An Introduction to Modern Astrophysics 2nd ed - SOLUTIONS by B. Carroll, D. Ostlie

By B. Carroll, D. Ostlie

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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. 5, we find N3 =N2 D 5:14 10 2 and N3 =N1 D 2:82 10 10. That is, for every neutral hydrogen atom in the n D 3 state, there are 19 in the n D 2 state and 3:55 109 in the n D 1 state. The number of neutral hydrogen atoms per unit area is then N1 hNa i D 5:35 1029 m 2 : N3 The ratio of the number of ionized to neutral atoms is obtained from the Saha equation, Eq. 4). The result shows that NII =NI D 1:33 10 4, so nearly all of the hydrogen atoms are neutral.

1 C X / D 0:034, and LEd D 4:6 106 Lˇ . 5 times greater than the actual luminosity. Clearly radiation pressure plays a significant role in this case. 24 (a) The Lane–Emden equation is a second-order differential equation. One very powerful tool in numerical analysis is to write second-order equations as two first-order equations, one of which is the derivative of the function that you are trying to solve. Beginning with Eq. 110) and taking the derivatives leads to d 2 Dn 2 dDn C C Dnn D 0: d 2 d Letting fn .

B) Âmoon D 2R=d D 9:0 10 3 rad D 190000, implying Âmoon=Âeye D 33:7. When Jupiter is at opposition (closest approach to Earth), ÂJupiter D 2:3 10 4 rad D 4700, implying ÂJupiter =Âeye D 1:7. The resolution is worse during all other relative positions of Jupiter and Earth. 8 (c) It is possible for the disk of the Moon to be resolved with the naked eye. However, the disk of Jupiter is comparable to the eye’s resolution limit; to distinguish any features on the surface of Jupiter, they would need to be nearly the size of the planet’s disk.

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