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Two satellites \mathrm{S_1 \: and \: S_2} revolve round a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 hour and 8 hour respectively. The radius of the orbit of \mathrm{S_1 \: is\, 10^4 \mathrm{~km}}. The speed of  \mathrm{S_2} relative to \mathrm{S_1} when they are closet (in \mathrm{\mathrm{kmh}^{-1}} ) is
 

Option: 1

10^4 \pi

 


Option: 2

2 \times 10^4 \pi
 


Option: 3

\frac{1}{2} \times 10^4 \pi
 


Option: 4

4 \times 10^4 \pi


Answers (1)

best_answer

Since \mathrm{T^2 \propto r^3}

\mathrm{ \Rightarrow \frac{r_1^3}{r_2^3}=\frac{T_1^2}{T_2^2}=\frac{1}{64} }

\mathrm{ \Rightarrow \frac{r_1}{r_2}=\frac{1}{4} }

\mathrm{ \Rightarrow r_2=4 \times 10^4 \mathrm{~km} }

\mathrm{ \quad v_1=\frac{2 \pi r_1}{T_1} \mathrm{andv}_2=\frac{2 \pi r_2}{\mathrm{~T}_2} }

\mathrm{ \Rightarrow \mathrm{v}_1= \frac{2 \pi \times 10^4}{1} \mathrm{andv}_2=\frac{2 \pi \times 4 \times 10^4}{8}(\mathrm{in} \mathrm{kmph}) }

\mathrm{\Rightarrow \mathrm{v}_1=2 \pi \times 10^4 \mathrm{kmh}^{-1} \mathrm{andv}_2=\pi \times 10^4 \mathrm{kmh}^{-1}}

So, speed of \mathrm{S_2} with respect to \mathrm{S_1\: is \: \pi \times 10^4 \mathrm{kmh}^{-1}.}

Hence option 1 is correct.

Posted by

Sayak

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