Question 7: Gravitation

Question

Two satellites $S_{1}$ and $S_{2}$ are revolving round a planet in coplanar circular orbits of radii $r_{1}$ and $r_{2}$ in the same direction, respectively. Their respective periods of revolution are 1$\mathrm{~h}$ and 8$\mathrm{~h}$. The radius of orbit of satellite $S_{1}$ is equal to $10^4 \mathrm{~km}$. What will be their relative speed (in $\mathrm{km} / \mathrm{h}$) when they are closest?

Accepted Answer

If $r_{1}$ and $r_{2}$ are radii of orbits of satellites $S_{1}$ and $S_{2}$ and $T_{1}, T_{2}$ are their respective time periods of revolution, then from Kepler's third law,
$\frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3} \Rightarrow r_2=\left(\frac{T_2}{T_1}ight)^{2 / 3} 	imes r_1$
$\begin{aligned}
	ext { Here, } \quad T_1 & =1 \mathrm{~h} \
T_2 & =8 \mathrm{~h} \
r_1 & =10^4 \mathrm{~km} \
\Rightarrow \quad\left(\frac{8 h}{1 h}ight)^{2 / 3} 	imes 10^4 & =4 	imes 10^4 \mathrm{~km}
\end{aligned} \$
The speed of satellite $S_{2}$ relative to $S_{1} \ $
$\begin{aligned}
\left|v_2-v_1ight| & =\frac{2 \pi r_2}{T_2}-\frac{2 \pi r_1}{T_1} \
& =2 \pi\left[\frac{r_2}{T_2}-\frac{r_1}{T_1}ight] \
& =2 \pi\left[\frac{4 	imes 10^4}{8}-\frac{10^4}{1}ight] \mathrm{km} / \mathrm{h} \
& =2 \pi 	imes \frac{1}{2} 	imes 10^4 \
& =\pi 	imes 10^4 \mathrm{~km} / \mathrm{h}
\end{aligned}$

Answer

$\pi / 2 	imes 10^4$

Answer

$\pi 	imes 10^4$

Answer

$2 \pi 	imes 10^4$

Answer

$4 \pi 	imes 10^4$

Gravitation

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