Okay, thanks to Ash, I know how to calculate the percent power of light hitting a planet at various angles. It is equal to cos(θ)2, where θ is the incident angle. I have a gas giant with a radius of 11.2 Earth radii that orbits its sun at a distance of 1 AU, and is orbited by a moon that has a radius of .9 Earth radii and a orbital radius of .00351 AU. There are eight different points on the moon where I'm testing for how much solar radiation bounces off the gas giant and strikes its moon, with one being the exact center of the side that’s tidally locked to the gas giant.

Each of the eight points are 45 degrees from each other. Using this model, can someone figure out the power percentage of radiation reflected off the gas giant and onto its moon?

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    $\begingroup$ Why not asking this on Physics.SE? It boils down to calculating some integrals, and has nothing to do with worlbuilding $\endgroup$
    – L.Dutch
    Apr 24, 2021 at 6:33
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    $\begingroup$ Maybe few mad drawings in paint or something, and not exactly that clear - what u actualy asking - power of moonshine around the globe in lumens or? $\endgroup$
    – MolbOrg
    Apr 24, 2021 at 6:51
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    $\begingroup$ It depends on the optical parameters of the gas giants atmosphere, and will change drastically throughout the orbit. That said it could be a fun one to draw some diagrams of! $\endgroup$
    – sdfgeoff
    Sep 21, 2021 at 19:44

2 Answers 2


Here are a few notes about your planet and moon.

I note that Earth has a mean radius of about 6,371.0 kilometers, while Jupiter has a mean radius of about 69,911 kilometers, which is 10.973316 times the radius of Earth. A radius 11.2 times the radius of Earth would be 1.0206592 times the radius of Jupiter.

Planets more than just a little more massive than Jupiter will stop epanding as mass is added. Planets more massive than that would tend to have smaller radii as their greater gravity compresses their matter more and more. A planet with a radius of 1.0206592 the radius of jupiter should be possible, but I don'tknow how to calculate what mass it would have. And you need to calculate the mass of the planet to know how fast the moon will have to orbit at a distance of .00351 AU.

Actually, if a radius 1.0206592 times the radius of Jupiter is less than the largest possible radius of a planet, there should be two different masses of planets with such a radius. One mass would be for a planet with a mass in the range where the radius is still increasing with increasing mass, and another mass would be for a mass in the range where the radius is decreasing with increasing mass.

Those two different possible masses would result in different orbital speeds and different orbital periods for the moon.

I note that 0.00351 AU is 525,088.5262 kilometers. I see that is within the Hill sphere of the planet Jupiter at a distance of 5.2044 AU from the Sun and even within the much smaller Hill sphere of Earth at a distance of 1 AU from the Sun. So the moon's orbit should be stable.

If the planet has a radius of 11.2 times the radius of Earth it should have a radius of 6,371.0 kilometers times 11.2, or 71,355.2 kilometers. Thus the orbit of the moon, with a radius of 525,088.5262 kilometers, should be 7.3587 times the radius of the planet.

Io orbits Jupiter at a distance of 421,700 kilometers and has an orbital period, and thus a day, 1.7196 Earth days long. Europa orbits Jupiter at a distance of 671,034 kilometers and has an orbital period, and thus a day, 3.5512 Earth days long.

If your moon orbited Jupiter at a distance of 525,088 kilometers, between the orbits of Io and Europa, it would have a day between 1.7196 and 3.5512 Earth days long. Since the planet would be more massive than Jupiter, the orital period of the moon, and thus its day, would be somewhat shorter. Thus it seems like the days and nights would not bee too long for native lifeforms or visiting Earth humans.


When light hits an atmosphere, there are a couple things that happen: enter image description here

  • The yellow lines represent incoming light - probably from the sun.
  • The orange lines are reflection - the light bounces off at an angle (roughly) equal to the reflection about the surface normal.
  • The blue lines are where the light is handled diffusely. This includes:
    • Transmission - light that passes through in a forwards direction
    • Back scattering - light that is "reflected" back in the direction the light came from.

There is another category not shown and that is the light absorbed (and re-emitted as infrared eventually)

The important thing to note is that all of these light "outputs" must have the same energy as the light that hits the surface.

What I've described here is roughly the same as a Bidirectional Reflectance Distribution Function.

So when you are trying to figure out how much light is reaching the moon, you are summing/integrating all of these different modes of interaction - which is (at least for me) nontrivial. Fortunately people have written software that does this for us! Such as ... blender3d! Blender3D contains a renderer (cycles) which tries to simulate lighting as accurately as possible.

So lets set up a your solar system in blender. Here it is. Completely to scale (it always amazes me that blender can handle distances this huge) lit with a star equivalent to our sun: enter image description here On the left is the gas giant, on the right is the moon.

And here we are in a high orbit around the moon looking at the gas giant: enter image description here You may notice some white speckles on the dark side of the gas giant. That's as the program starts to calculate some of the scattering from the moon onto the gas giant!

Looking at the moon from the top (the gas giant is off on the right) and you can clearly see some light scattering onto it: enter image description here If we let blender churn at it for longer, the speckles would smooth out into an even lighting distribution.

Clearly here I've used a single white sphere as the model for the surface, but with blender you can set it to whatever you want! Unfortunately I have to go to work now, but maybe I'll fiddle with this some more this evening.

But hopefully this answer has shown you that you don't have to calculate it yourself!


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