exam1solutions - AST 3722C Spring 2008 Midterm Solutions...

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AST 3722C - Spring 2008 Midterm - Solutions Part A. Download and view the movie at this URL: http : // www . theastronomer . org / asteroids / 2008 / 2007tu24 20080202 ndj . wmv . (If you have problems, see me.) An amateur astronomer took a sequence of digital pictures of the same piece of sky and strung them together to make this movie. The moving object is the asteroid 2007 TU 24 . The stationary dots are background stars, and you can see a couple of galaxies too. The UTC of each frame is given at the top, as is the orientation of the frame (north-up, east-left) and its field-of-view (“FOV”). The frame is centered on J2000 coordinates α = 12 h 00 m 37 . 5 s , δ = +33 21 15 ′′ . 1. The asteroid enters the field of view at about UTC 00:56 and leaves at about UTC 02:53. Assuming these are the exact times, what is the asteroid’s average rate of motion in Declination? Express your answer in arcseconds per hour. UTC uses the SI-second, but over the course of a few hours, the difference between that and the Earth rotation-second is negligible. The asteroid takes 1 hour and 57 minutes = 1.95 hours to move through the FOV which is 12.4 arcminutes. That FOV is (12 . 4 × 60 =) 744 arcseconds. So the asteroid is moving at a rate of 744 ′′ ÷ 1 . 95 hr = 381.5 ′′ / hr. However it’s moving south, so the motion is actually -381.54 ′′ / hr . 2. The asteroid’s motion in R.A. is 75 arcsec per hour. How many arcsec in R.A. does it move while it is in the FOV? In 1.95 hours, the asteroid moves (75 arcsec per hour × 1.95 hours =) 146.3 arcsec . 3. When the asteroid enters the FOV, its coordinates are α = 12 h 00 m 30 . 7 s , δ = +33 27 31 ′′ (J2000). What are the coordinates of the asteroid when it leaves the FOV? We now know from parts 1 and 2 that when the asteroid leaves the FOV that it has moved 744 arcsec south in declination and 146.3 arcsec east in R.A. Let the coordinates when it leaves the FOV be α l and δ l . First let’s handle the declination: δ l = δ + offset = δ 744 ′′ = δ 0 12 24 ′′ = +33 27 31 ′′ 0 12 24 ′′ = + 33 15 07 ′′ . Now what about the R.A.? We need to convert that offset into R.A. seconds: Note that yes indeed the initial δ is not quite exactly consistent with my statement in the preamble that the FOV is centered near +33 21 15 ′′ . There should be exactly 6.2 arcminutes of declination separation between them. Sorry! We’ll be lenient with credit here. 1
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offset in arcsec = offset in seconds × 15 × cos δ mean 146 . 3 ′′ = offset in seconds × 15 × cos(+33 21 19 ′′ ) 146 . 3 ′′ = offset in seconds × 12 . 5292 offset in seconds = 11 . 7 s . So now we can do the offsets: α l = α + offset = α + 11 . 7 s = 12 h 00 m 30 . 7 s + 11 . 7 s = 12 h 00 m 42 . 4 s . 4. The movie’s creator used a telescope in Chelmsford, northeast of London, England. The latitude there is φ = 51 33 . When the asteroid entered the FOV, the LST there was 09 h 44 m 25 s . What was the LST when the asteroid left the FOV?
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