Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 16 - Exercises and Problems - Page 301: 68


$R=4.1\times 10^9m$

Work Step by Step

As we know that $P=e\sigma AT^4$ but surface area A is $A=4\pi R^2$ $\implies P=e\sigma 4\pi R^2T^4 $ This simplifies to: $R=(\frac{P}{e\sigma 4\pi T^4})^\frac{1}{2}$ We plug in the known values to obtain: $R=(\frac{3.4\times 10^{30}}{1\times 5.67\times 10^{-8}\times 4(3.1416) (23000)^4})^\frac{1}{2}$ $R=4.1\times 10^9m$
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