Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 13 - Newton's Theory of Gravity - Exercises and Problems: 4

Answer

The ratio of the sphere's gravitational force on the dust particle and the earth's gravitational force on the dust particle is $1.61\times 10^{-7}$.

Work Step by Step

Let $M_s$ be mass of the lead sphere. Let $R_s$ be the distance from the dust to the sphere's center. Let $M_d$ be the mass of the dust particle. We can write an expression for the force of gravity $F_s$ of the sphere on the dust particle. $F_s = \frac{G~M_s~M_d}{R_s^2}$ 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 the dust particle. $F_e = \frac{G~M_e~M_d}{R_e^2}$ We can find the ratio of $\frac{F_s}{F_e}$. $\frac{F_s}{F_e} = \frac{(\frac{G~M_s~M_d}{R_s^2})}{(\frac{G~M_e~M_d}{R_e^2})}$ $\frac{F_s}{F_e} = \frac{M_s~R_e^2}{M_e~R_s^2}$ $\frac{F_s}{F_e} = \frac{(5900~kg)(6.38\times 10^6~m)^2}{(5.98\times 10^{24}~kg)(0.50~m)^2}$ $\frac{F_s}{F_e} = 1.61\times 10^{-7}$ The ratio of the sphere's gravitational force on the dust particle and the earth's gravitational force on the dust particle is $1.61\times 10^{-7}$.
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