Two satellites $S_1$ and $S_2$ are revolving around the earth in coplanar concentric orbits in the opposite sense. At $t=0$ , the positions of
satellites are shown in the diagram. The period of $S_1$ and $S_2$ are 4 h and 24 h, respectively. The radius of orbit of $S_1$ is $1.28\times {10}^4$ km. For
this situation mark the correct statement(s).

 

From Kepler’s law$T^2\propto R^3$     So,$\frac{T_1}{T_2}={\left(\frac{R_1}{R_2}\right)}^{\frac{3}{2}}$

$R_2={\left(\frac{T_2}{T_1}\right)}^{\frac{2}{3}}\times R_1={\left(\frac{24}{6}\right)}^{\frac{2}{3}}\times 1.28 \times {10}^4 km$   =$3.22 \times {10}^4 km$

Orbital velocity of $S_1$  is $v_1=\frac{2{\mathrm{\pi R}}_1}{T_1}= \frac{2\mathrm{\pi }\times 1.28\times {10}^4}{4}=0.64\mathrm{\pi }\times {10}^4$ km/h

Orbital velocity of $S_2$ is $v_2=\frac{2{\mathrm{\pi R}}_2}{T_2}=\frac{2\mathrm{\pi }\times 3.22\times {10}^4}{24}=0.2683\mathrm{\pi }\times {10}^4 km/h$  

At $t=12h$ the two satellites are closest to each other and after every 24 h they come at same position relative to each other. It is clear that $v_2$ w.r.t. $v_1$  is changing continuously both in magnitude and  direction.

Angular velocity of $S_2$ w.r.t $S_1$ at $t=12h$ is   $\omega =(v_1+v_2)/(R_2-R_1)=0.468\mathrm{\pi } \mathrm{rad} h^{-1}$  This can be easily proved by writing the basic equations.