Would a 50 mile asteroid, with the same albedo as the Moon, be visible from 250000 Miles (400000 KM) away in a dark sky?. And if so what would it's magnitude be?.
Sorry if this post doesn't make sense, i couldn't think of a way to put it.
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Topic authorCaptain-insane
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I'm not crazy, just ask my camel Steve!.
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Hungry4info
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I think we would need to know the exact dimensions of the asteroid, so we can get it's total anguar size/area whatever. In addition to that, it's position in the sky would be a big factor. If the sun - observer - asteroid angle is smaller, the asteroid will appear dimmer. Whereas if the sun - observer - asteroid angle is greater, the asteroid will appear brighter.
Well, other than that, I don't really know how else to help you. I havn't had physics class yet (10th grade... blah).
Edit:
(assuming your asteroid is spherical)
You could probably try taking the angular radius of the asteroid, squaring it, and multiplying it by pi to get the area. Then doing the same for the moon. Then compare them as a fraction (area of asteroid/area of moon) and that should give you a percentage of the moon's brightness that the asteroid will be. This should work since they have equal albedos.
As for translating that into magnitude, heck if I know. I wasn't really all to good at logarithms (that and never learned about it in school yet).
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|| That math is probably not the way to go about doing it, it's just what some uneducated fool like myself would try. I'd certainly wait on some of the other people on this site with far more knowledge about this to help you.
Well, other than that, I don't really know how else to help you. I havn't had physics class yet (10th grade... blah).
Edit:
(assuming your asteroid is spherical)
You could probably try taking the angular radius of the asteroid, squaring it, and multiplying it by pi to get the area. Then doing the same for the moon. Then compare them as a fraction (area of asteroid/area of moon) and that should give you a percentage of the moon's brightness that the asteroid will be. This should work since they have equal albedos.
As for translating that into magnitude, heck if I know. I wasn't really all to good at logarithms (that and never learned about it in school yet).
/\
|| That math is probably not the way to go about doing it, it's just what some uneducated fool like myself would try. I'd certainly wait on some of the other people on this site with far more knowledge about this to help you.
Current Setup:
Windows 7 64 bit. Celestia 1.6.0.
AMD Athlon Processor, 1.6 Ghz, 3 Gb RAM
ATI Radeon HD 3200 Graphics
Windows 7 64 bit. Celestia 1.6.0.
AMD Athlon Processor, 1.6 Ghz, 3 Gb RAM
ATI Radeon HD 3200 Graphics
You can try using the planmag script (see my sig) - you can either use the generic phase function there (which isn't hugely accurate, but better than nothing) or maybe you could try and modify the code so the asteroid uses Mercury's phase function (I don't have the Moon's in there, unfortunately)...
The easiest way would be to write an ssc for the asteroid, then run the script while targetting it to get a ballpark figure for the magnitude.
The easiest way would be to write an ssc for the asteroid, then run the script while targetting it to get a ballpark figure for the magnitude.
My Celestia page: Spica system, planetary magnitudes script, updated demo.cel, Quad system