2007-04-26 18:07:36 · 4 answers · asked by 4morewarspd 1 in Science & Mathematics ➔ Astronomy & Space
Not much different from on Earth. At the zenith on Earth, atmospheric extinction reduces apparent magnitude by about 0.3, so I'd go for 6.3. It'd be the same in all directions though. Given that an increase of 0.5 mag doubles the number of visible stars, there'd be a lot to look at!
2007-04-26 22:53:38 · answer #1 · answered by Iridflare 7 · 0⤊ 0⤋
Your eye views about equal with a 50mm lens. It will view equally here or on the Moon. Is that to which you refer?
Or do you mean how many times smaller does something appear if you're looking at it some 240,000 miles away on the Moon?
2007-04-27 01:30:51 · answer #2 · answered by Stratman 4 · 0⤊ 0⤋
Assuming light at 500nm and a 10mm lens aperture at the eye (its night after all) the angular resolution of the eye is:
theta = sin^-1(500nm/10mm) = 5 x 10^-5 radians.
At a distance of 384,400 to the moon this means the smallest object that can be resolved on its surfance is theta x distance = 19.2 km.
2007-04-27 04:50:22 · answer #3 · answered by Anonymous · 0⤊ 0⤋
I think your question is incorrect; rather, how small an object in the moon can be viewed by the naked eye.
2007-04-27 01:12:58 · answer #4 · answered by cattbarf 7 · 0⤊ 0⤋