Kepler's Third Law, also known as the Law of Periods, is a fundamental principle in astronomy that describes the relationship between the orbital period of a planet and its average distance from the sun. This law applies to any system where one body orbits another under the influence of gravity.
Formula
Kepler's Third Law is expressed mathematically as:
\[ \frac{T^2}{a^3} = \frac{4\pi^2}{GM} \]
Where:
\( T \) is the orbital period of the planet (in seconds)
\( a \) is the semi-major axis of the orbit (in meters)
\( G \) is the gravitational constant (\( 6.67430 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2} \))
\( M \) is the mass of the central body (in kilograms)
Calculation Steps
Let's calculate the orbital period of Earth around the Sun:
Given:
Semi-major axis of Earth's orbit (\( a \)) = \( 1.496 \times 10^{11} \text{ m} \)
Mass of the Sun (\( M \)) = \( 1.989 \times 10^{30} \text{ kg} \)
Rearrange Kepler's Third Law to solve for \( T \):
\[ T = \sqrt{\frac{4\pi^2a^3}{GM}} \]
Substitute the known values:
\[ T = \sqrt{\frac{4\pi^2(1.496 \times 10^{11} \text{ m})^3}{(6.67430 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2})(1.989 \times 10^{30} \text{ kg})}} \]
Perform the calculation:
\[ T \approx 3.156 \times 10^7 \text{ s} \]
Convert to years:
\[ T \approx 1.000 \text{ year} \]
Example and Visual Representation
Let's visualize Kepler's Third Law for the inner planets of our solar system:
This diagram illustrates:
The Sun at the center (yellow)
The orbits of Mercury (red), Venus (green), Earth (blue), and Mars (orange)
As the orbital distance increases, the orbital period increases more rapidly (proportional to a³/²)
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