**B.2 Planet distance averaged in time**

Average in time distance *dt* is defined by the integral for the period of a planet revolution T:

$$
\overline{d}\_t = \frac{1}{T} \int\_0^T ddt,\tag{13}
$$

Since according to the second Kepler's law, the radial arm of a given planet sweeps out an area at a constant rate h:

$$h = \frac{1}{2}d^2d\theta/dt,\tag{14}$$

or

$$d\mathbf{t}/d\boldsymbol{\theta} = d^2/2\boldsymbol{h}.\tag{15}$$

We know that the area of the ellipse is *Th* ¼ *πab*. Hence, the average distance in time can be defined as:

$$\overline{d}\_{t} = \frac{1}{T} \int\_{0}^{T} d\cdot dt = \frac{h}{\pi ab} \int\_{0}^{2\pi} d\cdot (dt/d\theta) d\theta = \frac{h}{\pi ab} \int\_{0}^{2\pi} d\frac{d^2}{2h} d\theta = \frac{1}{2\pi ab} \int\_{0}^{2\pi} d^3 d\theta,\tag{16}$$

where

$$\frac{1}{2\pi ab} \int\_0^{2\pi} d^3 d\theta = \frac{p^3 e^3 (2 + e^2) \pi}{1 - e^2)^{(5/2)} = b \left(3a^2 - b^2\right) \pi. \tag{17}$$

Then the average in time distance is

$$\overline{d}\_t = \frac{1}{T} \int\_0^T ddt = \frac{b\left(3a^2 - b^2\right)\pi}{2\pi ab} = 3a/2 - b^2/2a = \frac{3a}{2} - \frac{a^2(1 - e^2)}{2a} = a\left(1 + \frac{e^2}{2}\right). \tag{18}$$

or the averaged by time distance is

$$
\overline{d}\_t = a \left( 1 + \frac{e^2}{2} \right),
\tag{19}
$$

which is larger than the semi-major axis a by a factor of (1 <sup>þ</sup> *<sup>e</sup>*2*=*2).
