University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 32 - Electromagnetic Waves - Problems - Exercises - Page 1074: 32.25


$P_{av}=8.5\times10^{5}W $.

Work Step by Step

For a perfectly absorbing surface, we can relate the radiation pressure to the intensity. $$p_{rad}=\frac{I}{c}$$ $$I = c (p_{rad})=2.70\times10^3W/m^2$$ At a distance r from the source, the total average power from the star is spread over a spherical surface, of area $4 \pi r^2$. This equals the intensity at that distance. $$P_{av}=I4\pi r^2=(2.70\times10^3W/m^2)(4 \pi)(5.0m)^2=8.5\times10^{5}W $$
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