Question

Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are $1$ hour and $8$ hours respectively. The radius of the orbit of nearer satellite is $2\times 10^{3}km$. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $\frac{\pi }{x}\text{rad}{h}^{-1}$ where $x$ is .....

Answer 1

Answer: 3

$T_1=1$ hour
$\Rightarrow {\omega }_1=2\pi rad/$ hour
$T_2=8$ hours
$\Rightarrow {\omega }_2=\frac{\pi }{4}rad/$ hour
$R_1=2\times 10^{3}km$
As $T^2\propto R^3$
$\Rightarrow {\left(\frac{R_2}{R_1}\right)}^3={\left(\frac{T_2}{T_1}\right)}^2$
$\Rightarrow \frac{R_2}{R_1}={\left(\frac{8}{1}\right)}^{2/3}=4$
$\Rightarrow R_2=8\times 10^{3}km$

$V_1={\omega }_1{R}_1=4\pi \times 10^{3}km/h$
$V_2={\omega }_2{R}_2=2\pi \times 10^{3}km/h$
Relative $\omega =\frac{V_1-V_2}{R_2-R_1}=\frac{2\pi \times 10^3}{6\times 10^3}$
$=\frac{\pi }{3}rad/$ hour
$x=3$