The satellites $S_{1}$ and $S_{2}$ are revolving round a planet in coplanar and concentric circular orbits of radii $R_{1}$ and $R_{2}$ in the same sense respectively. Their respective periods of revolution are $1 h$ and $8 h$ . The radius of satellite $S_{1}$ is equal to $10^{4} k m$ . Find the relative speed in $k m / h$ when they are closet.

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$π \times 10^{4} km / h$

$2 π \times 10^{4} km / h$

$3 π \times 10^{4} km / h$

$5 π \times 10^{4} km / h$

answer is A.

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Detailed Solution

By Kepler's $3^{r d}$ 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 $R_{2} = 4 \times 10^{4} k m$
Distance travelled in one revolution, $S_{1} = 2 {πR}_{1} = 2 π \times 10^{4}$ and $S_{2} = 2 {πR}_{2} = 2 π \times 4 \times 10^{4}$
$v_{1} = \frac{S_{1}}{t_{1}} = \frac{2 π \times 10^{4}}{1} = 2 π \times 10^{4} km / h$ and $v_{2} = \frac{S_{2}}{t_{2}} = \frac{2 π \times 4 \times 10^{4}}{8} = π \times 10^{4} km / h$
$∴$ $Relative velocity$ $= v_{1} - v_{2} = 2 π \times 10^{4} - π \times 10^{4} = π \times 10^{4} km / h$