<blockquote>
<p>There are four moons that are closer to Jupiter than Io with higher eccentricities, yet they don't seem to have any volcanism at their surface.</p>
</blockquote>
<p>Only one of those innermost moons (Thebe) has an eccentricity higher than that of Io. The other three have lower eccentricities.</p>
<p>The reason they don't exhibit volcanism is because they are too small. The largest of the four, Amalthea, has a mean radius of 83.5 km. Compare that with Io's mean radius of 1821.6 km.</p>
<p>Tidal heating results in part from a tension between the tidal forces exerted by the planet that tends to pull a moon out of round and the moon's self-gravitational force that tends to pull a moon into a roundish shape. Small bodies such as those four innermost moons of Jupiter are too small to pull themselves into a roundish shape. The fuzzy boundary between small bodies such as those innermost moons and dwarf planetary bodies such as Pluto is the <a href="https://arxiv.org/abs/1004.1091" rel="noreferrer">potato radius</a>, somewhere between 200 and 300 km.</p>
<p>All of those four innermost moons of Jupiter are well below the potato radius, too small to pull themselves into a roundish shape via self-gravitation. The tidal forces on those small bodies are also smallish. (Tidal forces are proportional to body radius). This makes the tension between self-gravitation and tidal forces negligible for those small bodies.</p>
<p>Another factor that comes into play is the square-cube law. Cooling is proportional to surface area and hence the radius squared while mass is proportional to radius cubed. This means cooling per unit mass is inversely proportional to radius. Heating induced by those negligible tidal forces and negligible self-gravitation forces dissipates quickly in small bodies, not so quickly in larger bodies. This makes the interiors of larger bodies such as Io partially liquid. Small bodies such as those innermost moons of Jupiter are solid throughout. They don't flex as much as do those larger moons, and hence there is significantly less heat production in those small moons.</p>

CC BY-SA 4.0
2020-07-20T12:32:37.293Z
2020-07-21T08:16:43.757Z
2020-07-21T08:16:43.757Z
David Hammen
<p>tl;dr</p>
<p>The inner moons are much to small (and too stiff) to experience significant tidal work. They also cool down much faster though I don't think this is relevant in this case.</p>
<p>There're two things to look at, here:</p>
<ol>
<li>The total amount of tidal work that is done on a moon</li>
<li>The amount of energy that is radiated away over time</li>
</ol>
<p>When looking at the list of Jupiter's moon, one can easily see that especially the Galilean moons are more than a 1.000 times bigger than anything else that's orbiting Jupiter including the inner moons.</p>
<p>As we know work is force multiplied by distance, so it's not enough to <strong>push</strong> something - in order for work to be done, it also has to <strong>move</strong>.</p>
<p>For simplicity's sake let's think of a moon as a piece of rock. If you squeeze a piece of rock with your bare hands, you're just applying a force to it but not over a distance as the rock won't change shape. Thus you won't do any work and the rock won't heat up.</p>
<p>As with moons, the tidal force not only depends on the distance to the object it orbits, but also on of the mass of the object itself (among other things such as size), so a heavier object experiences higher tidal forces than a light one and there's a certain point at which those forces are strong enough to actually cause a change of shape, which means that's there's actually work being done that turns into heat.</p>
<p>The other aspect is probably negligible, though it's there: Larger objects cool down much more slowly, because they have a lower surface area compared to their volume.</p>

CC BY-SA 4.0
2020-07-23T17:40:00.827Z
Re: "the Galilean moons are more than a 1.000 times bigger than anything else that's orbiting Jupiter including the inner moons." By what measure? [Amalthea has a diameter of 170 km and Europa 3100 km](https://en.wikipedia.org/wiki/Moons_of_Jupiter#List) (a factor of 18)
Peter Mortensen
CC BY-SA 4.0
2021-02-20T10:30:10.073Z
@PeterMortensen by any measure that make sense, Europa is 23076 times as big as Amalthea.
PcMan
CC BY-SA 4.0
2020-07-20T11:56:23.460Z
2020-07-20T11:56:23.460Z
Thomas
<p>It’s because they are much smaller than Io.  Tidal forces are <em>differential</em> forces, that is, they result from the difference in gravitational pull on one side of a body compared to the other.  When an object is small, the difference in distance to the two sides of it is necessarily small as well.</p>
<p><a href="https://en.wikipedia.org/wiki/Moons_of_Jupiter" rel="noreferrer">According to Wikipedia</a>, Amalthea, the largest of those four innermost moons has a long axis that is only 250 km, and the others are smaller yet. The strength of the tidal heating <a href="https://en.wikipedia.org/wiki/Tidal_heating" rel="noreferrer">scales as the body’s radius to the 5th power</a>, quite a strong dependence.   The small mass plays a role, too, but the dependence on radius is stronger.</p>
<p>For this same reason, small objects on Earth don’t feel tidal stretching from our Moon, yet the Earth as a whole does.</p>
<p>On a related note, the inner three of those four satellites are inside Jupiter’s <a href="https://en.wikipedia.org/wiki/Roche_limit" rel="noreferrer">Roche limit</a>, the orbital distance at which a purely self-gravitating body should be pulled apart by tidal forces.  But they survive because most small bodies aren’t held together primarily by gravity (just as your own body isn’t) - the internal electromagnetic forces between atoms (which manifests macroscopically as the rigidity of the rock) is the main source of the bodies’ structural strength.</p>

