For a nearly radial Kepler orbit, , so modest changes in specific angular momentum can strongly change pericentre. A planetary torque acting for a reference orbital time gives . Coherent kicks change by order itself after orbital-scale encounters, whereas uncorrelated kicks require an order- random-walk count. Both estimates are conditional on encounter geometry: a close encounter with a smaller impact parameter can give a much larger kick.
For unit mass, incoming speed , impact parameter , and potential energy , put . At closest approach, conservation of energy and conservation of angular momentum give the displayed radii and tangential speeds . Consequently and . These relations compare equal force magnitudes. The attractive radial case is a singular collision, so the speed-product statement does not extend to that endpoint.
For an azimuthally symmetric classical scattering map from impact parameter to scattering angle ,When several impact parameters produce the same angle, their nonnegative contributions are summed.
Equatorial null orbit in Schwarzschild spacetime Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2010 ia Paper 4 10B b Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 64 1 Solution Created 2026-10-03 Updated 2026-10-07
Put , and . To leading order the mean-motion resonance fixes . The specific angular momentum isThe asteroid's instantaneous angular speed is . Thus the high-eccentricity angular-speed crossover satisfies , givingHere , so a parabolic Kepler orbit gives the local motion accurately. Its polar equation is . Write . Since ,This is the outward crossing; the inward crossing has true anomaly modulo .
For the pericentre-to-crossover flight time use . The parabolic Kepler orbit has and , whence direct integration gives the Barker equationAt the crossover . Consequently the planet's angular displacement isThe error tends to zero as at fixed ; it includes the finite binding energy of the elliptic Kepler orbit as well as the subleading term in the Barker equation.
At the th pericentre passage the asteroid's unwrapped mean longitude is . The resonant argument therefore givesThus specifies the planet's direction relative to the asteroid's longitude of pericentre when the asteroid passes pericentre, with branches separated by . It is not the instantaneous true-longitude separation throughout the orbit. At high orbital eccentricity the asteroid quickly sweeps to a direction nearly opposite pericentre, while the planet scarcely moves. During the long outer excursion the asteroid's direction changes slowly and the planet advances substantially. This makes an astronomical conjunction during the outer excursion much more dangerous than an astronomical conjunction close to pericentre.
At the following apocentre, the planet has advanced . The asteroid's direction in the rotating reference frame is consequentlyThese apocentre directions are equally spaced. Phase protection of interior integer resonances places the planet in the largest gap between them, rather than at an apocentre direction. For , puts an apocentre toward the planet, whereas puts the apocentres at . For , puts an apocentre toward the planet, whereas puts the apocentres at . Hence the geometrically favored resonant-argument libration centres in this regime of high orbital eccentricity areThis is an encounter-avoidance argument for likely stability, not a calculation of a resonant Hamiltonian or a claim that every orbit near either centre is stable at arbitrary orbital eccentricity.
The following rotating-frame drawing uses , , and . It follows three successive Kepler orbits, which fill one planet orbital period and close the pattern. The blue segment is just the first outward half-orbit. The marked crossover is a maximum of its rotating-frame polar angle, since changes from positive to negative there. The planet stays at ; the three apocentres avoid that direction.
To estimate angular-momentum kicks to nearly radial orbits, note that , so an order-one fractional change in requires of order . A characteristic distant encounter at radii of order has tangential acceleration of order and duration of order . Its torque per unit asteroid mass therefore producesWith the most favorable coherent signs, the crude encounter count isOrder-one factors depend on and encounter geometry. Uncorrelated signs instead give a random walk count of order when this is large. A librating phase-protected orbit can suppress the kicks even further.
There is an important limit to calling this a minimum. The question does not specify an encounter impact parameter. For a weak close encounter with impact parameter and relative speed , the impulse estimate gives and hence , capped below by one. In the 2:1 case the Kepler orbit can cross the planet's orbit, so one suitably close encounter can suffice: there is no impact-parameter-independent large lower bound. For , bounds the separation below by , and supplies a geometric factor at fixed . The boxed count is the usual orbital-scale, maximally coherent estimate; the assumptions are essential.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 63 2 Solution Created 2026-10-03 Updated 2026-10-06
Let , , and restrict first to . The radiation-pressure coefficient reduces the grain's central gravitational parameter to , whereas the planet's is approximately . Their mean motions obey . Setting this ratio to gives the radiation-pressure shift of a mean-motion resonance:It lies interior to the planet exactly when , or . Equality puts the frequency commensurability at the planet's radius. For every such commensurability is exterior.
There is a frame qualification in the printed encounter formula. The circular inertial speed of the grain at is . Subtracting the planet's inertial velocity at conjunction gives the local planet-relative inertial speedThis is the velocity used in the question's two-body gravitational assist calculation. In a frame actually rotating about the star at , the grain's tangential velocity is instead , with magnitude . In particular its nonradiating shear is , rather than . Thus the requested expression uses a local translating, inertial encounter convention despite the source's rotating-frame wording. The following derivation consistently uses that convention, and its coefficients are an encounter estimate rather than an exact rotating-frame result.
