Kepler's Equation

Kepler's equation connects the mean and the eccentric anomalies and is given by

$$E-e\sin E=M$$ (23)

To derive this equation, recall that the elliptic orbit is given by the formula

$$r=a\left(1-e\cos E).$$

Differentiating this with respect to time we get:

$$r'=ae\sin E E'$$ (24)

On the other hand we have that

$$r=\frac{a(1-e^2)}{1+e\cos f},$$

and hence

$$\frac1r=\frac{1+e\cos f}{a(1-e^2)}.$$

Differentiating this term we get

$$\frac{r'}{r^2}=\frac{e\sin f f'}{a(1-e^20}.$$ (25)

Because of the equal area law we have $r^2f'=na^2\sqrt{1-e^2}$, where $n$ denotes the mean angular velocity $2\pi/T$. Using this we get from the last equation (25)

$$r'=\frac{na\sin f}{\sqrt{1-e^2}}$$

Comparing this with (24), we get

$$E'=\frac{n\sin f}{\sqrt{1-e^2}\sin E}.$$ (26)

However, from the last section we know that

$$r\sin f= a\sqrt{1-e^2}\sin E,$$

and therefore

$$rE'=na.$$

since $r=a(1-\cos E)$, this becomes

$$\left(1-e\cos E\right)E'= n.$$

Integration over time yields

$$E-e\sin E =nt+c.$$

However, since $E=0$ at $t=\tau$, the constant of integration $c=-n\tau$ and

$$E-e\sin E=n(t-\tau)=M$$ (27)

which is Kepler's equation.

Kepler's equation makes it easy to compute the mean anomaly, whenever the eccentric anomaly is given, however, the other way round is difficult. The equation is a transcendental equation and astronomers have devised many numerical and analytic schemes to solve this equation. When the eccentricity $e$ is very small, the left hand side of (27) can be reasonably well approximated by $E$ itself, and therefore $M\approx E$. In the next section we investigate a few methods that will give numerical solutions of Kepler's equation.