In Chapter 3 we solved the Kepler problem to the extent that we were able to determine the geometric trajectory of a body moving in a gravitational field. However, as yet we do not know the position of the body as a function of time. In this chapter we show that the trajectory can be parametrised by a suitable angular variable, the so-called eccentric anomaly, which can be given at least implicitly as a function of time. In the elliptic and hyperbolic case, this so-called Kepler equation is transcendental and cannot be solved explicitly in terms of elementary functions.

In Section 4.1 we derive Kepler's equation in the elliptic case. The analogous hyperbolic case is relegated to the exercises. In Section 4.2 I present Newton's original solution of Kepler's equation (in the elliptic case) by means of a geometric construction involving a cycloid.

In the parabolic case, the relevant equation is cubic. I take this as an excuse to make a small detour, in Section 4.3, to the algebra of cubic equations and their explicit solution by radicals. This relies on an elementary decomposition of the cube, very much analogous to the solution of quadratic equations by quadratic extension.

Anomalies and Kepler's equation

In this section we concentrate on the elliptic case. Thus, let r: ℝ → ℝ3 \ {0} be a solution of the Kepler problem with h < 0 and c ≠ 0. As in Section 3.3, we may take r to describe an ellipse ε in the xy-plane, traversed positively. Moreover, we can choose our coordinates such that the eccentricity vector e points in the positive x-direction. Then the angle ω, as defined in Figure 3.1, is zero, and the true anomaly f describes the angle between the positive x-axis and r. Hence, writing r as a function of f - which is permissible since f > 0 -we have

Now draw a circle C with centre Z = (-ea, 0), which equals the centre of ε, and radius a.