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I want to ask what color is the Venusian atmosphere without the presence of dense clouds for our human vision? I searched for this question, and did not find a suitable answer. Everywhere they write about the color of the sky Venus already in the presence of clouds. Would we be able to see the surface of Venus through the dense atmosphere without clouds?

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    $\begingroup$ Have a look at the images sent back by the Soviet Venera landers in the 70s? That should be as close as you get to a cloud-less atmosphere, as those landers made it below the cloud cover. $\endgroup$ Mar 12, 2023 at 19:09
  • $\begingroup$ @AtmosphericPrisonEscape While I don' think the Soviet Venera landers in the 70s had color cameras and color-calibration targets, they certainly might speak to distance visibility. $\endgroup$
    – uhoh
    Mar 13, 2023 at 8:08
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    $\begingroup$ Does this answer your question? What is the color of Venus if it has no atmosphere? $\endgroup$ Mar 15, 2023 at 3:08

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The Wikipedia page for the atmosphere of Venus has two useful graphs due to Darekk2, reproduced below, showing the absorption spectra based on simulations. Simulated Earth spectrum Simulated Venus spectrum Note that there are fewer peaks in the visible range due to the lack of water and oxygen, while there is much more absorption in the infrared range due to all the CO2.

So, if there were no clouds or dust, the Venus atmosphere would be clearer than Earth's.

But at 93 bar, there is much more air to affect the light on the surface. Given that we lose about 30% of the sunlight on the way down to the ground on Earth (mostly scattered back into space than absorbed), we have about $\lambda\approx 0.3$ optical depth. The intensity reaching the ground is $I_0 e^{-m \lambda$}$ where $m$ is the air mass factor. Plugging in $\approx 93 m$ gives just $7\cdot 10^{-13} I_0$: practically all the light would disappear. Even a much lower optical depth is unlikely to change this much. At the same time this does not mean that the atmosphere is dark, it just scatters the light evenly.

So the impression I get is that, no, we would not be able to see the ground. The color of the air would be dominated by the scattered sunlight, making everything yellow.

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  • $\begingroup$ As I understand it, the color of the atmosphere of the second planet from the Sun will be yellowish. I think the color depends on atmospheric pressure. On Venus, the pressure is enormous, and therefore the light waves in it must be scattered more strongly. Thus, our eyes will see the Venusian atmosphere as yellow, and the clouds give Venus a bright white color, as they reflect the Sun's light well. $\endgroup$
    – Mikee
    Mar 15, 2023 at 15:39
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    $\begingroup$ @Mikee - How is this different from my answer? There is not much pressure dependence in Rayleigh scattering, the key factor is the total amount of atmospheric mass. opg.optica.org/josa/abstract.cfm?uri=josa-48-12-1018_1 Pressure broadening and temperature broadening of absorption lines matter, but since there are few in the optical spectrum for the Venusian atmosphere this doesn't matter much. $\endgroup$ Mar 15, 2023 at 18:18
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As far as I know, the color of the atmosphere of any planet depends primarily on its density and the temperature of the star that heats this planet. Thus, for a star of spectral type G2, 1 bar of the atmosphere will look blue to our eyes (if the star is at the zenith). If you increase atmospheric pressure to 10 bar, the color will turn turquoise. The sky of the planet at an atmospheric pressure of 40 bar will appear yellow-white. Thus, I dare to suggest that Venus without clouds could have such an atmosphere color, yellow-white. I don't know exactly how true this statement is.

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    $\begingroup$ I forgot to add that such colors will be visible at the zenith of the planet's atmosphere, and not at the horizon. I understand that near the horizon, the light in the atmosphere behaves differently. Thus, with increasing pressure, the color of the atmosphere for us will be perceived differently. It depends on the scattering of light rays. That's what I wanted to say. $\endgroup$
    – Mikee
    Mar 13, 2023 at 9:52

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