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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 a point projectile approaching an absorbing spherical body of radius $R$ in a prescribed exterior inverse-square force, conservation of energy and angular momentum give $b^2=r_p^2+2GM{r}_p/v_{\infty }^2$. This is increasing in the periapsis radius $r_p$, so $r_p=R$ gives the collision threshold displayed above. The gravitational scattering angle obeys $\tan (\chi /2)=GM/(b{v}_{\infty }^2)$. On the grazing family its deflection approaches zero at high asymptotic speed and $\pi$ as $v_{\infty }\to 0^{+}$. Strict noncollision requires $b>b_{\mathrm{crit}}$; the equality is the tangency limit.

For equal masses and relative position $\mathbf{r}$, the individual positions are ${\mathbf{r}}_A=\mathbf{R}-\mathbf{r}/2$ and ${\mathbf{r}}_B=\mathbf{R}+\mathbf{r}/2$, where the center of mass has uniform velocity in an isolated system. Bound inverse-square force motion makes $\mathbf{r}$ an ellipse; each mass follows a half-size ellipse in the center-of-mass frame. Adding a nonzero center-of-mass drift generally makes their laboratory trajectories nonclosed. The relative orbital period is not necessarily a period of either laboratory position.

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 a repulsive inverse-square force, $V(r)=+\kappa /r$. The effective potential is positive and decreases from infinity to zero, so a trajectory reaching a finite radius has $E>0$. For nonzero angular momentum, it is the accessible branch of a hyperbola, with $e^2=1+2E{J}^2/(m{\kappa }^2)>1$. There are no bound ellipses or finite-distance zero-energy parabolas. At zero angular momentum, the motion is straight inward followed by straight outward after reflection at a radial turning point.

For an attractive inverse-square force, write $V(r)=-\kappa /r$ with $\kappa >0$, and take nonzero conserved angular momentum $J$. The conic section orbit hasThe physically allowed cases are ellipse for $-m{\kappa }^2/(2J^2)\le E<0$, parabola for $E=0$, and hyperbola for $E>0$. The minimum allowed energy gives the circular special case $e=0$. Here $E$ is measured relative to infinity. Zero angular momentum instead gives radial collision or escape trajectories, interpreted as degenerate limits rather than regular closed ellipses.

For the attractive inverse-square force, choose zero potential energy at infinity, so $U=-A/r$. At either radial turning point $r_p,r_a$, the kinetic energy is purely tangential, with conserved angular momentum $J$. ThusIts two roots are the periapsis and apoapsis distances. Their sum is $r_p+r_a=-A/E$, whereas the ellipse has $r_p+r_a=2a$. Therefore $a=-A/(2E)$, with $E<0$ for a nondegenerate bound ellipse. The circular case follows by the double-root limit.

For the two-body problem, define $\mathbf{r}={\mathbf{r}}_B-{\mathbf{r}}_A$, total mass $M=m_A+m_B$, reduced mass $\mu =m_A{m}_B/M$, and center of mass $\mathbf{R}$. The relative equation is $\mu \ddot{\mathbf{r}}=\mathbf{F}$, while the isolated center of mass has $\ddot{\mathbf{R}}=0$. Once the relative orbit is known,The relative energy uses $\mu$ in its kinetic energy and $A=G{m}_A{m}_B$ in the gravitational potential energy.

In the equal-mass case the initial relative speed is $u$, $\mu =m/2$, and the center-of-mass velocity is half B's initial velocity. ThusNegative relative energy requires $u^2<4Gm/a_1$. The escape threshold is $u_E=2\sqrt{Gm/a_1}$; equality is marginal parabolic escape, so a strictly bound pair has $u<u_E$.

For $u=u_E/2=\sqrt{Gm/a_1}$, $E_{\mathrm{rel}}=-3G{m}^2/(4a_1)$ and the relative semimajor axis is $a=2a_1/3$. The initial transverse speed is below the relative circular value $\sqrt{2Gm/a_1}$, so the initial turning point is apoapsis. HenceChoose the initial separation along $+x$ from A to B, with B's initial velocity $+u\hat{\mathbf{y}}$. Conservation of angular momentum gives relative periapsis speed $3u$. At the opposite end of the ellipse the direction is $-\hat{\mathbf{y}}$, soThus B has the same speed as initially but the opposite velocity. A has $2u\hat{\mathbf{y}}$ at that instant. A value $3u/2$ for B would be its center-of-mass-frame speed, not the requested initial laboratory frame result.

The relative orbital period follows from Kepler's third law with total gravitating mass $2m$:In the center-of-mass frame each star follows an ellipse of semimajor axis $a_1/3$ and eccentricity $1/2$, with the center of mass at a focus; they remain opposite on their simultaneous position vectors. In the laboratory frame add the uniform drift $ut/2$ along $y$. The laboratory paths of an equal-mass binary are therefore not closed ellipses, even though their relative motion repeats.

For two isolated point masses, introduce total mass $M_T=M+m$, reduced mass $\mu =Mm/M_T$ and relative separation $r$. The relative acceleration is $\ddot{\mathbf{r}}=-G{M}_T\mathbf{r}/r^3$. Conserved specific angular momentum $h=|\mathbf{r}\times \dot{\mathbf{r}}|$ fixes a plane and a constant areal velocity $h/2$. The specific orbital energy isFor $h\ne 0$, the trajectory is a conic:Negative energy gives an ellipse, including the circular limit $e=0$; zero energy gives a parabola; positive energy gives a hyperbola. Bound ellipses obey $\epsilon =-G{M}_T/(2a_o)$ andThus Kepler's laws of planetary motion follow from angular momentum conservation and the inverse-square force. For a circular orbit of radius $r$, $v_c=\sqrt{G{M}_T/r}$, while escape from that radius requires $v_{\mathrm{esc}}=\sqrt{2G{M}_T/r}$. Energy determines whether the orbit is bound; angular momentum also determines its shape. These ideal results neglect extended-body perturbations, drag and relativistic corrections.

Use the relative star–planet separation and specific angular momentum per unit reduced mass,For $r=l/(1+e\cos \theta )$, differentiation gives $\dot{r}=Je\sin \theta /l$, while the transverse speed is $r\dot{\theta }=J/r=J(1+e\cos \theta )/l$. Adding the squared orthogonal components yieldsThe Binet equation for the relative two-body inverse-square force is $u^{\prime \prime }+u=G(M+m)/J^2$, with $u=1/r$. The conic has $u^{\prime \prime }+u=1/l$, soThe printed division by $l$ is an error: $J^2$ has dimensions $\mathrm{lengt}{\mathrm{h}}^4/tim{\mathrm{e}}^2$, while $G(M+m)/l$ has dimensions of speed squared. A circular orbit independently checks the correction: $J^2=r^2{v}^2=G(M+m)r$, with $l=r$.

For a closed ellipse, $l=a(1-e^2)$ and $1+2e\cos \theta +e^2=2l/r-(1-e^2)$. Substitution now gives the vis-viva equation,This is relative orbital speed; either body's speed about the barycentre follows by the corresponding mass-ratio factor.

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.