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Write $w=1/r$ and $h=r^2\dot{\theta }=b{v}_{\infty }$. Since $\dot{r}=-hdw/d\theta$, the radial equation becomes the Binet equationwhose solution isChoose the incoming asymptote at $\theta =\pi$ and the downstream axis at $\theta =0$. Then $w\to 0$ as $\theta \to \pi$, while $r\sin \theta \to b$. These two conditions giveThe mirror-image streamlines meet on the downstream axis atAt that point each streamline has radial velocity $-v_{\infty }$ and equal and opposite azimuthal velocity. An inelastic collision cancels the latter, so the specific energy afterwards isThe gas is bound when $E_{\mathrm{after}}<0$, orSweeping the corresponding capture cylinder through gas of density ${\rho }_{\infty }$ gives the Bondi--Hoyle--Lyttleton accretion rateUnlike stationary spherical Bondi accretion, this is a directed, supersonic flow with a downstream focusing wake; bulk speed replaces sound speed as the main resistance to capture.

A central force conserves angular momentumPut $u=1/r$. Since $\dot{\theta }=\ell u^2/m$,where primes now denote differentiation with respect to $\theta$. Substitution into the radial equation gives the Binet equation

In polar coordinates in the orbital plane, $h=r^2\dot{\theta }$. Setting $u=1/r$, the Binet equation for the effective inverse-square force isIts general solution isand henceThis is a branch of a hyperbolic Kepler orbit; because $1-\beta <0$, its physical branch has a negative denominator.

Put $u=1/r$. Since $h=r^2\dot{\theta }$, the radial equation becomes the Binet equationwhere primes denote derivatives with respect to the polar angle. Choosing the angular origin at periapsis givesWriting the semi-latus rectum as $h^2/\mu =a(1-e^2)$ yields the Kepler orbitHence the specific angular momentum and specific orbital energy are

For polar coordinates, the rotating unit vectors satisfy ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$ and ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$. Applying the product rule twice to $\mathbf{r}=r{\mathbf{e}}_r$ gives the velocity and accelerationNewton's second law for a central force now givesMultiplying the azimuthal force equation by $r/m$ yieldsThis is conservation of specific angular momentum. Here $F(r)$ is the signed radial component of the force; a negative value points inward.

When $h\ne 0$, $\theta$ is a valid local parameter along the trajectory. Put $u=1/r$ and let primes denote derivatives with respect to $\theta$. The chain rule yieldsSubstituting into the radial Newton's second law equation and dividing by $-m{h}^2{u}^2$ proves the Binet equation:

Here $m{h}^2=b$, so the Binet equation reduces toThe initial conditions becomeMultiply the differential equation by $2u^{\prime }$ and use the chain rule:The initial conditions give $C=0$. Since $e^{-u}>0$, $u^{\prime }$ cannot change sign at a finite value of $u$. Thus $u^{\prime }=e^{-u/2}$ on the physical branch. Separation of variables givesThereforeThe natural logarithm must be positive because $r>0$. For the forward trajectory, $\theta \ge 0$ and $h=\sqrt{b/m}>0$, so $\dot{\theta }=h/r^2>0$: the particle travels counterclockwise while $r$ strictly decreases.

For $a=1$, it starts at $(x,y)=(1,0)$, with tangent vector proportional to $(-e^{-1/2},1)$. Its successive crossings of the positive $x$ coordinate axis have radiiIt makes infinitely many inward turns, approaching the origin as $\theta \to \infty$. There is no finite-time arrival at the origin: $dt/d\theta =r^2/h$ behaves as $1/(4h{\log }^2\theta )$, whose improper integral diverges. If extended backwards, it comes from infinity as $\theta \downarrow 2(1-\sqrt{e})$.

Divide Newton's second law by the mass and resolve along the radial and transverse unit vectors in polar coordinates. A central force has no transverse component, soMultiplication of the second equation by $r$ shows that $d(r^2\dot{\theta })/dt=0$. Thuswhere $h$ is the signed specific angular momentum perpendicular to the orbital plane.

If $h\ne 0$ and $r>0$, the polar angle is locally a valid independent variable. For $u=1/r$, use primes for derivatives with respect to $\theta$. The chain rule givesSubstitution into the radial equation gives the Binet equationIf $h=0$, the orbit is radial and division by $h$ is unavailable; the first radial equation still applies with $\dot{\theta }=0$ away from the origin.

The initial transverse velocity gives $h=a(L/a)=L$. From $\dot{r}=-h{u}^{\prime }$, the initial conditions are $u(0)=1/a$ and $u^{\prime }(0)=1/a$. DefineThe Binet equation becomes the constant-coefficient linear ordinary differential equationIts particular solution is $({\nu }^2-1)/(a{\nu }^2)$. Applying the two initial conditions to the complementary solution givesThe physical branch is the connected range containing $\theta =0$ on which the right-hand side is positive. A zero corresponds to escape to infinite radius, rather than continuation to negative $r$.

The source calls $F$ an inward force magnitude, but here $F(r)=(r-a)/(a{r}^3)$ changes sign at $r=a$. The stated initial condition has inward radial velocity, so the particle immediately enters $r<a$, where the force is outward. The consistent interpretation is that $F$ is the signed inward radial component of the force per unit mass.

