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?

  • $\begingroup$ What do yo mean with atmosphere density? Maybe you want to say atmospheric pressure. $\endgroup$
    – Ender Look
    Jun 13, 2017 at 1:43
  • 9
    $\begingroup$ The shade of blue varies, and it's not monochromatic. It's a collection of wavelengths from Rayleigh and Mie scattering. $\endgroup$
    – Cort Ammon
    Jun 13, 2017 at 1:46
  • $\begingroup$ All the little particles coming into contact with sunlight is what gives our sky its colors. Thus, it really depends on the color of the particles. Larger objects will give more complexity to the color as well, such as clouds, damp air, etc. $\endgroup$ Jun 13, 2017 at 2:13
  • 6
    $\begingroup$ Even on this planet, the exact shade of blue varies. Sometimes that shade of blue is pink, or orange, or even greenish. $\endgroup$ Jun 13, 2017 at 2:16
  • 2
    $\begingroup$ This feels relevant xkcd.com/1145 $\endgroup$
    – Raditz_35
    Jun 13, 2017 at 12:22

2 Answers 2


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.

  • 3
    $\begingroup$ Whether Rayleigh dominates strongly depends on weather. In some places you can often see silvery-blue to almost white sky due to very thin clouds — with the Sun shining at the same time from the same location in the sky. $\endgroup$
    – Ruslan
    Jun 13, 2017 at 9:54

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.

  • $\begingroup$ I must be doing something wrong...when I ran the equation (because I would like to do this for myself) my answers came up COMPLETELY different. Could you walk me through it? Thank you for your answers, though. I must admit I was a little surprised by the final result. But, then again, real life contains many surprises itself... $\endgroup$
    – Mike
    Jun 14, 2017 at 11:41
  • $\begingroup$ @Mike Which result was different for you ($n_{new}$ or $I_{new}$)? I've run it a couple of times and get the same answer. $\endgroup$ Jun 14, 2017 at 14:49
  • $\begingroup$ Inew (sorry, no italics on this thing)...came up .99966958 (it was actually a bit longer, but I just wanted to give the general sense). As far as nnew, I came up with .00000000006256. Clearly, I messed up somewhere. (This was usually my problem in math class...I THOUGHT I was doing the equation right.) Thanks for your patience. $\endgroup$
    – Mike
    Jun 15, 2017 at 11:21
  • $\begingroup$ What was your equation for $n_{new}$? The problem does seem to be there. $\endgroup$ Jun 15, 2017 at 14:27
  • $\begingroup$ Honestly, I entered what I saw there in your post. $\endgroup$
    – Mike
    Jun 15, 2017 at 22:15

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