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The time period of a satellite in a circular orbit of radius $\mathrm{R}$ is $\mathrm{T},$ the period of another satellite in a circular orbit of radius $4 \mathrm{R}$ is
A.$4~\text{T}$
B.$\dfrac{\text{T}}{4}$
C.$8~\text{T}$
D.$\dfrac{\text{T}}{8}$

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Last updated date: 14th Jun 2024
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Answer
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Hint: Kepler's law gives the relationship between the time period (T) and the radius (r) of the circular path. The square of the time period is according to Kepler, directly proportional to the cube of the circular path radius.

Formula used:
$T^{2}=k r^{3}$

Complete Step-by-Step solution:
Objects rotating under the influence of a larger planet or star in circular orbits have a constant velocity. You may know that when a body is in a constant-speed circular motion, its motion is called a uniform circular motion. A periodic motion is a uniform circular motion.

Therefore, its motion has a time period. Here the satellite has a revolutionary time period.
The amount of time an object (satellite) rotates in circular orbits depends on the radius of the circular path.

By Kepler's law, the relationship between the time period (T) and the radius (r) of the circular path is given. The square of the time period is according to Kepler, directly proportional to the cube of the circular path radius.
$T^{2} \propto r^{3}$

Remove the proportionality sign and add a proportionality constant $\mathrm{k}$.
Therefore, we get, $T^{2}=k r^{3}$.
We can write the above equation as $\dfrac{T^{2}}{r^{3}}=k$
Let the time period of satellite 1 be $\mathrm{T}$ and let the radius be $\mathrm{r}$.
Let the time period of satellite 2 be $T^{\prime}$ and its radius be $r$.

Therefore, with respect to statement 1 we get:
$\dfrac{{{T}^{2}}}{{{r}^{3}}}=\dfrac{{{\left( {{T}^{\prime }} \right)}^{2}}}{{{\left( {{r}^{\prime }} \right)}^{3}}}$- (I)
It is given that $T=T, r=R$ and $r^{\prime}=4 R$.

 Substitute these values in equation (1) $\Rightarrow \dfrac{T^{2}}{R^{3}}=\dfrac{\left(T^{\prime}\right)^{2}}{(4 R)^{3}}$
$\Rightarrow \dfrac{T^{2}}{R^{3}}=\dfrac{\left(T^{\prime}\right)^{2}}{4^{3} R^{3}}$
$\Rightarrow T^{2}=\dfrac{\left(T^{\prime}\right)^{2}}{4^{3}}$
$\Rightarrow 4^{3} T^{2}=\left(T^{\prime}\right)^{2}$

Take square roots on both sides.
Therefore, $T^{\prime}=\sqrt{4^{3} T^{2}}=8 T$

Therefore, the time period of the satellite in a circular orbit of radius $4 \mathrm{R}$ is equal to $4 \mathrm{~T}$.

Hence, the correct option is $\mathrm{C}$.

Note:
The time period square of a body revolving in a circular orbit is directly proportional to the radius cube of the circular orbit. The orbit of circles. Satellites or any other celestial body rotating in circular paths of the same radius have the same duration of time.