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(A) Given:$T_1 = 1\, h$,$T_2 = 8\, h$$R_1 = 2 	imes 10^3\, km$Angular velocities are $\omega_1 = \frac{2\pi}{T_1} = 2\pi\, rad/h$ and $\omega_2 = \fr

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$3$

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(A) Given:$T_1 = 1\, h$,$T_2 = 8\, h$$R_1 = 2 	imes 10^3\, km$Angular velocities are $\omega_1 = \frac{2\pi}{T_1} = 2\pi\, rad/h$ and $\omega_2 = \frac{2\pi}{T_2} = \frac{\pi}{4}\, rad/h$.Using Kepler's third law,$\frac{R_2^3}{R_1^3} = \frac{T_2^2}{T_1^2} \Rightarrow \frac{R_2}{R_1} = (\frac{8}{1})^{2/3} = 4$.So,$R_2 = 4 	imes R_1 = 8 	imes 10^3\, km$.The linear velocities are $v_1 = \omega_1 R_1 = 2\pi 	imes 2 	imes 10^3 = 4\pi 	imes 10^3\, km/h$ and $v_2 = \omega_2 R_2 = \frac{\pi}{4} 	imes 8 	imes 10^3 = 2\pi 	imes 10^3\, km/h$.When the satellites are closest,the relative angular velocity $\omega_{rel}$ as observed from the nearer satellite is given by $\omega_{rel} = \frac{v_1 - v_2}{R_2 - R_1}$.$\omega_{rel} = \frac{4\pi 	imes 10^3 - 2\pi 	imes 10^3}{8 	imes 10^3 - 2 	imes 10^3} = \frac{2\pi 	imes 10^3}{6 	imes 10^3} = \frac{\pi}{3}\, rad/h$.Comparing this with $\frac{\pi}{x}$,we get $x = 3$.

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