Kepler–isochrone radial transformation ($r=\sqrt{\bar{E}/(\lambda E)}(\sqrt{{\bar{r}}^2+b^2}-b)$)

With $a=(1-\lambda )GM$, $h=\lambda E>0$ and $\bar{E}>0$, set $\bar{E}{q}^2=h{r}^2+ar$. If $a\ge 0$, its positive-radius root is $r=\sqrt{\bar{E}/h}(\sqrt{q^2+b^2}-b)$, with $b=a/(2\sqrt{h\bar{E}})$. The transformed Newtonian gravitational potential is an additive constant plus $-G\bar{M}/(b+\sqrt{q^2+b^2})$, the spherical isochrone model. The mass sign and branch conditions must be checked.

For bound old Kepler orbits, $E<0$, $\lambda <0$ and $\bar{E}>0$ give a positive radius, positive isochrone scale $b$, and positive transformed mass. The shifted energy is $\bar{E}-k=\bar{E}/\lambda <0$. For unbound old orbits with $E>0$, $1/2<\lambda \le 1$ gives an attractive unbound isochrone branch. The case $\lambda =0$ is handled by the separate Kepler–harmonic radial duality.