Between a moon and the primary, the equation for tidal heating is:

$$\dot E_\mathit{Tidal} = - Im(k_2) \frac{21}{2} \frac{GM_h^2 R^5 n e^2}{a^6}$$

But how does one calculate the tidal heating between moons?

Simplifying assumptions I'm fine with:

  • The moons are coplanar, $I_\mathit{affected} = I_\mathit{perturbing} = 0$
  • Both orbits have no eccentricity, $e_\mathit{affected} = e_\mathit{perturbing} = 0$ (which also implies no heating from the primary)
  • The affected moon is tidally locked to the primary
  • The perturbing moon can be treated as a point mass

While I'm not able to come up with any formula, I suspect the following properties hold:

  • The tidal heating is still proportional to $Im(k_2)$, as this seems to only be an internal property of the moon.
  • It's still proportional to $R^5$
  • Since tidal forces are inversely proportional to distance cubed, I think the overall heating is proportional to ${(a_\mathit{affected} - a_\mathit{perturbing})^{-3}}$, due to most heating happening while they are in close proximity.
  • It's inversely proportional to the relative synodic period of the two moons.
  • 1
    $\begingroup$ Why do you think the formula only applies when mass of primary >> mass of moon? $\endgroup$ Jul 23, 2020 at 14:40
  • $\begingroup$ @CarlWitthoft I don't see I have made such an assumption, and I can't see such an assumption in the linked article either. If you are suggesting adapting the equation by treating the second moon as the "primary", I'm not sure that's valid since the tidal locking condition no longer holds. $\endgroup$ Jul 23, 2020 at 14:50
  • $\begingroup$ OK, why wouldn't tidal locking apply? $\endgroup$ Jul 23, 2020 at 17:14
  • 1
    $\begingroup$ Re Both orbits have no eccentricity, $e_\text{affected}=e_\text{perturbing}=0$ (which also implies no heating from the primary) This is not true. Enceladus has an eccentricity of 0.0047, greater than that of the 0.0041 eccentricity of Io's orbit about Jupiter. Just as Jupiter causes significant tidal heating in Io, Saturn causes significant tidal heating in Enceladus. $\endgroup$ Jul 23, 2020 at 19:15
  • $\begingroup$ @DavidHammen $e = 0$ is the model I'm looking for, I'm not claiming Enceladus has zero eccentricity. May I also point out that Dione is not a point mass. $\endgroup$ Jul 23, 2020 at 19:20

2 Answers 2


This is for instance the case for the Saturn moons Enceladus and Dione.

This is not the case for Enceladus and Dione. Enceladus's orbit about Saturn has an eccentricity of 0.0047, which while low, is more than enough to result in tidal heating. Dione does play a role in the tidal heating of Enceladus, but that role is secondary. Dione is what keeps Enceladus's eccentricity non-zero via Enceladus's 2:1 mean motion orbital resonance with Dione.

  • $\begingroup$ This removes some of the motivation (different mechanism for example chosen), but it doesn't really address what I'm asking for. $\endgroup$ Jul 23, 2020 at 19:17
  • $\begingroup$ What you are asking for effectively does not exist. $\endgroup$ Jul 23, 2020 at 19:19
  • 2
    $\begingroup$ Well, that would be the answer then, if you have some way to show it. $\endgroup$ Jul 23, 2020 at 19:21

You have asked a very reasonable question.

Although moon-moon tidal heating is usually very weak, it may in principle be of interest in tightly packed systems. A detailed theory of this effect (motivated by TRAPPIST-1) was developed in 2021 by Michaela Walterova in her outstanding Ph.D. thesis. Also see page 39 in her defence slides.

I begged Michaela to publish the solution, but she keeps procrastinating.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .