Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems: 15

Answer

The free-fall acceleration toward the sun at the distance of the earth's orbit is $5.9\times 10^{-3}~m/s^2$.

Work Step by Step

Let $g_s$ be the free-fall acceleration toward the sun. $g_s = \frac{G~M_s}{r^2}$ $g_s = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(2.0\times 10^{30}~kg)}{(1.50\times 10^{11}~m)^2}$ $g_s = 5.9\times 10^{-3}~m/s^2$ The free-fall acceleration toward the sun at the distance of the earth's orbit is $5.9\times 10^{-3}~m/s^2$.
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