Question

Two satellites $S_1$ and $S_2$ are revolving around a planet in the opposite sense in coplanar circular concentric orbits. At time $t=0$, the satellites are farthest apart. The periods of revolution of $S_1$ ad $S_2$ and $3$h and $24$h respectively. The radius of the orbit of $S_1$ is $3\times {10}^4$km. Then the orbital speed of $S_2$ as observed from.

• A. The planet is $4\pi \times {10}^4$km $h^{-1}$ when $S_2$ is closest from $S_1$
• B. The planet is $2\pi \times {10}^4$km $h^{-1}$ when $S_2$ is farthest from $S_1$
• C. $S_1$ is $\pi \times {10}^4$km $h^{-1}$ when $S_2$ is closest from $S_1$
• D. $S_1$ is $3\pi \times {10}^4$km $h^{-1}$ when $S_2$ is closest to $S_1$

Answer 1

The correct option is D $S_1$ is $3\pi \times {10}^4$km $h^{-1}$ when $S_2$ is closest to $S_1$

From Kepler's Laws,

$\frac{R_1^3}{R_2^3}=\frac{T_1^2}{T_2^2}$

$\frac{R_1}{R_2}=\frac{1}{4}$

$R_2=12\times {10}^{4}km$

Relative to the planet,

$v_1=\frac{2\pi R_1}{T_1}=\frac{2\pi \times 3\times {10}^4}{3}=2\pi \times {10}^{4}km/h$

$v_2=\frac{2\pi R_2}{T_2}=\frac{2\pi \times 12\times {10}^4}{24}=\pi \times {10}^{4}km/h$

At closest approach distance between $S_1$ & $S_2$, relative orbital speed of $S_2$ as seen from $S_1$ will be $v_1+v_2$ as they are moving in the opposite directions.

Hence, the above orbital speed is :

$v=v_1+v_2=3\pi \times {10}^{4}km/hr$