By Carroll B.W., Ostlie D.A.

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**Additional resources for An introduction to modern astrophysics: Solution manual**

**Example text**

Using Fig. 10, we choose log10 D 0 kg m 3 and draw the best-fitting straight line to the curve. Two wellseparated points on this line are log10 Ä 1 D 1:19 at log10 T1 D 5:55, and log10 Ä 2 D 0:867 at log10 T2 D 6:15. For these two points, log10 Ä 2 log10 Ä 1 D 3:43 ' 3:5; log10 T2 log10 T1 as expected for a Kramers opacity law. 1:5 105 kg m 3 / D3 10 5 m: (b) From Eq. 29), the number of random-walk steps of this size from the center to the surface of the Sun is Â Ã2 Â Ã d Rˇ 2 N D D D 5 1026: ` ` The time for a photon to cover this many steps of size ` is tD N` D5 c 1013 s; almost two million years!

The amount of light blocked by the transiting Jupiter is R2J Te4 . Thus the relative change becomes R2 F D 2J D 0:0106 F Rˇ or just over 1%. 14 The fractional decrease in the flux appears to be approximately 1 0:983 D 0:017. 13, the ratio of the radius of the planet to the radius of the star is approximately Rp =Rs D 0:017 D 0:13. The radius of the star is estimated to be Rs D 1:1 Rˇ , implying that the radius of the planet is approximately Rp D 0:13Rs D 1:27 RJ , in agreement with the value found in the text.

Mp C me / ' V =mp . Here, D 10 6 kg m 3 is the density. The mass of the 44 Chapter 8 The Classification of Stellar Spectra electron is much less than the mass of the proton, and may be safely ignored in the expression for N t . Combining these expressions produces NII ne D N t mp for use in the Saha equation. With these substitutions, the Saha equation becomes Â NII N t mp D NI NII 2 me kT h2 Substituting this into the above equation produces " Ã Â N t mp 2 me kT 3=2 NII e 1C Nt NII h2 # I = kT Ã3=2 I = kT e N t mp NII Â : 2 me kT h2 Ã3=2 e I = kT : Multiplying each side by NII =N t and rearranging terms gives Â NII Nt Ã2 Â C NII Nt Ã mp Â 2 me kT h2 Ã3=2 e mp I = kT Â 2 me kT h2 Ã3=2 e I = kT D 0: (b) This is a quadratic equation of the form ax 2 C bx C c D 0, where aD1 Â bD mp cD b: 2 me kT h2 Ã3=2 e I = kT It therefore has the solution NII 1 D Nt 2a bC p b2 Á b p 4ac D 1 C 4=b 2 Á 1 : Evaluating b for a range of temperatures between 5000 K and 25,000 K produces the graph shown in Fig.