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An attractive inverse-square force together with radial drag $-k\dot{r}/r^2$ preserves angular momentum $h=r^2\dot{\theta }$ but modifies the Binet equation for $u=1/r$ by a first derivative. For $k/|h|<2$ and $h\ne 0$, the reciprocal radius is a decaying oscillation about $p/h^2$. Forward angular distance $\mathrm{sgn}(h)(\theta -{\theta }_0)$ makes its damping sign explicit for either orbital orientation.

For the signed inverse-square force $\mathbf{F}=-\kappa \mathbf{r}/r^3$ on a particle of mass $m$, with Hamiltonian $H=p^2/(2m)-\kappa /r$, the Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{p}\times \mathbf{L}-m\kappa \hat{\mathbf{r}}$ is conserved. Attraction has $\kappa >0$ and repulsion has $\kappa <0$. It obeys $\mathbf{A}\cdot \mathbf{L}=0$ and $A^2=m^2{\kappa }^2+2mH{L}^2$. For unit mass in a Kepler orbit, $\kappa =GM$.

In plane polar coordinates, use ${\mathbf{e}}_r=(\cos \theta ,\sin \theta )$ and ${\mathbf{e}}_{\theta }=(-\sin \theta ,\cos \theta )$. Their derivatives are ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$ and ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$. Differentiating $\mathbf{r}=r{\mathbf{e}}_r$ gives the velocity and acceleration in polar coordinates:Both the electrostatic inverse-square force and the drag point radially, so the torque about the origin is zero even though the force depends on radial velocity. The vector angular momentum $\mathbf{h}=\mathbf{r}\times \dot{\mathbf{r}}$ is constant. For $\mathbf{h}\ne 0$, every position lies in the fixed plane perpendicular to it; if $\mathbf{h}=0$, the trajectory is radial and still lies in a plane. In the nonzero case the transverse equation is $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$, givingThe signed $h$ depends on the plane orientation. The stipulated ratio $k/|h|$ requires $h\ne 0$.

Set $u=1/r$ and use primes for differentiation with respect to $\theta$. Then $\dot{\theta }=h{u}^2$, $\dot{r}=-h{u}^{\prime }$ and $\ddot{r}=-h^2{u}^2{u}^{\prime \prime }$. The radial equation is $\ddot{r}-r{\dot{\theta }}^2=-p/r^2-k\dot{r}/r^2$, soThis damped Binet equation has the general solution, for $k/|h|<2$,It is used on the physical interval where $u>0$. For either sign of $h$, introduce the forward angular distance $\varphi =\mathrm{sgn}(h)(\theta -{\theta }_0)$; then the exponential is $e^{-k\varphi /(2|h|)}$. Thus damping acts forward in time for both orientations, not just when $h>0$.

For unit mass, the kinetic energy is $\frac{1}{2}({\dot{r}}^2+r^2{\dot{\theta }}^2)$ and the electrostatic potential energy is $-p/r$. ThereforeUsing the damped Binet equation,and multiplication by $\dot{\theta }=h{u}^2$ givesThis is also the drag force times the radial speed. Energy is nonincreasing in time, strictly decreasing while radial motion occurs; a circular trajectory already has constant energy. The sign of $dE/d\theta$ alone would be misleading for negative $h$.

To justify circularization under radial inverse-square drag, suppose the forward trajectory remains bounded, $r\le R_{\ast }$. Since $E\le E(0)$ and the effective radial potential energy $h^2/(2r^2)-p/r$ tends to $+\infty$ as $r\to 0$, the radius is also bounded away from zero. Velocity then stays bounded, so the solution continues for all forward time without collision. Moreover $\dot{\varphi }=|h|/r^2\ge |h|/R_{\ast }^2$, so $\varphi \to \infty$. The decaying general solution gives $u\to p/h^2$ and $u^{\prime }\to 0$. ConsequentlyThe tangential motion remains, so the limiting orbit is circular. The convergence is asymptotic; boundedness is needed because a physical solution could otherwise reach $u=0$ and escape rather than complete indefinitely many turns.