CC BY-SA 4.0
2020-07-20T21:51:08.357Z
To take this to a limit, artificial satellites around Earth don't experience measurable tidal heating, and don't disintegrate from tidal stresses, even though they're well inside the Roche limit.
jamesqf
CC BY-SA 4.0
2020-07-23T17:28:40.883Z
Re *"...as the rigidity of the rock"*: But the inner three includes Amalthea which *"[consist of porous water ice with unknown amounts of other materials.](https://en.wikipedia.org/wiki/Amalthea_(moon))"*.
Peter Mortensen
CC BY-SA 4.0
2020-07-23T17:34:43.423Z
Only inside for the fluid-body Roche limit?
Peter Mortensen
CC BY-SA 4.0
2020-07-23T18:56:33.910Z
Water ice has rigidity - one surmises, in these specific instances, enough to withstand tidal forces. Presumably the moons for which that was not the case are all the ones that are no longer there.
Jonathan Hartley
CC BY-SA 4.0
2020-07-20T12:16:25.147Z
2020-07-20T12:16:25.147Z
Eric Jensen
<p>Volcanism on Io is caused by the fact that it is <a href="https://en.wikipedia.org/wiki/Tidal_heating_of_Io" rel="noreferrer">tidally heated</a>. There are four moons that are closer to Jupiter than Io with higher eccentricities, yet they don't seem to have any volcanism at their surface. Why isn't the tidal heating enough to generate volcanism on any other moons of Jupiter?</p>

CC BY-SA 4.0
2020-07-21T08:44:48.757Z
There are no spherical moons between Jupiter and Io. All the satellites inbetween are asteroids, the largest of them are Amalthea and Thebe having diameters of about 250 km (150 mi) and 110 km (70 mi). So there is no interior differentiation and thus any volcanism impossible.
Ioannes
<jupiter><natural-satellites><tidal-forces><io>
Why don't the inner moons of Jupiter have tidally-induced volcanism?
usernumber
# Why don't the inner moons of Jupiter have tidally-induced volcanism?

(Asked by: username_0 on 2020-07-20)

Volcanism on Io is caused by the fact that it is [tidally heated](https://en.wikipedia.org/wiki/Tidal_heating_of_Io). There are four moons that are closer to Jupiter than Io with higher eccentricities, yet they don't seem to have any volcanism at their surface. Why isn't the tidal heating enough to generate volcanism on any other moons of Jupiter?

**Tags:** jupiter, natural-satellites, tidal-forces, io
> **Comments:**
> **username_1**: There are no spherical moons between Jupiter and Io. All the satellites inbetween are asteroids, the largest of them are Amalthea and Thebe having diameters of about 250 km (150 mi) and 110 km (70 mi). So there is no interior differentiation and thus any volcanism impossible.  

---

## 3 answers

### Answer by username_6 on 2020-07-20. Score: 48 | Accepted answer

It’s because they are much smaller than Io. Tidal forces are *differential* forces, that is, they result from the difference in gravitational pull on one side of a body compared to the other. When an object is small, the difference in distance to the two sides of it is necessarily small as well.

[According to Wikipedia](https://en.wikipedia.org/wiki/Moons_of_Jupiter), Amalthea, the largest of those four innermost moons has a long axis that is only 250 km, and the others are smaller yet. The strength of the tidal heating [scales as the body’s radius to the 5th power](https://en.wikipedia.org/wiki/Tidal_heating), quite a strong dependence. The small mass plays a role, too, but the dependence on radius is stronger.

For this same reason, small objects on Earth don’t feel tidal stretching from our Moon, yet the Earth as a whole does.

