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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