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: 88

Answer

The period of one revolution would be 9.0 Earth days.

Work Step by Step

Let $r$ be the average distance from the Earth to the sun. Then, $r = 1.5 \times 10^{11} ~m$ Then, $\frac{v^2}{r} = g$ $v= \sqrt{gr}$ We can use the velocity to find the period of one rotation. $T = \frac{2\pi r}{\sqrt{gr}} = 2\pi \sqrt{\frac{r}{g}}$ $T = 2 \pi \sqrt{\frac{1.5 \times 10^{11} ~m}{9.80 ~m/s^2}}$ $T = 7.77 \times 10^5 ~s$ We can convert this time to Earth days. $T = (7.77 \times 10^5 ~s)(\frac{1 ~h}{3600 ~s})(\frac{1 ~day}{24 ~h}) = 9.0 ~days$ The period of one revolution would be 9.0 Earth days.
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