#### Answer

(a) The ratio $\frac{F_1}{F_2}$ is $\frac{1}{2}$
(b) The ratio $\frac{a_1}{a_2}$ is 1

#### Work Step by Step

(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.