Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction.
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 $\frac{\pi}{x} \operatorname{rad~h}^{-1}$ where $x$ is ……….

Solution:

$\mathrm{T}_{1}=1$ hour

$\Rightarrow \omega_{1}=2 \pi \mathrm{rad} /$ hour

$T_{2}=8$ hours

$\Rightarrow \omega_{2}=\frac{\pi}{4} \mathrm{rad} /$ hour

$\mathrm{R}_{1}=2 \times 10^{3} \mathrm{~km}$

As $\mathrm{T}^{2} \propto \mathrm{R}^{3}$

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

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

$V_{1}=\omega_{1} R_{1}=4 \pi \times 10^{3} \mathrm{~km} / \mathrm{h}$

$V_{2}=\omega_{2} R_{2}=2 \pi \times 10^{3} \mathrm{~km} / \mathrm{h}$

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

$=\frac{\pi}{3} \mathrm{rad} /$ hour

$x=3$

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