Let's say you've calculated the tidal heating of a moon in Watts using the method described in How does one calculate the tidal heating of a satellite?, how do you then convert this to surface temperature?

In my case, the moon orbits a gas giant which orbits a star.

  • $\begingroup$ The answer for this type of question is going to depend heavily on the parameters of the problem and what physics you care about. Ie- what is the heating? What material is the moon made of, what's its mass/radius, does it have an atmosphere? Do you care about the daily variation of temperatures or just the time average over an orbit? Either way, for this you would need to know the orbital period of the planet and moon and their inclinations. These types of questions are important to answer because they change what physical processes are going to determine the answer. $\endgroup$ Commented Oct 30, 2020 at 14:29
  • $\begingroup$ (cont) For instance, LDutch's answer assumes that heat transport is dominated by conduction. In reality, depending on the mass of the planet, the age of it, and what it's made of, the inner layers may be fluid enough to be dominated by convective transfer instead. I might be able to write some code to answer this in a more satisfactory way but that depends strongly on all the questions I listed above. If you don't have answers to these things, then MarvinKitfox's comment details a good way to get a rough upper bound for planets with no atmosphere. $\endgroup$ Commented Oct 30, 2020 at 14:36

3 Answers 3


Equilibrium assumption. Once you have the heating power generated inside, $\dot E$, a reasonable way to calculate (or at least approximate) the surface temperature, is to simply assume that: all the power generated by the tidal heating, is radiated outwards and lost to space.

Why assumption is reasonable. Note such assumption is indeed reasonable: assume the opposite: assume that the power generated is greater than radiated power, $\dot E > P$. This means that more heat is generated than the system is able to dump, meaning, an increase in temperature, meaning, $P$ will get larger because $P$ increases with $T$, and this will proceed until $P = \dot E$. Assume the opposite, assume $\dot E < P$, in this case, more power is being radiated outwards than being generated, meaning, temperature will decrease, and thus $P$ will decrease, until $P = \dot E$. Indeed, $P = \dot E$ is the equilibrium case.

Calculating surface temperature. Using Stefan-Boltzmann Law, the power radiated by a surface of temperature $T$ is: $$ P = A\epsilon\sigma T^4 $$

where $A$ is the surface area of the object, $\epsilon$ is the emissivity of the object [for perfect blackbodies, $\epsilon=1$], $\sigma$ is a constant, known as Stefan–Boltzmann constant, and $T$ is the temperature of the surface (after all, power is being radiated outwards from the surface).

Since you claim you have $\dot E$, then just make said assumption above: $P = \dot E$. That said, temperature becomes trivial to find:

$$ T = \left(\frac{\dot E}{A\epsilon\sigma}\right)^{\frac{1}{4}} $$

You might also want to include $P_0$, the power radiated inwards towards the satellite [say, by a star, or whatever]. In such a case, the equation would be: $P - P_0 = A\epsilon\sigma T^4$. The calculation of $P_0$ is not complicated and can be done simply using geometrical reasoning.

This is just an approximation: A similar calculation than this is used to estimate temperature of planets, and to calculate Goldilocks zone (or habitable zones) around a star: the power received by the planet [Stefan-Boltzmann] + generated inside [tidal locking + etc] = power radiated outwards [Stefan-Boltzmann].

Above procedure is also used to calculate temperature of stars based on the radiated power [it is reasonable to assume stars are perfect blackbodies].

However, this calculation complete ignores absorption and re-emission of thermal radiation by atmospheric gasses [like, the greenhouse effect]. The more dense and atmosphere of a planet, the more this calculation risks being in error.


This article:

"Exomoon Habitability Constrained by Illumination and Tidal Heating", Rene Heller and Roy Barnes, Astrobiology 2013 - https://arxiv.org/vc/arxiv/papers/1209/1209.5323v2.pdf - has a lot info for anyone interested in the potential habitability of hypothetical giant exomoons orbiting giant exoplanets.

  • $\begingroup$ This paper was a fascinating read! It also made me realise that tidal heating is not the only thing affecting the global flux of a gas giant's moon. For instance: you also have reflected light from the gas giant and of course: eclipses! Also, side note: your link is broken, the [1] should not be included. $\endgroup$
    – Astavie
    Commented Nov 1, 2020 at 16:20
  • $\begingroup$ @Astavie - I fixed the link. $\endgroup$ Commented Nov 19, 2020 at 2:59

What you are looking for is called heat equation, and is a well established way to calculate the spatial and temporal variation of the temperature of an extended body knowing the energy flow.

$\partial u \over \partial t$$-a \nabla^2u=0$

In your case it can be assimilated to the case of internal heat generation.

$1 \over \alpha$$\partial u \over \partial t$$=($$\partial^2u \over {\partial x^2}$$+$$\partial^2u \over {\partial y^2}$$+$$\partial^2u \over {\partial z^2}$$)$$+ $$1\over k$$q$

where $\alpha$ is the thermal diffusivity $\alpha = $$k \over {c_p \rho}$

By solving that equation after having assigned the boundary conditions you can get the temperature distribution in the entire body and thus also on its surface.

  • $\begingroup$ I think I'm going to need a little more help here... How does one calculate the thermal diffusivity (in the equation, the only familiar thing I see is the density)? How does one solve this equation? How does one then combine it with the energy coming from the star? $\endgroup$
    – Astavie
    Commented Oct 30, 2020 at 11:58
  • 1
    $\begingroup$ @Astavie, that's an entire exam of engineering, plus calculus to know how to solve differential equations. I cannot summarize all of these in one answer here. $\endgroup$
    – L.Dutch
    Commented Oct 30, 2020 at 11:59
  • $\begingroup$ Right, but this is worldbuilding, where approximations cutting corners are hopefully allowed. Can't I just add the tidal energy (divided by the planet's surface area) to the planetary equilibrium temperature equation? (Fsun + Ftidal = σT^4) $\endgroup$
    – Astavie
    Commented Oct 30, 2020 at 12:05
  • $\begingroup$ @Astavie: No you cannot. Your problem, stripped of its worldbuilding disguise, asks to calculate the external temperature of an oven (or of a stove etc.) given the thermal power produced inside, the material of which the oven (or the stove etc.) is made, the glazing (or lack of) on the surface, and the surrounding environment. It is not trivial. Just an example: most of the heat will be dissipated as infrared radiation; now, the surrounding environment (in your case, the atmosphere) may or may not couple with the infrared radiation, producing wildly different surface temperatures. $\endgroup$
    – AlexP
    Commented Oct 30, 2020 at 12:09
  • 2
    $\begingroup$ The TRUE answer is very complex, but the approximation is simple. You have tidal heating. You have incoming sunlight heating. So you know sum heat generation. Cheat and assume the whole moon is the same temp(infinite thermal conductivity). Cheat some more and assume the rest of the universe is absolute zero. plug albedo, plug in blackbody radiation to match this heat rate. Hey presto. $\endgroup$
    – user79911
    Commented Oct 30, 2020 at 12:35

You must log in to answer this question.

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