Kepler's laws from Newtonian mechanics

Gravity on a planet points along the line to the Sun, so its torque about the Sun, τ = r × F, is zero and the angular momentum L = r × mv cannot change. A fixed direction for L keeps r and v in one plane, so the orbit is flat. A fixed magnitude is the second law. In a short time dt the radius line turns through dθ and sweeps a thin triangle of area dA = ½r² dθ, so dA/dt = ½r² dθ/dt = L/(2m), a constant. Nothing here used the inverse square: equal areas hold for any central force. Two consequences follow. The angular speed dθ/dt = L/(mr²) falls as 1/r², and the part of the velocity perpendicular to r, v⊥ = L/(mr), falls as 1/r. At perihelion and aphelion the velocity is entirely perpendicular, so rₚvₚ = rₐvₐ. Earth is 1.471 × 10¹¹ m from the Sun in early January and 1.521 × 10¹¹ m in early July, so it moves 1.034 times faster in January, 30.3 against 29.3 km s⁻¹, and its angular speed is 1.069 times higher.

An ellipse is the set of points whose distances from two fixed foci add to the same total, r + r′ = 2a, where the semi-major axis a is half the longest diameter. The foci lie a distance ae either side of the centre, with eccentricity 0 ≤ e < 1. The Sun occupies one focus; the other focus and the centre are empty. Seen from the Sun, the nearest and farthest points are rₚ = a(1 − e) and rₐ = a(1 + e), so two observed distances fix everything: a = (rₚ + rₐ)/2, e = (rₐ − rₚ)/(rₐ + rₚ), and the semi-minor axis b = a√(1 − e²) = √(rₚrₐ). Solving Newton's second law for a bound inverse-square orbit gives the same curve in polar form, r = a(1 − e²)/(1 + e cos θ), with θ measured from perihelion. An asteroid ranging from 1.20 AU to 3.60 AU has a = 2.40 AU, e = 0.500 and b = 2.08 AU. Its centre is 1.20 AU from the Sun, it is 1.80 AU away at θ = 90°, and at the ends of the minor axis, where r = r′, it is exactly a = 2.40 AU away.

The areal rate needs L, and the two apsides supply it. There the velocity is perpendicular to the radius, so angular momentum gives rₚvₚ = rₐvₐ, and energy conservation gives ½vₚ² − GM/rₚ = ½vₐ² − GM/rₐ. Substituting vₐ = vₚrₚ/rₐ and solving gives vₚ² = 2GMrₐ/(rₚ(rₚ + rₐ)). Multiply by rₚ² and use rₚ + rₐ = 2a and rₚrₐ = a²(1 − e²): L/m = rₚvₚ = √(GMa(1 − e²)). For the 1.20 to 3.60 AU asteroid, with GM = 1.327 × 10²⁰ m³ s⁻² for the Sun and a = 3.590 × 10¹¹ m, L/m = 5.98 × 10¹⁵ m² s⁻¹. Dividing by each distance gives 33.3 km s⁻¹ at perihelion and 11.1 km s⁻¹ at aphelion, a factor of 3.00, exactly rₐ/rₚ. At fixed a, a circular orbit carries the most angular momentum; as e approaches 1 the ellipse collapses toward a straight line through the Sun and L falls to zero, the orbit of a body released from rest.

One period sweeps the whole ellipse, of area πab, at the constant rate L/(2m), so T = πab ÷ (L/(2m)) = 2πab/(L/m). Insert b = a√(1 − e²) and L/m = √(GMa(1 − e²)): T = 2πa²√(1 − e²)/√(GMa(1 − e²)) = 2π√(a³/GM). The factor √(1 − e²) that shrinks the area shrinks the areal rate by exactly as much, so T² = 4π²a³/GM holds for every ellipse, with a standing where the radius stood for a circular orbit. Orbits with the same semi-major axis share a period, however different their shapes. The 2.40 AU asteroid takes 2π√((3.590 × 10¹¹)³/(1.327 × 10²⁰)) = 1.173 × 10⁸ s = 3.72 yr, the same as a circular orbit of radius 2.40 AU. Taking Earth's orbit as the unit, a in AU and T in years, turns the law into T² = a³ for anything orbiting the Sun, and the check costs one line: 2.40³ = 13.8 and 3.718² = 13.8.

Rearranged, T² = 4π²a³/GM becomes a balance, M = 4π²a³/(GT²), in which the orbiting body's own mass never appears. Europa circles Jupiter at a = 6.709 × 10⁸ m every 3.551 days, 3.068 × 10⁵ s, which gives M = 1.90 × 10²⁷ kg, about 318 Earth masses, from one distance and one time. The same law tests data. Plot ln T against ln a for all the moons of one planet and they fall on a straight line of slope 3/2, with the intercept carrying GM; a moon twice as far out takes √(2³) = 2.83 times as long. Precision exposes the one-body assumption. The central body is pulled too, so both bodies orbit their common centre of mass, and with a now the semi-major axis of their separation the law becomes T² = 4π²a³/(G(M + m)). For Earth and the Moon, m/M = 0.0123, so leaving m out overstates the Moon's period by 0.61%; for two stars of similar mass, the orbit gives only their total.

The three laws test different properties of the force. Equal areas needs only a central force, so it survives any force aimed at the Sun. The focus and the power 3/2 need the inverse square: a planet tied to the Sun by an ideal spring, F = −kr, would also move on ellipses, but centred on the Sun, with the same period 2π√(m/k) for every size of orbit. Among central forces, only the inverse-square law and the spring law make every bound orbit close on itself; almost any other law leaves a rosette whose perihelion turns a little each lap. That is what makes a turning ellipse a measurement. The other planets' pulls do not point at the Sun, and Mercury's perihelion advances about 575 arcseconds per century; Newtonian tugs from the other planets account for all but 43 of them, and general relativity supplies the rest. Kepler's laws are the exact solution of an idealised problem, one fixed point mass and one planet, and each measured departure from them names physics the ideal left out.