homework2solutions

# homework2solutions - AST 3722C Spring 2008 Homework#2...

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AST 3722C - Spring 2008 Homework #2 – Solutions (a) [0.5 points] Suppose we’re in Orlando so φ = φ 0 = 28 . 6 . Find the upper and lower bounds on δ such that a can be zero. I.e. what is the range of declinations of objects that are neither circumpolar nor permanently below the horizon? This is similar to a problem from homework #1. Remember that you found that the southernmost declination that you can see from your location is φ 90 . In this case, that’s 61 . 4 . See the solution set for that problem if you don’t see why this is. As for the maximum declination, think about extremes. If you were at the North Pole, δ = 0 at your due north horizon. As you walk south, Polaris would drop in the sky toward the due north horizon. At the Equator, Polaris would have dropped all the way down and so δ = 90 at your due north horizon. You should be able to visualize that therefore the maximum declination you can see is 90 φ , or in our case 61 . 4 . The inelegant way to do this is to use the formula on Handout #5. A = 0 and a = 0, so: sin δ = sin a sin φ + cos a cos φ cos A = sin 0 sin φ + cos 0 cos φ cos0 = cos φ = sin(90 φ ) δ = 90 φ = 61 . 4 . (b) [0.5 points] Let t up ( δ ) be the time that an object at declination δ is above the horizon over the course of a day. What is t up (90 ) and t up ( 90 ) for us in Orlando? At δ = 90 , we’re looking at the North Celestial Pole, basically Polaris. Polaris is always up – it’s circumpolar! Therefore t up (90 ) must be 24 hours . At δ = 90 , we’re looking at the South Celestial Pole. We can never see that far down; remember from the last problem that the minimum δ we can see is 61 . 4 . Therefore t up ( 90 ) must be 0 hours . (c)

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homework2solutions - AST 3722C Spring 2008 Homework#2...

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