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I have a world with a moon, much like Earth and the Moon, except that the planet is almost tidally locked to the moon. The moon therefore appears to move only very slowly through the sky, and takes, say, 7 years to do a full revolution.

Across the equator of the planet runs a fresh water sea. My goal is to create an environment like ancient Egypt with the Nile, where there's periodic and very dramatic floods and ebbs. Also, I want a relatively easy way to circumnavigate the world following this sea/river current.

I just want have some idea that this setting makes sense and that I've correctly predicted likely behaviour, since I don't know much about seas.

  1. Would the difference in high tide and low tide be more dramatic than on Earth for a similarly sized moon, because the sea would have more time to "catch up" to the moon? Or would they be about the same?
  2. On earth the two high tides are roughly equal in size. Would that also be true for this system?
  3. Supposing you were in a boat, and you wanted to follow the tides around the world, where in the cycle would you want to sail? I would think you'd want to lag about half way between a high tide and low tide bulge, chasing after high tide. I would think that would be when the currents are strongest. Which would put the Moon at about a 45 degree angle ahead of you in the sky if I'm right (you'd basically be chasing the moon, and from your perspective it wouldn't move in the sky).
  4. Could you just drift on the tidal currents around this world, or would you need a motor or a sail?
  5. How would the currents work relative to the high tide bulges? I'm thinking that there are two options. Either the tidal currents always point towards the high tide bulge, and there's basically two convection cells on either side of the high tide slack current, or there's a single convection cell, and the tidal bulge works like a raindrop sliding down a window. Which would mean there's actually a current that flows away from the tidal bulge ahead of it before dropping down to the sea floor/river bed and reversing direction.

Any insight would be appreciated. If there are any striking features that I haven't thought of that would also be interesting to know.

Jay Lemmon
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    I think this confuses the concept of tidal lock where an orbiting body always directs it's same "side" towards the other object (in the most known form of tidal locking, there are other possibilities) with the concept of geosynchronous orbit - where an orbiting body seems to occupy the same position in the sky. – G0BLiN Jul 09 '17 at 15:50
  • @G0BLiN - As I understand it, if you are tidally locked to a celestial body, it's position in the sky is approximately constant. See https://starchild.gsfc.nasa.gov/docs/StarChild/questions/question58.html – Jay Lemmon Jul 09 '17 at 16:16
  • That's a lot of questions to be asked in one post. Pick one for this post, and create another post linking to this question, asking the different question. – Vylix Jul 09 '17 at 17:37
  • I'm not an expert, but as I understand it, tides are affected only by lunar phases - or to be precise the relative position of the moon, earth, and sun. If the moon, earth, and sun is in a straight line, then it's high tide, and if the moon, earth, and sun makes a right angle, then it's low tide. For more information http://science.howstuffworks.com/environmental/earth/geophysics/tide-cause.htm – Vylix Jul 09 '17 at 17:44
  • And I don't understand the planet is tidally locked to the moon, as I know only body with less mass that can be tidally locked to the higher mass. – Vylix Jul 09 '17 at 17:50
  • @vylix. No the main daily tide is due to the position of the moon alone not the angle it makes with the sun. You are thinking of spring/neap tides. And it is possible for the planet to be locked to the moon. See Pluto/Charon. – James K Jul 09 '17 at 22:37
  • Definitely confusing tidally locked with geosynchronous. The articlereferenced is from the moon's point of view not the earth's point of view. The earth does stay fixed in the moon sky when the moon is tidally locked, but the orbital radius determines how rapidly a satellite moves across the sky. – Jim Jul 10 '17 at 01:54
  • @Jim - Okay, enough people mentioned this that I looked in to it. As it turns out, if you are tidally locked to a celestial body, that body is in your geosynchronous orbit. Makes sense? So the Earth is in geosynchronous orbit of the Moon, which is why the same side of the Moon always faces the Earth, and why the Earth doesn't appear to change position in the Lunar sky. So if the Earth were tidally locked to the Moon as well, the Moon would be in geosynchronous orbit of the Earth, and it's position in the sky wouldn't change. – Jay Lemmon Jul 10 '17 at 09:23
  • @JayLemmon The Earth can't be in geosynchronous orbit of the Moon, because the geo- prefix means the Earth, not the Moon. It's in selenosynchronous orbit. – Mike Scott Jul 10 '17 at 10:19
  • @MikeScott - Yes, fair point :) – Jay Lemmon Jul 10 '17 at 10:28
  • It doesn't take 7 years to sail anywhere in the world (https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0ahUKEwiYjt3m3_7UAhVEPz4KHRTTB5YQFgguMAI&url=http%3A%2F%2Fwww.worldrecordacademy.com%2Fsports%2Ffastest_round_the_world_sailing_world_record_set_by_Francis_Joyon_80127.htm&usg=AFQjCNGFZ2g91S1xY8n0JK6rERj9xMWGsA), so sailing by the currents caused by a 7 year tide would be pretty unnecessary. – bendl Jul 10 '17 at 12:48
  • Sailing is a bad term. Drifting? – Jay Lemmon Jul 10 '17 at 17:11

4 Answers4

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If the moon takes 7 years to do an orbit, relative to the surface, then it must be doing one orbit of the planet each day, relative to the centre of the planet (as it is nearly in geosynchronous orbit)

If the planet has about 86400 seconds in a day (like Earth), then the moon is orbiting at 36000km about the equator, much much closer than the moon really is. This would potentially lead to much bigger tides. However the way tides work is not simple bulges. There are tidal flows, the moon generates a flowing wave that moves around the Earth, and as this wave meets land it can be pushed up and that gives us large tides at the coast. The tidal range mid-ocean is much smaller (about a metre). If the moon isn't moving quickly, relative to the surface, then these flows will stop, and the coastal tide will be less.

I don't think that there would be significant tidal flows. The moon is moving so slowly, and the tide would rise so slowly that the required flow of water would be very little. You couldn't surf the world's tidal wave.

Tidal bulges are an idealisation, assuming a world in which there is no land. In reality the tidal flows are strongly determined by the shape of the land https://www.youtube.com/watch?v=ZEhm_ONTQKc

There would be two tidal bulges, just as on Earth. Except on Earth, tidal flows mean that in some places one tide is bigger than the other.

So I would expect the mid-ocean tide to be much larger, but the coastal effect is less, and there are no significant tidal flows. Also the tidal heating by the moon of the Planet's interior is much greater: I would expect lots more tectonic activity as the Planet bends and creaks with the nearby moon. The moon would also be massive: ten times larger than it appears in the sky. and eclipses would be commonplace.

James K
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  • Thanks, I hadn't thought about it in terms of where that necessarily places the moon. From https://www.npl.washington.edu/av/altvw63.html it sounds like the tidal effect of a celestial body is more or less proportional to its apparent size in the sky, so even if the distance is fixed I can still play with the size of the actual moon to get reasonable tidal forces. – Jay Lemmon Jul 09 '17 at 22:11
  • Also, a note on your numbers, from https://www.lhup.edu/~dsimanek/scenario/tides.htm, the tidal bulge in the mid ocean is ~ 1 meter. – Jay Lemmon Jul 09 '17 at 22:14
  • Er, sorry, I should have said "cubicly proportional to apparent size". – Jay Lemmon Jul 09 '17 at 22:21
  • @JayLemmon 1m is bigger than I'd remembered. corrected. – James K Jul 09 '17 at 22:33
  • A geosynchronous orbit has nothing to do with a tidal lock. If Earth and Moon one day end up in a mutual lock they will be much further apart than today, and the Moon is already ten times more distant from Earth than a GSO. – pablodf76 Jul 10 '17 at 01:39
  • @pablodf76 If Earth and Moon one day end up in a mutual lock they will be much further apart than today, THe rotation period of the the earth will be much slower, and the height of a gso will be at the distance of the moon, since in a mutual lock the orbit period of the moon = rotation period of Earth = rotation period of moon. If the Earth has a 24 hour rotation, and in mutual lock, then the moon would be 36000km high. – James K Jul 10 '17 at 05:45
  • @JamesK The Earth cannot become tidally locked to the Moon and remain rotating at the same speed as today. Indeed that's why the Earth's spin is slowing down right now while the Moon recedes at about an inch per year (I think). – pablodf76 Jul 10 '17 at 10:15
  • Earth - Moon distance is around 384,400km, this is an order of magnitude larger than the 36,000km of this scenario. Since gravity's force diminishes in a square proportion to distance, you'll need to re-scale the moon's mass by two orders of magnitude (diminish it to 0.0088 of it's current mass). If you assume homogeneous density for simplicity, it means diminishing the moon's radius by around 0.2 (compared to the distance, which was diminished by around 0.1) - all this means that you'll get a much closer, much much less massive but quite bigger (around twice it's angular size) moon... – G0BLiN Jul 10 '17 at 11:41
  • @pablodf I agree entirely. But if we make the moon and the planet mutually locked, and the Planet is the same mass as the Earth, then the moon has to be closer. The question isn't about the Earth. – James K Jul 10 '17 at 15:15
  • @JamesK Yes of course. If the planet is Earth-sized and the mass of the moon is negligible compared to it and you want the lock to happen at 24 hours (which is not specified in the question), then yes, the moon should be at today's GSO. – pablodf76 Jul 10 '17 at 15:31
  • Tidal force is proportional to the cube of the distance. With the Moon at GSO (an order of magnitude closer than IRL) the tides would be three orders of magnitude stronger (i.e. ~1000 times as strong as IRL). – pablodf76 Jul 10 '17 at 15:34
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From the numbers you gave (planet and satellite of sizes comparable to Earth and Moon respectively, 7 years synodic month for the satellite) you cannot really infer the distance between the planet and the satellite and thence the magnitude of the tides.

