Chapter 13-14 - Therefore Mass = a^3/P^2 = 10^27/4x10^16 =...

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Astronomy –odd- numbered -Homework Problem 13 1./ M = -2.81*log(P) - 1.43 = -2.81*log(30) - 1.43 = -5.58 If m = 20, then the distance modulus μ = 20 - (-5.58) = 25.58 d = 10^(0.2μ + 1) = 1.31 Mpc 3./ d = 10 (m v - M v + 5)/5 M v = -19 m v = 17 d = 10 (m v - M v + 5)/5 = 10 (17 - (19) + 5)/5 = 10 8.2 d = 158 Mpc (1.58 X 10 8 pc.] 7./ Actual diameter(km)= 150,000,000 x distance(pc) x angular diameter(arc seconds) 22 arc minutes= 1,320 arc seconds. 9 mpc= 9,000,000 parsecs actual diameter= 150,000,000 x 9,000,000 x 1,320 actual diameter= 1.78200 × 10^18 kilometers= 57,750.6 Parsecs= 188,361.5 light years 9./ We have: a ^3 = P^2 In this case d=5,000 parsec = 5,000 parsec = 2 x 10^5 AU/parsec = 10^9 AU
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Unformatted text preview: Therefore Mass = a^3/P^2 = 10^27/4x10^16 = 2.5 x 10^10 M_Sun. Problem 14 1./ E = m c 2 = (2 X 10 30 kg) X (3 X 10 8 m/s) 2 = 1.8 X 10 47 J. Dividing this into 10 53 J, yields 10 53 J / 1.8 X 10 47 J = 56,000. So, it would take approximately 56,000 solar masses to produce this energy 5./ M = a^3 / P^2 so, orbital period = P = (a^3 / M)^ where M is in solar masses, a in AU and P in years P= (0.33^3/ 10^6) ^ yrs= (3.59 x 10^-8)^ yrs= 1.9 x 10^-4yrs= 1.66 hrs . 9./ z=(observed wavelength/rest wavelength-1) z= (563.9/486.1-1) Z= .16...
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Chapter 13-14 - Therefore Mass = a^3/P^2 = 10^27/4x10^16 =...

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