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Astronomy related angle question (more of a Trig Q)

  1. Nov 16, 2006 #1
    For an astronomy lab I am looking at pictures of the moon, measuring pixel distances and converting to km. Easy enough for the diameter of a crater, but the depth gets more complicated. To find the wall height of the crater, you measure the shadow length and the angle of altitude (theta in diagram). I was told that theta is equal to the distance to the terminator(D in diagram) divided by the radius of the moon, which is also easy to calculate. But to use the trig tangent(theta)=height/shadow, doesn't theta have to be in radians? the dist. to terminator over the radius is a dimensionless number (km/km after conversions)... should I multiply by a factor of 2pi to get theta, or is it okay for theta to be dimensionless?

    Attached Files:

  2. jcsd
  3. Nov 16, 2006 #2
    well... looks like I'm handing in my lab report as is whether it's right or wrong... just printed it.

    Still would like to get an answer though!
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