Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 316 2 iii Solution Created 2026-10-03 Updated 2026-10-05
Parameterize the trajectory by the decreasing orbital eccentricity. The Poynting–Robertson drag invariant givesFor , . Initially , and the orbit moves almost vertically down a plot of against : the apocentre distance shrinks rapidly while the pericentre distance changes little. Integrating the high-eccentricity slope givesOnce the orbit has moderate orbital eccentricity, both distances change appreciably. For example gives and in the limit. Eventually , , and the trajectory approaches the diagonal before reaching the origin. Thus the two approximate phases are apocentre contraction at nearly fixed pericentre, followed by nearly circular inward migration. They are a smooth crossover, not two separate exact solutions.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 316 3 ii Solution Created 2026-10-03 Updated 2026-10-05
The planetary expansion is a decomposition into secular eigenmodes of conservative linear Laplace-Lagrange secular theory. It assumes small orbital eccentricities, coplanarity, approximately constant semi-major axis values, and averaging over fast orbital phases away from important mean-motion resonances and close encounters. Dissipation, migration, and nonlinear secular effects are absent from this constant-coefficient representation.
The are real eigenfrequencies of the Laplace-Lagrange secular matrix, describing apsidal precession. For mode , the signed real coefficients give its amplitude in planet ; their relative magnitudes are fixed by an eigenvector, and their signs distinguish aligned and anti-aligned apsides. The common phase specifies that mode's orientation at the chosen time origin. A planet's actual orbital eccentricity is , which generally varies through interference of modes; it is not a sum of the positive mode amplitudes.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 316 3 v Solution Created 2026-10-03 Updated 2026-10-05
To first order in orbital eccentricity, the semi-minor axis is , while the ellipse's center is displaced from the star by . Thus each orbit is a circle of radius about , in complex spatial coordinates.
Write at one secular epoch. The centers of the individual circles lie on a circle of radius about . Their union is the proper-eccentricity annulusIts full radial width is and its center is displaced by toward the forced apoapsis direction. Relative to the star, if the forced longitude of periapsis is zero,The middle panel of the preceding figure shows this construction. For fully sampled orbital phases and proper phases, the annulus is occupied throughout, but its density need not be uniform; turning points in the proper radial excursion give integrable edge enhancements. The construction requires and is an instantaneous secular geometry.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 316 1 ii Solution Created 2026-10-03 Updated 2026-10-05
In resonant trapping of dust, the resonant torque must compensate the inward Poynting–Robertson drag. At the nearly fixed resonant semi-major axis, requiresSince , a necessary torque-balance condition is . When the limiting orbital eccentricity is small, this givesEquality allows only the extreme phase , with no margin for restoring torque; a robust trapped state needs . This is a necessary condition for resonance capture, not a sufficient probability-of-capture criterion for a migrating particle.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 316 1 i Solution Created 2026-10-03 Updated 2026-10-05
Expand the secular Poynting–Robertson drag rates using and . Through second order in orbital eccentricity,The second formula follows from . The cubic correction in contributes only at fourth order in .
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 316 1 vi Solution Created 2026-10-03 Updated 2026-10-05
The specific angular momentum is , so the periapsis angular speed isAt the exterior mean-motion resonance, neglecting radiation pressure, . Thus at the limiting orbital eccentricity ,The condition for overtaking the planet in angle isFor the formal second-order equilibrium, it holds for every physical bound value . To see this, rewrite , so . For , , whileThe instantaneous angular motion is therefore prograde relative to the planet near periapsis, although its mean relative motion is retrograde. The range is the regime where the eccentricity expansion is most reliable.
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 316 4 iii Solution 2026-09-28
The expansion separates naturally into three timescales:
- A short-period orbital perturbation contains rapidly varying mean longitudes. It describes individual conjunction kicks and usually averages to zero away from a commensurability.
- A resonant orbital perturbation has a nearly stationary integer combination of mean longitudes near a mean-motion resonance. Successive conjunctions then occur at correlated phases, so their effects accumulate and the resonant argument may librate.
- A secular perturbation survives averaging over both mean longitudes. It drives slow apsidal precession, nodal precession, and exchange of orbital eccentricity or orbital inclination over many orbital periods.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 322 2 Solution 2026-09-28
In centre-of-mass and relative coordinates, the kinetic energy separates into centre-of-mass motion plus , while the mutual potential is . In the centre-of-mass frame the orbital energy is therefore
The force is central, so the specific relative angular momentum is conserved. With , the radial equation becomes the Binet equationwhose solution isHere is the true anomaly measured from periapsis and is the orbital eccentricity. Substitution into the energy, or evaluation at an apsis, givesand hence
Immediately before the supernova the circular relative speed obeys . The impulsive supernova kick in a binary star changes it toApplying the energy formula just after the explosion at the unchanged separation givesUsing the original circular-speed relation,The post-explosion system is bound exactly when . The left side ranges from to , so every kick direction remains bound ifwhereas every direction unbinds the stars if
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 349 2 Solution 2026-09-28
Direct evidence for stellar feedback includes expanding ionized shells and superbubbles around young associations, hot X-ray-emitting gas in supernova remnants, broad or split emission lines, and blueshifted absorption showing cool and warm outflows. P-Cygni profiles reveal massive-star winds, while extraplanar filaments and metal-enriched gas demonstrate transport away from star-forming disks. Indirect evidence includes the galaxy mass--metallicity relation, low baryon fractions and suppressed star formation in dwarf galaxies, chemically enriched circumgalactic gas, and correlations of outflow speed and mass loading with the star formation rate.
