By Carroll B.W., Ostlie D.A.
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Extra resources for An introduction to modern astrophysics: Solution manual
From Kepler’s third law, the total mass of the system is M D 0:99 Mˇ . Since m sin i D 0:45 MJ M , the mass of 51 Peg is approximately 0:99 Mˇ . (b) For the planet orbiting HD 168443c, P D 1770 d, and a D 2:87 AU. Again, from Kepler’s third law, M D 1:01 Mˇ . We can again neglect the mass of the planet. 13 Considering the integrated flux over the projected solar disk, we have R2ˇ Te4 . 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%.
B) For M D 120 Mˇ , we may consider electron scattering as an appropriate opacity (the surface is ionized at that temperature). 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.
From Eq. 25 Heisenberg’s uncertainty principle, Eq. 20), relates E, the uncertainty in the energy of an atomic orbital, to t, the time an electron occupies the orbital before making a downward transition: E t D h 2 t : When an electron makes a downward transition from an initial to a final orbital, the energy of the emitted photon is (Eq. W = /. 594 nm line: These values can then be used with the general curve of growth for the Sun, Fig. 22, to obtain a value of log10 Œf Na . =500 nm/ for each wavelength.