Take impact parameter and a small gravitational scattering angle . Expanding the supplied hyperbolic scattering relation gives . Initially the relative tangential velocity is ; after the deflection it has tangential component . The relative speed is unchanged in the local hyperbolic Kepler orbit, but the stellar-frame kinetic energy changes. Taking the dot product with the planet's velocity givesThe specific orbital energy of the grain is . Comparing it immediately before and after the encounter at the same position givesConsequentlyThe encounter increases the grain's semi-major axis within this geometry. These steps assume , , and an orbital kick small enough for the subsequent linearization.
Choose the initial conjunction longitude as zero. The synodic period is , and the planet's longitude accumulated by the next conjunction isThese are unwrapped longitudes, counting complete revolutions. The second uses the new mean orbital period while neglecting the encounter's immediate phase offset and the difference between true and mean conjunction. A comparison reduced modulo could not implement the stated one-turn sensitivity test.
For , , so and . Conjunction-longitude sensitivity to an encounter givesRequiring this to exceed gives . At a high-index first-order mean-motion resonance, , henceThe interpretation of the resonance-overlap encounter estimate is loss of coherent encounter phases: a small kick changes the next synodic encounter time by more than one planetary revolution. Resonant protection by repeatedly meeting at controlled orbital phases can then fail, allowing irregular kicks and diffusion. This sensitivity criterion motivates an unstable region; it is not a proof that every orbit or resonant phase in it must escape.
For a quantitative sketch with radiation, put and . Differentiating the conjunction formula and substituting the kick givesEvaluate this at the positive and take as the critical-index curve. At small , increased encounter speed weakens planetary scattering, moving the threshold to larger . As rises further, the radiation-pressure shift of a mean-motion resonance brings each commensurability closer to the planet, strengthening the close encounters. The curve turns over and approaches the exterior-limit boundary . Larger planetary mass produces stronger kicks and a lower critical index. These competing effects are shown below; the curves retain the factors in , so their intercepts differ slightly from the leading high- formula. The boundary for disappearing exterior commensurabilities is geometric; divergence of the encounter approximation near should not be read as a precise physical scattering rate.
Encounter-estimate critical resonance index versus radiation pressure for three planet-to-star mass ratios
. Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 322 3 Solution Created 2026-10-03 Updated 2026-10-06
For constant component masses, the center of mass frame has and . Newton's law of universal gravitation gives the relative equation . Substitution in the component angular momenta yieldsHere is the reduced mass, while is specific angular momentum.
The printed total-energy expression lacks a factor of in its kinetic term. With the stated physical separation coordinate, the dimensionally consistent total two-body orbital energy isThe second expression is specific orbital energy; these two conventions must not be mixed. Its conservation follows directly:For the eccentricity vector, differentiate . Since is constant,The cancellation is the vector triple product identity. Thus all three corrected Kepler integrals are conserved. Dotting the eccentricity relation with also gives the useful orbit identity
For ballistic streamline focusing by a moving star, work in the star's rest frame and take the incoming gas to have velocity , with downstream behind the star at . A streamline with impact parameter approaches from . Its upstream specific angular momentum and eccentricity vector areAt its downstream axis crossing, . The orbit identity gives , hence the collision distance isAt that point, , while the component of gives . ThereforeA symmetric streamline with opposite impact parameter has the opposite transverse velocity. Their shock wave removes the opposing transverse motion while preserving the common downstream component. Thus the remaining velocity is in the star frame. In the frame where the undisturbed medium is stationary, adding the star's velocity gives zero velocity immediately after this idealized collision.
After transverse kinetic energy is dissipated, the remaining specific orbital energy is . It is negative if , or equivalently if the upstream impact parameter satisfies . This is the post-shock capture criterion in ballistic accretion; bound axial streams can return to the star. The captured incident mass flux through the corresponding disk gives the ballistic Bondi--Hoyle--Lyttleton accretion rate:This is the pressureless, dissipative capture model, with the shock's transverse energy unavailable to unbind the wake.
For wind mass transfer in a binary star, steady isotropic donor mass loss gives the local wind mass density at the companion:A fast wind has and a capture scale small compared with , so its relative incident speed is to leading order. Apply the same Bondi--Hoyle--Lyttleton accretion rate with accretor mass to obtainFinally, Kepler's third law eliminates the unquoted separation, , giving the rate in terms of the stated orbital period:Equivalently, the fast-wind accretion fraction in a circular binary is , with the relative circular orbital speed. Its smallness is consistent with using an almost undisturbed isotropic stellar wind.

