This answer to the question Is Io a magic energy machine? suggests that the energy from the internal heating of Io due to tidal "squishing" as it moves cyclically closer and farther from Jupiter in its elliptical orbit will come from the energy of Io's orbit. A lower energy orbit is necessarily smaller, and that actually means that the velocity will be larger. (When you want to raise a satellite's orbit to a higher altitude, you actually use thrust in the direction of motion to slow it down.)

Given that tidal forces are a little complicated (cf. Why is the Moon receding from the Earth due to tides? Is this typical for other moons?), is it a priori certain that the heating will lower Io's orbit, causing it to speed up? (Consider that Earth's Moon's receding is due in part to Earth's liquid ocean, and Jupiter is a gas giant.) Is it just the perijove that will decrease, or the semi-major axis?

How can a (seemingly, naively, on average) radial force cause a tangential acceleration? Io is tidally locked to Jupiter so its rotation around its own axis is synchronous to its rotation around Jupiter.

edit: fwiw if the gravitational interactions between Io and Jupiter's other moons causes the problem to be too complex to answer easily, I'm more interested in the basic dynamics of tidal heating and effects on the orbit of a single moon, rather than specifically Io's situation.

  • $\begingroup$ Please check " ctually means that the velocity will be larger. (When you want to raise a satellite's orbit to a higher altitude, you actually use thrust in the direction of motion to slow it down.)". $\endgroup$
    – Alchimista
    Feb 15, 2018 at 9:52

2 Answers 2


How can tidal heating lower Io's orbit?

It doesn't, at least not to first order. The first order effect is that tidal heating acts to circularize Io's orbit. Counter to that, orbital resonances with Europa and Ganymede act to make Io's orbit more elliptical. This leads to a nice hysteresis loop.

Suppose Io is in a fairly circular orbit. This results in reduced tidal stresses, thereby making Io cool down. A cooler and hence more rigid Io is less susceptible to tidal deformations than is a warmer and hence more plastic Io. Given two bodies in the same orbit, one warm and plastic, the other cool and rigid, the warmer body will suffer more tidal deformations than will the cooler one. This is captured by the object's $k_2$ Love number. The inevitable lag in response means that the response will not be symmetric about periapsis/apoapsis for an elliptical orbit, and the greater the plasticity, the greater the greater asymmetry. This is captured by the object's tidal quality factor $Q$.

This cooling of Io as its orbit becomes close to circular enables the ever-present resonance effects to now dominate over the circularizing effects. Io's orbit slowly becomes more elliptical. That elliptical orbit increases tidal stresses on the cold, rigid Io, eventually making it start to warm up and become more plastic. The circularization effects grow as the orbit grows more elliptical and as Io's interior becomes more flexible and more plastic. Eventually the circularization effects dominate over the orbital resonance effect, making Io's orbit become more circular -- until the cycle repeats.

Rinse and repeat, at least as long as that three-way orbital resonance between Io, Europa, and Ganymede holds up. How long that three-way tidal resonance has existed and how long it will last is, as far as I know, unknown.

  • $\begingroup$ Well, apart from some residual (since Io is generally in tidal lock already) roation/oscillation contributions to the heating, the energy must come from Io's and other moons' orbital energy, hence lowering the orbit compared to an imagined non-deformable body's trajectory. (There simply is no other source.) In a thought experiment, two lone, deformable, massive celestial bodies passing each other closely would lose some of their kinetic energy to tidal heat from deformation and move apart slower than they approached. Io's situation is basically the same situation, just repetitious. $\endgroup$ Feb 14, 2018 at 11:36
  • $\begingroup$ (That Io's orbital energy is most likely increasing overall due to Jupiter's rotation, since Io is rotating prograde, i.e. in Jupiter's rotational direction, notwithstanding.) $\endgroup$ Feb 14, 2018 at 11:37
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    $\begingroup$ @PeterA.Schneider - There most certainly is another source, and that source is Jupiter's rotation. Moons that orbit a planet prograde slower than the planet's rotation rate (e.g., Jupiter's Galilean moons) tend to spiral out. How fast depends on the planet's and the moon's tidal quality factors $Q$ and $k_2$ Love numbers. There's a good deal of tidal dissipation in both the Jovian and Saturnian systems. $\endgroup$ Feb 14, 2018 at 12:09
  • $\begingroup$ This answer is extremely unclear to me. How does Io's temperature affect whether the resonance effects dominate over the circularizing ones? $\endgroup$ Feb 14, 2018 at 13:02
  • $\begingroup$ You are right, there is another potential souce, so that part was incorrect; but does Jupiters rotation contribute much to Io's heat? I don't think so -- it just drags Io around and ramps its orbit up. Io's vulcanism would be the same around a non-rotating Jupiter. $\endgroup$ Feb 14, 2018 at 13:05

The answer by David Hammen includes many of the interesting details of how Io's orbit evolves in time (and explains why Io can still be volcanic even though right now Io's orbit is extremely circular). It also explains that if Io was fully tidally locked, with no other moons, then it would not heat and its orbit would not change, which may be what the questioner was mostly wondering about. Perhaps the only remaining question might then be, why does a moon that is in a circular orbit, but not rotating at the correct rate, find its orbit changing?

For this, there is an interesting result that if the moon is spinning faster than its orbit, the delay in the response of the moon's shape to its tidal equipotential means that the "points" of its bulges will get out ahead of alignment with the planet. This produces a torque from gravity that tends to slow its spin. The opposite holds if it is spinning slower than its orbit. So that's how the spin gets tidally locked, and there is some heating associated with that. But the planet-moon system (ignoring other moons) must conserve angular momentum, so if the spin slows, that angular momentum must show up somewhere else-- it shows up in the orbit. So instead of thinking about the energy of the orbit (which is not conserved because heat is created and spins are changing), think about the angular momentum of spin plus orbit. If the spin slows, the orbit must raise up to conserve angular momentum, and the opposite if the spin increases.

Since Io is not tidally locked, it isn't doing either, but in its history before it was locked, it would have done one or the other. As for the Earth and Moon, the Moon is tidally locked but the Earth is spinning faster than the Moon's orbit, so the Earth's bulges get out ahead of the Moon and so the Moon is torquing down our spin. That loss of angular momentum must go into the Moon's orbit, so that's why the Moon is getting farther away.

If you think in terms of energy, then you see that the Earth is being heated by the Moon's gravity. Also, the Moon's orbit is increasing in energy. So there has to be a source for both of those, and it is the energy in the Earth's spin. Here there is no question about how energy lost as heat radiated by the Earth can come out of the Moon's orbit, because in fact the orbital energy is increasing. It's more clear how spin energy can go into both the heating and the orbit, because it is the spin that is creating the forces that are causing both the heating and the orbital effects. Similarly, if a moon is spinning faster than its own orbit, then that spin creates forces on the moon that slow its spin, and some of that energy goes into heating the moon, and some goes into lifting its own orbit (to conserve angular momentum). If the moon is spinning slower than its orbit, then energy is going into both heat, and its spin, and this will lower the orbit because of the forces that are doing both those things also must conserve angular momentum.

  • $\begingroup$ Thanks for your answer and insight! Jupiter-Io is quite an interesting system and provides us a lot to think about. $\endgroup$
    – uhoh
    Mar 27, 2019 at 22:22

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