On a tidally locked earthlike moon of a gas giant, would the side that faces towards the gas giant be noticeably warmer than the side that faces away? Would the gas giant generate enough heat to have an impact on climate?

The moon is earthlike and orbits at 932,000 km.

Gas giant properties:

  • Radius: 60,613 km
  • Rings radius: 141,432 km
  • Mass: 108 earths
  • Density: 687 kg/m3
  • Temperature: -36.4°C
  • Albedo: 0.42
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    $\begingroup$ I know that the gravitational pull of the gas giant definitely can cause international heating of a planet moon. But i don't know if it could warm up one side. $\endgroup$ Commented Mar 22 at 13:22
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    $\begingroup$ If the relationship between planet and moon is correct. Compare Io (volcanic due to Jupiter) to Europa (which isn't). Get the size of your moon, its distance to the planet, and the size of the planet, to roughly match the ratios found with Jupiter and you've exceeded suspension of disbelief. Frankly, a larger mass than Io would be plausible so long as the ratio of planet mass to distance is met. $\endgroup$
    – JBH
    Commented Mar 22 at 16:24
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    $\begingroup$ @Pelinore, Jupiter is still producing heat from gravitational collapse, with the result that it's a source of radiated heat. $\endgroup$
    – Mark
    Commented Mar 22 at 22:14
  • $\begingroup$ @Mark Huh? [checks link] "Jupiter radiates more heat than it receives through solar radiation, due to the Kelvin–Helmholtz mechanism within its contracting interior" nice! is that going to be enough for any real effect on its moons though? guess that's what this question is asking. $\endgroup$
    – Pelinore
    Commented Mar 22 at 22:32

1 Answer 1


You can apply the Stefan-Boltzmann law to estimate the luminosity ($L$) due to thermal radiation. The law is $L = \sigma A T^4$, where $\sigma = 5.67 \times 10^{-8} \, \text{W}\,\text{m}^{-2}\,\text{K}^{-4}$ is the Stefan-Boltzmann constant, $A$ is the surface area of the gas giant, and $T$ is its temperature in Kelvin.

Using the radius $R = 6.0613 \times 10^7 \, \text{m}$, $T = 236.75 \, \text{K}$, and calculating the surface area $A$ using $A = 4\pi R^2$, yields $L = 8.22 \times 10^{18} \, \text{W}$.

To determine the intensity ($I$) of this radiation upon reaching the moon, the formula $I = L/(4\pi d^2)$ can be used, where $d$ is the distance from the gas giant to the moon. This yields $I = 0.75 \, \text{W}/\text{m}^{2}$. For comparison, the intensity of radiation from the Sun is about $I = 1000 \, \text{W}/\text{m}^{2}$.

Moving on to the high albedo ($a$) of $0.42$, assuming that this gas giant receives $2000 \, \text{W}/\text{m}^{2}$ from its Star, then the total luminosity it'll reflect due to albedo is $2000 \, \text{W}/\text{m}^{2}$ times its surface area times $0.42$, or $L= 3.88 \times 10^{19} \, \text{W}$ which is higher than the heat from the Stefan-Boltzmann Law. But due to the large distance, only $I= 3.88 \times 10^{19} \, \text{W}/(4\pi d^{2})$ will reach the moon or $3.55 \, \text{W}/\text{m}^{2}$, which is still not a lot. It'll definitely receive more from the star directly.

Other heating mechanisms are the Kelvin-Helmholtz mechanism, which generates internal heat within the gas giant, but has a negligible impact on the moon and tidal heating which may affect the moon's internal geology but not its surface climate.

To summarize, I think that the radiative heating from the gas giant and the albedo effect does not make the side of the moon facing it noticeably warmer.


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