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,
$\therefore v_{2} = \frac{2\pi r_{2}}{T_{2}} = \frac{2\pi\left(4 \times10^{4}\right)}{8}$
$= \pi \times 10^{4}\,km / hour$
Speed of $S_{2}$ relative to $S_{1}= \left(v_{2} - v_{1}\right)$
$= \pi \times 10^{4}\,\left(1-2\right)$
$=-\pi \times 10^{4}\,km/h$