Question

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) \(\mathrm{T}^{2}=4 \pi \mathrm{Mm} \mathrm{GR}^{2}\)

Short answer

The short answer is: (A) \(T^2 = \frac{4\pi^2R^3}{GMm}\).

Step-by-step solution

Identify the main rules and formulas

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: \[F = G \frac{Mm}{R^2}\]

Combine Kepler's Third Law with Newton's Law

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: \[GM = \frac{v^2 R}{R^2}\]

Calculate the Square of the Time Period

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).

Key concepts

Newton's Law of Gravitation

Newton's Law of Gravitation is a cornerstone principle in physics. It explains how two masses attract each other with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between their centers. This universal law is crucial to understanding planetary motion and many other gravitational phenomena. The gravitational force can be mathematically described using the formula:
\[F = G \frac{Mm}{R^2}\]
Here:

• \(F\) is the gravitational force.
• \(G\) is the gravitational constant.
• \(M\) and \(m\) are the masses of the two objects involved (e.g., a planet and its moon).
• \(R\) is the distance between the centers of the two masses.

Understanding this equation helps explain why planets orbit stars in a predictable manner and provides insights into the forces keeping moons in orbit around their planets.

Orbital Radius

The orbital radius is the average distance between an object in orbit (like a moon) and the body it is orbiting (like a planet). In the context of celestial mechanics, the orbital radius plays an integral role in determining various dynamic properties of the orbit.
Kepler's Third Law of Planetary Motion states that the square of the orbital period \(T^2\) is proportional to the cube of the semi-major axis of its orbit, commonly referred to as the 'orbital radius' for circular orbits:
\[T^2 \propto R^3\]
This law highlights the dependency of the orbit duration on its size:

• A larger orbital radius means a longer period.
• A smaller orbital radius results in a shorter orbital period.

This principle, blended with Newton's Law of Gravitation, helps us understand the precise motion of natural satellites and planets within their respective celestial systems.

Centripetal Force

Centripetal force is critical in keeping an object moving in a circular path. It acts perpendicular to the velocity of the object and is directed towards the center around which the object is moving. In the planetary motion context, it is this force that balances the gravitational pull keeping a moon in orbit around its planet.
This can be represented mathematically by:
\[F_{c} = \frac{mv^2}{R}\]
Where:

• \(F_{c}\) is the centripetal force.
• \(m\) is the mass of the orbiting object (e.g., a moon).
• \(v\) is the orbital velocity.
• \(R\) is the orbital radius.

By equating the gravitational pull with the centripetal force, it becomes possible to derive a relationship between the orbital velocity, radius, and period of the orbiting object. This interconnection is at the heart of understanding planetary motion and celestial dynamics.