On a related note, the inner three of those four satellites are inside Jupiter’s [Roche limit](https://en.wikipedia.org/wiki/Roche_limit), the orbital distance at which a purely self-gravitating body should be pulled apart by tidal forces. But they survive because most small bodies aren’t held together primarily by gravity (just as your own body isn’t) - the internal electromagnetic forces between atoms (which manifests macroscopically as the rigidity of the rock) is the main source of the bodies’ structural strength.

> **Comments:**
> **username_7**: To take this to a limit, artificial satellites around Earth don't experience measurable tidal heating, and don't disintegrate from tidal stresses, even though they're well inside the Roche limit.  
> **username_4**: Re *"...as the rigidity of the rock"*: But the inner three includes Amalthea which *"[consist of porous water ice with unknown amounts of other materials.](https://en.wikipedia.org/wiki/Amalthea_(moon))"*.  
> **username_4**: Only inside for the fluid-body Roche limit?  
> **username_8**: Water ice has rigidity - one surmises, in these specific instances, enough to withstand tidal forces. Presumably the moons for which that was not the case are all the ones that are no longer there.  

---

### Answer by username_2 on 2020-07-20. Score: 18

> There are four moons that are closer to Jupiter than Io with higher eccentricities, yet they don't seem to have any volcanism at their surface.

Only one of those innermost moons (Thebe) has an eccentricity higher than that of Io. The other three have lower eccentricities.

The reason they don't exhibit volcanism is because they are too small. The largest of the four, Amalthea, has a mean radius of 83.5 km. Compare that with Io's mean radius of 1821.6 km.

Tidal heating results in part from a tension between the tidal forces exerted by the planet that tends to pull a moon out of round and the moon's self-gravitational force that tends to pull a moon into a roundish shape. Small bodies such as those four innermost moons of Jupiter are too small to pull themselves into a roundish shape. The fuzzy boundary between small bodies such as those innermost moons and dwarf planetary bodies such as Pluto is the [potato radius](https://arxiv.org/abs/1004.1091), somewhere between 200 and 300 km.

All of those four innermost moons of Jupiter are well below the potato radius, too small to pull themselves into a roundish shape via self-gravitation. The tidal forces on those small bodies are also smallish. (Tidal forces are proportional to body radius). This makes the tension between self-gravitation and tidal forces negligible for those small bodies.

Another factor that comes into play is the square-cube law. Cooling is proportional to surface area and hence the radius squared while mass is proportional to radius cubed. This means cooling per unit mass is inversely proportional to radius. Heating induced by those negligible tidal forces and negligible self-gravitation forces dissipates quickly in small bodies, not so quickly in larger bodies. This makes the interiors of larger bodies such as Io partially liquid. Small bodies such as those innermost moons of Jupiter are solid throughout. They don't flex as much as do those larger moons, and hence there is significantly less heat production in those small moons.


---

### Answer by username_3 on 2020-07-20. Score: 6

tl;dr

The inner moons are much to small (and too stiff) to experience significant tidal work. They also cool down much faster though I don't think this is relevant in this case.

There're two things to look at, here:

1. The total amount of tidal work that is done on a moon
2. The amount of energy that is radiated away over time

When looking at the list of Jupiter's moon, one can easily see that especially the Galilean moons are more than a 1.000 times bigger than anything else that's orbiting Jupiter including the inner moons.

As we know work is force multiplied by distance, so it's not enough to **push** something - in order for work to be done, it also has to **move**.

For simplicity's sake let's think of a moon as a piece of rock. If you squeeze a piece of rock with your bare hands, you're just applying a force to it but not over a distance as the rock won't change shape. Thus you won't do any work and the rock won't heat up.

As with moons, the tidal force not only depends on the distance to the object it orbits, but also on of the mass of the object itself (among other things such as size), so a heavier object experiences higher tidal forces than a light one and there's a certain point at which those forces are strong enough to actually cause a change of shape, which means that's there's actually work being done that turns into heat.

The other aspect is probably negligible, though it's there: Larger objects cool down much more slowly, because they have a lower surface area compared to their volume.

> **Comments:**
> **username_4**: Re: "the Galilean moons are more than a 1.000 times bigger than anything else that's orbiting Jupiter including the inner moons." By what measure? [Amalthea has a diameter of 170 km and Europa 3100 km](https://en.wikipedia.org/wiki/Moons_of_Jupiter#List) (a factor of 18)  
> **username_5**: @PeterMortensen by any measure that make sense, Europa is 23076 times as big as Amalthea.  

---

