The time period \(\mathrm{T}\) of the moon of planet Mars \((\mathrm{Mm})\) is related to its orbital radius \(\mathrm{R}\) as \((\mathrm{G}=\) Gravitational constant \()\) (A) $\mathrm{T}^{2}=\left[\left(4 \pi^{2} \mathrm{R}^{3}\right) /(\mathrm{GMm})\right]$ (B) $\mathrm{T}^{2}=\left[\left(4 \pi^{2} \mathrm{GR}^{3}\right) /(\mathrm{Mm})\right]$ (C) \(T^{2}=\left[\left(2 \pi R^{2} G\right) /(M m)\right]\) (D) \(\mathrm{T}^{2}=4 \pi \mathrm{Mm} \mathrm{GR}^{2}\)

Since we are dealing with the motion of a moon orbiting a planet, we need to use two main principles: Kepler's Third Law of Planetary Motion and Newton's Law of Gravitation. Kepler's Third Law states that the square of the orbital period of an object in orbit is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, it can be represented as: \[T^2 \propto R^3\] Newton's Law of Gravitation states that the force between two objects is equal to the product of their masses divided by the square of the distance between their centers. The formula for gravitational force is: \[F = G \frac{Mm}{R^2}\]

In order to find the relationship between the time period (T) and the orbital radius (R), we need to combine Kepler's Third Law with Newton's Law of Gravitation. This can be done by equating the gravitational force to the centripetal force on the moon, which is given by: \[F_{c} = \frac{m v^2}{R}\] Now, we have: \[\frac{GMm}{R^2} = \frac{m v^2}{R}\] Notice that the mass of the moon (m) cancels out on both sides: \[GM = \frac{v^2 R}{R^2}\]

In order to find the relationship between T^2 and R, we need to first express the velocity (v) in terms of the time period (T). Since the circumference of the orbit is given by \(2\pi R\), the average orbital speed is: \[v = \frac{2\pi R}{T}\] Now, square the above equation to get the relationship between T^2 and R: \[\frac{4\pi^2R^2}{T^2} = v^2\] From Step 2, substitute the expression for v^2 in the equation: \[GM = \frac{4\pi^2R^2}{T^2}\cdot \frac{R}{R^2}\] Now, solve for T^2 to get the relationship between the time period and the orbital radius: \[T^2 = \frac{4\pi^2R^3}{GM}\] This equation shows the correct relationship between the time period (T) and the orbital radius (R). Comparing the given options, we see that: (A) \(T^2 = \frac{4\pi^2R^3}{GMm}\) (B) \(T^2 = \frac{4\pi^2GR^3}{Mm}\) (C) \(T^2 = \frac{2\pi R^2G}{Mm}\) (D) \(T^2 = 4\pi MmGR^2\) The closest option to our derived formula is option (A).