Radial orbit time transformation ($d\tau /dt=(q/r)dq/dr$)

For a central-force radial energy equation, choose a monotone radius $q=q(r)$ and clock $\tau$ such that $d\tau /dt=(q/r)dq/dr$. Then $q^{\prime }=(r/q)\dot{r}$, and multiplying the old energy equation by $r^2/q^2$ preserves the centrifugal form $L^2/(2q^2)$. A split of the remaining terms into constant energy and a new Newtonian gravitational potential produces a transformed radial orbit.

A transformed radial equation containing $L^2/(2q^2)$ becomes a full planar central-force orbit by choosing $d\bar{\theta }/d\tau =L/q^2$. For $d\tau /dt=(q/r)dq/dr$ and old angular velocity $L/r^2$, this gives $d\bar{\theta }/d\theta =(r/q)dq/dr$. Keeping the old angle generally fails to preserve the stated transformed angular momentum.

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.

Taking $\bar{E}=GMr/q^2>0$ in a radial orbit time transformation gives $r=(\bar{E}/GM)q^2$ and $\bar{\Phi }(q)=-E(\bar{E}/GM{)}^2{q}^2$. Bound Kepler orbits with $E<0$ map to a confining harmonic oscillator; positive energy maps to an inverted oscillator. Full angular reconstruction halves the original orbital angle.