First of all, I'd like to point out that I'm a worldbuilder and I like my worlds to be as physically possible as... possible.

I am in the process of building a world with a habitable moon orbiting around a gas giant in a similar fashion to Triton's orbits Neptune. I had originally intended it to be highly eccentric as well, but then I found out that Triton has a nearly perfect circular orbit. I looked into it a little and found something about tidal circularization.

What I'm not sure about is if my imaginary moon would undergo the process as well. Also, if it didn't, I'm not sure if it's even possible for life (and if, intelligent life,) to develop on a moon with such eccentric orbit.

Is it possible to explain what the parameters are that determine how quickly tidal circularization happen for systems similar to the Triton-Neptune system? How would the conditions need to be different for the system to remain in an elliptical orbit and avoid circularization?

  • 4
    $\begingroup$ Note that there is also a "worldbuilding" StackExchange, where this would (also?) be on tipic. $\endgroup$
    – James K
    Mar 22, 2020 at 17:14
  • $\begingroup$ I've modified the wording of your question to make it a better fit for Astronomy SE. I think this question can be answered here. If you are unfamiliar with World Building SE mentioned in the previous comment then have a look! However please don't post the same question in multiple SE sites. $\endgroup$
    – uhoh
    Mar 23, 2020 at 0:42
  • $\begingroup$ Did you look at astronomy.stackexchange.com/questions/19639/… $\endgroup$
    – ProfRob
    Jun 20 at 8:35

1 Answer 1


Tidal circularisation is produced by a combination of tidal dissipation in the planet and that in the satellite. Calculations performed for the Jovian and Saturnian moons inducate that for them the dissipation in the planet dominates, and my guess is that this should be so also for the Neptune-Triton system.

The eccentricity is reducing its value exponentially: $$ e\;=\;e_o\,\exp(-t/\tau)~~, $$
where the circularisation timescale $\tau$ obeys $$ \tau^{-1}\,=~A~\frac{M^{\,\prime}}{M}\;\left(\frac{R}{a}\right)^5\,K_2~~. $$ Here $A$ is a positive factor not very different from unity; $M^{\,\prime}$ and $M$ are the masses of the moon and the planet, correspondingly; $R$ is the planet's radius, $a$ is the semimajor axis of the moon, while $K_2$ is the the so-called ``quality function'' of the planet. It is equal to the ratio $k_2/Q$, where $k_2$ is the quadrupole Love number, while $Q$ is the quality factor.

For rapidly rotating planets, $A$ may be negative, in which case the eccentricity grows, the orbit elongates, and the moon may eventually be hit the planet (and get swallowed, if this is a gas or liquid planet). This is not the case of Neptune and Triton.

Nobody knows the value of Neptune's $K_2=k_2/Q$. My wild guess is that it should be somewhere between that of the (solid part of the) Earth (about $1.8\times 10^{-3}$) and that of Jupiter (about $10^{-5}$).

The values of $M^{\,\prime}$, $M$, $R$, and $a$ can be found in Wikipedia.

To have life, you need liquid water, either on the surface or not too deep in the crust. So the body should be sufficiently warm. This warmth may be provided by radioactivity, provided Triton has somehow preserved radioactive material in its depths. Another source of heat is tidal dissipation (so-called obliquity tides).

  • $\begingroup$ Not following where the factor $A$ arises and how if it is of order unity then it can be negative in some cases. Presumably then it could be close to zero? Or is it bimodal? $\endgroup$
    – ProfRob
    Jun 20 at 8:34
  • $\begingroup$ @ProfRob The circularisation rate is calculated via eqn (154) in arxiv.org/abs/1904.02253 where the 3rd and 4th lines should be omitted when we ignore the tides in the moon. In neglect of the inclination of the moon's orbit on the planet's equator, eqn (154) may be approximated with (156) where we should ignore the 2nd line. Usually, this approximation works -- but not for Triton whose orbit is inclined. If we nevertheless accept (156) as a very crude approximation, and if we also apply the CTL tidal model, we shall get the first term in (157). $\endgroup$ Jun 20 at 13:55
  • $\begingroup$ @ProfRob From the first term in (157), we easily derive the exponential formula given in my answer, and we see that the sign of A depends on the interrelation of the moon's mean motion $n$ and the planet's rotation rate $\dot{theta}$. Roughly speaking, an orbit tents to circularise when the planet's spin is not too fast. A swiftly spinning primary boosts the eccentricity of the secondary. Let me also reitarate that my approximation is very crude. First, Triton's orbit is inclined, so the neglect of $i$ in (156) causes big errors. Second, the CTL model is not necessarily applicable to Neptune $\endgroup$ Jun 20 at 14:04

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