|MadSci Network: Astronomy|
In the Earth's atmosphere the thing that makes the sky blue is a process called Rayleigh scattering, where light of shorter wavelengths is preferentially scattered by the molecules in the atmosphere. Shorter wavelength visible light is blue to the eye. Having the blue light scattered all over to a much greater degree than the red means that the diffuse wash of light from the sky, which is due almost entirely to scattered sunlight, is most definitely blue. You can find items in the MAD Scientist archives mentioning this process here, here, here, and here, in rough order of relevance and complexity.
You're clearly asking a new question, taking the Earth's atmosphere as it now is and asking what spectrum of radiation input would be needed to make the sky look reddish despite the Rayleigh scattering mechanism's natural tendency to make for a bluish background. I think it would take a real calculation to get the answer to this one right. You'd need to do what is called the "radiative transfer" correctly (that is, get a correct computation of the amount of scattered light as a function of wavelength) and fold in knowledge of the human eye's response to color (which is a complicated question of itself; see here for some leads about that). The latter, in effect, is equivalent to choosing what "paint chip" is the least red that could be called pink, and then figuring out what input spectrum is needed to match that color. At some level, of course, that's a subjective choice, as anyone who has chosen a paint color for a living room knows.
To a physicist it might seem tempting to take a "black-body" spectrum (the radiation given off by a hot object; it's a simple spectrum that depends only on the temperature of the radiating object -- see here about black-body radiation), and fold it into some simple assumptions about the atmosphere and Rayleigh scattering (which has a simple wavelength dependence), and get a fast answer that way. You could get an answer that way, but I don't think it'd be a reliable one. For stars, the spectrum on the short-wavelength side of the black-body peak departs strongly from the black-body curve, so that describing the spectrum in this way turns out to be a remarkably bad approximation. The Sun's black-body peak is in the visible, and for cooler stars it is in the red or infrared, so reality will depart more and more from the physicist's simple solution as you investigate cooler and cooler stars' spectra.
The other thing to be wary of is whether another physical effect jumps up and takes over. I don't think that'd be the case, but here's an example to indicate what I mean. On Mars the Sun's spectrum is obviously the same as we see here on Earth, but the daytime sky is pink; this is due to the reddish dust in the Martian atmosphere. Mars's surface atmospheric pressure is 0.006 of Earth's (there's a lot fewer molecules there for Rayleigh scattering), so that the "filtering" (wavelength-selective absorption of the blue light) by the reddish dust wins over the blue-favoring scattering of light off gas molecules. This can happen on Earth under abnormal conditions (when the atmosphere is full of smoke particles from a fire, or after a volcano has put lots of dust and droplets at high altitudes in the atmosphere by an explosive eruption, to give two examples), but you're not asking about those.
In short, I don't see a good way to get the answer to your question without a few days' solid work. The atmospheric science people who study things like the thermal and chemical balance in the Earth's atmosphere (things like greenhouse warming, the ozone hole, etc.) are more likely to have the appropriate radiative transfer tools handy than an astrophysicist who works on stellar atmospheres: there's a lot about the two problems that are different in important ways.
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