homework4solutions - AST 3722C Spring 2008 Homework#4...

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AST 3722C - Spring 2008 Homework #4 – solutions corrected version, for problem 6 1. [2 points] Sirius (a.k.a. α CMa) is the brightest star in the sky (aside from the Sun). Its equatorial coordinates in the B1950 equinox and the B1950 epoch were α = 06 h 42 m 56 . 7 s and δ = - 16 38 45 . 4 . (a) Use the formulae in Handout #11 to precess these coordinates to the J2000 equinox. For this use Handout #11. First step, calculate T : T = ( t - 2000 . 0) / 100 = (1950 . 0 - 2000 . 0) / 100 = - 0 . 5 . Next, calculate M and N : M = 1 . 2812323 T + 0 . 0003879 T 2 + 0 . 0000101 T 3 , N = 0 . 5567530 T - 0 . 0001185 T 2 - 0 . 0000116 T 3 , and you should get M = - 0 . 640520 and N = - 0 . 278405 . Next step is to calculate α m and δ m . For that we’re going to need α in decimal degrees: α = 06 h 42 m 56 . 7 s = 6 . 715750 h = (6 . 715750 × 15) = 100 . 736250 . Now we get α m first: α m = α - 1 2 ( M + N sin α tan δ ) = 100 . 736250 - 1 2 ( M + N sin 100 . 736250 tan( - 16 38 45 . 4 )) = 100 . 736250 - 1 2 ( - 0 . 558738 ) = 101 . 015619 . Next δ m , which depends on α m : δ m = δ - 1 2 N cos α m = - 16 38 45 . 4 - 1 2 ( - 0 . 278405 ) cos 101 . 015619 = - 16 . 672542 . Then finally we can apply the actual equations to get the J2000 coordinates. First do α 0 : 1
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α 0 = α - ( M + N sin α m tan δ m ) = 100 . 736250 - ( M + N sin 101 . 015619 tan( - 16 . 672542 )) = 100 . 736250 - ( - 0 . 558676 ) = 101 . 294926 = 6 . 752995 h = 06 h 45 m 10 . 8 s . And then we do δ 0 : δ 0 = δ - N cos α m = - 16 38 45 . 4 - ( - 0 . 278405 ) cos 101 . 015619 = - 16 . 699141 = - 16 41 56 . 9 . (b) The proper motion of Sirius is - 0 . 54605 /year in R.A. and - 1 . 22314 /year in Dec. What are the coordinates of Sirius at epoch J2000? In 50 years, Sirius has moved ( - 0 . 54065 /year) × 50 years = -27.0325 in R.A. and ( - 1 . 22314 /year) × 50 years = -61.1570 in Dec. So it’s moved west and south. The J2000 epoch coordinates are these offsets from the J2000 equinox (B1950 epoch) coordinates. Let α n and δ n be the coordinates at J2000 epoch. So: δ n = δ 0 + offset = - 16 41 56 . 9 + ( - 61 . 1570 ) = - 16 42 58 . 1 , α n = α 0 + offset = 6 . 752995 h - 27 . 0325 = 06 h 45 m 10 . 8 s - (27 . 0325 / (15 cos δ mean )) s = 06 h 45 m 10 . 8 s - 1 . 9 s = 06 h 45 m 08 . 9 s . For δ mean , I used the average of δ n and δ 0 . 2
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2. [1 point] Suppose you have a friend that was born in London on February 15, 1988, at 6:00 a.m.GMT = 6:00 a.m. UT. Calculate how many SI-standard seconds your friend will have been alive at the time this homework is due. (Hint: how many leap seconds have been inserted?) This is basically asking you to understand how leap seconds come into play. First let’s calculate how many days have elapsed since your friend’s birthday. From to is this many days Feb 15 1988 6am Feb 15 1989 6am 366 days Feb 15 1989 6am Feb 15 1990 6am 365 days Feb 15 1990 6am Feb 15 1991 6am 365 days Feb 15 1991 6am Feb 15 1992 6am 365 days Feb 15 1992 6am Feb 15 1993 6am 366 days . . . . . . . . . Feb 15 2007 6am Feb 15 2008 6am 365 days Feb 15 2008 6am Feb 19 2008 6am 4 days Total 7309 days You could also do this straightaway by thinking that your friend is 20 years old, so: 20 × 365 + 5 leap days + 4 days past birthday = 7309 days . This homework was due at 7 pm EST, which is 12 midnight GMT/UT, and that’s 18 hours past the birth time, which is 0.75 days. So your friend is 7309.75 days old, and since there are 86,400 seconds in a day, that gives you 631,562,400 seconds.
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