Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 9 - Work and Kinetic Energy - Exercises and Problems - Page 230: 65


The distance to the supernova is $1.24\times 10^8~ly$

Work Step by Step

Let's assume that the total energy $E$ emitted from the supernova each second is spread uniformly around the area of a sphere centered on the supernova. Let $d$ be the distance to the supernova. Then the amount of energy per unit area is $\frac{E}{4\pi ~d^2}$ We can calculate the amount of energy detected $E_0$ on the earth per unit area of the telescope. Note that the energy observed each second $E_0$ is $9.1\times 10^{-12}~J$ $\frac{E_0}{\pi ~r^2} = \frac{9.1\times 10^{-12}~J}{(\pi)(1.0~m)^2} = 2.90\times 10^{-12}~J/m^2$ We can find the distance $d$ to the supernova. Note that the energy emitted each second by the supernova is $E = 5.0\times 10^{37}~J$ $\frac{E}{4\pi ~d^2} = 2.90\times 10^{-12}~J/m^2$ $d^2 = \frac{E}{(4\pi)(2.90\times 10^{-12}~J/m^2)}$ $d = \sqrt{\frac{E}{(4\pi)(2.90\times 10^{-12}~J/m^2)}}$ $d = \sqrt{\frac{5.0\times 10^{37}~J}{(4\pi)(2.90\times 10^{-12}~J/m^2)}}$ $d = 1.17\times 10^{24}~m$ We can find the number of meters in one light-year. $1~ly = (3.0\times 10^8~m/s)(365~days/yr)(24~hr/day)(3600~s/hr)$ $1~ly = 9.46\times 10^{15}~m$ We can convert the distance $d$ to units of light-years. $d = (1.17\times 10^{24}~m)(\frac{1~ly}{9.46\times 10^{15}~m})$ $d = 1.24\times 10^8~ly$ The distance to the supernova is $1.24\times 10^8~ly$
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