The example uses $p=h=1$, $k=0.6$ and $u(\theta )=1+0.35e^{-0.3\theta }\cos (\sqrt{0.91}\theta )$, which stays positive. Radial oscillations decay while angular momentum stays fixed.

Outside a spherical Earth the gravitational field is the field of its total mass at the center, by the shell theorem. Since $G{M}_E=g{R}^2$, the inverse-square force on the upward radial trajectory givesThe independent dimensional inputs are $R,V,g$, of dimensions $L,L{T}^{-1},L{T}^{-2}$. A dimensionless monomial in them must be a power of $V^2/(gR)$. Thus dimensional analysis givesFor a finite turning height and time, assume $0<V<\sqrt{2gR}$, equivalently $0<\lambda <2$. This escape velocity restriction is implicit in the requested maximum-height formulas.

The conservation of energy, per unit particle mass, issoThe numerator vanishes at the turning point, givingOn ascent $\dot{z}>0$, and integrating $dt=dz/\dot{z}$ givesThe upper endpoint has an integrable inverse-square-root singularity. Set $z=Hx$, use $H=R\lambda /(2-\lambda )$ and $V^{2}R=(2gR-V^2)H$, and note $R\lambda /V=V/g$. The result isFor small $\lambda$, $F(\lambda )\sim \lambda /2$ and $G(\lambda )\to 1$, recovering $H\sim V^2/(2g)$ and $T\sim V/g$. At $\lambda =2$ the particle escapes marginally and has no finite maximum; for $\lambda >2$ it escapes with nonzero asymptotic speed. The finite-turning formulas must not be extended to those cases.

For unit mass, the inverse-square force has potential energy $-k/r$, with zero potential at infinity. Conservation of angular momentum gives $r^2\dot{\theta }=h$. Differentiate the polar equation of the Kepler orbit:Consequently the total mechanical energy iswhere $k=h^2/\ell$. This also explains the sign distinction between an elliptic orbit, a parabolic Kepler orbit, and a hyperbolic Kepler orbit.

At the original periapsis, the parabolic Kepler orbit has $e=1$, $r=d=\ell /2$, and purely tangential speed $V$. ThusThe outward radial impulse has zero moment about the Sun, so it preserves angular momentum and hence $\ell$. Its additional velocity is perpendicular to the original velocity, so the new mechanical energy is $E^{\prime }={\alpha }^2{V}^2/2$. Applying the energy formula givesFor $\alpha >0$, the new orbit is a hyperbolic Kepler orbit; $\alpha =0$ leaves the parabola unchanged.

To fix the orientation of the sketch, put the Sun at the origin, the impulse point at $(d,0)$, and take the original motion there upwards. The new velocity is $(\alpha V,V)$. The eccentricity vector isIt points towards the new periapsis, which is rotated clockwise through $\beta =\arctan (2\alpha )$ from the original periapsis direction. Thus, measured using the original polar angle $\theta$, the new orbit isIn particular, the impulse point is already on the outgoing part of the new orbit and is not its new periapsis. The old parabola satisfies $x=d-y^2/(4d)$. Both orbits pass through $(d,0)$ and have the same focus, but the hyperbola's new periapsis axis is tilted.

For the proposed circular motion,The separation is ${\mathbf{x}}_1-{\mathbf{x}}_2=2a(\cos \omega t,\sin \omega t,0)$, so the inverse-square force on particle 1 isThe equation $m{\ddot{\mathbf{x}}}_1={\mathbf{F}}_{12}$ holds exactly whenThe equation for particle 2 follows by symmetry.

For an attractive inverse-square force, $r=p/(1+e\cos f)$, where $f$ is true anomaly, $e$ is orbital eccentricity, and $p=h^2/\mu$ is the semilatus rectum. The specific angular momentum relation $\dot{f}=h/r^2$ converts this geometric equation into an angular evolution law.