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What I'm trying to achieve

For a series of realistic renders I am currently working on I am trying to calculate the light level on different celestial bodies, in terms of the luminous flux (in lumens), as this is the parameter used in the rendering software. (as it's made with earth in mind). Specifically, I am now looking to find the luminous flux on Saturn's moon Enceladus.

Thanks in advance for any help provided :)

What I've tried/found so far

Now the solar irradiance is easy to find as lying somewhere inbetween 16.7 and 13.4 W/m^2. However, in order to derive the luminous flux the frequency contents must be known as the luminous flux weights the light by a sensitivity function for human sight.

Wikipedia gives a formula for the luminous flux as

$$\Phi_v = 683.002 \int_0^{\inf}V(\lambda)\Phi_{e,\lambda}(\lambda)d\lambda $$

Where $\Phi_v$ is the luminous flux in lumen, $\Phi_{e,\lambda}$ is the spectral radiant flux in W/nm and $V(\lambda)$ is the luminosity function describing the sensitivity of the human eye to different wavelengths of light (dimensionless).

$V(\lambda)$ was relatively easy to find, but I have difficulty finding the spectral radiant flux $\Phi_{e,\lambda}$ of sunlight at this distance. Can this be calculated from the irradiance?

I did manage to find the solar irradiance spectrum in space at the earth in (W/m^2)/nm, and I assume this could simply be scaled by $R^{-2}$ ($R$ being the distance from enceladus to the sun) to get the same spectrum at Enceladus, but I am not sure how this relates to the spectral radiant flux.

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You don't need to do it that complicated. The sunlight's spectrum is the same everywhere in the solar system. And in order to get the light intensity, you can simply apply the inverse square law that intensity drops to 1/4 if you double the distance to the sun - like you did for getting the irradiance. The spectral sensitivity of the human eye is also the same everywhere in the solar system.

What is different, is the local conditions: On the surface of bodies it is, of course, influenced by the atmosphere - and that is likely the most important aspect when you want to model different colour; this does include possible aerosol content (like you see for instance on Mars during dust storms). The 2nd most important impact probably is the colour of the light reflected from the body you are "standing" on, thus a feature of the surface material.

So in order to get the colour impression you have to account for these two spectral filters, applied to the (nearly) black-body distribution of the sunlight. Thus in your equation you need to insert these factors as well - or just look at these separately as your equation states the default. Unfortunately this opens a VERY wide field as you can have a variety of atmospheric compositions - just think of the different light conditions you can have here on Earth - and now add the factor of completely different chemestry in the atmosphere. Modelling that is definitely an ongoing scientific endeavour.

If your quest is "just" to get some artists impression, play around with different materials and adjust the spectral transmission and scatter properties of the atmospheres (especially that) - and the ground material.

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  • $\begingroup$ Atmospheric effects could probably be incorporated by multiplying the spectrum with a specific filter. Luckily, for moons such as Enceladus the atmosphere is sparse enough that it can likely be neglected. In this case, can I then just take the black-body spectrum at the appropriate distance for $\Phi_{e,\lambda}$? $\endgroup$ – Mark Marketing May 5 at 12:23
  • $\begingroup$ yes, that's the essence of my probably unnecessary long answer. yes, in the absence of atmosphere: absolutely. You might try to have a look at Stellarium to get a ready-made look from other solar system bodies; it does an excellent job in that. Enceladus is one of the objects in the shortlist. But you have to select the appropriate ground separately. $\endgroup$ – planetmaker May 5 at 12:39
  • $\begingroup$ Thank you. This seems to work pretty well. Additionally, Stellarium is an amazing piece of software! Really helps to put planets into scale :) $\endgroup$ – Mark Marketing May 5 at 12:59
  • $\begingroup$ What color is the Sky on other Planets? $\endgroup$ – uhoh May 6 at 3:21
  • $\begingroup$ Depends. See apod.nasa.gov/apod/ap140302.html for an example $\endgroup$ – planetmaker May 6 at 5:44

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