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 $1 \,h$ and $8 \,h$, 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 $kmh ^{-1}$	1	$\pi / 3$
B	Speed of $S_{1}$ in $kmh ^{-1}$	2	$2 \pi \times 10^{4}$
C	Velocity of $S_{2}$ relative to $S_{1}$ when $S_{2}$ is closest to $S_{1}$ in $kmh ^{-1}$	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 $T_{1}(1 h )$ and $T_{2}(8 h )$, respectively.
As the square of the time period is proportional to the cube of the radius,
$\left(\frac{R_{2}}{R_{1}}\right)^{3}=\left(\frac{T_{2}}{T_{1}}\right)^{2}=\left(\frac{8 h }{1 h }\right)^{2}=64$
$\Rightarrow \frac{R_{2}}{R_{1}}=4 \Rightarrow R_{2}=4 R_{1}=4 \times 10^{4} km$
Now, the time-period of $S_{1}$ is $1 h$. So, $\frac{2 \pi R_{1}}{v_{1}}=1$
$\Rightarrow$ Speed of $S_{1}, v_{1}=\frac{2 \pi R_{1}}{1}=2 \pi \times 10^{4} km h ^{-1} \ldots$ (i)
Similarly, speed of
$S_{2}, v_{2}=\frac{2 \pi R_{2}}{8}=\pi \times 10^{4} kmh ^{-1}$ ....(ii)
At the closest separation, they are moving in the same direction. Hence, the velocity of $S_{2}$ with respect to $S_{1}$ is
$v_{2}-v_{1} =\pi \times 10^{4} km h ^{-1}-2 \pi \times 10^{4} kmh ^{-1}$
$=-\pi \times 10^{4} km h ^{-1}$ ....(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, $A \rightarrow 3, B \rightarrow 2, C \rightarrow 4$ and $D \rightarrow 1$.