On Earth, tides generally tie to a 12 or 24 hour cycle depending on the location. The exact times and heights are generally based on the shape of the sea, and the positions of the moon and sun.

What I'm wondering is if there's a planetary system setup that would give very long cycles - say at least a month - between high and low tides.

Additional requirements:

  1. The planet needs to be habitable by humans (habitable zone of the star, or otherwise kept warm/cold enough).
  2. Tides should be at least as extreme (in terms of height between high/low tides) as on Earth. More extreme is better.
  3. The planet should be of roughly the same composition as Earth - no making it out of diamond and then sticking it inside the Roche limit of a gas giant or the like.
  4. No magic.

Note: It seems like you could just change the period of the moon's orbit to get this, but I believe that would also involve changing the moon's orbit and mass, which would change tides in other ways. So it's not quite as simple as that.

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    $\begingroup$ Wouldn't this just be dictated by the rotation speed of the plant? A shorter day would give you a shorter cycle and vise versa? $\endgroup$
    – James
    Mar 12, 2015 at 21:13
  • $\begingroup$ @James: To some extent, but a really long day might not be conductive to human life - it would get much hotter during the day, and extremely cold at night. Also at some point you have to take into account the moon's orbital period - if your day gets too long that becomes the determining factor, and I'm not sure how short or long you can make those cycles and keep everything kosher for the inhabitants. I would be fine with a long day though, as long as it's not too crazy. $\endgroup$ Mar 12, 2015 at 21:16
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    $\begingroup$ I've tried a scenario where the moons rotation around earth is similar to the earths rotation, keeping the moon over the same portion of the planet longer (not quite tidal locked, it leaves the moon visible in the sky for half the month), which should increase the tide as the moon is over that part of the earth longer and influences the tide gravitationally for longer (end result is kind of a double tide, one from the suns affect and the other from the moon). Unfortunately the speed required for the moon to do this launches it off into space. $\endgroup$
    – Twelfth
    Mar 12, 2015 at 21:41
  • $\begingroup$ What about a highly eccentric orbit? $\endgroup$ Mar 13, 2015 at 11:37
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    $\begingroup$ @Pureferret: That's an interesting thought. My only concern is stability - I think it might be difficult to have a long-lasting eccentric orbit between a planet, a moon and the parent star. $\endgroup$ Mar 13, 2015 at 14:25

6 Answers 6


There are two formulas at play here. Since you want the tides to be similar or more extreme in magnitude than on Earth, we need to consider the tidal force (thanks to celtschk for the correction) between the planet and moon: $$\overrightarrow{a}_t\rm{(axial)}\approx\pm\hat{r} 2\Delta rG\frac{m}{r^3}$$ Also, since you want to change the tide cycles, we need to consider the orbital period of the planet and moon: $$T=2\pi\sqrt{\frac{a^3}{G(m_1+m_2)}}$$ Here, $r$ is the distance between the two objects, and $a$ is the semi-major axis of the orbit. If we assume an approximately circular orbit, $a=r$, so we get $$T=2\pi\frac{r^{3/2}}{\sqrt{G(m_1+m_2)}}$$ In order to get a longer tidal cycle, we need to have an orbital period closer to the planet's rotational period, so we need to change the moon to decrease $T$ while keeping $a$ constant (or increasing).

For a simple case, let's cut the radius of the orbit in half. In order to keep the force the same, we'll need to divide the moon's mass by 8. What effect does this have on the orbital period? Well, if we assume the moon's mass is relatively negligible (our moon is about 1% of the mass of Earth) we have $$T_1=2\pi\frac{(\frac12r)^{3/2}}{\sqrt{G(m_1+\frac18m_2)}} =2\pi\frac{\frac{1}{\sqrt{8}}r^{3/2}}{\sqrt{G(m_1+m_2)}}=\frac{1}{\sqrt{8}}T$$ That means the moon orbits about $2.8$ times faster.

