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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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