Question

Two satellites ${{S}_{1}}$ and ${{S}_{2}}$ revolve around a planet in coplanar circular orbits in the same sense. Their periods of revolution are $1\text{ hour}$ and $8\text{ hours}$ respectively. The radius of the orbit of ${{S}_{1}}$ is ${{10}^{4}}km$. When ${{S}_{1}}$ is closest to ${{S}_{2}}$, the angular speed of ${{S}_{2}}$ as observed by an astronaut on ${{S}_{1}}$ is
A. $\pi \text{ }rad/hr$
B. $\dfrac{7\pi }{8}\text{ }rad/hr$
C. $2\pi \text{ }rad/hr$
D. $\dfrac{\pi }{3}\text{ }rad/hr$

Answer 1

Hint: To solve this problem, we have to find out the relative velocity of ${{S}_{2}}$ with respect to ${{S}_{1}}$. The individual speeds of the satellites can be found out by finding out the radii of the orbits and the time periods of motion. The radii and time periods of motion are related by Kepler’s third law for the motion of planetary bodies.

Formula used:
For a body moving in a circular orbit of radius $R$ with time period $T$, the speed $v$ is given by
$v=\dfrac{2\pi R}{T}$
For a body moving with speed $v$ in a circular orbit of radius $R$, the angular velocity $\omega$ is given by,
$\omega =\dfrac{v}{R}$
According to Kepler’s third law of motion of planetary bodies,
${{T}^{2}}\propto {{R}^{3}}$
Where $T$ is the time period and $R$ is the radius of orbit.

Complete step by step answer:
In essence, to get the relative angular velocities between the satellites, we have to first find their individual speeds with respect to the ground and then using this find the relative velocity of ${{S}_{2}}$ with respect to ${{S}_{1}}$. Then by using the relative distance between them, we can get the relative angular velocity.
Now, to get the individual velocities, we can use the radii of orbit and time period of the satellites. To get the radii of orbit and time period of the satellites, we can use Kepler’s third law and use the information given in the question.
Hence, let us analyze the question.
Let the radius of orbit, speed and time period of ${{S}_{1}}$ be ${{R}_{1}},{{v}_{1}}$ and ${{T}_{1}}$ respectively.
Let the radius of orbit, speed and time period of ${{S}_{2}}$ be ${{R}_{2}},{{v}_{2}}$ and ${{T}_{2}}$ respectively.
According to Kepler’s third law of motion of planetary bodies,
${{T}^{2}}\propto {{R}^{3}}$ --(1)
Where $T$ is the time period and $R$ is the radius of orbit.
Hence, using (1), we get,
$\dfrac{{{T}_{2}}^{2}}{{{T}_{1}}^{2}}=\dfrac{{{R}_{2}}^{3}}{{{R}_{1}}^{3}}$
Now, according to the question, ${{T}_{2}}=8hr$, ${{T}_{1}}=1hr$, ${{R}_{1}}={{10}^{4}}km$. Putting this information in the above equation, we get,
$\dfrac{{{8}^{2}}}{{{1}^{2}}}=\dfrac{{{R}_{2}}^{3}}{{{\left( {{10}^{4}} \right)}^{3}}}$
$\therefore {{R}_{2}}^{3}=64\times {{\left( {{10}^{4}} \right)}^{3}}$
Cube rooting both sides we get,
$\sqrt[3]{{{R}_{2}}^{3}}=\sqrt[3]{64\times {{\left( {{10}^{4}} \right)}^{3}}}=4\times {{10}^{4}}km$
$\therefore {{R}_{2}}=4\times {{10}^{4}}km$ --(2)
For a body moving in a circular orbit of radius $R$ with time period $T$, the speed $v$ is given by
$v=\dfrac{2\pi R}{T}$ --(3)
Now, using (3) and the information given in the question, we get,
${{v}_{1}}=\dfrac{2\pi \times {{10}^{4}}}{1}=2\pi \times {{10}^{4}}km/hr$ --(4)
Now, using (2),(3) and the information given in the question, we get,
${{v}_{2}}=\dfrac{2\pi \times 4\times {{10}^{4}}}{8}=\pi \times {{10}^{4}}km/hr$ --(5)
Since, in the question, it is mentioned that the satellites go about in the same sense, therefore, the relative speed of ${{S}_{2}}$ with respect to ${{S}_{1}}$ will be $\left| {{v}_{2}}-{{v}_{1}} \right|$.
Therefore, using (4) and (5), we get,
${{v}_{21}}=\left| \pi \times {{10}^{4}}-\left( 2\pi \times {{10}^{-4}} \right) \right|=\left| -\pi \times {{10}^{4}} \right|=\pi \times {{10}^{4}}km/hr$ --(6)
This relative velocity is perpendicular to the line joining the two satellites, when they are the closest.
When the satellites are closest, the relative distance $R$ between them will be the difference of their radii of orbit. That is,
$R=\left| {{R}_{2}}-{{R}_{1}} \right|$
Using (2) and the information in the question, we get,
$R=\left| 4\times {{10}^{4}}-{{10}^{4}} \right|=3\times {{10}^{4}}km$ --(7)
For a body moving with speed $v$ in a circular orbit of radius $R$, the angular velocity $\omega$ is given by,
$\omega =\dfrac{v}{R}$ --(8)
Putting (6) and (7) in (8), we get the required relative angular velocity ${{\omega }_{21}}$ as
${{\omega }_{21}}=\dfrac{{{v}_{21}}}{R}=\dfrac{\pi \times {{10}^{4}}}{3\pi \times {{10}^{4}}}=\dfrac{\pi }{3}rad/hr$
Hence, the answer is D) $\dfrac{\pi }{3}rad/hr$.

Note: Students can make the mistake of directly relating the angular velocity with the time periods and finding out the individual angular velocities and then finding their difference to get the relative angular velocity. However, this is wrong since this method does not take into account that the distance is not the radii of the individual orbits but the difference between the radii of the individual orbits which gives the actual perpendicular distance between the satellites, which can then be used along with the relative speeds to find out the relative angular velocity required.