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  • 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 <img alt="2 \times 10^{3} \mathrm{~km}" src="htt

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{E}{x}$ rad $\mathrm{h}^{-1} where x  is_________.

 

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best_answer


S_{1}\rightarrow satellite\, \left ( 1 \right )
S_{2}\rightarrow satellite\, \left ( 2 \right )
r_{1}= 2\times 10^{6}\, m
T_{1}= 1\, hour
T_{2}= 3\, hour
T^{2}\alpha r^{3}
\frac{T_{1}^{2}}{T_{2}^{2}}= \frac{r_{1}^{3}}{r_{2}^{3}};r_{2}^{3}= 64\times 8\times 10^{18}
r_{2}= 8\times 10^{6}m

V_{1}= \omega _{1}r_{1}= 2\pi \times 2\times 10^{3}\frac{km}{hr}

V_{2}= \omega _{2}r_{2}= 8\times \frac{2\pi }{8}\times 10^{3}\frac{km}{hr}
\omega _{rel}= \frac{V_{2}-V_{1}}{r_{2}-r_{1}}= \frac{2\pi \times 10^{3}\frac{km}{hr}}{6\times 10^{6}m}
= \frac{2\pi }{6}= \frac{\pi }{3}
x=3

 

Posted by

Deependra Verma

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