Question

Two satellites $S_1$ and $S_2$ revolve round a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 hour and 8 hours respectively. The radius of the orbit of $S_1={10}^4$km When $S_2$ is closest to $S_1$ find speed of $S_2$ relative to $S_2$ and the angular speed of $S_1$ actually observed by an astronaut at $S_1$.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$\because T^2\propto r^3$

Given $r_1={10}^4$km = ${10}^7$m,$T_1$ = 1 hour and $T_2$ = 8 hours.

Therefore, $r_2^3=({10}^7{)}^3\times (\frac{8}{1}{)}^2=64\times {10}^{21}$

or $r_2=4\times {10}^{7}m=4\times {10}^{4}km$

(a) The orbital speeds of satellites $S_1$ and $S_2$ respectively are

$v_1=\frac{2\pi r_1}{T_1}=\frac{2\pi \times {10}^4}{1}=2\pi \times {10}^{4}km/h$ our

And $v_2=\frac{2\pi r_2}{T_2}=\frac{2\pi \times 4\times {10}^4}{8}=\pi \times {10}^4$km/h our

The magnitude the speed of $S_2$ relative to

$S_1$ is $|v_2-v_1|=\pi \times {10}^4=3.14\times {10}^4$ km/h our

The angular speed of $S_2$ relative to $S_1$ is

$\omega =\frac{v_2-v_1}{(r_2-r_1)}=-\frac{3.14\times {10}^4}{(4\times {10}^4-{10}^4)}=-\frac{3.14}{3}$

= $-\frac{3.14}{3}=\frac{3.14}{3}=-\frac{1}{60}\times \frac{1}{60}$

= -2.91 $\times {10}^{-4}$ rad $s^{-1}$

The negative sign indicates that the sense of rotation of opposite to the sense of revolution of the satellites.