Two satellites S_{1} and S_{2} are revolving | vedclass

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(D) From Kepler's third law,we have $T^{2} \propto R^{3}$.For satellites $S_{1}$ and $S_{2}$,we have $\left(\frac{T_{1}}{T_{2}}\right)^{2} = \left(\fr

Vedclass · Accepted Answer

$S_{1}$ is $3 \pi 	imes 10^{4} \,km \,h^{-1}$,when $S_{2}$ is closest to $S_{1}$

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(D) From Kepler's third law,we have $T^{2} \propto R^{3}$.For satellites $S_{1}$ and $S_{2}$,we have $\left(\frac{T_{1}}{T_{2}}ight)^{2} = \left(\frac{R_{1}}{R_{2}}ight)^{3}$.Given $T_{1} = 3 \,h$,$T_{2} = 24 \,h$,and $R_{1} = 3 	imes 10^{4} \,km$.The radius of the orbit of satellite $S_{2}$ is $R_{2} = R_{1} \left(\frac{T_{2}}{T_{1}}ight)^{2/3} = 3 	imes 10^{4} 	imes \left(\frac{24}{3}ight)^{2/3} = 3 	imes 10^{4} 	imes (8)^{2/3} = 3 	imes 10^{4} 	imes 4 = 12 	imes 10^{4} \,km$.Since the satellites revolve in opposite senses,when they are closest to each other,their velocity vectors are in opposite directions relative to the planet,meaning their relative speed is the sum of their orbital speeds.The orbital speeds are $v_{1} = \frac{2 \pi R_{1}}{T_{1}} = \frac{2 \pi 	imes 3 	imes 10^{4}}{3} = 2 \pi 	imes 10^{4} \,km \,h^{-1}$ and $v_{2} = \frac{2 \pi R_{2}}{T_{2}} = \frac{2 \pi 	imes 12 	imes 10^{4}}{24} = \pi 	imes 10^{4} \,km \,h^{-1}$.The relative speed of $S_{2}$ as observed from $S_{1}$ is $v_{rel} = v_{1} + v_{2} = 2 \pi 	imes 10^{4} + \pi 	imes 10^{4} = 3 \pi 	imes 10^{4} \,km \,h^{-1}$.

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