Question

Two satellites $S_1$ and $S_2$ revolve around a planet in coplanar circular orbits in the same sense. Their periods of revolutions are $1h$ and $8h$, respectively. The radius of the orbit of $S_1$ is $10^{4}km$. With reference to the above situation, match the Column I (quantities) with Column II (approximate values) and select the correct answer from the codes given below.

Column I		Column II
A	Speed of $S_2$ in $km{h}^{-1}$	1	$\pi /3$
B	Speed of $S_1$ in $km{h}^{-1}$	2	$2\pi \times 10^4$
C	Velocity of $S_2$ relative to $S_1$ when $S_2$ is closest to $S_1$ in $km{h}^{-1}$	3	$\pi \times 10^4$
D	Angular speed of $S_2$ as observed by an astronaut in$S_1$ when $S_2$ is closest to $S_1$in radh ${}^{-1}$	4	$-\pi \times 10^4$

• A. $A-(3),B-(2),C-(4),D-(1)$
• B. $A-(2),B-(3),C-(4),D-(1)$
• C. $A-(2),B-(3),C-(1),D-(4)$
• D. $A-(4),B-(2),C-(1),D-(3)$

Answer 1

Answer: (A)

Let the mass of the planet be $M$, that of $S_1$ be $m_1$ and of $S_2$ be $m_2$. Let the radius of the orbit of $S_1$ be $R_1$ $\left(=10^{4}km\right)$ and of $S_2$ be $R_2$. Let $v_1$ and $v_2$ be the linear speeds of $S_1$ and $S_2$ with respect to the planet. The figure shows the situation.

If the period of revolutions of satellites $S_1$ and $S_2$ are $T_1(1h)$ and $T_2(8h)$, respectively.
As the square of the time period is proportional to the cube of the radius,
${\left(\frac{R_2}{R_1}\right)}^3={\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{8h}{1h}\right)}^2=64$
$\Rightarrow \frac{R_2}{R_1}=4\Rightarrow R_2=4R_1=4\times 10^{4}km$
Now, the time-period of $S_1$ is $1h$. So, $\frac{2\pi R_1}{v_1}=1$
$\Rightarrow$ Speed of $S_1,v_1=\frac{2\pi R_1}{1}=2\pi \times 10^{4}km{h}^{-1}\dots$ (i)
Similarly, speed of
$S_2,v_2=\frac{2\pi R_2}{8}=\pi \times 10^{4}km{h}^{-1}$ ....(ii)
At the closest separation, they are moving in the same direction. Hence, the velocity of $S_2$ with respect to $S_1$ is
$v_2-v_1=\pi \times 10^{4}km{h}^{-1}-2\pi \times 10^{4}km{h}^{-1}$
$=-\pi \times 10^{4}km{h}^{-1}$ ....(iii)
As seen from $S_1$, the satellite $S_2$ is at a distance $R_2-R_1=3\times 10^{4}km$ at the closest separation. Also, it
is moving at $\pi \times 10^{4}km{h}^{-1}$ in a direction perpendicular to the line joining them. Thus, the angular speed of $S_2$ as observed by $S_1$ is
$\omega =\frac{v}{r}=\frac{\pi \times 10^{4}km{h}^{-1}}{3\times 10^{4}km}=\frac{\pi }{3}rad{h}^{-1}$
Hence, $A\to 3,B\to 2,C\to 4$ and $D\to 1$.