Question

Two satellites $S_1$ and $S_2$ revolve round a planet in coplanar circular orbits in the same sense. Their periods of revolution are $1h$ and $8h$ respectively. The radius of the orbit of $S_1$ is ${10}^{4}km$. When $S_2$ is closest to $S_1$, find (a) the speed of $S_2$ relative to $S_1$ and (b) the angular speed of $S_2$ as observed by an astronaut in $S_1$.

Answer 1

According to Kepler's law $T^2\times r^3$

${\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{r_2}{r_1}\right)}^3$

$r_2=r_1{\left(\frac{T_2}{T_1}\right)}^{2/3}$

$r_2=\left({10}^4\right){\left(\frac{8}{1}\right)}^{2/3}$

$r_2=4\times {10}^{4}km$

$\therefore V_1=\frac{{2\pi r}_1}{T_1}=\frac{2\pi \left({10}^4\right)}{1}=2\pi \times {10}^{4}Km/hr$

Similarly,

$V_2=\frac{{2\pi r}_2}{T_2}=\frac{2\pi (4\times {10}^4)}{8}=\pi \times {10}^{4}Km/hr$

(a) Speed of $S_2$ relative to $S_1=(V_2-V_1)$

$=\pi \times {10}^4(1-2)=-\pi \times {10}^{4}Km/hr$

(b) Angular speed of $S_2$ as observed by astronaut in $S_1$

$W_r=\frac{V_1-V_2}{r_2-r_1}$

$=\frac{\pi \times {10}^{4}Km/hr}{4\times {10}^4-{10}^4}$

$=\frac{\pi }{3}$ rad/hr