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There are some great explanations on StackExchange of why we see a blue sky on Earth, eg here.

Namely, that it's a combination of :

And, that when the Sun is low and thus light going through ~2000km of atmosphere rather than ~10km (do the trig), Rayleigh scattering means there's very little blue light left, so the end effect is a reddish sky. (A little counterintuitive, but this is a good explanation of how that works.)

My question is why is sky red on Mars.

I've seen two explanations, both dependent on the fact that the much thinner Martian atmosphere means the effect of Rayleigh scattering of light by gas is dominated by the effect of airborne dust.

Explanation one: the airborne dust scatters out all the blue light, so that the sky we'd see is reddish for same reason our sunrise/sunset sky is reddish. (Because there's so much dust?)

Or, simply that because the dust is red (because the surface of Mars is red - ie, lots of iron oxide), we would see a reddish sky.

Can anyone help with this?

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With great thanks to the wonderful Dr Eriita Jones of the University of South Australia...

Mar's reddish sky is due to a combination of:

  • The amount of airborne dust (there's a lot of it - due to the dryness of the Martian surface, the very small size of the dust grains, and the planet's lower gravity).

  • The thinness of the atmosphere, which means there is less scattering of photons by the atmospheric gases (which means less Rayleigh scattering, though it does occur).

Because all that airborne dust is more similar in size to the incident visible photons, more Mie scattering occurs than Rayleigh scattering.

Mars dust is about 10—100 micrometres, visible wavelengths are approx 0.35—0.7 micrometres ...ish) while the $\rm{N}_2$ and $\rm{O}_2$ molecules or our atmosphre, or the $\rm{CO}_2$ molecules of the Martian atmosphere, are ~200—300 picometres, or a thousand times smaller.

Mie scattering (which happens when particle size approaches wavelength) preferentially scatters larger wavelength (redder) light, or more completely the angle of Mie scattering is more uniform for longer wavelengths, while shorter wavelengths (blues) tend to scatter at slight angles. This means that under Mie scattering, blue light tends to be deflected less than red light (thanks to Brian K for this explanation).

Conversely, Rayleigh scattering (when particle size is much smaller than wavelength, eg from gas molecules), which predominates in our own atmosphere, strongly preferentially scatters smaller wavelength (bluer) light (inversely proportional to $\lambda^4$).

So the key difference with Earth is there is less of the Rayleigh scattering (because of Mars' thin atmosphere) and more of the Mie scattering (caused by tiny Mars dust particles), so the Mie scattering dominates.

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    $\begingroup$ $\propto^{-1}$ seems to not really be a common notation for inverse proportionality. Wouldn't it be better to simply say $\propto \lambda^{-4}$ instead? $\endgroup$ – Ruslan Feb 15 '17 at 15:24
  • $\begingroup$ Totally agree Ruslan. And plain English even better. Edited. $\endgroup$ – Errol Hunt Feb 15 '17 at 22:35
  • $\begingroup$ So, is it the done thing to tick my own (Eriita's) answer as the best answer? Or is that a little too self congratulatory? I assume I can't claim my own bounty! $\endgroup$ – Errol Hunt Feb 21 '17 at 0:01
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    $\begingroup$ I didn't wrote my answer "just for the bounty", so don't bother. I Just thank you for a very good question. And even for this "mainstream" answer, which is a very good approach too. Yet I think it should be tested in the lab, if this dust can cause any observable redness. I doubt it can, but then my doubts are nothing, and experiments/true observations are everything. $\endgroup$ – Jokela Feb 21 '17 at 13:20
  • $\begingroup$ Where did you take that Mie mechanism results in preferential scattering of redder light? It's actually very non-uniform: see e.g. this image. And if you average over a range of size parameters, you'll get white scattered light (look outside at the clouds in the day time). $\endgroup$ – Ruslan Aug 11 '18 at 11:04
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Good and interesting question. I just recently realized that the phase transitions can be defined through the speed of light, and this explains the layers of all planets atmospheres. Most interesting was the Stratopause, which is the point where Gas starts to turn to plasma.

So as I was testing this theory, I noted that Mars Practically doesn't have any gaseous atmosphere. It's only few kilometers thick. And the rest of it is plasma.

This might cause the red colour become from the same source as the air glow, which is red also in Earth.

I want to point out with this lection, that Mie scattering merely influences the color, as then "all colours scatters equally." (33:10-> at video)

This is why the Clouds in earth are white. (explained up to 38:00 in the video) and means that if the dust is the cause, then the atmosphere of Mars should be white. (The Red is explained in up to 40:00 in the video, There is even an experience which shows how this happens in this video up to 43:30)

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    $\begingroup$ I believe this is not mainstream physics. $\endgroup$ – Rococo Feb 19 '17 at 23:55
  • $\begingroup$ @Rococo I believe Walter Lewin lectured Meinstream physics in MIT. Pls. look his video. It atleast explains why the other answer cant' be correct. $\endgroup$ – Jokela Feb 20 '17 at 0:00
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    $\begingroup$ I was referring specifically to your paper as not mainstream. Regarding the Levin lecture, although I am not terribly familiar with Mie scattering it looks like Levin's description is probably a simplification- see here for example: osa-opn.org/Content/ViewFile.aspx?id=10460 in which it is explained how Mie scattering can preferentially scatter either the red or blue end of the spectrum under different conditions. So I think it is at least possible that Errol Hunt's conclusion is correct. $\endgroup$ – Rococo Feb 20 '17 at 0:36
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    $\begingroup$ @Rococo Yes. it's possible. -lot of things are possible, especially when things are simplified. but is the amount enough, to really be the cause?? Levin's description at 41:00 -43:30 in the video can't be a simplification. He's showing the whole thing with an experiment. It shows, that the color doesn't depend only from the particles/light source, but also where you look, and from te point of view; at 41:23 -The sky is not everywhere, and it isn't blue to all viewers. Pls. look how the sun is turning red, but it's only the sun which is red. The scattering is also polarizing the light etc. $\endgroup$ – Jokela Feb 20 '17 at 7:01

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