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Unformatted text preview: Astronomy 210 Spring 2011 Homework Set #4: Solutions 1. The Cooling Nighttime Moon. (a) Combinign T (K) = T ( ◦ C) + 273 and T ( ◦ C) = 5 9 [ T ( ◦ F) − 32], we find T (K) = 5 9 [ T ( ◦ F) − 32] + 273 (1) and so T ( ◦ F) = 9 5 [ T (K) − 273] + 32 (2) and so the daytime Moon is at temperature T d ≈ 330 K = 135 ◦ F (3) This is hotter than the hottest summer day in a deserts on Earth. Thus moonbases have to have a good cooling mechanism during the day. (b) This is just a simple application of Wein’s law: λ peak = 2 . 9 × 10 − 3 m K /T and so λ peak = 2 . 9 × 10 − 3 m K T d = 8 . 8 × 10 − 6 m = 8 . 8 μ m (4) where here and below, we use absolute temperature, i.e., T in Kelvin. This is 8.8 microns, which is deeply into the infrared part of the spectrum. In fact, this is the heart of the mid infrared (as opposed to near and far IR). (c) Here we use the StefanBoltzmann law: F = σT 4 = 672 Watt / m 2 (5) (d) Recall that the definition of flux is F = light power area = LA (6) where L is the light power and A is the collecting area. Solving, we have L = FA , and for the case of a blackbody at temperature T , this gives L = σT 4 A = 672 Watt (7) for an area A = 1 m 2 . at temperature T d . A typical lightbulb gives off about 100 Watts of power, so every square meter of the nighttime Moon radiates off almost seven lightbulbs worth of power in the infrared (at least at first, when it is still nearly at daytime temperatures). Note that this is in the same general ballpark as the energy consumption in your home (actually this is a bit less).as the energy consumption in your home (actually this is a bit less)....
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 Spring '08
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 Astronomy, Thermodynamics, Energy, Entropy, Heat, Eheat

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