Two satellites $S_1$ and $S_2$ revolve round a planet in coplanar circular orbits in the same sense. Their periods of revolution are $1 h$and $8 h$ respectively. The radius of the orbit of $S_1$ is $104 km.$ When $S_2$ is closest to $S_1$ find The speed of $S_2$ relative to $S_1$

We have $T^2 \infty r^3$

Therefore  ${\left(\frac{T_1}{T_2}\right)}^2={\left(\frac{R_1}{R_2}\right)}^3$

or             ${\left(\frac{1}{8}\right)}^2={\left(\frac{R_1}{R_2}\right)}^3$

$\Rightarrow R_2=4R_1=4\times {10}^4 km.$

Let ${\nu }_1$ and ${\nu }_2$ be the linear speeds of $S_1$ and $S_2$ with respect to the planet. Then

${\nu }_1=\frac{2\pi R_1}{T_1}=2\pi \times {10}^4 km/h$

and ${\nu }_2=\frac{2\pi R_2}{T_2}=\pi \times {10}^4 km/h$

At the closest separation, they are moving in the same direction. Therefore the speed of $S_2$ with respect to $S_{1 }$ is   $\left|{\nu }_2-{\nu }_1\right|=\pi \times {10}^4 km/h.$