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I was reading an article on exomoon habitability constrained by illumination and tidal heating. The article imagines an Earth-like exomoon around Jupiter-like host planet. That got me thinking about ocean tides on the Earth-like exomoon, and according to my calculations using Newton equilibrium theory, they should be kilometers high, which I thought was a lot.

Deriving the tidal range from Newton I got : $$h = \frac{3}{2} \frac{M_J}{M_E} \frac{{R_E}^4}{d^3} $$ where $M_J$ is the mass of Jupiter, $M_E$ the mass of the Earth, $R_E$ the radius of the earth and $d$ the distance between the exomoon and the planet.

According to the article, the distance can be as low as $d = 5 R_J$ ($R_J$ radius of Jupiter). That gives $h \approx 18 km$, which is higher than the Earth's troposphere. Even with $d = 10 R_J$, the value the article gives for slightly higher eccentricities, we get $h \approx 2 km$. That still seems quite high to me compared to the result we get when computing for the Earth-Moon system ($h \approx 54cm$).

More generally, this is about tidal behavior with different orbital parameters and mass ratios. I was able to find very little on that topic, but I thought that example might be one of those edge cases/extreme scenarios where the formulae don't apply because something we normally neglect isn't negligible anymore. So :

  • Firstly, I wonder whether there might be some other parameters than gravity and hydrostatic pressure to take into account in Newton's formula. If that's the case, I'm not sure what those are.
  • Secondly, I know that there is no tidal bulge on Earth. Then, since the moon is expected to have some form of tectonic activity, one might assume continents and wonder how the cyclic forcing interacts with them, which I'm aware might be a very complicated question requiring precise topographic information, but maybe some sort of qualitative argument can be made. If there is a nice way to do that using Laplace's Tidal Equations, I haven't been able to find it.
  • Thirdly, I'm know the article assumes synchronous rotation, which makes the forcing from the planet static in the moon's referential and renders the wave analysis irrelevant. But, because it's more general and I like wave mechanics, I'm still interested in the case of the non-tidally locked moon.

In summary:

  • Is there a reason why the tidal range derived from Newton is wrong, in such extreme conditions, even for a tidally locked moon? If so, how do we correct it?
  • If the exomoon isn't tidally locked and thus submitted to cyclical forcing, is there a similar, corrected way to estimate the tidal range?
  • Does "synchronous rotation" mean tidal locking? All I know is Io is tidally locked to Jupiter, and the 100m solid tides are due to the eccentricity of the orbit...which is not the usual expression of the tidal tensor: https://en.wikipedia.org/wiki/Tidal_tensor – JEB Nov 11 '23 at 05:04
  • You appear to be neglecting the solid tide, so your calculation is giving the variation in water height relative to the centre of the exomoon, rather than relative to the (raised) land surface. – PM 2Ring Nov 11 '23 at 10:00
  • @JEB Hi, thank you for answering. Synchronous rotation is a form of tidal locking, yes, but for some eccentricities, tidal locking gives rise to spin-orbit resonance instead (as is the case with mercury). I will study the question of whether or not the case of Io is made special by its proximity to Jupiter, thank you for this idea. (Although the paper I'm referring to suggests such an effect shouldn't happen with the wider orbits I chose.) – Boreal Stars Nov 16 '23 at 06:33
  • @PM2Ring Hi, thank you for your answer. I derived the tidal range formula from "Modern Theory and Practice of Tide Analysis and Tidal Power", of which you can a PDF find online (pages 13-14, equations 39-35.) I realize this might indeed neglect the solid tide but I didn't think that would be too relevant. I can try and compute a correction for that when I have the time, but the calculation should already give the tidal range/amplitude, from lowest to highest point, not from the center of the exomoon. (It does for the Earth) – Boreal Stars Nov 16 '23 at 06:36

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