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: 11

Answer

Please see the work below.

Work Step by Step

We know that $W=mg=\frac{GmM}{r^2}$ Thus, the weight is inversely proportional to $r^2$ $\implies \frac{W_{planet}}{W_{Earth}}=(\frac{r_{Earth}}{r_{planet}})^2$ This can be rearranged as: $r_{planet}=\sqrt{\frac{W_{Earth}}{W_{planet}}}r_{Earth}$ $\implies r_{planet}=\sqrt{\frac{1}{2}}r_{Earth}$ We plug in the known values to obtain: $r_{planet}=(0.7071)(6370Km)$ $r_{planet}=4500Km$
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