[NMSU, N. Vogt]

Consider the positions of a planet, P₁, P₂, P₃, and P₄, at times t₁, t₂, t₃, and t₄. Between times t₁ and t₂ the planet's radius vector sweeps out the area bounded by the radius vectors SP₁, SP₂ and the arc P₁P₂. Similarly, the area swept out by the radius vector in the time interval (t₄-t₃) is the area SP₃P₄. Then, Kepler's second law states that:

If t₂-t₁ = t₄-t₃, then area [SP₁P₂] = area [SP₃P₄]. If the area is the area of the ellipse itself (which is given by ab), the radius vector will be back to its original position. Kepler's second law thus implies that the planet's orbital period is constant.

It is possible to derive a mathematical expression for Kepler's second law by supposing that the time interval (t₂-t₁) is very small and equal to interval (t₄-t₃). Position P₂ will be very close to P₁, just as P₄ will be close to P₃. The area SP₁P₂ is then approximately the area of triangle SP₁P₂, or

Using the small angle approximation,

sin (P₁SP₂) = P₁SP₂ = ₁.

Also,

so that the area SP₁P₂ is given by

½ r²_{1 1}.

Similarly, area SP₃P₄ is given by

½ r²_{2 2},

where SP₃ = r₂ and angle P₃SP₄ = ₂. Let t₄-t₃ = t₂-t₁ = t. Then, from Kepler's second law,

½ r²₁(₁ / t) = ½ r²₂(₂ / t) = constant.

However, /t is the angular velocity, , in the limit when t tends to zero. Hence

½ r²₁₁ = ½ r²₂₂ = constant,

is the mathematical expression of Kepler's second law. In order for this law to be obeyed, the planet has to move fastest when its radius vector is shortest, at perihelion, and slowest when it is at aphelion, as shown here.