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Two satellites $S_{1}$ and $S_{2}$ are revolving around a planet in the opposite sense in coplanar circular concentric orbits. At time $t=0$, the satellites are farthest apart. The periods of revolution of $S_{1}$ and $S_{2}$ are $3 \,h$ and $24 \,h$, respectively. The radius of the orbit of $S_{1}$ is $3 \times 10^{4} \,km$. Then, the orbital speed of $S_{2}$ as observed from
the planet is $4 \pi \times 10^{4} \,km h ^{-1}$, when $S_{2}$ is closest from $S_{1}$
the planet is $2 \pi \times 10^{4} \,km h ^{-1}$, when $S_{2}$ is farthest from $S_{1}$
$S_{1}$ is $\pi \times 10^{4} \,km h ^{-1}$, when $S_{2}$ is closest from $S_{1}$
$S_{1}$ is $3 \pi \times 10^{4} \,km h ^{-1}$, when $S_{2}$ is closest to $S_{1}$
Solution
$(d)$ From Kepler's law, we have
$T^{2} \propto R^{3}$
So, for satellites $S_{1}$ and $S_{2}$ is
$\Rightarrow \quad\left(\frac{T_{1}}{T_{2}}\right)^{2}=\left(\frac{R_{1}}{R_{2}}\right)^{3}$
Radius of orbit of satellite $S_{2}$ around planet is
$R_{2}=\left(\frac{T_{2}^{2} \times R_{1}^{3}}{T_{1}^{2}}\right)^{\frac{1}{3}}$
$=\left(\frac{24 \times 24 \times\left(3 \times 10^{4}\right)^{3}}{3 \times 3}\right)^{\frac{1}{3}}$
$=4 \times 3 \times 10^{4} \,km$
$=12 \times 10^{4} \,km$
When satellites are closest to each other, orbital speed of $S_{2}$ as observed from $S_{1}$ is
$v_{\text {relative }} =v_{1}+v_{2}$
$=R_{1} \omega_{1}+R_{2} \omega_{2}$
$=R_{1} \times \frac{2 \pi}{T_{1}}+R_{2} \times \frac{2 \pi}{T_{2}}$
$=3 \times 10^{4} \times \frac{2 \pi}{3}+12 \times 10^{4} \times \frac{2 \pi}{24}$
$=2 \pi \times 10^{4}+\pi \times 10^{4}$
$=3 \pi \times 10^{4} \,kmh ^{-1}$
