Question

Two satellites $S_1$ and $S_2$ are revolving round a planet in coplanar and concentric circular orbit of radii $R_1$ and $R_2$ in the same direction respectively. Their respective periods of revolution are $1hr$ and $8hr$. The radius of the orbit of satellite $S_1$ is equal to $10^{4}km$. Find the relative speed in kmph when they are closest.

• A. $2\pi \times 10^{4}kmph$
• B. $\pi \times 10^{4}kmph$
• C. $\frac{\pi }{2}\times 10^{4}kmph$
• D. $\frac{\pi }{3}\times 10^{4}kmph$

Answer 1

Answer: (B)

By Kepler’s $3^{rd}$ law, $\frac{T^2}{R^3}=$ constant

$\therefore \frac{T_1^2}{R_1^3}=\frac{T_2^2}{R_2^3}$ or $\frac{1}{{\left(10^4\right)}^3}=\frac{64}{R_2^3}$ or

$R_2=4\times 10^{4}km$

Distance travelled in one revolution,

$S_1=2\pi R_1=2\pi \times 10^4$

and $S_2=2\pi R_2=2\pi \times 4\times 10^4$

$v_1=\frac{S_1}{t_1}=\frac{2\pi \times 10^4}{1}=2\pi \times 10^4$ kmph

and $v_2=\frac{S_2}{t_2}=\frac{2\pi \times 4\times 10^4}{8}$

$=\pi \times 10^{4}kmph$

$\therefore$ Relative velocity

$v_1=v_2=2\pi \times 10^4-\pi \times 10^4$

$=\pi \times 10^4$ kmph