Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 12 - Gravity - Problems and Conceptual Exercises - Page 412: 59

Answer

$2.96Km$

Work Step by Step

We know that $V_{esc}=\sqrt{\frac{2GM_S}{R}}$ as $V_{esc}=c$, where c is speed of light $\implies c=\sqrt{\frac{2GM_S}{R}}$ This can be rearranged as: $R=\frac{2GM_S}{c^2}$ We plug in the known values to obtain: $R=\frac{2(6.67\times 10^{-11})(2.0\times 10^{30})}{(3.0\times 10^8)^2}=2.96Km$
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