Two satellites $S_{1}$ and $S_{2}$ revolve round a planet in coplanar circular orbits in same sense. Their periods of revolution are 1 hr and 8 hrs respectively. Radius of orbit of $S_{1}$ is $10^{4}$ km. When $S_{1}$ is closest to $S_{2}$ , relative speed of $S_{1}$ with respect to $S_{2}$ is $n \times 10^{4}$ km/h, here n is (Answer upto two digits after the decimal point.)

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Solution

Since $T^{2} \propto r^{3} \Rightarrow {(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{r_{1}}{r_{2}})}^{3}$

${(\frac{1}{8})}^{2} = {(\frac{10^{4}}{r_{2}})}^{3} \Rightarrow r_{2} = 4 \times 10^{4}$ km

$V_{1} = r_{1} w_{1} = 10^{4} \times \frac{2 π}{1} = 2 π \times 10^{4}$ km/h
$V_{2} = r_{2} w_{2} = 4 \times 10^{4} \times \frac{2 π}{8} = π \times 10^{4}$ km/h

$V_{1} - V_{2} = π \times 10^{4}$ km/h

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Two satellites $S_{1}$ and $S_{2}$ revolve round a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 hour and 8 hours respectively. The radius of the orbit of $S_{1} = 10^{4}$ km When $S_{2}$ is closest to $S_{1}$ find speed of $S_{2}$ relative to $S_{2}$ and the angular speed of $S_{1}$ actually observed by an astronaut at $S_{1}$ .