College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 5 - Problems - Page 187: 34

Answer

The magnitude of the force exerted by the Sun on the Earth is $3.6\times 10^{22}~N$

Work Step by Step

The gravitational force provides the centripetal force to keep the Earth moving in a near-circular orbit around the Sun. We can find the gravitational force, that is, the magnitude of the force exerted by the Sun on the Earth: $F = \frac{mv^2}{r}$ $F = \frac{(6.0\times 10^{24}~kg)(3.0\times 10^4~m/s)^2}{1.5\times 10^{11}~m}$ $F = 3.6\times 10^{22}~N$ The magnitude of the force exerted by the Sun on the Earth is $3.6\times 10^{22}~N$
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