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What's the equation for moon-moon tidal heating?

Astronomy Asked by SE - stop firing the good guys on February 9, 2021

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.

One Answer

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.

Answered by David Hammen on February 9, 2021

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