The Moon is currently about 384000 km from Earth on average and is tidally locked to Earth; for a mutual tidal lock to take place the Earth would have to decelerate its rotation and the Moon would have to recede a lot, a process that would take tens of billions of years. The Moon is obviously not on a geosynchronous orbit and as it recedes from Earth it will be even less so (if you take the value of today's GSO, of course!). As Earth's (or any planet's) rotation slows down due to tidal braking, the GSO will get farther from the planet.

The distance between two mutually tidally locked bodies depends on the sum of their angular momentum, which cannot increase or decrease. You can start with any value within a broadly reasonable range. Angular momentum depends on mass and rotational speed, and a planet could conceivably end up with almost zero rotational speed after it has formed.

On to your question: I would think that, irrespective of the magnitude of the tides, their extremely low frequency would make them almost unnoticeable. We're talking about an acceleration vector that takes seven years to go around an Earth-sized planet.

pablodf76
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  • re: the tidal frequency, if the tidal forces were much stronger than on Earth, so that the tidal swells were ~15meters, I think you would notice, even if it took almost 2 years to rise and another 2 years to fall. – Jay Lemmon Jul 10 '17 at 09:58
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    You'd certainly notice it, but not on a timescale needed to sail a boat by it. – Salda007 Jul 10 '17 at 10:10
  • @JayLemmon You would notice the sea levels going up and down predictably. You wouldn't experience those changes as ocean waves (as we think of them on Earth). You wouldn't be able to ride a "swell" like that. – pablodf76 Jul 10 '17 at 10:18
  • If you're moving around the planet following the tides once every 7 years, you only need to achieve an average speed of .18 meters/second. That's not very fast. I guess it depends what the surface currents would look like whether you could just rely on drifting or if you'd need some power like sails. If this sea acted like a river my guess is you'd be able to achieve at least that speed if you stuck to the thalweg. – Jay Lemmon Jul 10 '17 at 10:27
  • @JayLemmon That's out of my league, really. I think the wind would be the major factor in navigation, not the tides or the currents they produce. But then winds and ocean currents on Earth are heavily influenced by Earth's spin. It'd really take a sophisticated model to guess what the outcome would be. – pablodf76 Jul 10 '17 at 10:32
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To expand on pablodf76's last sentence:

If your planet has an Earthlike circumference of 40,000 km, and your moon is orbiting once every seven years (relative to the surface), then, relative to the surface, the peak of your tides (i.e. the "groundspeed" of the moon) only travels at <1 km/hr. By comparison, the peak of Earth's tides (which effectively circle the planet each day) moves closer to 1700 km/hr. So, while local topography will of course cause variations on-the ground, in general, no, you're not going to see any appreciable tidal flow.

In fact, if the moon's orbital plane wasn't aligned with the plane of your planet's equator (geosynchronous but not geostationary), you'd probably see a stronger north-south tidal movement than east-west.

Salda007
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Your Tides would be HUGE, hundreds or thousands of feet high, but VERY slow. More like "Don't build anything permanent or expensive here, in three years it will be under water. Your moon will have to be about 10 times closer, all things being equal. Give the diminishing square laws, inverted, your moon, all things bing equal, will have 100 times the influence on the surface liquid. It would have so much influence that its gravity would have to be taken into account designing very tall structures. It also could very well have destructive effects on core heating and crust/mantle tectonics.

Other things to consider, IF you had something that you were not willing to leave every 2 years or so (figuring a tide every 3.5 years that is a year or or so long) Any mining could only be done for a couple of years at a time then all equipment pulled out and the mines allowed to flood till next 'low' tide. Any city would have to be built on towers tall enough to be higher than the water level at 'high' tide. OR you could build a city of inter connected structures, designed to float, tethered or moored on cables, thousands of feet long that would ride the tide up every time, or make a 'walking' city that would continue to move to stay ahead of the tide. You can also adjust the mass of your moon to adjust the tide to a manageable level. Your scenario, as given, would result is a single tide thousands of feet high (or deep as the case may be) every 7 years so explore how to make a civilization the avoids their equator (the place the gigantic blob of surface liquid would collect and move) or stay ahead of it. Do your equatorial latitudes have topography that would prevent a city sized machine from rolling/walking across it every 7 years? Do your movers have contingency plans for when a wheel/axle/leg goes down. How many, straight number or percentage can go down before speed or forward movement is impeded ect. have fin with this world.

Supertankerm60a3
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  • Actually tidal forces follow an inverse cube law, so a moon the size of the Moon 10x closer would have 1000x the influence on the surface liquid. See https://www.npl.washington.edu/av/altvw63.html. Of course, you could just scale the moon down so that it has the same apparent size in the sky. Then it would have the same tides as on Earth. – Jay Lemmon Jul 15 '17 at 17:06
  • YES! (I was close, only a factor of 10 off.) Thanks. – Supertankerm60a3 Jul 15 '17 at 20:50