College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 26 - Problems - Page 1010: 22


$L =6.5\times 10^{10}~m$

Work Step by Step

Let $L_0 = 0.75\times 10^{11}~m$ We can find the distance $L$ according to the astronaut: $L = L_0~\sqrt{1-\frac{v^2}{c^2}}$ $L = (0.75\times 10^{11}~m)~\sqrt{1-\frac{(0.50~c)^2}{c^2}}$ $L =6.5\times 10^{10}~m$
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