Chapter 13-14

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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)^½ where M is in solar masses, a in AU and P in years P= (0.33^3/ 10^6) ^ ½ yrs= (3.59 x 10^-8)^ ½ yrs= 1.9 x 10^-4yrs= 1.66 hrs . 9./ z=(observed wavelength/rest wavelength-1) z= (563.9/486.1-1) Z= .16...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online