If we go with 1/9th the radius and 1/729th of the mass, we get an orbit $27$ times faster, or about $1.1$ days. This gives you a tidal cycle of about $10$ days, as opposed to $30$, but the ratios to get a $1.03$ day orbit aren't as nice. (Edit: it's about 1/9.35 for the radius, and 1/817.4 for the mass.) You can go with a little more mass to increase the magnitude of the tides.

Also, while this moon is a lot closer than ours, it is still well outside the Earth's Roche limit. Also, it has a faster orbit but is below escape velocity for that distance. So it won't break apart or go flying off into space.

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    $\begingroup$ The height of the tides is not determined by the total force, but by the tidal force (inhomogeneity of the gravitational field), which goes with 1/r^3. $\endgroup$
    – celtschk
    Mar 13, 2015 at 1:25
  • $\begingroup$ Yay, thank you for pulling in the math that I didn't have time to write in yesterday. +1 $\endgroup$ Mar 13, 2015 at 12:44
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    $\begingroup$ @celtschk Really? Well, crud. My math is all wrong. At least the general concept still works. I'll redo that math and edit this answer later. Hopefully this doesn't put the moon above escape velocity. $\endgroup$
    – KSmarts
    Mar 13, 2015 at 13:39
  • $\begingroup$ Now it looks much better, +1. Interestingly, this means that, assuming an unchanged density of the moon, the apparent size of the moon would remain the same (the moon would be smaller, but also closer; since the volume goes with $r_{\text{Moon}}^3$, the ration $r_{\text{Moon}}/r$, which is responsible for the apparent size, remains constant). $\endgroup$
    – celtschk
    Mar 13, 2015 at 23:16

I'm sorry that my knowledge of how this would work is a bit limited...but here is a possibility.

You can have your day/night cycles however you'd like...but the planet would have to be almost tidally locked to its moon. The mechanics of this dictate that it would only stay that way for so long before the planet and moon became truly tidally locked to each other...but it'd still be a pretty darn long time.

In short, if a planet's day and lunar month were almost the same (which, given enough time, would happen to any planet/moon system), then you could have a very slow tidal cycle. The catch here, is that the planet's day would be about the same length as the tidal cycle, which leads to a whole host of problems for any inhabitants of the planet.

To have a stable orbit, the distance from the planet, mass of the orbiting bodies, and speed they are moving all must fit together. For a moon to move faster (thus catching up with a planet's rotation speed) and not escape the planet, it must either be more massive, or closer to the surface of the planet. Either of which also increase the strength of the tides, possibly to the point of leaving the planet uninhabitable.

The balance to this is to slow the rotation of the planet, lengthening the day, rather than decreasing the lunar month. This would lead to greater extremes in temperature between day and night (just look at the temperature variance between day and night on Earth).

So, ultimately, you'd need to play a balancing act of a moon that was massive enough and close enough to effect the tides, and then speed it up or slow down planetary rotation to reach an equilibrium where the lunar month was just a little bit longer than a planetary day.

Other than that...the near-synching of the day/lunar month period...there is no way to produce a long-lasting tide cycle.

In order to maintain a near-Earth-like day cycle...here would be a good balance: A Super-Earth to increase the gravity well of the planet, a fairly massive moon, orbiting at high speeds, at fairly close proximity to the planet. Unfortunately, this may cause the tidal effects to mess with the landscape as well as the water. I don't currently have the time to crunch numbers to try to come up with a solid set of results...but that's a base to start on.

