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homework4sol - Key for AST3018 HW 4 27 Points Leah Simon 16...

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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 ( θ 1 ) - v 2 cos ( θ 2 ). Using v 1 = 80 km/s, v 2 = 10 km/s, θ 1 = 30 , and θ 2 = 20 we find v = 59 . 9 km/s . b.) Therefore using Δ λ = λv/c we find Δ λ = 1 . 31 ˚ A for ( λ = 6562 . 8 ˚ A. c.) Along similar lines, if we are moving away, then θ 2 = 180 + θ 2 , or v = v 1 cos ( θ 1 ) + v 2 cos ( θ 2 ) we find v = 78 . 7 km/s and Δ λ = 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 ( θ ) θ = 100 AU/d ( pc ). 5.19 (5 pts) a.) Using v = Δ λc/λ we find v 1 = 1 . 002 × 10 4 m/s and v 2 = 2 . 01 × 10 4 m/s. Then using P 2 πG ( 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 M Sun 5.20 (2 pts) This problem is not solvable in its given form (the author of the book actually wanted you to solve for M tot ). However, to solve for M
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