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Ch45Word - Ch 45 Page 1 CHAPTER 45 Astrophysics and...

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Ch. 45 Page 1 CHAPTER 45 – Astrophysics and Cosmology Note: A factor that appears in the analysis of energies is e 2 /4p Å 0 = (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 / φ ; 1 pc = (1.50 × 10 11 m)/[(1.000 )/(3600 /°)(180°/p rad)] = 3.094 × 10 16 m. From the definition of the light-year, 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/ φ = [1/(0.38 )](3.26 ly/pc) = 8.6 ly . 3. We find the distance from D = 1/ φ = [1/(0.00019°)(3600 /°)](3.26 ly/pc) = 4.8 ly . 4. ( a ) We find the parallax angle from D = 1/ φ ; 36 pc = 1/ φ , which gives φ = 0.028 . ( b ) We convert this to degrees: φ = (0.028 )/(3600 /°) = (7.7 × 10 –6 )°. 5. We find the parallax angle from D = 1/ φ ; (55 ly)/(3.26 ly/pc) = 1/ φ , which gives φ = 0.059 . The distance in parsecs is D = (55 ly)/(3.26 ly/pc) = 17 pc . 6. We find the distance in light-years: 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/ φ , we see that D 1 / D 2 = 2 = φ 2 / φ 1 , or φ 1 / φ 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 .
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Ch. 45 Page 2 10. The diameter of our Galaxy is 100,000 ly so the angle subtended by our Galaxy is φ 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 φ 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 φ Galaxy ˜ 6 φ Moon . 11. If we assume negligible mass change, as a red giant, the density of the Sun will be ρ = 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 φ = 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 ρ dwarf = M / V = M / ) p r 3 ; = (2.0 × 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: ρ dwarf / ρ Earth = M dwarf / M Earth = (2.0 × 10 30 kg)/(5.98 × 10 24 kg) = 3.3 × 10 5 . 14. The density of the neutron star is ρ ns = M ns / V ns = M ns / ) p r ns 3 ; = (1.5)(2 × 10 30 kg)/ ) p(11 × 10 3 m) 3 = 5.4 × 10 17 kg/m 3 . From the result for Problem 13 we see that ρ ns = (3 × 10 8 ) ρ dwarf . For the density of nuclear matter we calculate the density of a proton: ρ p = m p / ) p r p 3 = (1.67 × 10 –27 kg)/ ) p (1.2 × 10 –15 m) 3 = 2.3 × 10 17 kg/m 3 . Thus we see that ρ ns ˜ ρ nuclear matter . 15. For the reaction He 2 4 + He 2 4 Be 4 8 + γ , we determine the Q -value: Q = [ M ( 4 He) + M ( 4 He) – M ( 8 Be)] c 2 = [2(4.002603 u) – (8.005305 u)] c 2 (931.5 MeV/u c 2 ) = – 0.0922 MeV = – 92.2 keV .
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