Question
Answers

The time period of an earth satellite in a circular orbit of radius $R$ is $2$ days and its orbital velocity is ${{v}_{0}}$.If time period of another satellite in a circular orbit is 16 days then
A. Its radius of orbit is \[4R\] and orbital velocity is ${{v}_{0}}$.
B. Its radius of orbit is $4R$ and orbital velocity is $\dfrac{{{v}_{0}}}{2}$
C. Its radius of orbit is \[2R\] and orbital velocity is ${{v}_{0}}$
D. Its radius of orbit is \[2R\] and orbital velocity is $\dfrac{{{v}_{0}}}{2}$

Answer Verified Verified
Hint: Kepler’s third law relates the time period of a planetary object with the radius of orbit. Using this Kepler’s third law calculates the radius of the orbit of the 2nd satellite. Using this radius calculate the orbital velocity of the satellite.

Formula used:
The square time period $T$ of a satellite is directly proportional to the cube of the radius $R$ of the orbit and is given by ${{T}^{2}}\alpha {{R}^{3}}$.
For two different objects. Of time periods ${{T}_{1}}\text{ and }{{T}_{2}}$ of radius ${{R}_{1}}\text{ and }{{R}_{2}}$ it is given by
$\dfrac{{{T}_{1}}^{2}}{{{T}_{2}}^{2}}=\dfrac{R_{1}^{3}}{R_{2}^{3}}$
The orbital velocity of the satellite of radius $R$of time period $T$is
$v=\left( \dfrac{\text{circumference of the orbit}}{\text{time period}} \right)$
So, $v=\dfrac{2\pi R}{T}$

Complete answer:
According to Kepler’s third law of planetary motion, the square of time period of revolution of any object around the sun or the earth is directly proportional to the cube of the radius i.e.
${{T}^{2}}\alpha {{R}^{3}}$
For two different objects. Of time periods ${{T}_{1}}\text{ and }{{T}_{2}}$ of radius ${{R}_{1}}\text{ and }{{R}_{2}}$ it is given by
$\dfrac{{{T}_{1}}^{2}}{{{T}_{2}}^{2}}=\dfrac{R_{1}^{3}}{R_{2}^{3}}$
For the first satellite. ${{R}_{1}}=R$,${{T}_{1}}=2\text{ days}$
For the second satellite. ${{T}_{2}}=16\text{ days}$
So
 $\begin{align}
  & \dfrac{{{T}_{1}}^{2}}{{{T}_{2}}^{2}}=\dfrac{R_{1}^{3}}{R_{2}^{3}} \\
 & \Rightarrow R_{2}^{3}=\dfrac{{{T}_{2}}^{2}}{{{T}_{1}}^{2}}R_{1}^{3} \\
\end{align}$
Putting the values of ${{T}_{1}},{{T}_{2}}\text{ and }{{R}_{2}}$we get.
$\begin{align}
  & R_{2}^{3}={{\left( \dfrac{16}{2} \right)}^{2}}{{R}^{3}}=64{{R}^{3}} \\
 & \Rightarrow {{R}_{2}}=\sqrt[3]{64{{R}^{3}}}=4R \\
\end{align}$
So the radius of second satellite is $4R$
Now the orbital velocity of this satellite is given by
$\begin{align}
  & v=\left( \dfrac{\text{circumference of the orbit}}{\text{time period}} \right)=\dfrac{2\pi {{R}_{2}}}{{{T}_{2}}} \\
 & \Rightarrow v=\dfrac{2\pi (4R)}{16}=\dfrac{2\pi R}{4}=\dfrac{1}{2}{{v}_{0}} \\
\end{align}$
Where ${{v}_{0}}=\dfrac{2\pi R}{2}$ is the orbital velocity of the first satellite.
So the radius of 2nd satellite is $4R$ and its orbital velocity will be $\dfrac{{{v}_{0}}}{2}$

So, the correct answer is “Option B”.

Additional Information:
Kepler’s first law: Each planet revolves around the sun in an elliptical orbit with the sun situated at one of its foci.
Kepler’s second law(law of areas): The area vector drawn from the sun to a planet sweeps out an equal area in an equal interval of time.
Kepler’s third law: the square of period of revolution of a planet around the sun is proportional to the cube of the semi major axis of its elliptical orbit.

Note:
Note that Kepler’s law is valid for any object revolving around any object. It is also valid for circular orbit. A planet revolves around the sun due to the centripetal force of the sun which is directed towards the sun. You can also calculate the angular momentum of revolution if you know orbital velocity and radius by the formula
$L=mvr$