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JEE MAIN 2021
01-09-21 S2
Question
Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hours respectively. The radius of the orbit of nearer satellite is $2 \times 10^3 \mathrm{~km}$. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $\frac{\pi}{x} \mathrm{rad} \mathrm{h}^{-1}$ where x is $\_\_\_\_$
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Solution
$\begin{aligned} & \mathrm{T}_1=1 \text { hour } \\ & \Rightarrow \omega_1=2 \pi \mathrm{rad} / \text { hour } \\ & \mathrm{T}_2=8 \text { hours } \\ & \Rightarrow \omega_2=\frac{\pi}{4} \mathrm{rad} / \text { hour } \\ & \mathrm{R}_1=2 \times 10^3 \mathrm{~km} \\ & \text { As } \mathrm{T}^2 \propto \mathrm{R}^3 \\ & \Rightarrow \frac{\mathrm{R}_2}{\mathrm{R}_1}=\left(\frac{8}{1}\right)^{210}=4 \Rightarrow \mathrm{R}_2=8 \times 10^3 \mathrm{~km}\end{aligned}$
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