Tides on a nearly tidally locked planet

In summary, a two body system with one body being close to tidally locked would have larger tides due to the slower rotation and the longer time for the water and ground to follow the varying potential across the planetary surface. The tides would scale with M/R3 where M is the mass of the other object and R is the separation (for circular orbits). Once the planet becomes tidally locked, there would be a permanent tidal bulge, but it would not be noticeable like the bulge on Earth's equator. The size of the bulge would depend on the objects involved, but it could be significant in the case of two Earth-like planets in a 30 hour orbit. The rotation speed of the planet would also affect the
  • #1
Althistorybuff
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Non-Scientist here.

Assume that you had a two body system (planet-moon, double-planet) where one of the bodies was very close to tidally locked.

Say they are closely co-orbiting at 6 to ten diameters in a relatively quick 20 to 50 hour orbit.

One of the planets is close to being tidally locked but not quite there.

Say the "month" is 1/10th to 1/25th longer than the day. The companion moon/planet would be seen from the sky for extended periods on one hemisphere (maybe weeks), then disappears for quite some time.

Question 1: Would this affect the tides in any meaningful way if there was a slow motion "high" or "low" tide? Would they be weaker in any way, meaning not rise as high? Or would they just rise and stay there for a longer period?

As the moon/double-planet would be very close, I would assume they would be very high tides in the perhaps hundreds of meters.

Does speed of planetary rotation effect the tides in any way other than how high they remain at their peak?

Question 2: Once the planet becomes tidally locked, is there a permanent "High Tide" excluding any solar tides, or are there other factors in play? With the moon's/other planet's gravity create a permanent "bulge" pulling the tidally locked face of the planet to the other body?

I have this idea for a double-planet system, one of which is tidally locked but the other is close to being tidally locked. I'm just not sure if there is something I'm missing.

Thanks.
 
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  • #2
Althistorybuff said:
Question 1: Would this affect the tides in any meaningful way if there was a slow motion "high" or "low" tide? Would they be weaker in any way, meaning not rise as high? Or would they just rise and stay there for a longer period?
Tides would be larger than with a faster rotation. The water and the ground have more time to follow the varying potential across the planetary surface.

Tides scale with M/R3 where M is the mass of the other object and R is the separation (for circular orbits).
Althistorybuff said:
Once the planet becomes tidally locked, is there a permanent "High Tide" excluding any solar tides, or are there other factors in play?
That "high tide" is just a permanent deviation from a spherical shape. Tides don't only affect water, they deform the solid ground as well. You won't notice such a permanent tide, in the same way as you don't notice that the radius of Earth at the equator is 20 km larger than the radius at the poles.
 
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  • #3
mfb said:
Tides scale with M/R3 where M is the mass of the other object and R is the separation (for circular orbits).
That's how tidal acceleration scales, right? Does the equipotential surface displacement (i.e. the height of the tidal bulge) also scale the same way? I could never quite visualise why it should or shouldn't.
 
  • #4
It scales in the same way. It also scales with properties of the planet where the tides are on, but that is a different topic.
 
  • #5
mfb said:
Tides would be larger than with a faster rotation. The water and the ground have more time to follow the varying potential across the planetary surface.

Tides scale with M/R3 where M is the mass of the other object and R is the separation (for circular orbits).That "high tide" is just a permanent deviation from a spherical shape. Tides don't only affect water, they deform the solid ground as well. You won't notice such a permanent tide, in the same way as you don't notice that the radius of Earth at the equator is 20 km larger than the radius at the poles.
Thank.

Any idea how large the bulge/permanent tide would be?

As for the lesser tides of a slowly rotating / very close to tidally locked world, how much weaker would b they be with this lesser rotation?

Interestingly, this scenario may have a much longer "day" than "month".

Assume two closely orbiting bodies may orbit in, say, 30 hours, but the one not-quite-tidally locked planet would have a rotation of 300 hours. You would have 10 "months" to a "day".
 
  • #6
Althistorybuff said:
Any idea how large the bulge/permanent tide would be?
Depends on the objects.
Let's take two Earth-like planets in a 30 hour orbit, that leads to a separation of 62000 km. 6370 km (radius of Earth) closer to the other object, we are 600 kJ/kg deeper in the gravitational potential, but 1470 kJ/kg higher in terms of effective centrifugal potential. Sum: 870 kJ/kg. That corresponds to 87 km tidal bulge. Yeah... quite significant. Your planets probably won't get a cold solid surface until they are tidally locked.

