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Eclipse problem

  1. Nov 16, 2003 #1

    Math Is Hard

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    This is an astronomy homework question that my classmates and I are having an awful time with.

    If the penumbra of the Earth’s shadow is 16,000 km across, and if the Moon moves 3400 km/hr with respect to the shadow, why does it take 6 hours instead of only 5 hours to get completely through the penumbra?

    We first thought that the extra hour was due to the Earth's revolution around the Sun pushing the edge of the penumbra ahead so the moon would have to play "catch up" with that.
    But now that I am re-reading the question, it says that the moon moves "with respect to" the shadow so I am thinking that the Earth's rotation has already been factored in.

    The other theory is that the extra hour would be due to the moon travelling in an arc instead of a straight line through the penumbra, and that would account for the time delta.

    Are either of these close? Thanks in advance for your help.
     
  2. jcsd
  3. Nov 16, 2003 #2
    Have you considered how the diameter of the moon figures into this?
     
  4. Nov 16, 2003 #3

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    Not at all - but I think what you're saying is that one edge of the moon has to enter and the other edge has to completely pass through before the transit is complete? I guess the problem is we have been treating it like a speck starting on one edge and finishing at the other edge instead of calculating additional time for entry and exit.Which I think would be radius in and radius out - or a diameter's length of additional space to cross?
     
  5. Nov 16, 2003 #4
    Yes, that's right. If you just use time = speed over length, then you can calculate the time for the leading edge of the Moon to go from one side of the penumbra to the other --- but then the trailing edge of the Moon still has to pass out of the penumbra as well.
     
  6. Nov 16, 2003 #5

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    Thank you both so much. Worked like a charm - I have about 5.73 hours for the moon to get in, get through, and get out.
     
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