Kepler's Laws from Angular Momentum & Energy

You’ll be able to

Derive Kepler's second law from conservation of angular momentum
Relate Kepler's third law to Newtonian gravitation for a circular orbit
Analyze an elliptical orbit using conservation of energy and angular momentum together

The equal-area law is angular momentum conservation

Kepler's second law states that the line joining a planet to the sun sweeps out equal areas in equal times. This is not an independent empirical fact but a direct consequence of a central force. The area swept in a short time dt is dA = ½r²dθ, so dA/dt = ½r²(dθ/dt) = L/2m. Since gravity is central it exerts no torque about the sun, L is constant, and therefore dA/dt is constant. Kepler discovered from data what Newton later showed follows from conservation.

The three laws in Newtonian form

ellipse with sun at a focus dA/dt = L/2m = constant T² = (4π²/GM)a³

In the third law a is the semi-major axis of the ellipse, which reduces to the radius for a circular orbit. The constant depends only on the mass of the central body, so it is the same for every planet in a system.

Deriving the third law for a circular orbit

For a circular orbit gravity supplies the centripetal force: GMm/r² = mv²/r, so v² = GM/r and the orbiting mass cancels. Substituting v = 2πr/T gives 4π²r²/T² = GM/r, and rearranging yields T² = (4π²/GM)r³. The satellite's own mass is absent, which is why a communications satellite and a spent bolt at the same altitude orbit with the same period. The full derivation for an ellipse replaces r with the semi-major axis and is beyond the scope of the course, but the circular case carries the physics.

Using both conservation laws on an ellipse

An elliptical orbit is fully handled by applying two conservation laws at once. Energy gives ½mv² − GMm/r = constant, evaluated most conveniently at perihelion and aphelion. Angular momentum gives mv_p r_p = mv_a r_a, which is especially simple at those two points because the velocity is perpendicular to the radius there. Two equations, two unknown speeds. This pairing is the standard structure of orbital free-response questions, and recognizing that perihelion and aphelion are the points where L takes its simplest form is the key step.

Worked example

A comet has speed v_p at perihelion distance r_p and speed v_a at aphelion distance r_a. Write the two conservation statements linking them, and state what happens to the comet's speed as it moves outward.

1.Angular momentum about the sun is conserved, and at perihelion and aphelion the velocity is perpendicular to r, so L = mv_p r_p = mv_a r_a.
2.Rearranged: v_a = v_p(r_p/r_a). Since r_a > r_p, the aphelion speed is smaller.
3.Energy is conserved: ½mv_p² − GMm/r_p = ½mv_a² − GMm/r_a.
4.Physically, moving outward increases the gravitational potential energy — which is negative and rising toward zero — so kinetic energy must fall and the comet slows.

Answer: v_p r_p = v_a r_a from angular momentum, together with the energy equation; the comet is fastest at perihelion and slowest at aphelion, which is the equal-area law expressed as speeds

Checkpoint

A planet in an elliptical orbit moves fastest at perihelion. The most fundamental reason is —

Tip

Kepler's three laws map onto three different pieces of physics: the ellipse comes from solving the inverse-square force law, equal areas from conservation of angular momentum, and T² ∝ a³ from equating gravitational and centripetal force. Naming which principle produces which law is a frequent short-answer question.

Checkpoint

Two satellites of very different mass orbit Earth at the same altitude. Their orbital periods are —

Answer the 2 checkpoints as you read.