By Carroll B.W., Ostlie D.A.
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For T D 5777 K, u=n D 1:34 eV. 4 For blackbody radiation, I D B , and so Eq. T / d D Z 8 k 4T 4 D 3h3 c 3 Prad 4 3c 1 0 Z 1 0 2hc 2= e hc= x3 ex 1 5 kT 1 d : dx: The integral is equal to 4=15. 14 for the Stefan– Boltzmann constant, D 2 5k 4 =15c 2h3 , produces the desired result, Prad D 4 T4 1 1 D aT 4 D u; 3c 3 3 via the definition of the radiation constant, a D 4 =c, and Eq. 7) for the blackbody energy density. 5 We start by integrating Eq. 8) with I D B over all outward directions (0 Ä Â Ä =2).
For dish 2 there are 48 additional baselines (we don’t want to recount the baseline between 1 and 2). For dish 3 there are 47 additional baselines, and so on. The total number becomes 49 C 48 C C 1 D 1125. 15 The projected resolution of SIM PlanetQuest is 4 arcsec D 0:00000400 D 2 (a) 10 11 radians. If grass grows at a rate of 2 cm per week, this is equivalent to 3:3 10 8 m s 1 . From a distance of 10 km, SIM PlanetQuest can resolve a width of D D Âd D 0:2 m. It would take the grass approximately 6 s to grow the necessary length to be measured!
At a room 105 K. 3 The peak of the Maxwell–Boltzmann distribution, Fig. 6, shows that nv =n ' 6:5 10 5 s m 1 where v D vmp . nv =n/ v ' 0:13. 4 The most probable speed, vmp , occurs at the peak of the Maxwell–Boltzmann distribution, Eq. 1). Setting d nv =dv D 0, we find Á d nv m Á3=2 d 2 e mv =2kT v 2 D 0: D4 n dv 2 kT dv This leads to Â Ã mv 3 2 C 2v e mv =2kT D 0; kT p so that vmp D 2kT =m, which is Eq. 2). 5 Solving the Boltzmann equation, Eq. 2/2 D 8, E1 D 13:6 eV, and E2 D 3:40 eV. Then, with N2 =N1 D 0:01, T D 1:97 104 K.