Two satellites A and B revolve round the same planet in coplanar circular orbits lying in the same plane. Their periods of revolutions are 1 h and 8 h, respectively. The radius of the orbit of A is 10⁴ km. The speed of B relative to A, when they are close in km/h is [AIIMS 2018]

Given,   $T_A=1 \mathrm{h},T_B=8 \mathrm{h}$

and                  $r_A={10}^4 \mathrm{km}$

From Kepler's law,   $\frac{T_A^2}{T_B^2}=\frac{r_A^3}{r_B^3}\Rightarrow \frac{1^2}{8^2}=\frac{{\left({10}^4\right)}^3}{r_B^3}$

$\Rightarrow r_B^3=64\times {\left({10}^4\right)}^3$

$\therefore r_B=4\times {10}^4 \mathrm{km}$

Speed of satellIte  $A,v_A=\frac{2\pi r_A}{T_A}=\frac{2\pi \times {10}^4}{1}$

                                      $=2\pi \times {10}^4 \mathrm{km}/\mathrm{h}$

Speed of satellIte   $B,v_B=\frac{2\pi r_B}{T_B}=\frac{2\pi \times 4\times {10}^4}{8}$

                                       $=\pi \times {10}^4 \mathrm{km}/\mathrm{h}$

The speed of B relative to A when they are close,

$v_{BA}=v_A-v_R=2\pi \times {10}^4-\pi \times {10}^4$

      $=\pi \times {10}^4 \mathrm{km}/\mathrm{h}$