The Gravitational Two-Body Problem

A complete derivation — from Newton's law of gravitation to Kepler's three empirical laws, via the orbit equation, conic sections, and conservation laws.

1. Introduction

The gravitational two-body problem asks: given two point masses interacting solely through Newtonian gravity, what are their trajectories? This is one of the few problems in classical mechanics that admits an exact, closed-form solution.

The key insight is a mathematical sleight of hand: by switching to centre-of-mass and relative coordinates, the coupled six-dimensional system separates into (i) a trivially solved centre-of-mass translation and (ii) the motion of a single fictitious particle of reduced mass orbiting a fixed centre. The relative-motion problem then has two conserved quantities — energy and angular momentum — that together determine the complete orbit.

2. Problem Setup

Consider two point masses $m_1$ and $m_2$ with position vectors $\mathbf{r}_1$ and $\mathbf{r}_2$ measured from an inertial origin. The gravitational force exerted by $m_2$ on $m_1$ is:

By Newton's third law, $\mathbf{F}_{21} = -\mathbf{F}_{12}$. Writing $\mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2$ for the relative separation, the equations of motion are:

3. Centre-of-Mass Reduction

Define the total mass $M$ and the centre-of-mass (CM) position $\mathbf{R}$:

Adding the equations of motion gives $M\ddot{\mathbf{R}} = 0$, so:

The CM moves at constant velocity — it is an inertial frame. Working in the CM frame ($\mathbf{R} = 0$) decouples the problem: we only need to solve for the relative coordinate $\mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2$.

4. The Reduced Mass and the One-Body Problem

Subtracting the scaled equations of motion yields:

where $\hat{\mathbf{r}} = \mathbf{r}/r$ and $r = |\mathbf{r}|$. This is Newton's equation for a single particle orbiting a fixed mass $M$. Introducing the reduced mass:

5. Conservation Laws

5.1 Angular Momentum

Because the force is purely radial (central), the torque is zero and $\mathbf{L}$ is conserved:

$\mathbf{L}$ is perpendicular to the plane of motion, so the orbit is confined to a fixed plane. In polar coordinates $(r, \theta)$:

5.2 Energy

The total mechanical energy is also conserved:

5.3 Effective Potential

Using (5.2) to eliminate $\dot{\theta}$, we can define the effective potential:

The first term is the centrifugal barrier (repulsive) and the second is the gravitational well (attractive). Bound orbits require $E < 0$; $E = 0$ corresponds to escape.

6. Deriving the Orbit Equation

To find $r(\theta)$ we use the Binet substitution $u \equiv 1/r$, giving:

where $\ell = L^2/(\mu^2 GM)$ is the semi-latus rectum. The general solution, choosing the periapsis as the reference direction, is:

This is the polar equation of a conic section with focus at the origin. The eccentricity $e$ is determined by the energy and angular momentum:

7. Classification of Orbits

The shape of the orbit is entirely determined by the eccentricity $e$:

8. Elliptical Orbits in Detail  $(0 \le e < 1)$

For bound orbits ($E < 0$) the trajectory is an ellipse with one focus at the CM. The standard orbital parameters are:

The orbital period follows from $dA/dt = L/(2\mu) = \text{const}$ (Kepler's second law) integrated over the full ellipse of area $\pi ab$:

This is Kepler's Third Law: the square of the orbital period is proportional to the cube of the semi-major axis.

9. Kepler's Three Laws

The solution to the two-body problem reproduces and explains all three of Kepler's empirical laws:

10. Summary of Key Results