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# 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 0 45 . 4 00 . (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 0 45 . 4 00 )) = 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 0 45 . 4 00 - 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 0 45 . 4 00 - ( - 0 . 278405 ) cos 101 . 015619 = - 16 . 699141 = - 16 41 0 56 . 9 00 . (b) The proper motion of Sirius is - 0 . 54605 00 /year in R.A. and - 1 . 22314 00 /year in Dec. What are the coordinates of Sirius at epoch J2000? In 50 years, Sirius has moved ( - 0 . 54065 00 /year) × 50 years = -27.0325 00 in R.A. and ( - 1 . 22314 00 /year) × 50 years = -61.1570 00 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 0 56 . 9 00 + ( - 61 . 1570 00 ) = - 16 42 0 58 . 1 00 , α n = α 0 + oﬀset = 6 . 752995 h - 27 . 0325 00 = 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
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

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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.

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

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