Lecture 10-26 (11,12,13,14,15,16,17,18,19,20,21,22,23,24,25) Problem Solutions (Ansewrs)

Lecture 10-26 (11,12,13,14,15,16,17,18,19,20,21,22,23,24,25) Problem Solutions (Ansewrs)

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Astro 301 5/11/10 1 End-of-Lecture Problems: Answers (Lecture 10–26) Lecture 10. The net result of the p-p reaction, can be summarized as 4 × 1 H 4 He+ energy , so 4 H nuclei are converted to He in each reaction. The energy released is Δ E = Δ mc 2 , where Δ m is the mass difference between the 4 H nuclei and the produce He nucleus. The number of p-p reactions/sec required to sustain the Sun’s luminosity is N ~ L / Δ E . The mass of H converted to He/sec is then 4 m H N . On the other hand, the amount of mass converted to energy/sec is Δ mN (alternatively, we can get the mass-energy conversion rate by taking the time derivative of Einstein’s equation dE/dt = ( dm/dt ) c 2 = L ). We can now calculate, respectively, the total mass converted to He and the total mass converted to energy over Sun’s lifetime to date: M (H He) ~9 × 10 28 kg (0.04 M ); M ( Δ E )~6.5 × 10 26 kg (3 × 10 -4 M ). Use the mass loss rate given in the lecture (~10 –14 M /yr ) to estimate the total mass lost due to the solar wind: ~4.6 × 10 –5 M . Lecture 11. The ratio of luminosities is given by L Sirius / L = 100 [M(Sun)–M(Sirius)]/5 , where the “M”’s refer to the absolute magnitudes (NB: NOT masses). Thus, Sirius is about 23 × more luminous. Similarly, the flux ratio is F Sirius / F = 100 [m(Sun)–m(Sirius)]/5 , where the “m”’s are the apparent magnitudes. For the Sun, m = –26.7 (slide 4), hence F Sirius / F 8 × 10 –11 . Flux is related to luminosity via the inverse-square law, hence F Sirius / F = ( L Sirius / L ) × (1 AU/ r Sirius ) 2 . Solving for the distance, r Sirius 5.37 × 10 5 AU or 2.6 pc. Luminosity, radius, and surface temperature are related by the blackbody equation L = 4 π R 2 σ T 4 . Recasting in terms of Solar values and solving for the radius, R / R = ( L / L ) –1/2 × ( T / T ) –2 . Thus, for Sirius, R 1.65 R . Lecture 12. First, use the small angle formula to estimate the linear diameter of the nebular: D = " d 206 265 . With d = 2000 pc and θ = 30 × 60 = 1800”, we have D 17.5 pc. The size of the nebula is related to the number of ionizing photons/sec, Q , produced by the central star cluster. Assuming that it is spherical, the Stromgren radius is R S 8.7 pc = 2.7 × 10 17 m. Balancing photionizations and recombinations within the nebula gives Q = 4 3 R S 3 n e n H # H . Within the Strömgren sphere, the gas is fully ionized so
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Lecture 10-26 (11,12,13,14,15,16,17,18,19,20,21,22,23,24,25) Problem Solutions (Ansewrs)

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