Two satellites revolve around a planet in coplanar circular orbit… - Vidyadip

MCQ

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 ..... .

✓
$3$
B
$30$
C
$0.3$
D
$4$

✓

Answer

Correct option: A.

$3$

a
${T}_{1}=1$ $hour$

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

${T}_{2}=8$ $hours$

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

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

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

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

${x}=3$

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