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Math Help - Moonlight over my head?

  1. #1
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    Moonlight over my head?

    Hi.

    It's been waaaay to long since I've been to university, and I need a little help with a problem. Hope this is the right forum.

    I would like to determine the time prior to moonrise that the rays of the moon begin to illuminate the sky above me. I have selected a height of 6 miles above me as my area of interest.

    I am an amateur astronomer, and when we want to study faint deep sky objects, the light from the moon is a major impediment. It's easy to look up the time of moonrise on various web sites, but I am interested to know how much in advance the light of the moon will be affecting the dark sky visibility at my location.

    I have sketched it out by drawing a tangent line at point A, and it seems what I need to calculate is point B, where the surface of the earth has dropped 6 miles away from the tangent line. Then I need to determine the angle between A & B (from the center of the earth), from which I can determine the amount of time it takes the earth to rotate that much.

    I'm a little fuzzy on this, but I think the answer would vary depending on my latitude, so let's just keep it simple and assume I'm at the equator. According to Google, the radius of the earth is approximately 3,959 miles. And that is about as far as I can get! Anybody care to help me out a bit?
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  2. #2
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    Re: Moonlight over my head?

    After looking at my sketch some more, I realized that it is a simple right angle triangle! After some basic trig calculations I found my answer.
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  3. #3
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    Re: Moonlight over my head?

    Quote Originally Posted by russdirks View Post
    After looking at my sketch some more, I realized that it is a simple right angle triangle! After some basic trig calculations I found my answer.
    GOOD! Hope the NHL settles as quickly!!
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