Expert reviewed • 04 March 2025 • 10 minute read
In physics, few principles have had as profound an impact on our understanding of the cosmos as Kepler's Laws of Planetary Motion. These laws, formulated by Johannes Kepler in the early 17th century, provide a mathematical framework for describing the motion of planets around the Sun and, more broadly, any celestial body orbiting another due to gravitational attraction.
Kepler's First Law states that the orbit of each planet around the Sun is an ellipse, with the Sun located at one of the two foci of the ellipse.
The shape of an elliptical orbit is characterised by its eccentricity , which ranges from to :
For example, Earth's orbit has an eccentricity of approximately 0.0167, making it nearly circular.
Kepler's Second Law states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
Kepler's Third Law, also known as the Law of Harmonies, establishes a relationship between the orbital period of a planet and its average distance from the Sun.
Mathematically, it states that the square of the orbital period is directly proportional to the cube of the semi-major axis of the orbit:
where,
For circular orbits, which are a good approximation for many planetary and satellite orbits, the semi-major axis is equal to the radius of the orbit. In this case, we can derive a more specific form of Kepler's Third Law:
where:
This equation allows us to relate the orbital period to the radius and the mass of the central body.
Let's apply Kepler's Laws to solve some practical problems.
Problem: Earth's Orbital Radius:
Given that Earth's orbital period around the Sun is approximately 365.25 days, calculate Earth's average distance from the Sun.
Given:
Using the equation , we can solve for :
Problem: Satellite Orbits
A satellite orbits Earth in a circular orbit with a radius of 7,000 km (measured from Earth's centre). Calculate its orbital period.
Given:
We'll use the equation :
Problem: Comparing Orbital Periods
Jupiter's moon Io has an orbital radius of 421,700 km and an orbital period of 1.77 days. Another moon, Europa, has an orbital radius of 671,100 km. Calculate Europa's orbital period.
We can use the fact that is constant for all moons orbiting Jupiter:
Where subscript 1 refers to Io and 2 refers to Europa.
Solving for :