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If not then which colour would the sky be and why?

Could you determine which elements the atmosphere is made of just by looking at the colour of the sky?

Is it possible to know the density of a planet by knowing the colour of its sky?

Thanks in advance for the brilliant mind that sees this and answers my questions.

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  • $\begingroup$ Even on Earth, the sky's colour differs depending on a range of factors including humidity, pollution, latitude and elevation, and of course time of day. Compare the vivid blue seen from outback Australia, the pastel blue of northern England, and the deep blue (almost indigo) from the Bolivian plateau (elevation 5000m, extremely dry atmosphere). Or take a high-altitude balloon to the stratosphere to see what colour the sky is at 1% of sea-level atmospheric density. $\endgroup$ Apr 9, 2021 at 0:19

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The color of the sky doesn't just depend on the atmospheric density, but also on aerosols and particulate matter. Mars has about 1% of Earth's atmospheric density, just as you specify in the question. But its sky looks pale orange during the day, whereas sunsets look bluish. So that's almost opposite to what we see on Earth. The reason for these colors is scattering mostly by fine-grained dust particles. A nice picture is linked here (NASA).

The more general question is if you can tell a planet's composition by looking at its atmosphere. But from the context it appears that you imagine standing on the planet and actually looking. In that case, you can tell from the example of our own planet that this is not enough to deduce the composition of the planet. On the other hand, spectroscopic studies of the atmosphere do indeed help pin down its composition, especially if combined with other measurements. Most importantly, you need a way to deduce the planet's mass. This can be done if you find some other object gravitationally interacting with the planet (either a moon, or a spacecraft, etc.).

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  • $\begingroup$ Oh is Mars only 1% of Earth's density? I didn't realize it was that low. $\endgroup$
    – DKNguyen
    Apr 6, 2021 at 19:37
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    $\begingroup$ @DKNguyen I didn't realize it either, until I had to teach an astronomy course last year :) $\endgroup$
    – Jens
    Apr 6, 2021 at 19:43
  • $\begingroup$ Is the blue colour of the sky in broad daylight for example is not enough to determine the density of its atmosphere? If a planet has a blue sky with the sun being the same as ours what information can we deduce from this fact? $\endgroup$ Apr 6, 2021 at 19:49
  • $\begingroup$ @MohammedAli Rayleigh scattering, the cause of the blue sky, is a single-particle effect that has a specific frequency dependence (increasing with the fourth power of the frequency). To read off the atmospheric density from the color alone, you need to figure out how multiple scattering events accumulate to change the color. In principle, this could be done, but it won't give you enough information about all the constituents of the atmosphere to deduce its density. $\endgroup$
    – Jens
    Apr 6, 2021 at 19:59
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    $\begingroup$ @DKNguyen It’s actually about 0.6% of Earth’s sea-level atmospheric pressure at the datum, but varies from 1.2% at the bottom of the Hellas basin to 0.07% at the top of Olympus Mons. $\endgroup$
    – Mike Scott
    Apr 7, 2021 at 12:55
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If there is not enough atmosphere or stuff in the atmosphere to scatter incoming light, it will appear blac or very dark.. Think about how you can't see a laser beam middle mid-air unless it bounces off something like fog to

Our sky appears blue because blue wavelengths are small relative to the air molecules so bounce off them and get scattered more than longer wavelengths. At sunset the sky looks red because the sun being on the horizon causes sunlight has to travel through more atmosphere. So much blue light is scattered along the way that there is very little left by the time it reaches your eye. Longer wavelengths like red are larger than the air molecules don't scatter as much and so can reach your eye. All this is called Rayleigh scattering.

The particles in clouds are huge relative to visible wavelengths so scatter the wavelengths more evenly and so appear white. This is Mie scattering.

And this is why long-infrared (not near-IR) can penetrate clouds (think satellites).

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  • $\begingroup$ So let's say hypothetically we find a planet near a star that is exactly the same as our sun. We land a rover and take a bunch of pictures and we observe a blue sky, is it fair to conclude that their atmospheric density is similar to ours? $\endgroup$ Apr 6, 2021 at 19:09
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    $\begingroup$ The chances are fairly good I think as long as the composition of the air is of similar sized particles. Note that without dust in the air, the Martian sky is blue closest to the sun and red farthest from the sun. $\endgroup$
    – DKNguyen
    Apr 6, 2021 at 19:10
  • $\begingroup$ @DKNguyen The Martian sky is that color because of dust in the air. $\endgroup$
    – nick012000
    Apr 7, 2021 at 4:33
  • $\begingroup$ @nick012000 Wait, so you're saying 1% is thin enough to appear as black? $\endgroup$
    – DKNguyen
    Apr 7, 2021 at 5:11
  • $\begingroup$ @DKNguyen I'm saying that the Martian atmosphere has enough ambient dust particles floating in it to give it a red color, even outside of things like dust storms. The reason why it's that color is because of the difference between the size of the dust particles and the wavelengths of light waves. $\endgroup$
    – nick012000
    Apr 7, 2021 at 6:28
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The sky would look as if you were at a height of near 26 km, because that is where the pressure has dropped to 0.01 atmosphere.

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    $\begingroup$ Not quite. At 26 km altitude above the Earth your eyes will likely still be adapted to the brilliant blue glow below your aircraft and even more brilliant white glow near the horizon. This will make you think that zenith is black. But in fact, it will have brightness similar to what you'd observe on the ground at twilight when the Sun is about 2° below horizon. $\endgroup$
    – Ruslan
    Apr 6, 2021 at 21:14
  • $\begingroup$ @Ruslan Indeed, don't look down. $\endgroup$
    – my2cts
    Apr 6, 2021 at 21:48
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Why not make it simple? Let's say, that here on Earth nothing else changes, just that the atmospheric density decreases to your required level (even if it is impossible for the sake of argument let's say that). What do you see when you look up? Good old blue. Why?

The answer is Rayleigh scattering. If nothing else changes, this is the dominant form of scattering when the particles (atmospheric molecules) are much smaller then the wavelength of (visible 400-700nm) light. This is why the sky is blue now, and atmospheric density alone will not change the color itself.

After all, you have relatively small thickness near zenith, which makes most of the light scattered to you not too extincted due to Beer-Lambert law, while near the horizon the thickness is much larger, and the light scattered into the observer, in addition to becoming bluer due to Rayleigh scattering depending on wavelength, becomes also redder due to extinction along this long path. The combination of this bluing and reddening effects gives a color closer to white (which you can see in the daytime simulation above), or reddish-orange (in the twilight).

Why is the sky *uniformly* blue?

Will it change the shade? I bet it will somewhat, maybe it will be a different shade of blue.

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  • $\begingroup$ -1. Árpád, this seems like an uninformed guess rather than a science-based response - especially the last line "maybe it will be a brighter shade of blue". It's quite easy to test this: visit the Atacama Large Millimeter Array in Chile, at 5000m, and you'll see a very deep blue sky (almost indigo overhead). Take a high-altitude balloon into the stratosphere, to 0.01 atm, and the sky looks almost black overhead - for the very obvious reason that the fewer atmospheric molecules there are, the less Rayleigh scattering there is. $\endgroup$ Apr 9, 2021 at 0:27
  • $\begingroup$ @ChappoHasn'tForgottenMonica I edited the last sentence. Though, the rest should be correct, because if nothing else changes, the dominant scattering is Rayleigh, and the main color remains blue. $\endgroup$ Apr 9, 2021 at 4:38

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