Cross-check: (mass of Earth)/(mass of Moon) * (distance to moon)^3 / (62000 km)^3 = 21000 times the tidal influence. Give or take a factor 4 because we have a different mass ratios between the objects. Plausible.

Althistorybuff said:
As for the lesser tides of a slowly rotating / very close to tidally locked world, how much weaker would b they be with this lesser rotation?
Wrong direction. The tides increase for a nearly tidally locked world. But the amount is hard to predict without a much more detailed model.
Althistorybuff said:
Assume two closely orbiting bodies may orbit in, say, 30 hours, but the one not-quite-tidally locked planet would have a rotation of 300 hours. You would have 10 "months" to a "day".
That is not nearly tidally locked. Nearly tidally locked needs one rotation in close to 30 hours. The day would be about as long as a month, probably a bit shorter.
 
  • #7
mfb said:
Wrong direction. The tides increase for a nearly tidally locked world.
I'm not sure that's correct. Taking into account that most of the tidal height we get on Earth comes from water sloshing about and piling up in various nooks and crannies of the shoreline, giving it more time to gradually seep towards the new equipotential surface should reduce the tidal height. IIRC the equipotential surface height difference of the lunar tides raised on Earth is merely 0.5m between low and high tide. If you slowed down Earth's rotation significantly, then I think that 0.5m is all we'd ever get.

Of course putting the Moon at ten Earth radii would raise that 0.5 m to a hundred, and switching the satellite for another Earth-like planet by another factor of 100. But just the slowing down of relative rotation should make the tides comparatively less severe.
 
  • #8
Well, you don't get the occasional resonance leading to extreme tides, okay. But you can get the true equipotential surfaces.
 
  • #9
Tidal locking necessarily requires mass asymmetry. In planet formation central limit theory suggests many uniform density/mass planets are formed around red dwarf stars as their rotation initially does not match their revolution. As such worlds become locked, their water mass will migrate from the hot side and freeze on the cold side in Everest mass mountains. This will disrupt the tidal locking and cause a reversal. A resonance of such cycles could cause very long day/night cycles on such 'tidally locked' worlds. Angular monentum tells us that surface mass is much more significant than mass near the core. This could have significant implications for life.
 
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  • #10
aaa_clem said:
As such worlds become locked, their water mass will migrate from the hot side and freeze on the cold side in Everest mass mountains. This will disrupt the tidal locking and cause a reversal.
I can buy the water migration as part of the water cycle, but I don't get why this would disrupt or reverse tidal locking. The Moon has asymmetrical mass distribution and is tidally locked. Please, explain.
 

What is a tidally locked planet?

A tidally locked planet is a type of planet where one side always faces its parent star, while the other side always faces away. This occurs due to the gravitational pull of the star on the planet, causing the planet's rotation to synchronize with its orbit around the star.

How do tides occur on a nearly tidally locked planet?

Tides on a nearly tidally locked planet are caused by the gravitational pull of the planet's parent star. As the planet orbits the star, the side facing the star experiences a stronger gravitational pull, resulting in a bulge of water on that side. This creates high tides. Meanwhile, the side facing away from the star experiences a weaker gravitational pull, resulting in a low tide on that side.

Do tides on a nearly tidally locked planet differ from those on Earth?

Yes, tides on a nearly tidally locked planet can differ from those on Earth. On Earth, tides are primarily caused by the moon's gravitational pull. On a nearly tidally locked planet, the tides are primarily caused by the planet's parent star. Additionally, the distance between the planet and its star, as well as the planet's rotation speed, can also affect the tides.

Can tides on a nearly tidally locked planet impact its habitability?

Yes, tides on a nearly tidally locked planet can impact its habitability. In extreme cases, the tides could cause large floods or even create a planet-wide ocean. However, the presence of tides can also help regulate the planet's temperature and distribute nutrients, which could potentially support life.

What are some other factors that can influence tides on a nearly tidally locked planet?

Other factors that can influence tides on a nearly tidally locked planet include the planet's distance from its star, the star's mass and size, and the planet's axial tilt. These factors can affect the strength and frequency of tides on the planet.

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