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Unformatted text preview: AST 3722C - Spring 2008 Homework #6 – Solutions • 1. (2 points.) One day you decide to observe a comet that is passing Δ = 0 . 8 AU away from the center of Earth. To make it easy, you’ll observe the comet when it is transiting. You look up the ephemeris and see that its J2000 coordinates at the time of transit are α = 05 h 10 m 20 s , δ = 20 ◦ 30 ′ 00 ′′ . However this is the position for a hypothetical person standing at the center of Earth. (a) Calculate the distance Δ ′ to the comet from your location on the surface of Earth at latitude φ = 28 . 6 ◦ . Assume that the radius of Earth is 6378 km, and that 1 AU is 149,597,871 km. Look at the figure below, which sets up the situation in this problem. “CoE” refers to the center of Earth. Let R E be the radius of Earth. The line extending horizontally from CoE is the line that marks the direction to the celestial equator. The horizontal line extending from you through point P and onward is parallel to that line; in other words, that line indicates the direction to the celestial equator as seen by you. The trick is to see that you know one of the angles in the triangle formed by comet-you- CoE. This angle is φ − δ . So then you can use law of cosines to figure out Δ ′ : Δ ′ 2 = Δ 2 + R 2 E − 2 R E Δ cos( φ − δ ) . Putting in the numbers gets you an answer of . 7999578 AU , or 119 , 671 , 984 km . So Δ ′ is not so different from Δ. (b) Calculate the apparent declination δ app of the comet when you observe it from your location. You need to reason this through, since it may not be obvious how to get δ app , but you just need geometry. The angle defined by you-P-CoE is actually equal to δ – interior angles to parallel lines are equal. So then you know two of the three angles in the triangle defined by you-P-CoE. That means that the third angle, call it ψ , is 180 ◦ − δ − ( φ − δ ) = 180 ◦ − φ . 1 But that’s not the end, we need δ app , which makes up part of the triangle defined by CoE-you-comet. And the angle defined by CoE-you-comet is ψ + δ app . So by the law of sines: sin( φ − δ ) Δ ′ = sin( ψ + δ app ) Δ . Which means: sin( φ − δ ) Δ ′ = sin(180 ◦ − φ + δ app ) Δ = sin( φ − δ app ) Δ ⇒ sin( φ − δ app ) = Δ Δ ′ sin( φ − δ ) = 0 . 140909 ⇒ δ app = 20 . 49957 ◦ = 20 ◦ 29 ′ 58 . 5 ′′ . (c) The time it would take for a photon to travel from the comet to the center of Earth would be Δ /c . How much less time does it take for a photon from the comet to get to you? The speed of light is exactly 299,792.458 km/s. So the time Δ /c is 399.204 seconds. However Δ ′ /c is 399.183 seconds. So it takes light 0.021 seconds less time to get to you than it does to get to the center of Earth....
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This note was uploaded on 11/09/2009 for the course AST 4700 taught by Professor Fernandez during the Spring '09 term at University of Central Florida.
- Spring '09