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Solar eclipse - magnitude and albedo

  1. Dec 1, 2013 #1
    Hello, everyone.
    On physics Olympiad in our country I get a interesting example... But, I didn't know how to solve it.
    Can you tell me how?
    It is known that the Moon when it is full has the apparent magnitude of approximately -12 mag and the Sun during the day has the apparent magnitude of -27 mag. Try to figure out what is the apparent magnitude of the Moon directly before a solar eclipse, if you know that the albedo of the Earth is approximately 0.36 and the albedo of the Moon 0.12. Presume that light after reflection disperes the same way on the surface of both the Moon and Earth.

    Hmm... I know Pogson's equation. So I know the I_Moon/I_Sun thanks for it. I also know, how looks position of solar eclipse. However, this is everything. I know what is albedo, ok, but any equation between magnitude and albedo? No, really no :-(
    Please, give me a advice.
    Thank you very much and sorry for my bad English :-)
     
  2. jcsd
  3. Dec 1, 2013 #2
    Apply Pogson´s equation to that.
    The given facts are:
    Moon is stated 15 magnitudes, that is by Pogson equation 1 million times dimmer than Sun. (The real difference is more like 14 magnitudes, but compute on the stated numbers).

    The albedo of Earth is 3 times that of Moon. The area of Earth is about 14 times that of Moon. Therefore Earth is about 42 times brighter than Moon - 4,05 magnitudes brighter than Moon. Thus 10,95 magnitudes dimmer than Sun.
    If Moon lit by Sun is magnitude -12, then Moon lit by full Earth that is 10,95 magnitudes dimmer than Sun should be 10,95 magnitudes dimmer than full Moon - therefore -1,05.
     
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