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If kT D 13:6 eV, then T D 1:58 10 5 eV K 1 . 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  à 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.

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! (b) Assuming a resolution of 4 as, and a baseline of 2 AU, the measureable distance is d D 2 AU=4 as D 500 kpc. (c) From Eq. 6) and using MV D 4:82, V D 28:3.

1) and a D a1 C a2 D 2:0 AU, we have a1 D 1:6 AU and a2 D 0:4 AU. Now, the orbital velocities of the two stars are v1 D 2 a1=P D 26:5 km s 1 and v2 D 2 a2=P D 6:6 km s 1 . Finally, taking into consideration the orbital inclination of 30ı, the maximum observable radial velocities are v1r; max D v1 sin i D 13:2 km s 1 and v2r; max D v2 sin i D 3:351 km s 1 . (c) The eccentricity can be determined by considering the asymmetries in the velocity and/or light curves. In particular, the velocity curves deviate from being perfectly sinusoidal (the amount of deviation increasing with increasing e).

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