Kepler's Second Law

Here again we follow the notations of (Katz, 1993; p. 378).
As before, let:
  • (rho, nu) be solar coordinates of a point R=(X, Y) on the ellipse,
  • the semi-major axis a=1,
  • the semi-minor axis be arbitrary, b < 1,
  • e denote the eccentricity, and
  • beta be the central polar coordinate of the point W=(X, Y0) on the circle, where Y=b*X0.
Then as we have seen, the first law may be expressed by the three equations:
[1] X = rho*cos(nu) = e + cos(beta)
[2] Y = rho*sin(nu) = b*sin(beta)
[3] rho ~= 1 + e*cos(beta)
We also want to make use of Kepler's 2nd Law in the infinitesimal form,
[4] rho2*nu-dot = K (a constant)
where nu-dot denotes the derivative of nu with respect to time. This follows directly from Kepler's Second Law, and the Lemma below.

Kepler's Second Law

Let DA denote the area swept out by the solar radius vector, rho, as the planet moves from position R1 to position R2 on the ellipse, during an interval of time, DT. Then the ratio DA/DT is a constant.

As DT approaches zero, we get A-dot = K, a constant.

Lemma. As e is small, rho2*nu-dot closely approximates the time rate of change of the area swept out by the solar radius vector, rho.

That is, A-dot ~= rho2*nu-dot. Combining with the 2nd law, rho2*nu-dot ~= K.

To prove the lemma, consider this figure. In the interval of time, DT, the planet moves from R1 to R2 along the red epplise, and the solar radius vector, rho, decreases from rho1 to rho2, sweeping out the blue shaded area, DA. At the same time, the solar angle coordinate increases from nu1 to nu2, a change of Dnu. The blue area is a bit larger than the portion of it, A2, enclosed by the smaller black circle, which passes through R2, and has radius rho2. The area A2 is given by

A2 = (PI*rho22) * (Dnu/2*PI)
= (rho22) * Dnu/2

That is, the area of the smaller black circle times the fraction of the circle covered by the blu wedge (with Dnu measured in radians). Note that the difference, DA-A2, is the area of the small triangle of DA outside the small black circle. And if DT shrinks to zero, this difference shrinks to zero, while DA gets closer and closer to A2. Dividing both sides of this by DT, we get the lemma,

A-dot ~= rho2 * nu-dot/2

Revised 30 dec 2001 by ralph abraham