  • $\begingroup$ I was thinking the same, but I keep having the moon launch off into space at the speeds required to keep it up to the earths rotation $\endgroup$
    – Twelfth
    Mar 12, 2015 at 21:42
  • $\begingroup$ Maybe try something like this, but with a true double planet scenario with both rotating around a shared center of gravity? That might let you disassociate the day/night cycle with the rotation and tides. $\endgroup$ Mar 12, 2015 at 21:44
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    $\begingroup$ Be aware that you have both solar tides and lunar tides. (Solar ones are much weaker on earth). Solar tides will always match the length of days. Spring and Neap Tides (big ones and small ones) happen when the Solar and Lunar Tides either reinforce or cancel each other out. $\endgroup$
    – Tim B
    Mar 12, 2015 at 22:14
  • $\begingroup$ Why would the planet's day length and the tidal cycle length have to be approximately the same? If the moon is very close to geosynchronous orbit, doesn't that give you very long tidal cycles regardless of day length? Also, I'm pretty sure that orbital period of a moon is determined (almost) entirely by the mass of the planet and the shape and size of the orbit; the mass of the moon is (almost) inconsequential to orbital period. $\endgroup$ Mar 13, 2015 at 2:11
  • $\begingroup$ Note that if you do nearly tidal lock the moon to the planet, even substantially larger tides would be less drastic, since they would rise and fall far slower. So a double or triple tide height would probably be acceptable. $\endgroup$ Mar 13, 2015 at 5:55

How multiple moons would act on the tides:

enter image description here

Ignore the frequency and times given here and look at how the waves add up. This is called a beat.

The important bit is that the sum can seem like it has a much longer wavelength and thus your moons can get away from being nearly geo-locked when your tides hit the once-a-month mark. Actually, when you're geo locked the tides hit the once-an-infinity mark.

But what about those little extremes inside the beat? Instead of this giving you half a month of eb then half a month of flow it's giving half a month of almost nothing going on then half a month of twice daily extreme tide changes.

Not exactly the same as what you asked for, but interesting.


For a simple earth-moon system, the moon would have to be near a geosync orbit - that way the moon's movement across the surface can be whatever you need it to be. Geosync seems rather close for a large moon, but it is still well outside the Roche limit, so I don't see anything preventing it from being reasonably stable as long as the planetary composition is uniform enough to still behave like a point source at that distance.

Multiple moons could also work - a bunch of small moons that have little effect on their own, but cause large tides when they all line up in the right way. The actual tide cycle might end up quite complicated, as you would also get smaller tides when only some of the moons are aligned.

If the moon was in a highly elliptical orbit, it could spend most of its time too far away to cause tides, but really make things slosh around when it gets close to earth. A month would probably be on the low end here - that would basically amount to stretching the moon's existing orbit, and likely put the closest approach a bit too close.

For something more extreme, try a co-orbital planet - no tides at all for centuries, and then anything short of tearing the planet apart.


Does it have to be the whole planet or perhaps only a localized phenomenon?

Imagine a narrow land mass (a bit like Panama) dividing a large ocean and a giant lake (the lake about the size of the Caribean Sea, or perhaps a little smaller).

At spring tide the tide runs over the the narrow land mass and into the lake resulting in a large tidal wave at the other edge of the lake when the tide gets there. The duration of the phenomenon will be a couple of days each time. Slowly the extra water will disappear through rivers/evapouration.

(Writing this it occurs to me that the narrow land mass would erode too quickly for this to be a phenomenon that occurs over millennia, maybe others can extend this idea).

  • $\begingroup$ Localized could work, although I'd like the area of the tide to be critical in some way - in other words, the tide should constrain trade and military movement. $\endgroup$ Mar 13, 2015 at 20:48

Yes, there are a few ways.

1. Slow the rotation of the earth

The tide is created twice a day as the earth rotates under the moon. The gravitational pull of the moon makes the aquasphere slightly elliptical.

To create a longer cycle, slow the rotation of the planet. This will also create a longer day/night cycle.

2. More than one moon

The tide is a stable waveform with a period of approximately 12 hours. It is created not only by the pull of the moon, but also by the inertia of the aquasphere. If you add a second moon you might potentially create an interference pattern which would flatten the waveform for most of the month, building to a giant tide once a month when the two moons were in phase.

3. Elliptical lunar orbit

An exaggerated elliptical lunar orbit would bring the moon close to the planet less frequently. The sligshot would create a very high tide once a month or so. Other tides would be much less noticeable.


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