Question

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

Answer 1

Hint: The relation between the time period (T) and the radius (r) of the circular path is given by Kepler’s law. According to Kepler, the square of the time period is directly proportional to the cube of the radius of circular path.

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

Complete step-by-step answer:
Objects revolving in circular orbits under the influence of a larger planet or star, have a constant speed. You may know that when a body is in a circular motion with a constant speed, then its motion is called a uniform circular motion.
A uniform circular motion is a periodic motion. Therefore, it has a time period of its motion. Here, the satellite has a time period of revolution.
Time period of an object (satellite) revolving in circular orbits depends on the radius of that circular path. The relation between the time period (T) and the radius (r) of the circular path is given by Kepler’s law. According to Kepler, the square of the time period is directly proportional to the cube of the radius of circular path. We can write it as,
${{T}^{2}}\propto {{r}^{3}}$.
Remove the proportionality sign and add a proportionality constant k.
Therefore we get, ${{T}^{2}}=k{{r}^{3}}$.
We can write the above equation as $\dfrac{{{T}^{2}}}{{{r}^{3}}}=k$.
Since, k is a constant, $\dfrac{{{T}^{2}}}{{{r}^{3}}}$ is also constant. ……….(statement 1)
Now, we can solve the given question. We can use statement 1.
Let the time period of satellite 1 be T and let the radius be r.
Let the time period of satellite 2 be T’ and its radius be r’.
Therefore, with respect to statement 1 we get,
 $\dfrac{{{T}^{2}}}{{{r}^{3}}}=\dfrac{{{\left( T' \right)}^{2}}}{{{\left( r' \right)}^{3}}}$ ……….(1).
It is given that T=T, r=R and r’ = 4R. Substitute these values in equation (1)
 $\Rightarrow \dfrac{{{T}^{2}}}{{{R}^{3}}}=\dfrac{{{\left( T' \right)}^{2}}}{{{\left( 4R \right)}^{3}}}$.
Open the brackets.
$\Rightarrow \dfrac{{{T}^{2}}}{{{R}^{3}}}=\dfrac{{{\left( T' \right)}^{2}}}{{{4}^{3}}{{R}^{3}}}$
$\Rightarrow {{T}^{2}}=\dfrac{{{\left( T' \right)}^{2}}}{{{4}^{3}}}$
$\Rightarrow {{4}^{3}}{{T}^{2}}={{\left( T' \right)}^{2}}$
Take square roots on both sides.
Therefore, $T'=\sqrt{{{4}^{3}}{{T}^{2}}}=8T$
Therefore, the time period of the satellite in a circular orbit of radius 4R is equal to 4T.
Hence, the correct option is C.

Note: Points to remember.
(i) The square of the time period of a body revolving in a circular orbit is directly proportional to the cube of the radius of the circular orbit.
(ii) Satellites or any other celestial bodies revolving in circular paths with the same radius have the same time period.