Lecture 1-9 (2, 3, 4, 5, 6 7, 8) Problem Solutions (Ansewrs)

Lecture 1-9 (2, 3, 4, 5, 6 7, 8) Problem Solutions...

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Astro 301 3/28/10 1 End-of-Lecture Problems: Answers Lecture 2. The problems make use of the small angle formula: D = " d 206 265 . In the first case, we are solving for the distance d , given the Moon’s angular size α = 0.5°. Converting to arcesconds, α = 0.5 × 3600 = 1800’’. We also need the linear diameter, which is D = 3476 km. We then get d 400 000 km. In the second problem, we solve for α , given the distance d (=1 AU) and the diameter D = 1.4 × 10 6 km, of the Sun: = 206 265 D d . D and d must be in the same units, hence = 206 265 # 1.4 # 10 6 km ( ) 1.5 # 10 8 km ( ) , giving α 1900’’.This is about the same size as the Moon, which is why we have eclipses. Lecture 3 Due to geometrical dilution, the solar flux at radius r is F = L ! 4 r 2 . Therefore, the flux reaching Mercury will be F (Mercury) = (1 AU/0.39 AU) 2 F (Earth) 6.6 F (Earth). Similarly, for Neptune F (Neptune) = (1 AU/30 AU) 2 F (Earth). The energy of a “typical” photon will be E = hc / λ , where h = Planck’s constant, c = speed of light and λ (wavelength) 500 nm. Looking up the values of the constants (and remembering to convert wavelength to meters), E 4 × 10 –19 J. Since the Sun’s luminosity is
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Lecture 1-9 (2, 3, 4, 5, 6 7, 8) Problem Solutions...

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