's by a very small amount -- and in the case of a moon of a gas giant this effect basically goes to zero with the increasing distance to the tide inducer, so the tides should be considered symmetric; (2) Powerful tides like this will tidally lock the moon very quickly, even more so with water sloshing around. Think <100 million years. So unless you want a very young system, this is a challenge. One possibility is to have had it locked in the recent past, and recently spun up by a collision. $\endgroup$
    – addaon
    May 19, 2022 at 22:38
  • $\begingroup$ Also, this answer is optimizing for size of tide. It's possible to get a normal-sized but slower tide in a more traditional planet/moon system if you're willing to relax constraints on the length of a day. $\endgroup$
    – addaon
    May 19, 2022 at 22:39
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    $\begingroup$ Tide speed is the difference between planet rotation (day length) and moon rotation speed ("month" length, if you want). On Earth, the moon orbits so slowly that the day term is dominant, and tides are roughly twice per day. If you want to slow down the tides, you have to either slow down the day to (nearly) match the month, or speed up the moon to make the month (nearly) match the day. Speeding up the moon brings it closer, and you can only go so far before you have a ring system instead of a moon. $\endgroup$
    – addaon
    May 19, 2022 at 22:44
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    $\begingroup$ @addon - I see. "Day" is shorthand for planet rotation speed, which because of the star brings day and night but also brings regions on the rotating planet underneath the moon. $\endgroup$
    – Willk
    May 19, 2022 at 23:19
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    $\begingroup$ One more idea that may help: As mentioned, any system with strong slow tides is going to want to be tidally locked. You'll have to spin up the planet at least a bit; and if you want day- or week- scale tides this is quite a bit of spin up. There is a solution that lets you have significant tides of choosable length in a tidally locked system, though! If you have a reasonably highly eccentric orbit the tidally-locked planet will have a tidal peak location facing the gas giant, but this peak will "slosh" during the orbit, in sync with the eccentricity seasons (which could be interesting). $\endgroup$
    – addaon
    May 20, 2022 at 0:52
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Just put the Moon really close to geostationary orbit.

If you have a moon exactly in a stationary orbit, the tides will not move... which means they might as well not exist. But if it's just a little bit higher or a little bit lower, then the tides will move very slowly--as fast as the moon appears to move relative to the ground. By adjusting the moon's altitude, and thus its orbital period, and more specifically the ratio of its orbital period to the planet's day, you can get the tides to move as slowly as you want (and thus for peaks and minima to last as long as you want), along with arbitrary day lengths.

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You could try something more earth-like, starting from Earth and Moon configuration, by playing with these factors:

  • Mass of moon: to increase or decrease the tidal range. Doubling the mass also doubles the pull. I don't how that translates into height of the water tides. I am not astrophysicist, but as far as I can tell, changing the mass of the moon without changing its orbital speed won't change its orbit.

  • Rotation speed of planet: Drawing on @addaon comments (if I understood it well), there will be two tides per "day", so increasing the length of the day makes the tides slower. If you want low tides of double the time in Earth, half its rotation speed.

  • Orbital speed and distance of moon: faster and closer for longer and stronger tides (what you are looking for), slower and further for the opposite. Halving the distance implies multiplying by four the orbital speed (four times longer tides) and the gravitational pull.

Corrections are welcome.

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