Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 8 - Exercises and Problems - Page 148: 67

Answer

$1.42 \times10^3 \ km$

Work Step by Step

The first rocket must go 10.5 orbits in the time that it takes the other rocket to go 10. Since radius is directly related to the period raised to the 2/3 power, we find: $(\frac{10}{10.5})^{2/3}=\frac{r}{45,800}$ $ r = 44,334\ km$ Thus, we find the difference is: $h = 45,800\ km - 44,334 \ km =1,465 \ km$
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