By B. Carroll, D. Ostlie

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**Additional info for An Introduction to Modern Astrophysics 2nd ed - SOLUTIONS MANUAL **

**Sample text**

28) was used for the integral of B . Finally, integrating the flux over the surface area of a sphere of radius R gives Z 2 Z FR2 sin Â dÂ d D 4 R2 T 4 ; LD D0 ÂD0 which is Eq. 17). 6 According to Eq. 2r /2 with Eq. 6 ns. 2 and 1 Ä500 `D 3 , the mean free path is D 21:8 m: D 2=3; that is, about 2=3 of a mean free path. Thus you You always look back to an optical depth of about could see to a distance of only 2`=3 D 18:5 m. 8 Using Eq. sec Â1 or Solving Eqs. 9 Consider the problems from the reference frame in which the electron is initially at rest.

2: A comparison between the limb darkening formulae of Van Hamme and the Eddington approximation (Eq. 58). 1 From Eq. 15), dr D d =Ä . Substitution into Eq. 6) leads immediately to the final result. 1 (see also Fig. 8). In this case however, r < r1 and r1 < R < r2 . R C r /2 . R r /2 After integrating over u, exact cancellation occurs, giving F D 0. 3 Assuming that the Sun is composed entirely of hydrogen atoms, the number of atoms in the Sun is approximately 1 Mˇ N ' D 1:2 1057: mH If each atom releases 10 eV of energy during a chemical reaction, the time scale for chemical reactions would be Echem 1:2 1058 eV tchem D D ' 5 1012 s D 1:6 105 yr: Lˇ 3:84 1026 W This is much shorter than the age of the solar system, and so the Sun’s energy cannot be chemical.

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. (d) Assuming a limit of V D 20, Eq. 6) implies that Betelgeuse would be detectable from a distance of d D 1 Mpc. 16 Given the number of telescopes the student is asked to research, it may be useful to have the class split into teams and have each team subdivide the research.