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Put $\mu =GM$ and use energy and angular momentum per unit particle mass. The Kepler orbit equation is $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$. HenceThus $E$ and $\mathbf{L}=\mathbf{r}\times \dot{\mathbf{r}}$ are integrals of motion. Splitting the velocity into radial and transverse parts, $|\dot{\mathbf{r}}{|}^2={\dot{r}}^2+L^2/r^2$, gives the Kepler radial energy equationHere $\dot{r}$ denotes the derivative of the scalar radius, unlike $\dot{\mathbf{r}}$.

For the Laplace-Runge-Lenz vector, differentiate directly:The vector triple-product identity yieldsThe terms cancel, proving every component of $\mathbf{A}$ is conserved.

Write $q=\bar{r}(r)$ and $\tau =\bar{t}$ for a monotone radial orbit time transformation. The chain rule givesImpose $q^{\prime }=(r/q)\dot{r}$, equivalentlyMultiplying the radial energy equation by $r^2/q^2$ then givesThis is a transformation along each trajectory; the new clock is obtained by integrating its radius-dependent rate.

For the Kepler–harmonic radial duality, take the old inverse-radius term as the new constant energy, $\bar{E}=\mu r/q^2$. On a positive-radius branch $\bar{E}>0$, soThe remaining term is minus the new Newtonian gravitational potential:For $E<0$ this is an attractive isotropic harmonic oscillator with ${\bar{\omega }}^2=-2E(\bar{E}/\mu {)}^2$. For $E>0$ it is an inverted oscillator, and for $E=0$ the Newtonian gravitational potential vanishes. Calling it a confining oscillator therefore requires a bound original Kepler orbit.

For the angular reconstruction of a radial orbit transformation, choose the new angle by $d\bar{\theta }/d\tau =L/q^2$. The same $L$ then supplies its centrifugal term. From $d\theta /dt=L/r^2$, the transformation is $d\bar{\theta }/d\theta =(r/q)(dq/dr)$; in the harmonic case this gives $\bar{\theta }=\theta /2$ up to a constant. Keeping the old angle unchanged would not in general preserve the angular momentum used in the displayed transformed radial equation.

For the second transformation use the explicit displayed split of the energy terms. It assigns weight $\lambda$ to $E$ and $1-\lambda$ to $\mu /r$; the surrounding printed prose interchanges these weights. Define the transformed quantities consistently byTheir difference is exactly $(r^2/q^2)(E+\mu /r)$, so they give the same transformed radial energy equation.

Let $a=(1-\lambda )\mu$, $h=\lambda E$, and $\epsilon =\bar{E}$. Constancy of the new energy imposesFor the standard regular isochrone branch take $h>0$, $\epsilon >0$, and $a\ge 0$. The positive-radius root isFor $q>0$ it is monotone. The corresponding clock is also regular there:With $\alpha =\sqrt{\epsilon /h}$ and $s=\sqrt{q^2+b^2}$, rationalization gives $r/q^2=\alpha /(s+b)$. Eliminate $r^2/q^2$ using $h{r}^2/q^2=\epsilon -ar/q^2$:where one possible identification isThis is the Kepler–isochrone radial transformation to the spherical isochrone model, up to the additive energy constant $k$.

The attractive branch of the Kepler–isochrone transformation requires a suitable parameter domain. The displayed square root needs $\lambda E\bar{E}>0$, and the positive-root branch used above needs $\lambda E>0$, $\bar{E}>0$. An attractive regular isochrone also needs $b\ge 0$ and $\bar{M}>0$. For bound original orbits a convenient domain is $E<0$, $\lambda <0$, $\bar{E}>0$: it satisfies all these requirements and gives $\bar{E}-k=\bar{E}/\lambda <0$. For $E>0$ one may instead take $1/2<\lambda \le 1$, with $\bar{E}>0$; these map to unbound attractive isochrone orbits. Other parameter choices can produce a repulsive Newtonian gravitational potential or a different branch, so no unrestricted real-parameter statement is justified. The singular algebraic case $\lambda =0$ is the harmonic transformation already handled separately. The angular reconstruction $d\bar{\theta }/d\tau =L/q^2$ likewise completes the isochrone radial solution into a central-force orbit.

Assume the station's mass remains fixed through an instantaneous kick, with subsequent motion governed only by the Earth's Newtonian gravitational field. Put $\kappa =G{M}_E$, and work per unit station mass. The initial circular orbit has specific angular momentum $j$ and specific orbital energy ${\epsilon }_0$ given byLet $\delta$ denote the fractional increase of kinetic energy. The position and potential energy are unchanged during the kick, soUnchanged angular momentum at unchanged position fixes the tangential velocity; the added kinetic energy must be radial. Thus this is a radial kick, not a purely tangential speed increase. The new orbital eccentricity follows fromAt either radial turning point, the Kepler radial energy equation gives $\epsilon =j^2/(2r^2)-\kappa /r$. Its two roots, equivalently the periapsis and apoapsis of the elliptic orbit, areFor $\delta =0.01$, $e=0.1$ andAn inward or outward kick gives the same pair of radial turning points, although it places the station at a different phase of the orbit. These two finite positive radii require $0\le \delta <1$.