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According to its Wikipedia page:

Rotation period: synchronous

Eccentricity: 0.0041

But also

...extreme geologic activity is the result of tidal heating...

How is it possible? It should not be heated, if its rotation is synchronous, no tidal waves should exist in it.

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2 Answers 2

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How can Io be tidally heated while it is in tidal lock?

It is tidally locked in a mean motion sense of "tidally locked". That Io is in an eccentric orbit rather than a circular orbit means that tidal stresses can and do build up. Lainey et al. claim that the global energy dissipation in a tidally-stressed moon is given by $$\dot E = -\frac{21}2 \frac{k_2}Q \frac{n^5R^5}G e^2$$ where

  • $\dot E$ is the rate at which tidal energy dissipates,
  • $k_2$ is the moon's second order tidal Love number,
  • $Q$ is the moon's tidal quality factor,
  • $n$ is the moon's mean motion,
  • $R$ is the moon's radius,
  • $G$ is the universal gravitational constant, and
  • $e$ is the eccentricity of the moon's orbit.

The ratio $k_2/Q$ strongly depends on the makeup of the moon's interior. Compared to a moon with a solid interior, a moon with a partially molten interior will have a slightly higher value of $k_2$ and a significantly lower value of $Q$. Io's volcanism is a sign of a moon with at least a partially molten interior.

The power of five on the mean motion and moon radius means that a large moon that orbits close to its parent planet will be subject to vastly more tidal stress than a small moon that orbits far from the parent planet. Io is a large moon (larger than our Moon) and it orbits fairly close to Jupiter.

Finally, even though Io's eccentricity is small, it is not zero. The fact of $e^2$ means that the tidal energy dissipation strongly depends on eccentricity. Those tidal stresses normally would act to circularize Io's orbit about Jupiter, thereby reducing the tidal stresses. However, Io is also in a 1:2:4 orbital resonance with Europa and Ganymede. These interactions tend to increase Io's eccentricity.

This has been hypothesized to lead to an interesting hysteresis loop (e.g., Yoder). Suppose Io's interior is cool and its eccentricity is very low. This makes tidal stresses very low. This reduces the impact of Jupiter's circularization effects on Io's orbit. The resonance effects now begin to dominate, making Io's orbit become more eccentric. Tidal stresses now become significant and Io's interior warms up. At some point, the tidal stresses that lead to circularization dominate over the effects of Europa and Ganymede. Io's orbit circularizes and Io's interior cools. Rinse and repeat.


References:

Lainey, et al. "Strong tidal dissipation in Io and Jupiter from astrometric observations," Nature 459.7249 (2009): 957-959.

Yoder, Charles F. "How tidal heating in Io drives the Galilean orbital resonance locks." Nature 279.5716 (1979): 767-770.

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    $\begingroup$ +1 I recall some discussion of yours from a few years ago of the "lather, rinse, repeat" variety involving this eccentricity and heating/cooling cycles (of Io?) but I can't remember if it's in another answer or only in comments. I think readers will benefit from having them linked, will try to find it. $\endgroup$
    – uhoh
    Commented Dec 1, 2021 at 22:45
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    $\begingroup$ @uhoh It was in an answer I wrote to your question How can tidal heating lower Io's orbit? I didn't add references to either answer. I'll fix that here. $\endgroup$ Commented Dec 2, 2021 at 0:14
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    $\begingroup$ FWIW, here's a typical plot of the eccentricity of Io over ~4 orbital periods, with a time step of 1 hour, produced using Horizons. $\endgroup$
    – PM 2Ring
    Commented Dec 2, 2021 at 10:10
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    $\begingroup$ And here's a graph of the Io-Jupiter distance for the same period, produced using my script linked here. The mean distance is ~421770 km. $\endgroup$
    – PM 2Ring
    Commented Dec 3, 2021 at 8:08
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    $\begingroup$ Thank you very much! So, the essence is: although it is roughly in tidal lock, the little ellipsiodness of its orbit still causes some swing. And so close to such a big planet, even this is enough to heat it. $\endgroup$
    – peterh
    Commented Dec 3, 2021 at 14:18
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That sentence on wikipedia continues "from friction generated within Io's interior as it is pulled between Jupiter and the other Galilean moons".

There's an orbital resonance (with the other Galilean moons) that prevents Io's orbit from circularizing, and also prevents Io from migrating away from Jupiter. If the other moons didn't exist, Io would be on a circular orbit much further away from Jupiter, and wouldn't be getting any appreciable tidal heating. Yes, the eccentricity of Io isn't particularly high. But consider that Io is mere 400kkm away from Jupiter, and Jupiter is rather... massive.

It's easy to underestimate just how massive Jupiter is. If you do the math, the tidal forces on Io are twenty thousand times the tidal forces the Moon causes on the Earth (and the Moon is pretty massive as moons go). The moon's tidal bulge stretches by a hundred meters between the apogee and perigee, despite the low eccentricity. Run the model, and you get about 0.6-1.6E14 W heat generated by this process, which is consistent with the observed heat loss from Io. This absolutely dwarfs both the heat produced through radioactive decay and insolation.

Jupiter is really, really massive.

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    $\begingroup$ I calculate that the tidal forces exerted by Jupiter on Io are "only" 250 times stronger than the tidal forces exerted by the Earth on the Moon. You need to use the distance from the center of the planet to the center of the moon in question rather than the distance between the surface of the planet to the center of the moon. $\endgroup$ Commented Dec 1, 2021 at 23:33
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    $\begingroup$ @DavidHammen You're right, but I was talking about the force the Moon exerts on the Earth. The figure comes from a quote from NASA (science.nasa.gov/science-news/science-at-nasa/2000/ast04may_1m). Doing the naïve math, I get "only" about 5k times the force, not 20k. I expect the 20k figure also includes the (maximal or average?) contribution from the other moons. $\endgroup$
    – Luaan
    Commented Dec 2, 2021 at 12:54

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