This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Ch. 45 Page 1 CHAPTER 45 Astrophysics and Cosmology Note: A factor that appears in the analysis of energies is e 2 /4p = (1.60 10 19 C) 2 /4p(8.85 10 12 C 2 /N m 2 ) = 2.30 10 28 J m = 1.44 MeV fm. 1. Using the definition of the parsec, we find the equivalent distance in m: D = d / ; 1 pc = (1.50 10 11 m)/[(1.000 )/(3600 /)(180/p rad)] = 3.094 10 16 m. From the definition of the lightyear, we have 1 pc = (3.094 10 16 m)/(3.00 10 8 m/s)(3.16 10 7 s/yr) = 3.26 ly. 2. We find the distance from D = 1/ = [1/(0.38 )](3.26 ly/pc) = 8.6 ly . 3. We find the distance from D = 1/ = [1/(0.00019)(3600 /)](3.26 ly/pc) = 4.8 ly . 4. ( a ) We find the parallax angle from D = 1/ ; 36 pc = 1/ , which gives = 0.028 . ( b ) We convert this to degrees: = (0.028 )/(3600 /) = (7.7 10 6 ). 5. We find the parallax angle from D = 1/ ; (55 ly)/(3.26 ly/pc) = 1/ , which gives = 0.059 . The distance in parsecs is D = (55 ly)/(3.26 ly/pc) = 17 pc . 6. We find the distance in lightyears: D = (35 pc)(3.26 ly/pc) = 114 ly. Thus the light takes 114 yr to reach us. 7. The star farther away will subtend a smaller angle, so the parallax angle will be less . From D = 1/ , we see that D 1 / D 2 = 2 = 2 / 1 , or 1 / 2 = ! . 8. ( a ) The apparent brightness of the Sun is = 1.3 10 3 W/m 2 . ( b ) The absolute luminosity of the Sun is L = A = (1.3 10 3 W/m 2 )4p(1.5 10 11 m) 2 = 3.7 10 26 W . 9. The apparent brightness is determined by the absolute luminosity and the distance: L = A = 4p D 2 . If we form the ratio for the apparent brightness at the Earth and at Jupiter, we get E / J = ( D J / D E ) 2 ; ( 1 . 3 10 3 W/m 2 )/ J = (5.2) 2 , which gives J = 48 W/m 2 . Ch. 45 Page 2 10. The diameter of our Galaxy is 100,000 ly so the angle subtended by our Galaxy is Galaxy = d 1 / D 1 = (100,000 ly)/(2 10 6 ly) = 0.05 rad = 2.9. The angle subtended by the Moon at the Earth is Moon = d 2 / D 2 = (3.48 10 6 m)/(3.84 10 8 m) = 9.06 10 3 rad = 0.52. Thus we have Galaxy 6 Moon . 11. If we assume negligible mass change, as a red giant, the density of the Sun will be = M / V = M / ) p r 3 ; = ( 2 10 30 kg)/ ) p(1.5 10 11 m) 3 = 1.4 10 4 kg/m 3 . 12. The angle subtended at the Earth by the Sun as a white dwarf will be = d / D = (3.48 10 6 m)/(1.5 10 11 m) = 2.3 10 5 rad = 4.8 . 13. The density of the white dwarf is dwarf = M / V = M / ) p r 3 ; = ( 2 . 10 30 kg)/ ) p(6.38 10 6 m) 3 = 1.8 10 9 kg/m 3 . The ratio of densities will be the ratio of masses: dwarf / Earth = M dwarf / M Earth = (2.0 10 30 kg)/(5.98 10 24 kg) = 3.3 10 5 ....
View
Full
Document
This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.
 Spring '09
 Fielding
 Physics

Click to edit the document details