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Unformatted text preview: Key for AST3018 HW 4 27 Points Leah Simon 16 Nov 2009 5.3 (3 pts) a.) Here we need to solve for the velocity along the line of sight. In this case, v = v 1 cos ( 1 ) v 2 cos ( 2 ). Using v 1 = 80 km/s, v 2 = 10 km/s, 1 = 30 , and 2 = 20 we find v = 59 . 9 km/s . b.) Therefore using = v/c we find = 1 . 31 A for H ( = 6562 . 8 A. c.) Along similar lines, if we are moving away, then 2 = 180 + 2 , or v = v 1 cos ( 1 ) + v 2 cos ( 2 ) we find v = 78 . 7 km/s and = 1 . 72 A. 5.16 (1 pt) Since 1 arcsec at 1 parsec subtends 1 AU, we find that 0.1 arcsec subtends 100 AU at 10,000 pc by simple division: tan ( ) = 100 AU/d ( pc ). 5.19 (5 pts) a.) Using v = c/ we find v 1 = 1 . 002 10 4 m/s and v 2 = 2 . 01 10 4 m/s. Then using P 2 G ( v 1 + v 2 ) 3 sin 3 i = m 1 + m 2 (1) we find M tot = 4 . 1 10 31 kg = 20 M Sun . Since we know v 1 v 2 = m 2 m 1 (2) We find m 2 = 1 . 37 10 31 kg = 6 . 9 M Sun and m 1 = 2 . 73 10 31 kg = 13 . 7 M Sun . b.) For i = 30 deg, use the same equations, but divide by sin 3 (30) to get M tot = 3 . 28 10 32 kg = 165 M Sun , and m 2 = 1 . 09 10 32 kg = 55 M Sun , m 1 = 2 . 19 10 32 kg = 110...
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This note was uploaded on 07/28/2011 for the course AST 3018 taught by Professor Lada during the Fall '08 term at University of Florida.
 Fall '08
 Lada

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