Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 8 - Exercises and Problems - Page 142: 63


$1.42 \times10^3 \ km$

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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