Let $h=J/m=b{v}_0$ be the specific angular momentum, and measure the orbital angle $\vartheta$ from periapsis. Set $w=1/r$. From $\dot{\vartheta }=h/r^2$, differentiating gives $\dot{r}=-hdw/d\vartheta$. The radial equation of motion $\ddot{r}-r{\dot{\vartheta }}^2=-{\mu }_J/r^2$ becomes the Binet equationIts solution symmetric about periapsis isFor example, the expression for the orbital eccentricity follows by substituting $w$ and $\dot{r}=-h{w}^{\prime }$ into the energy equation: $E/m={\mu }_J^2(e^2-1)/(2h^2)=v_0^2/2$. Since $e>1$, this is a hyperbolic Kepler orbit.

The outgoing asymptote occurs at ${\vartheta }_{\infty }$ with $\cos {\vartheta }_{\infty }=-1/e$, where $\pi /2<{\vartheta }_{\infty }<\pi$. The incoming velocity points inward along the incoming radial asymptote, whereas the outgoing velocity points outward. Therefore the change in velocity direction is $\varphi =2{\vartheta }_{\infty }-\pi$, not $2{\vartheta }_{\infty }$. It follows thatFor $b>0$, this gives $0<\varphi <\pi$: a large impact parameter produces a small deflection, and the ideal point-mass limit $b\to 0^{+}$ approaches a reversal.

Radial energy and effective potential. For a central force with potential energy $U(r)$, angular momentum conservation fixes $J=m{r}^2\dot{\varphi }$ and confines motion to a plane. The kinetic energy separates into radial and angular parts, givingThe second term is the centrifugal barrier. It converts the radial motion to a one-dimensional energy problem: allowed radii have $E\ge U_{\mathrm{eff}}$, turning points have equality, and a circular orbit occurs at a stationary point of $U_{\mathrm{eff}}$. Stability of that radial orbit requires a local minimum.

The first-order orbit equation. Write $A=GMm$. Since $f=-dU/dr=-A/r^2-A\gamma /r^3$, choose zero potential at infinity to getAlso $\dot{r}=(dr/d\varphi )J/(m{r}^2)$. Substitute this into the energy equation and multiply by $2m{r}^4/J^2$:Here $\gamma$ has units of length. A regular bound orbit with a centrifugal barrier requires $J^2>mA\gamma$ as well as the appropriate negative energy.

Solving in inverse radius. Let $u=1/r$ and defineDividing the first-order equation by $r^4$ givesDifferentiating, or directly using the Binet equation, gives $u^{\prime \prime }+qu=b$. The continuation through turning points follows from this smooth second-order equation. Its noncircular bound solution, with angular origin at a pericentre, isSubstitution into the first-order equation fixes the amplitude through this $e$. Therefore the requested representation holds withsince ${\alpha }_1/r-{\alpha }_2=\cos ({\alpha }_3\varphi )$. A general angular origin replaces $\varphi$ by $\varphi -{\varphi }_0$. This is the orbit equation for combined inverse-square and inverse-cube attraction. The circular limit $e=0$ is regular in $u=b/q$ but is not represented by finite ${\alpha }_1,{\alpha }_2$ in the source's normalized cosine form.

Precession and the inferred length. The last expression immediately givesSuccessive pericentres occur after angular advance $2\pi /{\alpha }_3$, so the apsidal precession per orbit isFor ${\alpha }_3$ near one, $1-{\alpha }_3^2\simeq \Delta \varpi /\pi$. Inferring an inverse-cube perturbation from apsidal precession consequently gives $\gamma \simeq [J^2/(mA)]\Delta \varpi /\pi$. For the stipulated nearly circular approximation, $J^2/(mA)\simeq r_0$, and Kepler's third law givesusing $P=88(86400)\mathrm{s}$, $M=2.0\times 10^{30}\mathrm{kg}$ and $G=6.6743\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$. A century contains about $36525/88=415.1$ orbits. Convert the observed advance to radians per orbit before using it:HenceThe positive sign strengthens the attraction and gives prograde precession. The fractional perturbation $\gamma /r_0\simeq 1.6\times 10^{-7}$ validates the small-deviation expansion. This is the parameter fitting the proposed phenomenological force law under the specified small-eccentricity approximation, not a derivation of that law as the physical explanation of Mercury's anomaly.

Put $f(r)=1-2\mu /r$, where $\mu =GM/c^2$, and work in the orbital plane. A dot denotes an affine-parameter derivative; for a material particle choose that parameter to be proper time.

First integrals and normalization. The geodesic Lagrangian isTime translation and axial rotation make $t$ and $\varphi$ cyclic, givingThe tangent norm is constant along an affine geodesic because the Levi-Civita connection preserves the metric. ThereforeFor proper-time parametrization of a timelike worldline this norm is $c^2$. For a null geodesic it is zero, for every affine normalization. Thus $\mathcal{E}=c^2$ for the chosen material-particle normalization and $\mathcal{E}=0$ for light. Proper time cannot itself parametrize a null ray, despite the same printed symbol $\tau$ being used for its affine parameter.

