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 \(2 \times 10^{3}\, {km}\). The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is \(\frac{\pi}{{x}}\, {rad} \,{h}^{-1}\) where \({x}\) is ..... .

\(\Rightarrow \omega_{1}=2 \pi {rad} / {hour}\)

\({T}_{2}=8\) \(hours\)

\(\Rightarrow \omega_{2}=\frac{\pi}{4} {rad} /\) \(hour\)

\({R}_{1}=2 \times 10^{3} {km}\)

As \({T}^{2} \propto {R}^{3}\)

\(\Rightarrow\left(\frac{{R}_{2}}{{R}_{1}}\right)^{3}=\left(\frac{{T}_{2}}{{T}_{1}}\right)^{2}\)

\(\Rightarrow \frac{{R}_{2}}{{R}_{1}}=\left(\frac{8}{1}\right)^{2 / 3}=4 \Rightarrow {R}_{2}=8 \times 10^{3} {km}\)

\({V}_{1}=\omega_{1} {R}_{1}=4 \pi \times 10^{3} {km} / {h}\)

\({V}_{2}=\omega_{2} {R}_{2}=2 \pi \times 10^{3} {km} / {h}\)

Relative \(\omega=\frac{{V}_{1}-{V}_{2}}{{R}_{2}-{R}_{1}}=\frac{2 \pi \times 10^{3}}{6 \times 10^{3}}\)

\(=\frac{\pi}{3} {rad} /\) \(hour\)

\({x}=3\)