Question

The 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). $8T$
(C). $\dfrac{T}{4}$
(D). $\dfrac{T}{8}$

Answer 1

Hint: According to Kepler’s third law of planetary motion, the square of the time period of a body in a circular orbit around another body is directly proportional to the cube of the distance from it, or the radius of orbit.
Formula used:
${{T}^{2}}\propto {{R}^{3}}$
Where $T$ is the time period of revolution and $R$ is the radius of the circular orbit.

Complete step by step answer:
This problem can be solved by making use of Kepler’s third law of planetary motion. Kepler’s third law states that the square of the time period of revolution of a body around another body is directly proportional to the cube of the radius of the circular orbit.
Hence,
${{T}^{2}}\propto {{R}^{3}}$ --(1)
Where $T$is the time period of revolution and $R$is the radius of the circular orbit.
Therefore, let us analyze the given question.
We are given that the period of a satellite in a circular orbit of radius $R$is $T$and we have to find out the period of another satellite in a circular orbit of radius $4R$.

Therefore let the radius of orbit and time period of the first satellite be ${{R}_{1}}$ and ${{T}_{1}}$ respectively.
Let the radius of orbit and time period of the second satellite be ${{R}_{2}}$ and ${{T}_{2}}$ respectively.
Therefore according to the question,
${{R}_{1}}=R$
${{T}_{1}}=T$
${{R}_{2}}=4R$
Therefore, using (1), we get,
$\dfrac{{{T}_{2}}^{2}}{{{T}_{1}}^{2}}=\dfrac{{{R}_{2}}^{3}}{{{R}_{1}}^{3}}$
Now, plugging in the values, we get,
$\dfrac{{{T}_{2}}^{2}}{{{T}^{2}}}=\dfrac{{{\left( 4R \right)}^{3}}}{{{\left( R \right)}^{3}}}$
$\therefore \dfrac{{{T}_{2}}^{2}}{{{T}^{2}}}={{\left( \dfrac{4R}{R} \right)}^{3}}={{4}^{3}}=64$
$\therefore {{\left( \dfrac{{{T}_{2}}}{T} \right)}^{2}}=64$
Square rooting both sides, we get,
$\sqrt{{{\left( \dfrac{{{T}_{2}}}{T} \right)}^{2}}}=\sqrt{64}$
$\therefore \dfrac{{{T}_{2}}}{T}=8$
$\therefore {{T}_{2}}=8T$
Hence, the required time period of the second satellite is $8T$. Therefore, the correct option is B) $8T$.

Note: In questions such as these, where the final answer is in the form of a ratio, it is better to find a relation between the required quantities and then use this ratio and consider all other variables which will anyway get cancelled at the end in the answer as constants that do not factor in the proportionality relation. For example, in this problem we found out a relation between $T$ and $R$. This method saves a lot of time, especially in competitive exams.
Actually, planetary orbits are elliptical in nature. However, the orbits can be considered as circular for all practical purposes and ease of calculations. Thus, the student must not get confused upon noticing in this topic that the problems are solved considering the orbits to be circular.