By Carroll B.W., Ostlie D.A.
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Additional info for An introduction to modern astrophysics: Solution manual
50. (b) In this binary system, the larger star is hotter. 0), the maximum amount of the larger star is covered, implying the least total integrated flux (or luminosity). The primary minimum is not flat because the proximity of the secondary star leads to tidal distortion of the shapes of the stars, thus removing spherical symmetry. 25, the two stars are seen in profile. 5. The secondary minimum is flat because the secondary (and the distortion effects) are occulted for a finite period of time. 9 See Fig.
26. log10 Œf hNa i. W = / 4:58 4:69 3:90 4:02 Now calculate log10 Œf . 9 nm line: The average value of log10 Na for this problem is hNa i D 20:18, so hNa i D 1:51 1020 m 2 . This is the value of the number of neutral hydrogen atoms per unit area with an electron in the n D 3 orbit. (b) The Boltzmann equation (Eq. 6) allows us to find the relative numbers of neutral hydrogen atoms in the n D 3, n D 2, and n D 1 states. From Eq. 14), E1 D 13:6 eV, E2 D 3:40 eV, and E3 D 1:51 eV, and gn D 2n2 for hydrogen.
16. 17(a). (b) See Fig. 5. 18 Using the data in the text and the problem, the mass of the star can be determined from Kepler’s third law to be approximately 1:1 Mˇ . Take the radius of the star to be roughly 1:1 Rˇ for a main-sequence star. The mass of the planet is estimated to by 0:00086 Mˇ . From the data, TwoStars generates a bolometric light curve shown in Fig. 6. 01 mag, in agreement with the value specified in the text. 17(b). 6: The synthetic light curve of OGLE-TR-56b for Prob. 18. 1 In the following, it is easiest to express Boltzmann’s constant as k D 8:6174 temperature of 300 K, 1 kT D 0:0259 eV eV: 40 If kT D 1 eV, then T D 1:16 104 K.