Kepler’s Laws of Planetary Motion

Expert reviewed • 22 November 2024 • 10 minute read

What are Kepler’s Laws of Planetary Motion?

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.

The First Law: The Elliptical Orbit

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.

Key points:

An ellipse is a closed curve where the sum of the distances from any point on the curve to two fixed points (foci) is constant.
The Sun’s position at one focus means that a planet’s distance from the Sun varies throughout its orbit.
The point of closest approach to the Sun is called perihelion, while the farthest point is aphelion.

Eccentricity

The shape of an elliptical orbit is characterised by its eccentricity $(e)$ , which ranges from $0$ to $1$ :

$e=0$ : A perfect circle (a special case of an ellipse)
$0<e<1$ $0 < e < 1$ : An ellipse
- $e$ approaches $1$ : The ellipse becomes more elongated

For example, Earth’s orbit has an eccentricity of approximately 0.0167, making it nearly circular.

The Second Law: Equal Areas in Equal Times

Kepler’s Second Law states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

Key implications:

Planets move faster when they are closer to the Sun (near perihelion) and slower when they are farther away (near aphelion).
This law is a consequence of the conservation of angular momentum.

The Third Law: The Harmonic Law

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 $(T)$ is directly proportional to the cube of the semi-major axis $(a)$ of the orbit:

T_2\propto a_3\\\text{or}\\T_2=ka_3

where,

$k$ is a constant of proportionality.

Application to Circular Orbits

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:

\frac{r^3}{T^2} = \frac{GM}{4\pi^2}

where:

$r$ is the orbital radius (m)
$T$ is the orbital period (s)
$G$ is the universal gravitational constant ( $6.67 × 10^{−11} \;\text{Nm²/kg²}$ )
$M$ is the mass of the central body (kg)

This equation allows us to relate the orbital period to the radius and the mass of the central body.

Problem-Solving Examples:

Let’s apply Kepler’s Laws to solve some practical problems.

Practice Question 1

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:

Orbital period, $T = 365.25 \;\text{days} = 3.156 × 10^7 s$
Mass of the Sun, $M = 1.989 × 10^{30}\; kg$
$G = 6.67 × 10^{−11} \;Nm²/kg²$

Using the equation $\frac{r^3}{T^2} = \frac{GM}{4\pi^2}$ , we can solve for $r$ :

r^3 = \frac{GMT^2}{4\pi^2}\\r = \sqrt[3]{\frac{GMT^2}{4\pi^2}}\\r = \sqrt[3]{\frac{(6.67 \times 10^{-11})(1.989 \times 10^{30})(3.156 \times 10^7)^2}{4\pi^2}}\\r ≈ 1.496 × 10^{11}\: m

Practice Question 2

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:

Orbital radius, $r = 7,000 \;km = 7 × 10^6\; m$
Mass of Earth, $M = 5.97 × 10^{24}\; kg$
$G = 6.67 × 10^{−11} \;Nm²/kg²$

We’ll use the equation $T^2 = \frac{4\pi^2r^3}{GM}$ :

T^2 = \frac{4\pi^2(7 \times 10^6)^3}{(6.67 \times 10^{-11})(5.97 \times 10^{24})}\\T^2 = 5.805 × 10 ^7 \\T = \sqrt{5.805 \times 10^7} \\= 7,619 s ≈ 2.12 \:\text{hours}

Practice Question 3

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 $\frac{r^3}{T^2}$ is constant for all moons orbiting Jupiter:

\frac{r_1^3}{T_1^2} = \frac{r_2^3}{T_2^2}

Where subscript 1 refers to Io and 2 refers to Europa.

\frac{(421,700)^3}{(1.77)^2} = \frac{(671,100)^3}{T_2^2}

Solving for $T_2$ :

T_2^2 = \frac{(671,100)^3 \times (1.77)^2}{(421,700)^3} \\T_2 = \sqrt{7.134} \\= 2.67 \:\text{days}

Key Takeaways

Kepler’s First Law describes the shape of planetary orbits as ellipses.
Kepler’s Second Law explains the varying speed of planets in their orbits.
Kepler’s Third Law relates orbital period to orbital size, applicable to any system of orbiting bodies.
These laws form the foundation for understanding orbital mechanics and have applications beyond our solar system.

Understanding Orbital Velocity

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Gravitational Potential Energy and Energy Changes in Space

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