Color of the sky

Question

I know the default will be blue, but just HOW blue, i.e., wavelength in nanometers? I have a world with the following stats:

• Radius: 4628 km.
• Mass: .35 Earth (2,09076 x 10^24 kg).
• Atmospheric Density: .88 Earths.
• Scale height: 14.63 km.
• Atm. Composition:
  • N2: 72%
  • O2: 19.6%
  • CH4 (Methane): .04%
  • CO2: .02%
  • Ar: 8.34%
  • Plus trace like ozone, water vapor, etc.
• Temperature: 299 K (25,85 °C)
• Solar Insolation: 1.04 (relative to Earth)
• It orbits a G3V star:
  • Temperature 5,756 K (5,482,85 °C)
  • Peak radiation: 503 nm.

What would be the formula for calculating the exact shade of blue? What factors do I need to keep in mind?

Answer 1

The answer to this depends on surprisingly few factors. In fact, the only things you will need to consider are your atmospheric depth and your sun's output spectrum.

The blue color of the sky is predominantly caused by Rayleigh Scattering, which is the deflection of electromagnetic waves as they travel through a medium of dipoles. To get the correct color, you draw a ray from the observer out into space. At each point along this line, you determine the intensity of the light hitting that point. The intensity of Rayleigh scattering is defined by the equation:

$$ I = I_0 \frac{ 1+\cos^2 \theta }{2 R^2} \left( \frac{ 2 \pi }{ \lambda } \right)^4 \left( \frac{ n^2-1}{ n^2+2 } \right)^2 \left( \frac{d}{2} \right)^6$$

Where R is the distance to the particle, theta is the scattering angle, lambda is the wavelength, n is the refractive index, and d is the diameter of the particle.

Of course, none of these are dependent on your medium except n, the refractive index, which will be roughly the same for your planet's atmosphere as is Earth's air.

What this means is that the effect of Rayleigh Scattering is going to be pretty much the same as you see on Earth. (There's Mie scattering as well, but Rayleigh dominates). The only difference will be the atmospheric depth, which affects the path lengths you will be integrating along (both the path length from sun to particle and path length from particle to ground observer).

The color of the sky will, of course, also be dependent on the star you choose. However, since the star you chose is really close to that of the sun, you should expect almost perfectly earth-like behaviors. The only difference will be from that path length difference. I believe that the result will be a darker blue with less scattering.

And, since you asked for a "shade" of blue and this scattering effect is multispectral, we can collapse it into a single shade as viewed by a human using the CIE standard observer.

Answer 2

Your atmospheric density

The refractive index of a material depends on the density. You've stated yours is $88\%$ of Earth's, the refractive index of Earth's air is $n_{air} = 1.000293$. For gasses $n-1$ is directly proportional to the density of the material, so: $$n_{air} - 1 \propto \rho_{air}$$ $$n_{new} - 1 \propto \rho_{new} = \rho_{air} \times 0.88 $$ $$\frac{n_{new} - 1}{n_{air} - 1} = \frac{\rho_{air} \times 0.88}{\rho_{air}} = 0.88 $$ $$ n_{new} = 1 + 0.88(1.000293-1) = 1.00025784$$

So your new refractive index is slightly lower than that on Earth.

If we look at rayleigh scattering, assuming all other constants are the same: $$ I \propto \left( \frac{ 1 }{ \lambda } \right)^4 \left( \frac{ n^2-1}{ n^2+2 } \right)^2$$

comparing the intensities for your new refractive index:

$$I_{new} = I_{Earth} \cfrac{\left( \cfrac{ n_{new}^2-1}{ n_{new}^2+2 } \right)^2}{\left( \cfrac{ n_{air}^2-1}{ n_{air}^2+2 } \right)^2} = I_{Earth} \times 0.774 $$

We get a $77\%$ reduction in the light you're scattering through the atmosphere across all wavelengths.

As a result your sky will be a more washed out blue - the lower refractive index will mean the light scatters less so the distinction will be lower - but the sky will be darker since the intensity of scattered light is so reduced.