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Jupiter appears to approximately follow Lambert's cosine law as it looks darker towards its limbs when viewed from the same direction as from where the Sun shines on it. Here an image from the article Hubble takes close-up portrait of Jupiter that shows it in opposition:

Now there is the Lommel-Seeliger law which is a good first approximation to diffuse reflection. Here is an image that shows its effects in the middle, while an approximately Lambertian surface is at the bottom:

The Lommel-Seeliger sphere appears flat at zero phase angle. The Lommel-Seeliger law is derived by considering what happens to a beam of light that enters a medium. Therefore, my assumption is that it should be a good approximation to atmospheres and gas giants, too. However, Jupiter apparently proves me wrong. Why does it not look flat?

An example of a celestial body that appears flat at zero phase angle is the Moon. It is covered by lunar regolith that is a medium of pulverized particles.

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  • $\begingroup$ I hope this one is on-topic here and does not belong in Physics SE. $\endgroup$
    – akuzminykh
    Aug 9, 2021 at 7:35
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    $\begingroup$ You don't link to the source - which camera took this picture at which distance with which field of view and apparent size of the planet? $\endgroup$
    – asdfex
    Aug 9, 2021 at 7:46
  • $\begingroup$ @asdfex It's from the Wikipedia article: "This image was taken by the Hubble Space Telescope, using the Wide Field Camera 3, on April 21, 2014. Jupiter's atmosphere and its appearance constantly changes, and hence its current appearance today may not resemble what it was when this image was taken. Depicted in this image, however, are a few features that remain consistent, such as the famous Great Red Spot, featured prominently in the lower right of the image, and the planet's recognizable banded appearance." $\endgroup$
    – akuzminykh
    Aug 9, 2021 at 7:56
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    $\begingroup$ Additional info: the angle Sun-Jupiter-Earth was about 11° at that time which explains the asymmetry between left and right. $\endgroup$
    – asdfex
    Aug 9, 2021 at 8:20
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    $\begingroup$ Question: do any of the models you describe account for the absorption of light passing through (literally) thousands to even tens of thousands of km of atmosphere? The approximations work for the Moon, because the Moon's atmosphere is slightly less dense than Jupiter's $\endgroup$
    – CuteKItty_pleaseStopBArking
    Aug 9, 2021 at 9:27

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Because the Lommel-Seeliger law and Lambertian surface reflection are both single scattering models, ignoring the optical properties of atmospheres.
That is a bad fit for a planet that is almost purely atmosphere and very little else.

The atmospheric effect observed here is reflective atmospheric limb darkening (not to be confused with radiative limb darkening).
This can intuitively be viewed as light having to pass through more atmosphere near the limb than in the centre.

Wildey and Traeton 1971 discusses the limb darkening of Jupiter specifically. While the observational data has been improved since then, the underlying theory remains the same.

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  • $\begingroup$ It might be implicit that a multiple-scattering atmospheric model wouldn't do this, but another issue is that those models assume a uniform medium, assuming the only thing to care about is the first layer of clouds. A multple-scattering model with the same assumption would still be a poor fit. $\endgroup$
    – Christopher James Huff
    Aug 9, 2021 at 12:45
  • $\begingroup$ @ChristopherJamesHuff doesn't multiple scattering imply taking into account the scattering between cloud layers? $\endgroup$
    – Ruslan
    Aug 9, 2021 at 15:24
  • $\begingroup$ @Ruslan multiple scattering in itself only implies taking multiple levels of scattering into account. A multiple scattering model could very well assume a uniform medium. In an atmosphere model, this would account for diffusion of light past the terminator, but not the layered structure. In the context of atmospheric models, I don't know whether anyone's bothered with multiple scattering without a more realistic atmosphere. $\endgroup$
    – Christopher James Huff
    Aug 9, 2021 at 15:42

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