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Two satellites $S_{1} and S_{2}$ are revolving round a planet in coplanar and concentric circular orbit of radii $R_{1} and R_{2}$ in the same direction respectively. Their respective periods of revolution are I hr and 8 hr. The radius of the orbit of satellite $S_{1}$ is equal to $10^{4}$ km. Find the relative speed in kmph when they are closest.

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$π \times 10^{4} kmph$

$\frac{π}{2} \times 10^{4} kmph$

$2 π \times 10^{4} kmph$

$\frac{π}{3} \times 10^{4} kmph$

answer is B.

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Detailed Solution

By Kepler's 3'd law, $\frac{T^{2}}{R^{3}} = constant$

$∴ \frac{{T_{1}}^{2}}{{R_{1}}^{3}} = \frac{{T_{2}}^{2}}{{R_{2}}^{3}} or \frac{1}{{(10^{4})}^{3}} = \frac{64}{{R_{2}}^{3}} or R_{2} = 4 \times 10^{4} km$

Distance travelled in one revolution, $S_{1} = 2 {πR}_{1} = 2 π \times 10^{4} and S_{2} = 2 {πR}_{2} = 2 π \times 4 \times 10^{4}$

$v_{1} = \frac{S_{1}}{t_{1}} = \frac{2 π \times 10^{4}}{1} = 2 π \times 10^{4} kmph$

and $v_{2} = \frac{S_{2}}{t_{2}} = \frac{2 π \times 4 \times 10^{4}}{8} = π \times 10^{4} kmph$

$∴$ Relative velocity

$v_{1} - v_{2} = 2 π \times 10^{4} - π \times 10^{4} = π \times 10^{4} kmph$

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