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SOLUTIONS TO EXERCISE 8
(a) Since a Megaparsec is about 3.1 * 10^22 m, this burst's luminosity is L = 4 * pi * r^2 * F = 4 * 3.14 * (3000 * 3.1 * 10^22 m)^2 * (5 * 10^-10 W/m^2) = 8.7 * 10^43 W or 8.7 * 10^43 Joule/sec For a 90 second period, the total energy release was E = (8.7 * 10^43 Joule/s) * (90 s) = 7.8 * 10^45 J. (b) The luminosity of a galaxy is usually about equal to the luminosity of all its stars put together. Since the Milky Way has about 10^11 stars, if we use the Sun as a typical star then our galaxy's luminosity is L = 10^11 * (4 * 10^26 W) = 4 * 10^37 W. The luminosity of the gamma-ray burst is (8.7 * 10^43 W) --------------- = 2.2 million times greater than an entire galaxy! (4 * 10^37 W) (c) The energy released by the burst was (7.8 * 10^45)/(10^44) = 78 times greater than that of a typical supernova. |