By Carroll B.W., Ostlie D.A.
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Extra resources for An introduction to modern astrophysics: Solution manual
8) with I D B over all outward directions (0 Ä Â Ä =2). Z F d DB d 2 Z D0 =2 ÂD0 cos Â sin Â dÂ d D B d : We now integrate over all wavelengths to find the total flux, Z F D 1 0 Z F d D 1 0 B d D T4 D T 4: where Eq. 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.
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.
M1 D 0:5 Mˇ , M2 D 2:0 Mˇ , P D 1:8 yr D 657:46 d, i D 30ı, D 90ı . 15 (a) See Fig. 2. ] (b) From Kepler’s third law (Eq. 37), with P D 1:8 yr, M1 D 0:5 Mˇ and M2 D 2:0 Mˇ , a D 2:0 AU. Also, since m1 =m2 D a2 =a1 (Eq. 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 .