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 13 - Newton's Theory of Gravity - Exercises and Problems: 1

Answer

The ratio of the sun's gravitational force on a person and the earth's gravitational force on a person is $6.02\times 10^{-4}$

Work Step by Step

Let $M_s$ be mass of the sun. Let $R_s$ be the distance from a person to the sun. Let $M_p$ be the mass of a person. We can write an expression for the force of gravity $F_s$ of the sun on a person. $F_s = \frac{G~M_s~M_p}{R_s^2}$ Let $M_e$ be the mass of the earth. Let $R_e$ be the earth's radius. We can write an expression for the force of gravity $F_e$ of the earth on a person. $F_e = \frac{G~M_e~M_p}{R_e^2}$ We can find the ratio of $\frac{F_s}{F_e}$. $\frac{F_s}{F_e} = \frac{(\frac{G~M_s~M_p}{R_s^2})}{(\frac{G~M_e~M_p}{R_e^2})}$ $\frac{F_s}{F_e} = \frac{M_s~R_e^2}{M_e~R_s^2}$ $\frac{F_s}{F_e} = \frac{(1.989\times 10^{30}~kg)(6.38\times 10^6~m)^2}{(5.98\times 10^{24}~kg)(1.50\times 10^{11}~m)^2}$ $\frac{F_s}{F_e} = 6.02\times 10^{-4}$ The ratio of the sun's gravitational force on a person and the earth's gravitational force on a person is $6.02\times 10^{-4}$.
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