Understanding Planetary Orbits and Periods

This question asks us to find the ratio of the orbital periods of two planets revolving around the Sun. We are given the radii of their circular orbits. This problem can be solved using Kepler's Third Law of planetary motion, which describes the relationship between the orbital period of a planet and the size of its orbit.

Applying Kepler's Third Law for Orbital Periods

Kepler's Third Law states that the square of the orbital period (\(T\)) of a planet is directly proportional to the cube of the semi-major axis of its orbit (\(a\)). For planets in circular orbits around the Sun, the semi-major axis is simply the radius of the orbit (\(R\)). Mathematically, this law can be written as:

\[T^2 \propto R^3\]

This proportionality can be written as an equation:

\[T^2 = k R^3\]

where \(k\) is a constant that is the same for all planets orbiting the same central body (in this case, the Sun). The constant \(k\) depends on the mass of the Sun.

Calculating the Period Ratio for Two Planets

Let \(T_1\) and \(R_1\) be the period and radius of orbit for the first planet, and \(T_2\) and \(R_2\) be the period and radius of orbit for the second planet. According to Kepler's Third Law:

• For Planet 1: \(T_1^2 = k R_1^3\)
• For Planet 2: \(T_2^2 = k R_2^3\)

We are given the radii of the orbits:

• \(R_1 = R\)
• \(R_2 = 4R\)

To find the ratio of their periods, \(T_1/T_2\), we can take the ratio of the squared periods:

\[\frac{T_1^2}{T_2^2} = \frac{k R_1^3}{k R_2^3}\]

The constant \(k\) cancels out:

\[\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}\]

Now, substitute the given values for \(R_1\) and \(R_2\):

\[\frac{T_1^2}{T_2^2} = \frac{(R)^3}{(4R)^3}\]

Calculate the cube of the radii:

\[\frac{T_1^2}{T_2^2} = \frac{R^3}{4^3 \times R^3}\]

\[\frac{T_1^2}{T_2^2} = \frac{R^3}{64 R^3}\]

The \(R^3\) terms cancel out:

\[\frac{T_1^2}{T_2^2} = \frac{1}{64}\]

We want to find the ratio \(T_1/T_2\), so we take the square root of both sides of the equation:

\[\sqrt{\frac{T_1^2}{T_2^2}} = \sqrt{\frac{1}{64}}\]

\[\frac{T_1}{T_2} = \frac{\sqrt{1}}{\sqrt{64}}\]

\[\frac{T_1}{T_2} = \frac{1}{8}\]

So, the ratio of the periods \(T_1/T_2\) is \(1/8\).

Summary of Calculation Steps

Here are the steps we followed to find the ratio of the orbital periods:

1. Identify the relationship between orbital period and radius using Kepler's Third Law: \(T^2 \propto R^3\).
2. Set up the equations for the two planets: \(T_1^2 = k R_1^3\) and \(T_2^2 = k R_2^3\).
3. Form the ratio of the squared periods: \(\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}\).
4. Substitute the given values \(R_1 = R\) and \(R_2 = 4R\).
5. Simplify the expression: \(\frac{T_1^2}{T_2^2} = \frac{R^3}{(4R)^3} = \frac{R^3}{64R^3} = \frac{1}{64}\).
6. Take the square root to find the ratio of the periods: \(\frac{T_1}{T_2} = \sqrt{\frac{1}{64}} = \frac{1}{8}\).

Conclusion on Orbital Period Ratio

The ratio of the periods of the two planets, \(T_1/T_2\), is \(1/8\). This result aligns with one of the provided options.

Revision Table: Kepler's Laws and Orbital Periods

Additional Information: Factors Affecting Orbital Periods

While Kepler's Third Law in the form \(T^2 = k R^3\) is very useful for comparing periods of objects orbiting the same central body, the constant \(k\) itself depends on the mass of the central body. The more massive the central body, the shorter the period for a given orbital radius. The full formula for the period \(T\) of a circular orbit is:

\[T = 2\pi \sqrt{\frac{R^3}{GM}}\]

Where:

• \(R\) is the radius of the orbit
• \(G\) is the gravitational constant
• \(M\) is the mass of the central body (the Sun, in this case)

From this formula, we can see that \(T^2 = \frac{4\pi^2}{GM} R^3\). Comparing this to \(T^2 = k R^3\), we see that \(k = \frac{4\pi^2}{GM}\). Since both planets orbit the same Sun, \(M\) is the same for both, and thus \(k\) is the same, justifying the method used to find the ratio.

This law applies not only to planets orbiting the Sun but also to moons orbiting planets or satellites orbiting Earth, provided they are orbiting the same central mass.