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(D) According to Kepler's third law of planetary motion,the square of the time period is proportional to the cube of the orbital radius: $\frac{T_{A}^

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$\pi 	imes 10^{4}$

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(D) According to Kepler's third law of planetary motion,the square of the time period is proportional to the cube of the orbital radius: $\frac{T_{A}^{2}}{T_{B}^{2}} = \frac{r_{A}^{3}}{r_{B}^{3}}$.Given $T_{A} = 1\, h$,$T_{B} = 8\, h$,and $r_{A} = 10^{4}\, km$.Substituting the values: $\frac{1^{2}}{8^{2}} = \frac{(10^{4})^{3}}{r_{B}^{3}} \Rightarrow \frac{1}{64} = \frac{(10^{4})^{3}}{r_{B}^{3}}$.Thus,$r_{B}^{3} = 64 	imes (10^{4})^{3}$,which gives $r_{B} = 4 	imes 10^{4}\, km$.The orbital speed of a satellite is given by $v = \frac{2 \pi r}{T}$.For satellite $A$: $v_{A} = \frac{2 \pi 	imes 10^{4}}{1} = 2 \pi 	imes 10^{4}\, km/h$.For satellite $B$: $v_{B} = \frac{2 \pi 	imes 4 	imes 10^{4}}{8} = \pi 	imes 10^{4}\, km/h$.Since both satellites revolve in the same direction in coplanar orbits,their relative speed when they are closest is the difference of their orbital speeds: $v_{rel} = v_{A} - v_{B} = 2 \pi 	imes 10^{4} - \pi 	imes 10^{4} = \pi 	imes 10^{4}\, km/h$.

Two satellites $A$ and $B$ revolve around the same planet in coplanar circular orbits. Their periods of revolution are $1\, h$ and $8\, h$,respectively. The radius of the orbit of $A$ is $10^{4}\, km$. The speed of $B$ relative to $A$ when they are closest is (in $km/h$):

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