This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 (northup, eastleft) and its fieldofview (“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 SIsecond, but over the course of a few hours, the difference between that and the Earth rotationsecond 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 actually381.54 ′′ / 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 − ◦ 12 ′ 24 ′′ = +33 ◦ 27 ′ 31 ′′ − ◦ 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 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 ....
View
Full
Document
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
 Fernandez
 Astronomy

Click to edit the document details