<p>That sentence on wikipedia continues &quot;from friction generated within Io's interior as it is pulled between Jupiter and the other Galilean moons&quot;.</p>
<p>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.</p>
<p>It's easy to underestimate just how massive Jupiter is. If you do the math, the tidal forces on Io are <em>twenty thousand</em> 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.</p>
<p>Jupiter is <em>really, really massive</em>.</p>

CC BY-SA 4.0
2021-12-01T23:33:07.390Z
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.
David Hammen
CC BY-SA 4.0
2021-12-02T12:54:06.123Z
@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 (https://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.
Luaan
CC BY-SA 4.0
2021-12-01T22:28:05.880Z
2021-12-01T22:28:05.880Z
Luaan
<blockquote>
<p>How can Io be tidally heated while it is in tidal lock?</p>
</blockquote>
<p>It is tidally locked in a mean motion sense of &quot;tidally locked&quot;. That Io is in an eccentric orbit rather than a circular orbit means that tidal stresses can and do build up. <em>Lainey et al.</em> claim that the global energy dissipation in a tidally-stressed moon is given by
<span class="math-container">$$\dot E = -\frac{21}2 \frac{k_2}Q \frac{n^5R^5}G e^2$$</span>
where</p>
<ul>
<li><span class="math-container">$\dot E$</span> is the rate at which tidal energy dissipates,</li>
<li><span class="math-container">$k_2$</span> is the moon's second order tidal Love number,</li>
<li><span class="math-container">$Q$</span> is the moon's tidal quality factor,</li>
<li><span class="math-container">$n$</span> is the moon's mean motion,</li>
<li><span class="math-container">$R$</span> is the moon's radius,</li>
<li><span class="math-container">$G$</span> is the universal gravitational constant, and</li>
<li><span class="math-container">$e$</span> is the eccentricity of the moon's orbit.</li>
</ul>
<p>The ratio <span class="math-container">$k_2/Q$</span> 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 <span class="math-container">$k_2$</span> and a significantly lower value of <span class="math-container">$Q$</span>. Io's volcanism is a sign of a moon with at least a partially molten interior.</p>
<p>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.</p>
<p>Finally, even though Io's eccentricity is small, it is not zero. The fact of <span class="math-container">$e^2$</span> 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.</p>
<p>This has been hypothesized to lead to an interesting hysteresis loop (e.g., <em>Yoder</em>). 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.</p>
<br>
References:
<p><a href="http://www.nature.com/nature/journal/v459/n7249/abs/nature08108.html" rel="noreferrer">Lainey, et al. &quot;Strong tidal dissipation in Io and Jupiter from astrometric observations,&quot; <em>Nature</em> 459.7249 (2009): 957-959.</a></p>
<p><a href="https://www.nature.com/articles/279767a0" rel="noreferrer">Yoder, Charles F. &quot;How tidal heating in Io drives the Galilean orbital resonance locks.&quot; Nature 279.5716 (1979): 767-770.</a></p>

CC BY-SA 4.0
2021-12-01T15:47:15.963Z
The eccentricity of the Io is very low, it is orbit is practically circular. Although maybe so close to the Jupiter might generate enough tidal forces even so.
peterh
CC BY-SA 4.0
2021-12-01T22:45:15.453Z
`+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.
uhoh
CC BY-SA 4.0
2021-12-02T00:14:32.387Z
@uhoh It was in an answer I wrote to your question [How can tidal heating lower Io's orbit?](https://astronomy.stackexchange.com/questions/25061) I didn't add references to either answer. I'll fix that here.
David Hammen
CC BY-SA 4.0
2021-12-02T00:36:14.413Z
@DavidHammen wow, almost four years ago; a testament to how memorable your answers are!
uhoh
CC BY-SA 4.0
2021-12-02T10:10:46.293Z
FWIW, here's a typical [plot of the eccentricity of Io](https://i.sstatic.net/eskyS.png) over ~4 orbital periods, with a time step of 1 hour, produced using Horizons.
PM 2Ring
CC BY-SA 4.0
2021-12-03T08:08:27.247Z
And here's a [graph of the Io-Jupiter distance](https://i.sstatic.net/MzEjY.png) for the same period, produced using my script linked [here](https://astronomy.stackexchange.com/a/47345/16685). The mean distance is ~421770 km.
PM 2Ring
CC BY-SA 4.0
2021-12-03T14:18:45.107Z
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.
peterh
CC BY-SA 4.0
2021-12-01T15:39:51.487Z
2021-12-02T01:13:52.623Z
2021-12-02T01:13:52.623Z
David Hammen
<p>According to its <a href="https://en.wikipedia.org/wiki/Io_(moon)" rel="noreferrer">Wikipedia page</a>:</p>
<blockquote>
<p>Rotation period: synchronous</p>
<p>Eccentricity: 0.0041</p>
</blockquote>
<p>But also</p>
<blockquote>
<p>...extreme geologic activity is the result of tidal heating...</p>
</blockquote>
<p>How is it possible? It should not be heated, if its rotation is synchronous, no tidal waves should exist in it.</p>

<solar-system><celestial-mechanics><tidal-forces><tidal-locking><io>
How can Io be tidally heated while it is in tidal lock?
peterh
# How can Io be tidally heated while it is in tidal lock?

(Asked by: username_0 on 2021-12-01)

According to its [Wikipedia page](https://en.wikipedia.org/wiki/Io_(moon)):

> 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.

**Tags:** solar-system, celestial-mechanics, tidal-forces, tidal-locking, io


---

## 2 answers

### Answer by username_2 on 2021-12-01. Score: 34 | Accepted answer

> 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.](http://www.nature.com/nature/journal/v459/n7249/abs/nature08108.html)

[Yoder, Charles F. "How tidal heating in Io drives the Galilean orbital resonance locks." Nature 279.5716 (1979): 767-770.](https://www.nature.com/articles/279767a0)

> **Comments:**
> **username_0**: The eccentricity of the Io is very low, it is orbit is practically circular. Although maybe so close to the Jupiter might generate enough tidal forces even so.  
> **username_3**: `+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.  
> **username_2**: @username_3 It was in an answer I wrote to your question [How can tidal heating lower Io's orbit?](https://astronomy.stackexchange.com/questions/25061) I didn't add references to either answer. I'll fix that here.  
> **username_3**: @DavidHammen wow, almost four years ago; a testament to how memorable your answers are!  
> **username_4**: FWIW, here's a typical [plot of the eccentricity of Io](https://i.sstatic.net/eskyS.png) over ~4 orbital periods, with a time step of 1 hour, produced using Horizons.  
> **username_4**: And here's a [graph of the Io-Jupiter distance](https://i.sstatic.net/MzEjY.png) for the same period, produced using my script linked [here](https://astronomy.stackexchange.com/a/47345/16685). The mean distance is ~421770 km.  
> **username_0**: 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.  

---

### Answer by username_1 on 2021-12-01. Score: 13

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*.

> **Comments:**
> **username_2**: 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.  
> **username_1**: @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 (https://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.  

---