Rapid gas removal changes the gravitational potential before stellar and dark-matter orbits can respond adiabatically. Positions and velocities are initially unchanged, but orbital binding energies rise; orbits expand and become more eccentric, and some particles escape. Repeated burst--outflow--reaccretion cycles can irreversibly transfer energy to collisionless matter and turn a central dark-matter cusp into a core. This matters because dwarf-galaxy rotation curves are used to test dark-matter microphysics: a feedback-made core can mimic a non-cold or self-interacting dark-matter signature.
Write the initial potential energy as . The virial theorem gives . If a well-mixed fraction remains after instantaneous mass loss, the immediate kinetic and potential energies are and , soAfter revirialization at , . Equating energies gives the impulsive mass-loss expansion lawThe remnant is bound only for ; loss of half or more of the gravitating mass disrupts this idealized system.
For many infinitesimal, individually revirialized losses, put in the impulsive result. To first order, . Integration yields the adiabatic mass-loss expansion lawSlow loss causes finite expansion for every positive remaining mass fraction and has no sharp disruption threshold.
Finally consider an initially circular orbit of radius around a point mass . Its speed and specific angular momentum obey and . Immediately after , these remain unchanged, while the new specific energy isUsing the orbital eccentricity relation givesIt is an ellipse for , parabolic at , and unbound for smaller , in agreement with the virial argument.
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 316 2 a Solution Created 2026-09-24 Updated 2026-09-25
Here is the planet's semi-major axis, while are the particle's semi-major axis, orbital eccentricity, and orbital inclination relative to the planet's plane. The formula assumes the circular restricted three-body problem: the planet-to-star mass ratio is small, the particle has negligible mass, and its motion is approximately heliocentric and Keplerian away from brief encounters. The planet's own orbit is circular.
The Jacobi constant is exactly conserved in that ideal rotating problem. Expressing it in heliocentric orbital elements away from the planet gives the approximately conserved Tisserand parameterThe first term measures the particle's normalized binding energy. The second is twice the component of its specific angular momentum normal to the planet's plane, normalized by .
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 316 4 a Solution Created 2026-09-24 Updated 2026-10-03
The radiation-pressure coefficient reduces the dust grain's effective stellar gravitational parameter to . At the exterior 5:4 mean-motion resonance, , so mean motion givesThis is a first-order mean-motion resonance. Its leading disturbing-function term is therefore linear in the small orbital eccentricity:Thus its dimensionless strength is of order , up to the Laplace-coefficient combination , and its resonant argument varies slowly near commensurability.
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 316 4 e Solution Created 2026-09-24 Updated 2026-09-25
During resonant trapping of dust, the mean semi-major axis is stationary. Setting the supplied to zero givesA real resonant phase exists only when . Hencewhere is the grain's orbital eccentricity when trapping begins. The inequality also has the expected sign , allowing the planetary torque to oppose Poynting–Robertson drag.
Keeping one resonant harmonic and treating orbital eccentricity as fixed reduces mean-motion resonance dynamics to a physical pendulum equation. For the inner first-order term , leading Lagrange planetary equations give . Around its stable phase, the small-amplitude libration frequency is . Evolution of eccentricity, other harmonics, and approach to a separatrix change this approximation.
In a high-orbital eccentricity interior mean-motion resonance, the resonant argument places successive rotating-frame apocentres at . Avoiding astronomical conjunction at these slow outer excursions favors for and for . This geometrical phase protection suggests stable resonant-argument libration centres; actual dynamical stability requires the resonant perturbation problem.
Planet-stationary orbit 2026-10-06
A prograde, equatorial circular orbit with orbital period equal to the planet's spin period. In the ideal model its satellite remains above one equatorial ground location. Nonzero orbital inclination or orbital eccentricity destroys exact stationarity even when the periods agree.
Polar equation of a Kepler orbit 2026-10-06
For an attractive inverse-square force, , where is true anomaly, is orbital eccentricity, and is the semilatus rectum. The specific angular momentum relation converts this geometric equation into an angular evolution law.
Poynting–Robertson drag invariant 2026-10-05
The orbit-averaged Poynting–Robertson drag equations preserve for . This relates shrinking semi-major axis to decreasing orbital eccentricity; a circular orbit is treated by its separate limiting solution.
Proper-eccentricity annulus 2026-10-05
For a common semi-major axis and proper eccentricity , randomly distributed proper longitudes of periapsis give an annulus of radii and about the point displaced from the star by minus times the forced eccentricity vector, to first order in orbital eccentricity.
Radiation pressure on circumplanetary dust 2026-10-06
The leading uniform stellar radiation pressure is a differential acceleration relative to a planet, of scale . Relative to planetocentric gravity at dust radius , its magnitude isThus lying inside a Hill sphere alone does not guarantee small radiative perturbations. Approximately circular tightly bound dust requires this ratio sufficiently small, as well as weak tides; the uniform radiative force can otherwise excite orbital eccentricity or unbind grains.
