considering Eq.(\[evol\_a\]) (with $\varepsilon=0$ for simplification) we can see that, for $\omega_\ii/n\leqslant N_a(e)/N(e)$, the tides raised on the body $i$ lead to a decrease of the semi-major axis, transferring the angular momentum from the orbit to the body’s internal rotation. It is easy to show that for a synchronous planet this condition is always fulfilled, since $\frac{\omega_\mathrm{eq}}{n}=\frac{N(e)}{\Omega(e)}\leqslant\frac{N_a(e)}{N(e)}$ for any eccentricity (respectively solid and dashed curves of Fig.\[fig:phasespace\]a). As a result, the semi-major axis of most short period planets is decreasing.

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On the other hand, truncating Eq.(\[evol\_a\]) at the order $e^2$ for the semi-major axis evolution yields $$\begin{aligned}
\frac{1}{a}\frac{\dd a}{\dd t}=\frac{4\,a}{GM_{\star}M_p}
&\Big\{&\Kp&\left[(1+\frac{27}{2}e^2)
\frac{\op}{n}-(1+23\,e^2)\right]  \nonumber\\
&+& \Ks&\left[(1+\frac{27}{2}e^2)
\frac{\os}{n}-(1+23\,e^2)\right]\Big\}\ .
\label{evol_ae2}\end{aligned}$$ and the previous condition becomes $\omega_\ii/n\leqslant(1+23\,e^2)/(1+\frac{27}{2}e^2)$. Up to $2^{\mathrm{nd}}$ order in eccentricity, the pseudo synchronization angular velocity is given by $\omega_\mathrm{eq}=(1+6\,e^2)n$ [^1]. One can see that $\omega_\mathrm{eq}/n=1+6\,e^2 \leqslant(1+23\,e^2)/(1+\frac{27}{2}e^2)$ only for $$e\leqslant\frac{1}{9}\sqrt{\frac{7}{2}}\approx0.208.$$ This means that even for a moderate eccentricity, $e\sim0.2$, the truncated model predicts that tides raised on a pseudo synchronous planet lead to a *growth* of the semi-major axis instead of a *decrease*, as obtained by the complete model. Therefore, truncating the tidal equations at the order $e^2$ for an eccentricity $e \geqslant 0.2$ not only predict quantitatively wrong but 