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 $1$ hour and $8$ hour respectively. The radius of the orbit of $S_1$ is $10^{4}km$. When $S_2$ is closest to $S_1$, the speed of $S_2$ relative to $S_1$ (in $km/h$)

• A. $\pi \times 10^4$
• B. $-\pi \times 10^4$
• C. $\pi \times 10^5$
• D. $-\pi \times 10^5$

Answer 1

Answer: (B)

According to Keplers law, $T^2\propto r^3$
$\therefore {\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{r_2}{r_1}\right)}^3$
or $\frac{r_2}{r_1}={\left(\frac{T_2}{T_1}\right)}^{2/3}$
or $r_2=r_1{\left(\frac{T_2}{T_1}\right)}^{2/3}$
or $r_2=\left(10^4\right){\left(\frac{8}{1}\right)}^{2/3}$
or $r_2=4\times 10^{4}km$
$\therefore v_1=\frac{2\pi r_1}{T_1}=\frac{2\pi \left(10^4\right)}{1}$
$=2\pi \times 10^{4}km/hour$
Similarly,
$\therefore v_2=\frac{2\pi r_2}{T_2}=\frac{2\pi \left(4\times 10^4\right)}{8}$
$=\pi \times 10^{4}km/hour$
Speed of $S_2$ relative to $S_1=\left(v_2-v_1\right)$
$=\pi \times 10^4\left(1-2\right)$
$=-\pi \times 10^{4}km/h$