Question

Two satellites $S_1$ and $S_2$ are revolving around the Earth in co-planar concentric orbits in the opposite sense. At $t=0$, the positions of satellites are shown in the diagram. The periods of $S_1$ and $S_2$ are 4h and 24h respectively. The radius of orbit of $S_1$ is $1.28\times {10}^{4}km$. For this situation, mark the correct statement(s) :

• A. The angular velocity of $S_2$ are observed by $S_1$ at $t=12$ h is $0.468\pi rad{s}^{-1}$.
• B. The two satellites are closest to each other at $t=12h$ and then after every 24 h they are closest to each other.
• C. The orbital velocity of $S_1$ is $0.64\pi \times {10}^{4}km$.
• D. The velocity of $S_1$ relative to $S_2$ is continuously changing in magnitude and direction both.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The angular velocity of $S_2$ are observed by $S_1$ at $t=12$ h is $0.468\pi rad{s}^{-1}$.
B The two satellites are closest to each other at $t=12h$ and then after every 24 h they are closest to each other.
C The orbital velocity of $S_1$ is $0.64\pi \times {10}^{4}km$.
D The velocity of $S_1$ relative to $S_2$ is continuously changing in magnitude and direction both.

From Kepler's law, $T^2\propto R^3$

So, $\frac{T_1}{T_2}=(\frac{R_1}{R_2}{)}^{\frac{3}{2}}$

$R_2=(\frac{T_2}{T_1}{)}^{\frac{2}{3}}\times R_1=(\frac{24}{6}{)}^{\frac{2}{3}}\times 1.28\times {10}^{4}km$

$=3.22\times {10}^{4}km$

Orbital velocity of $S_1$ is

$v_1=\frac{2\pi R_1}{T_1}=\frac{2\pi \times 1.28\times {10}^4}{4}$

$=0.64\pi \times {10}^{4}km$

Orbital velocity of $S_2$ is

$v_2=\frac{2\pi R_2}{T_2}=\frac{2\pi \times 1.28\times {10}^4}{4}$

$=0.64\pi \times {10}^{4}km$

At $t=12$ h the two satellites are closest to each other and after every 24 h they come at the same position relative to each other. It is clear that direction of $v_2$ w.r.t. $v_1$ is changing continuously in both magnitude and direction.

Angular velocity of $S_{2}w.r.t.s_1$ at $t=12h$ is

$\omega =v_1+v_2/R_2-R_1=0.468\pi rad{s}^{-1}$