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Unformatted text preview: AST 3722C  Spring 2008 Homework #3 – solutions • 1. [1 point] You want to observe Comet 17P/Holmes tonight and you look up online that it’s equatorial position at 9 p.m. (EST) will be α = 03 h 26 m 58 . 4 s and δ = +39 ◦ 12 ′ 19 . 7 ′′ (2000.0). You are also told that the comet is moving 31.9 arcsec per hour in R.A. and 12.1 arcsec per hour in Declination. Calculate the equatorial coordinates of the comet at 10 p.m. Since 10 p.m. is 1 hour after 9 p.m., the comet has moved 31.9 arcsec in R.A. and 12.1 arcsec in Dec. The latter is easier; you just have to remember that a negative Dec motion means the comet moves south. Let δ 10 be the Declination at 10 p.m. δ 10 = δ + offset = δ − 12 . 1 ′′ = +39 ◦ 12 ′ 19 . 7 ′′ − 12 . 1 ′′ = + 39 ◦ 12 ′ 07 . 6 ′′ R.A. is harder. We need to convert that 31.9 arcsec into an equivalent amount of change in R.A. You cannot just divide by 15; you have to account for the cos δ factor. Recall that R . A . distance in arcsec = R . A . distance in seconds × 15 × cos δ mean where δ mean is the average of the Declinations that are appropriate to the problem. In this case, δ mean = 1 2 × ( δ + δ 10 ). Therefore R . A . distance in seconds = R . A . distance in arcsec 15 × cos δ mean = 31 . 9 15 × cos(+39 ◦ 12 ′ 13 . 1 ′′ ) = 31 . 9 15 × . 7749 = 2 . 74 . The comet is moving east (since the rate is positive). Let α 10 be the R.A. at 10 p.m. So: α 10 = α + offset = α + 2 . 74 s = 03 h 26 m 58 . 4 s + 2 . 74 s = 03 h 26 m 61 . 1 s (now carry over) = 03 h 27 m 01 . 1 s . 1 • 2. [2 points] The nearest star system to the Sun is the α Centauri system, which has 3 stars – A, B, and C (a.k.a Proxima). A and B are easy to see, but C is pretty faint. A has equatorial coordinates (2000.0) of α = 14 h 39 m 36 . 5 s and δ = − 60 ◦ 50 ′ 02 . 3 ′′ . B has equatorial coordinates (2000.0) of α = 14 h 39 m 35 . 1 s and δ = − 60 ◦ 50 ′ 13 . 8 ′′ . (a) Given these coordinates, what was the angular separation of these stars? I didn’t put it in the problem but let a subscript X refer to star X. You should remember from previous homework that in situations like this, you can use the smallspherical triangle approximation. You just have to be careful of using cos δ factor. So: separation = radicalBig ( δ A − δ B ) 2 + (( α A − α B ) × 15 × cos δ mean ) 2 . In the second term above, the 15 and the cos δ mean allow us to convert the difference in R.A. from seconds to arcsec. In this case, δ mean is just 1 2 × ( δ A + δ B ). First we deal with the Declinations: δ A − δ B = − 60 ◦ 50 ′ 02 . 3 ′′ − ( − 60 ◦ 50 ′ 13 . 8 ′′ ) = − 60 ◦ 50 ′ 02 . 3 ′′ + 60 ◦ 50 ′ 13 . 8 ′′ = 11 . 5 ′′ ....
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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
 Fernandez
 Astronomy

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