Planetshine - need algorithms

[Sword7]

I had seen new photos from Artemis II lunar flyby. Astronauts took some photos during solar eclipse. I saw the dark crescent on northwestern moon. Sun shines on earth but it reflected 30% of sunrays (albedo .30) and shines on moon. I googled and found some information. Also, something shines on earth's night side. Where did it reflect from?

Click for more information - Planetshine but it does not have algorithms or formula.

Does anyone have any information about algorithms or formula for planetshine?

Tim

 

[Boxx]

as a very simplified low (linearization to avoid computation of complex integrals), I would consider the phase angle [math]\alpha[/math] (between the Sun and the target/moon, as seen from the reflecting body/planet) and define a light source in the graphic engine at the location of the reflecting body, whose emitted power will be: [math]\phi(\alpha)[/math]:

[math]\phi(\alpha)=\cos\left(\frac{\alpha}{2}\right)\,\phi_r[/math],

where [math]\phi_r[/math] is the "power" defined for the emitting body in the graphic engine (?). The problem is that the emitted power depends on the direction, which is bad if you want to code the graphic engine. You could also use a true "phase function" (more complex, maybe over-sized here). If we have to compute the re-emitted light of the Sun by the reflecting body (assumed as a dot source of light), then we need [math]\tau[/math] the albedo of the body, the distance to the Sun and the size of the body... warning, the conversion of the actual power (watts) into the intensity of light in the graphic engine is totally unknown by me.

[math]\phi_r = \tau\,\left(\frac{d_\text{Earth}}{d_\text{Body}}\right)^2 \, \pi(r_\text{Body})^2 \, \phi_E} [/math],

where [math]\phi_E[/math] is the flux received by the Earth from the Sun (1400W/m²), "d_" = distances to Sun, "r_" = radius. Again, it's an approximation of how much light is caught by the body and, then, partially reflected (as if the body were flat and only reflecting perpendicularly), but we're talking about weak reflections so it could probably look realistic.

The "haze" of the Sun behind the moon is, IMHO, another story, way more complex and quiet magnificent.
(edit: sorry for the math, it doesn't show up as expected.... copy/paste in Latex -> for instance here: https://quicklatex.com/ )

 

[GLS]

I might be wrong here, but if the albedo of the Moon is 0.12 and the Earth is 0.3, then the Earth-shine that we see on the Moon is 0.036% of the light that the Earth gets. If we consider this as level 255 (8 bits for light: 0-255), then this Earth-shine is 9, so it is more than the faintest level possible. But, a quick look comparison between black (level 0) and full blue at level 9, shows that they are barely distinguishable.
In addition, we then run into the "only one light source" limitation...

BTW, just to illustrate how hard it is to correctly show the light levels, if we again consider Earth as light level 255, Neptune would be about a quarter of the faintest light level possible, so full darkness. But the Earth shouldn't be at the brightest level, because Mercury and Venus get way more light, which means that Neptune should actually be even darker.

 

[Boxx]

in my understanding, the graphic engine is responsible to illuminate the Moon, and to compute the light spread on its surface (a sphere) assuming its albedo 0.12 (if you say so) that must be input, indeed, but not counted in light sources: there are 2 sources here, the Sun and the Earth. Ideally, does the graphic engine allow a "mask" on the light source (like a spotlight, for instance), then, it would be easier to apply a hemi-spherical mask on the night-side of the Earth...

 

[Sword7]

Yeah. Solution is using HDR (10-bit color depth).

 

[Sword7]

as a very simplified low (linearization to avoid computation of complex integrals), I would consider the phase angle [math]\alpha[/math] (between the Sun and the target/moon, as seen from the reflecting body/planet) and define a light source in the graphic engine at the location of the reflecting body, whose emitted power will be: [math]\phi(\alpha)[/math]:

[math]\phi(\alpha)=\cos\left(\frac{\alpha}{2}\right)\,\phi_r[/math], View attachment 48116

where [math]\phi_r[/math] is the "power" defined for the emitting body in the graphic engine (?). The problem is that the emitted power depends on the direction, which is bad if you want to code the graphic engine. You could also use a true "phase function" (more complex, maybe over-sized here). If we have to compute the re-emitted light of the Sun by the reflecting body (assumed as a dot source of light), then we need [math]\tau[/math] the albedo of the body, the distance to the Sun and the size of the body... warning, the conversion of the actual power (watts) into the intensity of light in the graphic engine is totally unknown by me.

[math]\phi_r = \tau\,\left(\frac{d_\text{Earth}}{d_\text{Body}}\right)^2 \, \pi(r_\text{Body})^2 \, \phi_E} [/math], View attachment 48117

where [math]\phi_E[/math] is the flux received by the Earth from the Sun (1400W/m²), "d_" = distances to Sun, "r_" = radius. Again, it's an approximation of how much light is caught by the body and, then, partially reflected (as if the body were flat and only reflecting perpendicularly), but we're talking about weak reflections so it could probably look realistic.

The "haze" of the Sun behind the moon is, IMHO, another story, way more complex and quiet magnificent.
(edit: sorry for the math, it doesn't show up as expected.... copy/paste in Latex -> for instance here: https://quicklatex.com/ )

Ok, got it now. Thanks for info. Yes, that called zodiacal light that was reflected from interplanetary dust.

Full moon reflected 12% (albedo .12) sunlight and shines on Earth. At night, I can see anything without flashlight! Astronauts were able taking a picture of night side of earth.