So you reckon it'd be less than x100... x100 is a difference of 5 magnitudes isn't it?
I wonder what the apparent magnitude of the earthshine seen on a new moon itself is...
Actually, I'll crosspost what I was asking about in the Users board here, because there are some astronomical questions here:
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Evil Dr Ganymede wrote:Right. That's intentional. Your planet orbits A with a semimajor axis of 0.4, so when B is 11au, A is still 27.5 closer. The 1/R^2 attenuation of the electromagnetic force means that B will be a dimmer by a factor of about 1/750. Also, A is intrinsically over 3.5 times brighter than B, so A is going to appear over 2500 times brighter than B. Celestia adjusts the apparent light source brightnesses with a power function; with the exponent I've chosen, light from B should be just barely visible. I could adjust the exponent to bring out the light from B more, but I'm not sure that this would be more realistic.
OK, just for ease of reference, here's the Prometheus CEL URL I'm referring to all the time. You'll need the code from
the thread on the Bugs board too.
cel://PhaseLock/Rigel%20Kentaurus%20A:Prometheus/Rigel%20Kentaurus%20B/2366-02-26T19:54:24.16683?x=AI5ZmJkqPKaB7+b//////w&y=oEJZ4FCPrA8sndL//////w&z=qtRYzZxDdzTDiCo&ow=-0.367940&ox=-0.121596&oy=0.674201&oz=-0.628719&select=Rigel%20Kentaurus%20B&fov=20.765589&ts=1.000000<d=0&rf=38711&lm=71
Now, the apparent magnitude of B over the range of 11 to 30 AU (or whatever the extremes are) is from -18 to -20. The full moon has a magnitude of only about -13. So it's what, at least 100 times bright than the full moon, even when it's furthest from the planet... and yet it provides no visible illumination? As it is the full moon casts noticeable, visible shadows when it's up. In this case B should certainly do the same if you were standing on the surface.
Because B isn't illuminating the planet at all though, I presume that B wouldn't cast ring shadows if Prometheus had a ring either. I could believe that it'd be hard to see the shadow while A was above the horizon, but you would surely see a shadow cast by B if it was the sole illuminator given what I just said.
Look at the crescent moon in the sky, and you can clearly see earthshine. Wouldn't B's illumination look similar to that? I wonder what the apparent magnitude of the full Earth is as seen from the moon?
Though the thing with earthshine is that you can see it when the moon is new/crescent, because the Earth is full as viewed from that hemisphere on the moon. Do we not see earthshine when the moon is "half full" because the illuminated side is too bright for us to see the dark side? Or is it because the Earth would also be "half full" as seen from the moon at that time, and that lowers the apparent magnitude of Earth enough that it doesn't illuminate the moon's surface enough to make it visible anymore? I suspect it's more the latter than the former...