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

(a) The ratio $\frac{F_1}{F_2}$ is $\frac{1}{2}$ (b) The ratio $\frac{a_1}{a_2}$ is 1
(a) Let $M_e$ be the mass of the earth. Let $R$ be the distance from the earth to the 1000-kg satellite. We can write an expression for the gravitational force of the earth on the 1000-kg satellite. $F_1 = \frac{G~M_e~(1000~kg)}{R^2}$ Note that distance from the earth to 2000-kg satellite is also $R$. We can write an expression for the gravitational force of the earth on the 2000-kg satellite. $F_2 = \frac{G~M_e~(2000~kg)}{R^2}$ $F_2 = 2~F_1$ We can divide $F_1$ by $F_2$. $\frac{F_1}{F_2} = \frac{F_1}{2~F_1} = \frac{1}{2}$ The ratio $\frac{F_1}{F_2}$ is $\frac{1}{2}$ (b) We can find an expression for $a_1$. $F_1 = (1000~kg)~a_1$ $a_1 = \frac{F_1}{1000~kg}$ We can find an expression for $a_2$. $F_2 = (2000~kg)~a_2$ $a_2 = \frac{F_2}{2000~kg}$ $a_2 = \frac{2~F_1}{2000~kg}$ $a_2 = \frac{F_1}{1000~kg}$ $a_2 = a_1$ We can divide $a_1$ by $a_2$. $\frac{a_1}{a_2} = \frac{a_1}{a_1} = 1$ The ratio $\frac{a_1}{a_2}$ is 1.