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: 44

Answer

$T_A=8T_B$

Work Step by Step

As we know that the time period for satellite A is $T_A=2\pi\sqrt{\frac{R_A^3}{GM}}$ Similarly for satellite B $T_B=2\pi\sqrt{\frac{R_B^3}{GM}}$ Now we can find the ratio of the two satellites time period as: $\frac{T_A}{T_B}=\sqrt{\frac{R_A^3}{R_B^3}}$ Substituting $R_A=4R_B$, we obtain: $T_A=8T_B$
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