Null normalization and impact parameter. Under $\tau \mapsto A\tau +B$ with $A>0$, both $k$ and $h$ scale as $1/A$, while the null path is unchanged. Their ratio is invariant. At large radius the spatial speed with respect to asymptotic time is $c$, and orbital angular momentum per affine tangent givesFor an orientation chosen with $h>0$, this is $h=bc$. Setting $k=1$ is a choice of affine scale, not a claim that all photons have equal measured energy. Once a tangent is identified with a physically normalized four-momentum, its energy fixes that scale.

Orbit shape. Substituting the constants in the norm and multiplying by $f$ givesFor $u=1/r$ and $h\ne 0$, $du/d\varphi =-\dot{r}/h$. HenceThe $h=0$ radial orbit must be treated with the radial equation, because azimuth is then constant.

Circular photons and stability. For a null ray define $W(r)=f(r)/r^2$. Its radial equation becomesThe circular orbit requires $\dot{r}=0$ and $W^{\prime }=0$, soIt is at a maximum of the null effective potential and is unstable. Equivalently the Binet equation $u^{\prime \prime }+u=3\mu u^2$ linearizes about $u_0=1/(3\mu )$ to $\delta u^{\prime \prime }=\delta u$, admitting exponentially growing radial deviations in azimuth.

Capture from infinity. Divide the null radial equation by $k^2{c}^2$. The quantity opposing inward motion is $b^{2}W(r)$, whose exterior maximum is $b^2/(27{\mu }^2)$. There is no outer radial turning point precisely whenAt equality an exactly critical ray from infinity asymptotes to the photon sphere; it does not cross the horizon at a finite affine parameter. Above the threshold an initially inward ray from infinity reaches an outer turning point and scatters. The radial case $b=0$ is also captured. These statements refer to inward rays launched from the asymptotically flat exterior, not arbitrary initial data inside the angular-momentum barrier.

Approach to the singularity. A captured nonradial ray bends inward, crosses $r=3\mu$ and then the event horizon $r=2\mu$, and continues toward the curvature singularity. A plot in $(r\cos \varphi ,r\sin \varphi )$ is only a coordinate projection, not a Euclidean spatial slice inside the horizon. For large $u$ the cubic term in the shape equation dominates:for the inward branch with $h>0$. Its integral isThus the azimuth tends to a finite limit at the singularity, rather than winding infinitely often there. Near-critical rays can accumulate many turns near $3\mu$ before capture, which is a separate phenomenon. A strictly radial ray has constant azimuth and does not admit this azimuthal asymptotic form. The radial equation also gives $r\propto ({\tau }_{\ast }-\tau {)}^{2/5}$ for a nonradial null ray, confirming finite remaining affine parameter.

Material capture and the condition at infinity. To recover the quoted threshold, take the usual marginally bound orbit, initially at rest at infinity: $k=1$. With $\mathcal{E}=c^2$,The turning-point quadratic has discriminant $h^2(h^2-16{\mu }^2{c}^2)$. It is positive for large $r$ and has no real positive root when $0<|h|<4\mu c$, while the radial case also falls in. ThereforeAt equality the quadratic is proportional to $(r-4\mu {)}^2$ and the orbit asymptotes to the unstable marginally bound circular orbit at $r=4\mu$. For larger $|h|$, an incoming marginally bound particle encounters the outer turning point and scatters.

The source says only “from infinity”, without explicitly fixing the asymptotic speed. For a general incoming speed $v_{\infty }$, $k=(1-v_{\infty }^2/c^2{)}^{-1/2}\ge 1$ andThe same bound $|h|<4\mu c$ remains sufficient for capture, since the added constant is nonnegative. It is the exact threshold only for $k=1$. At fixed $k>1$, the critical unstable orbit satisfiesThen $h_c>4\mu c$. This states both the printed sufficient condition and the additional assumption needed to call it a sharp capture boundary.

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.

Let $\theta$ be the scattering angle between incoming and outgoing velocities, so forward scattering is $\theta =0$ and exact reversal is $\theta =\pi$. Define the length $a=\kappa /(2E)$. ThenFor $b>0$, this gives $0<\theta <\pi$; $b=0$ is the head-on limiting case $\theta =\pi$. Large deflections select small impact parameters, rather than requiring every incident alpha particle to reverse.

A short derivation uses conservation of angular momentum and conservation of energy. Choose the polar angle $\varphi =0$ at closest approach, and write $u=1/r$. The Binet equation for the repulsive Coulomb potential givesThe last equality follows by substituting this orbit into $E=m{\dot{r}}^2/2+L^2/(2m{r}^2)+\kappa /r$. The hyperbola reaches infinity at $\varphi =\pm {\varphi }_{\infty }$, with $\cos {\varphi }_{\infty }=1/ϵ$. Its scattering angle is $\theta =\pi -2{\varphi }_{\infty }$, henceThis proves the relation without assuming that the deflection is small. The symbol $ϵ$ here is a dimensionless orbital parameter, distinct from ${\epsilon }_0$.