# I Tides on a nearly tidally locked planet

1. Apr 9, 2017

### Althistorybuff

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.

2. Apr 9, 2017

### Staff: Mentor

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.

3. Apr 9, 2017

### Bandersnatch

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. Apr 9, 2017

### Staff: Mentor

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. Apr 9, 2017

### Althistorybuff

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. Apr 9, 2017

### Staff: Mentor

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.

Wrong direction. The tides increase for a nearly tidally locked world. But the amount is hard to predict without a much more detailed model.
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. Apr 9, 2017

### Bandersnatch

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. Apr 9, 2017

### Staff: Mentor

Well, you don't get the occasional resonance leading to extreme tides, okay. But you can get the true equipotential surfaces.