Answer
(a) The orbital period of satellite 1 is 250 minutes.
The orbital period of satellite 3 is 459 minutes.
(b) The force on satellite 2 is 20,000 N
The force on satellite 3 is 4440 N
(c) $\frac{K_1}{K_3} = 1.5$
Work Step by Step
(a) We can write an expression for the orbital period $T_1$ of satellite 1. Let $M_p$ be the mass of the planet.
$T_1^2 = \frac{4\pi^2~(2R)^3}{G~M_p}$
$T_1^2 = 8\times ~\frac{4\pi^2~R^3}{G~M_p}$
$T_1 = \sqrt{8\times ~\frac{4\pi^2~R^3}{G~M_p}}$
Since the orbital period $T_2$ of satellite 2 has the same orbital radius as satellite 1, the orbital period of satellite 2 is also 250 minutes.
We can write an expression for the orbital period $T_3$ of satellite 3.
$T_3^2 = \frac{4\pi^2~(3R)^3}{G~M_p}$
$T_3^2 = 27\times ~\frac{4\pi^2~R^3}{G~M_p}$
$T_3 = \sqrt{27\times ~\frac{4\pi^2~R^3}{G~M_p}}$
We can divide $T_3$ by $T_1$.
$\frac{T_3}{T_1} = \sqrt{\frac{27}{8}}$
$T_3 = \sqrt{\frac{27}{8}}~T_1$
$T_3 = \sqrt{\frac{27}{8}}~(250~minutes)$
$T_3 = 459~minutes$
The orbital period of satellite 3 is 459 minutes.
(b) We can write an expression for the force on satellite 1.
$F_1 = \frac{G~M_p~m}{(2R)^2}$
$F_1 = \frac{1}{4}\times ~\frac{G~M_p~m}{R^2}$
We can write an expression for the force on satellite 2.
$F_2 = \frac{G~M_p~(2m)}{(2R)^2}$
$F_2 = \frac{1}{2}\times ~\frac{G~M_p~m}{R^2}$
$F_2 = 2F_1$
$F_2 = (2)(10,000~N)$
$F_2 = 20,000~N$
The force on satellite 2 is 20,000 N
We can write an expression for the force on satellite 3.
$F_3 = \frac{G~M_p~m}{(3R)^2}$
$F_3 = \frac{1}{9}\times ~\frac{G~M_p~m}{R^2}$
We can divide $F_3$ by $F_1$.
$\frac{F_3}{F_1} = \frac{1/9}{1/4}$
$\frac{F_3}{F_1} = \frac{4}{9}$
$F_3 = \frac{4}{9}~F_1$
$F_3 = \frac{4}{9}~(10,000~N)$
$F_3 = 4440~N$
The force on satellite 3 is 4440 N
(c) The orbital speed of an object around a planet is $v = \sqrt{\frac{G~M_p}{R_0}}$, where $R_0$ is the orbital radius. We can use the speed $v$ to find an expression for the kinetic energy of an object of mass $M_0$ in orbit around a planet.
$K = \frac{1}{2}M_0v^2$
$K = \frac{1}{2}M_0~(\frac{G~M_p}{R_0})$
$K = \frac{G~M_p~M_0}{2R_0}$
We can write an expression for the kinetic energy of satellite 1.
$K_1 = \frac{G~M_p~m}{(2)(2R)}$
$K_1 = \frac{1}{4}\times \frac{G~M_p~m}{R}$
We can write an expression for the kinetic energy of satellite 3.
$K_3 = \frac{G~M_p~m}{(2)(3R)}$
$K_3 = \frac{1}{6}\times \frac{G~M_p~m}{R}$
We can find the ratio of $\frac{K_1}{K_3}$.
$\frac{K_1}{K_3} = \frac{1/4}{1/6}$
$\frac{K_1}{K_3} = \frac{6}{4}$
$\frac{K_1}{K_3} = 1.5$
