Deriving Kepler's Laws from Newton

This page outlines how Kepler's three laws emerge mathematically from Newton's law of gravitation and Newton's second law of motion (F = ma).

Setup: The Two-Body Problem

Consider a planet of mass m orbiting a star of mass M, where M >> m. The gravitational force on the planet is:

F = −G·M·m / r² (directed toward the star)

We work in polar coordinates (r, θ) centered on the star. Newton's second law in the radial and tangential directions gives two coupled differential equations.

Deriving the Second Law (Easiest)

Since gravity is a central force (purely radial, no tangential component), there is no torque about the star. Therefore angular momentum is conserved:

L = m·r²·(dθ/dt) = constant

The area swept out per unit time is dA/dt = ½r²(dθ/dt) = L/(2m) = constant. This is Kepler's Second Law.

Deriving the First Law

Substituting u = 1/r and using the angular momentum relation, the radial equation of motion transforms into:

d²u/dθ² + u = G·M·m² / L²

This is a simple harmonic oscillator equation with a constant offset. Its general solution is:

u = (G·M·m² / L²)(1 + e·cos(θ))

Converting back to r:

r = (L² / G·M·m²) / (1 + e·cos(θ))

This is the equation of a conic section with eccentricity e. For 0 ≤ e < 1, it's an ellipse. This is Kepler's First Law.

Deriving the Third Law

The area of an ellipse is A = π·a·b. Since dA/dt = L/(2m), the period is T = 2m·A/L = 2π·m·a·b/L. Using b² = a²(1−e²) and the relation L² = G·M·m²·a·(1−e²), you can show after algebra:

T² = (4π² / G·M) · a³

This is Kepler's Third Law, with the proportionality constant revealed to depend on the central mass M.

Significance

This derivation was one of the great triumphs of Newton's Principia. It showed that a single force law explains all of Kepler's empirical rules. It also revealed that the two-body problem is exactly solvable, while real solar systems with multiple planets require the theory of perturbations.