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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.

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    $\begingroup$ 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. $\endgroup$ – G0BLiN Jul 9 '17 at 15:50
  • $\begingroup$ @G0BLiN - As I understand it, if you are tidally locked to a celestial body, it's position in the sky is approximately constant. See starchild.gsfc.nasa.gov/docs/StarChild/questions/… $\endgroup$ – Jay Lemmon Jul 9 '17 at 16:16
  • $\begingroup$ 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. $\endgroup$ – Vylix Jul 9 '17 at 17:37
  • $\begingroup$ 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 science.howstuffworks.com/environmental/earth/geophysics/… $\endgroup$ – Vylix Jul 9 '17 at 17:44
  • $\begingroup$ 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. $\endgroup$ – Vylix Jul 9 '17 at 17:50
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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.

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  • $\begingroup$ Thanks, I hadn't thought about it in terms of where that necessarily places the moon. From 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. $\endgroup$ – Jay Lemmon Jul 9 '17 at 22:11
  • $\begingroup$ Also, a note on your numbers, from lhup.edu/~dsimanek/scenario/tides.htm, the tidal bulge in the mid ocean is ~ 1 meter. $\endgroup$ – Jay Lemmon Jul 9 '17 at 22:14
  • $\begingroup$ Er, sorry, I should have said "cubicly proportional to apparent size". $\endgroup$ – Jay Lemmon Jul 9 '17 at 22:21
  • $\begingroup$ @JayLemmon 1m is bigger than I'd remembered. corrected. $\endgroup$ – James K Jul 9 '17 at 22:33
  • $\begingroup$ 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. $\endgroup$ – pablodf76 Jul 10 '17 at 1:39
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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.

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  • $\begingroup$ 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. $\endgroup$ – Jay Lemmon Jul 10 '17 at 9:58
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    $\begingroup$ You'd certainly notice it, but not on a timescale needed to sail a boat by it. $\endgroup$ – Salda007 Jul 10 '17 at 10:10
  • $\begingroup$ @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. $\endgroup$ – pablodf76 Jul 10 '17 at 10:18
  • $\begingroup$ 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. $\endgroup$ – Jay Lemmon Jul 10 '17 at 10:27
  • $\begingroup$ @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. $\endgroup$ – 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.

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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.

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  • $\begingroup$ 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 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. $\endgroup$ – Jay Lemmon Jul 15 '17 at 17:06
  • $\begingroup$ YES! (I was close, only a factor of 10 off.) Thanks. $\endgroup$ – Supertankerm60a3 Jul 15 '17 at 20:50

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