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Put $\mathcal{G}=G(M_{\star }+M_{\mathrm{pl}})$, $h=\sqrt{\mathcal{G}a(1-e^2)}$, $\theta =\omega +f$ and $L=n_{\mathrm{pl}}t$. Here $h$ is the specific angular momentum, $f$ is the true anomaly and $\omega$ is the planar longitude of periapsis. Differentiating the Kepler orbit equation with $\dot{f}=h/r^2$ givesResolve the velocity along the radial and tangential unit vectors, $\hat{\mathbf{r}}=(\cos \theta ,\sin \theta )$ and $\hat{\mathbf{\theta }}=(-\sin \theta ,\cos \theta )$. The identities $\sin f\cos \theta -\cos f\sin \theta =-\sin \omega$ and $\sin f\sin \theta +\cos f\cos \theta =\cos \omega$ yieldThis is the Kepler velocity hodograph: the velocity runs around a circle whose centre is displaced by the orbital eccentricity.

The centre of mass condition gives the star's barycentric position ${\mathbf{r}}_{\star }=-\mu a_{\mathrm{pl}}(\cos L,\sin L)$. Translating to the star without rotating the axes gives ${\mathbf{r}}^{\prime }=\mathbf{r}-{\mathbf{r}}_{\star }$ and ${\mathbf{v}}^{\prime }=\mathbf{v}-{\dot{\mathbf{r}}}_{\star }$. From Kepler's third law, $n_{\mathrm{pl}}^2{a}_{\mathrm{pl}}^3=\mathcal{G}$, soConsequentlyThe small displacement and the larger reflex velocity have different orders: $|{\mathbf{r}}^{\prime }-\mathbf{r}|/a=\mu \alpha$, but the reflex speed divided by the circular speed is $\mu {\alpha }^{-1/2}$. These are the source of the astrocentric osculating-element oscillations.

For the coordinate ratio, the exact result isFor fixed $e<1$, $r/a\ge 1-e$, so this is $O(\mu \alpha )$ on any angular region with $|\cos \theta |$ bounded away from zero. The printed coordinate claim is not uniform over a full orbit: at $x=0$ its ratio is undefined. The nonsingular statement is $|{\mathbf{r}}^{\prime }-\mathbf{r}|/r\le \mu \alpha /(1-e)$, assuming the pericentre distance remains much larger than the binary separation.

Squaring the two components of the velocity givesAt $\mu =0$, $\kappa =0$ and $\mathcal{G}=G{M}_{\star }$. Sincethis reduces to the vis-viva equation, $v^{\prime 2}=G{M}_{\star }(2/r-1/a)$.

Now let the barycentric Kepler orbit be circular, write $\Delta =L-\theta$ and $\epsilon =\mu {\alpha }^{-1/2}\ll 1$, and put $v_0=\sqrt{\mathcal{G}/a}$. The osculating orbital elements about the star use the gravitational parameter $G{M}_{\star }=(1-\mu )\mathcal{G}$. To the required order, $r^{\prime }=a[1+O(\mu \alpha )]$ and $v^{\prime 2}=v_0^2[1+2\epsilon \cos \Delta +O({\epsilon }^2)]$. The specific orbital energy therefore impliesHere $\mu =o(\epsilon )$ and ${\epsilon }^2=o(\epsilon )$ under the stated hierarchy. Even at a phase where the leading cosine vanishes, the absolute error bound remains valid.

Finally, use the eccentricity vector ${\mathbf{e}}^{\prime }=({\mathbf{v}}^{\prime }\times {\mathbf{h}}^{\prime })/(G{M}_{\star })-{\hat{\mathbf{r}}}^{\prime }$. In radial and tangential components it isThe translation changes the radial basis only by $O(\mu \alpha )$, whereas $v_r^{\prime }=-v_0\epsilon \sin \Delta +O(v_0\mu \alpha )$ and $v_{\theta }^{\prime }=v_0(1+\epsilon \cos \Delta )+O(v_0\mu \alpha )$. ThusFor the original circular Kepler orbit, $\omega$ and $f$ are separately arbitrary; their sum $\theta$ is well defined. Using the eccentricity vector avoids introducing an undefined pericentre for that circular Kepler orbit.