Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 5 - Circular Motion; Gravitation - General Problems - Page 136: 78

Answer

The length of one Earth day would be 84.5 minutes.

Work Step by Step

If the force of gravity was precisely equal to the centripetal force, then the normal force would be zero and a person would be apparently weightless. $m \frac{v^2}{r} = mg$ $v = \sqrt{gr}$ We can use this velocity to find the period of one rotation. $T = \frac{2\pi r}{\sqrt{gr}} = 2 \pi \sqrt{\frac{r}{g}} = 2 \pi \sqrt{\frac{6.38 \times 10^6 ~m}{9.80 ~m/s^2}}$ $T = 5070 ~s$ We can express this time in units of minutes. $T = (5070 ~s)(\frac{1 ~min}{60 ~s}) = 84.5 ~minutes$ The length of one Earth day would be 84.5 minutes.
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