From the Classical Mechanics Lecture Notes by Helmut Haberzettl, we know that in Newtonian Mechanics, the solution to Kepler's problem can be parametrized as a conic section equation $$r(\varphi)=\frac{r_{0}}{1-\varepsilon \cos (\varphi-\Phi)}\tag{1.329}$$ (It seems $\varphi$ and $\phi$ are rendered the same here, so I shall denote $\phi$ by $\Phi$)
In the Newton's cannonball problem, as illustrated below
Setting kinetic energy equal to potential energy gives the escape velocity $v_e$ $$\frac{m}{2} v_{\mathrm{e}}^{2}=\frac{\alpha}{r_{0}} \Rightarrow v_{\mathrm{e}}=\sqrt{\frac{2 G m_{E}}{r_{0}}} \quad \text { for } \quad \alpha=G m m_{E}\tag{1.350}$$ and hence $$\alpha=\frac{m}{2} v_{\mathrm{e}}^{2} r_{0} \quad \text { and } \quad \ell=r_{0}\left(m v_{\perp}\right)=m r_{0} v_{0} \sin \beta\tag{1.351}$$ where $\ell$ is the magnitude of the angular momentum.
The eccentricity is given by $$\varepsilon=\sqrt{1+\frac{2 \ell^{2}}{m \alpha^{2}} E}=\sqrt{1+4 \sin ^{2} \beta\left(\frac{v_{0}}{v_{\mathrm{e}}}\right)^{2}\left[\left(\frac{v_{0}}{v_{\mathrm{e}}}\right)^{2}-1\right]}\tag{1.327 & 1.352}$$ and, by the quadratic equation, we have
$$\left(\frac{v_{0}}{v_{\mathrm{e}}}\right)^{2}=\frac{1}{2}-s \frac{\sqrt{\varepsilon^{2}-\cos ^{2} \beta}}{2 \sin \beta}\tag{1.353}$$ for which $s=\pm 1$ depending on $\beta>\pi/2$ or $\beta<\pi/2$ respectively.
Now the books says, from the preceding results, $(1.329)$ can be rewritten as $$\frac{r_{0}}{r}=\left(\frac{v_{\mathrm{e}}}{v_{0}}\right)^{2} \frac{1-\varepsilon \cos (\varphi-\Phi)}{2 \sin ^{2} \beta}\tag{1.355}$$
I am not sure how this follows from the above discussion. Since if we merely rearrange $(1.329)$, we would get something like $r_0/r=1-\varepsilon\cos(\varphi-\Phi)$, I wonder how the additional factor $\left(\frac{v_{\mathrm{e}}}{v_{0}}\right)^{2} \frac{1}{2 \sin ^{2} \beta}$ arises in $(1.355)$.
