Understanding Planetary Motion and Kepler's Law

The question asks about the change in the time period of revolution for a planet when its distance from the sun changes. This problem can be solved using Kepler's Third Law of Planetary Motion.

Kepler's Third Law of Planetary Motion

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 (\(a\)) of its orbit. For a circular orbit, the semi-major axis is simply the radius or the distance from the sun.

Mathematically, this can be written as:

\(T^2 \propto a^3\)

Or, introducing a constant of proportionality, \(k\):

\(T^2 = k a^3\)

where \(k\) is a constant that depends on the mass of the central body (in this case, the sun).

Applying Kepler's Third Law to the Problem

We are given two scenarios for the planet's orbit around the sun:

1. Initial State: The planet is at a distance '\(a\)' from the sun, and its time period of revolution is '\(T\)' years.
2. New State: The planet suddenly comes to a distance '\(\frac{a}{2}\)' closer to the sun. This implies the new distance from the sun is \(a - \frac{a}{2} = \frac{a}{2}\). Let the new time period of revolution be \(T'\).

Using Kepler's Third Law for the initial state:

\(T^2 = k a^3 \quad \cdots (1)\)

Using Kepler's Third Law for the new state, where the new distance is \(\frac{a}{2}\) and the new period is \(T'\):

\((T')^2 = k \left(\frac{a}{2}\right)^3\)

\((T')^2 = k \frac{a^3}{8} \quad \cdots (2)\)

Calculating the New Time Period

To find the relationship between \(T'\) and \(T\), we can divide equation (2) by equation (1):

\(\frac{(T')^2}{T^2} = \frac{k \frac{a^3}{8}}{k a^3}\)

The constant \(k\) cancels out, and \(a^3\) also cancels out:

\(\frac{(T')^2}{T^2} = \frac{\frac{1}{8}}{1}\)

\(\frac{(T')^2}{T^2} = \frac{1}{8}\)

Now, take the square root of both sides to find the ratio of the periods:

\(\sqrt{\frac{(T')^2}{T^2}} = \sqrt{\frac{1}{8}}\)

\(\frac{T'}{T} = \frac{1}{\sqrt{8}}\)

Simplify the term \(\sqrt{8}\):

\(\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}\)

Substitute this back into the equation for the ratio of periods:

\(\frac{T'}{T} = \frac{1}{2\sqrt{2}}\)

Finally, solve for the new time period \(T'\):

\(T' = T \times \frac{1}{2\sqrt{2}}\)

\(T' = \frac{T}{2\sqrt{2}}\)

So, the new time period of revolution is \(\frac{T}{2\sqrt{2}}\) years.

Conclusion

When the planet's distance from the sun decreases from '\(a\)' to '\(\frac{a}{2}\)', its time period of revolution changes from '\(T\)' to '\(\frac{T}{2\sqrt{2}}\)'. This result is a direct application of Kepler's Third Law.

Revision Table: Kepler's Laws Summary

Additional Information: Factors Affecting Orbital Period

While our calculation focused on the change in distance based on Kepler's Third Law, it's important to note other factors in celestial mechanics:

• Mass of the Central Body (Sun): The constant \(k\) in Kepler's Third Law (\(T^2 = ka^3\)) is equal to \(\frac{4\pi^2}{GM}\), where \(G\) is the gravitational constant and \(M\) is the mass of the central body (sun). If the mass of the sun were to change, the orbital periods of all planets at a given distance would also change.
• Mass of the Planet: For orbits around a much larger body like the sun, the planet's mass is negligible compared to the sun's mass and does not significantly affect the orbital period according to this simplified form of Kepler's Law. However, in a more precise calculation considering the two-body problem, the constant depends on the sum of the masses of the two bodies (\(M_{sun} + M_{planet}\)).
• Orbital Shape: Kepler's Third Law applies to both circular and elliptical orbits, where '\(a\)' represents the semi-major axis. Our problem implicitly treats the orbit as changing from one state (distance \(a\)) to another (distance \(a/2\)). For perfectly circular orbits, distance is the radius, which is equal to the semi-major axis.