I'm looking for these data to apply them to the Kelvin-Helmholtz mechanism but I can't find the values. Or at least if there is an order of magnitude compared to the solar luminosity.


2 Answers 2


According to Li et al. (2018), the internal heat coming from Jupiter (and emerging almost entirely as infrared radiation) is $7.485 \pm 0.160$ W/m$^2$.

If we take an average radius for Jupiter of 70,000 km and assume this can be used to estimate a spherical surface area, then th total infrared luminosity (due only to internal heat) is about $4.6\times 10^{17}$ Watts. This is $1.2\times 10^{-9}$ solar luminosities.

A similar number for Saturn would be $2.01 \pm 0.14$ W/m$^2$ (Hanel et al. 1983, possibly there is something more recent). Given an average radius of 58,200 km, this gives an infrared luminosity of $8.6\times10^{16}$ Watts or $2.2\times10^{-10}$ solar luminosities.


The luminosity of an object is proportional to the square of its radius and the fourth power of its temperature, Jupiter has radius of 0.1 solar radii and a temperature of 160 or 0.0285 of the sun.

So its luminosity would be $0.1^2×0.0285^4=6.6×10^{-9}$ of the sun.

The luminosity calculator converts this to an absolute magnitude of 25.2. This would be a bolometric magnitude, since a negligible amount of that radiation is in visible wavelengths.

You can repeat the calculation for Saturn.

But Jupiter is still warmed by the sun, so it's temperature would be less if it were only warmed internally. I'm not sure on the temperature that Jupiter would be but guess it would be about half the temperature it is now, so 1/8 of the luminosity, and about two magnitudes dimmer.

  • 1
    $\begingroup$ Not only does Jupiter absorb heat from the Sun, but it also isn't anywhere near a blackbody - it has an albedo of around 0.5 and huge chunks of the spectrum are taken out by molecular absorption. $\endgroup$
    – ProfRob
    Oct 28, 2022 at 19:22
  • $\begingroup$ Indeed, which is why I'm pleased (and a little surprised) that this calculation gives very similar values to yours, without depending knowledge of the amount of heat produced in the interior, but only on the more easily measurable radius and temperature. Perhaps that is a co-incidence. Nevertheless I'm happy that I was within an order of magnitude with this "back-of-the-envelope" calculation. $\endgroup$
    – James K
    Oct 28, 2022 at 19:27
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    $\begingroup$ E.g. i.sstatic.net/qZVRJ.gif $\endgroup$
    – ProfRob
    Oct 28, 2022 at 19:29

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