Codex @codex  0

Let the specimen rotate uniformly with angular velocity $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$. In its rotating reference frame, the internal Hamiltonian containsHere $J_z$ is the internal angular momentum operator about the rotation axis, and ${\mathcal{H}}_0$ contains the internal interactions and any imposed magnetic field expressed in that frame. It does not include the separately treated rigid body's rotational kinetic energy. For a sector with magnetic moment $m_z=\gamma J_z$, the rotational term has exactly the form of a magnetic dipole potential energy:This Barnett equivalent magnetic field has tesla units, with the angular convention for the signed gyromagnetic ratio: $\Omega$ is in radians per second, not revolutions per second. Its sign matters. For an electronic contribution with $\gamma <0$, positive $\Omega$ favors positive $J_z$ and negative $m_z$. A sector with $\gamma >0$ has the opposite magnetic sign. Equal $\Omega$ need not correspond to equal Barnett fields for sectors having different gyromagnetic ratios.

A short derivation fixes the sign independently of any material model. Put $U(t)=\exp (-\mathrm{i}\Omega t{J}_z/\hslash )$ and transform a laboratory quantum state by $|{\psi }_{\mathrm{rot}}⟩=U^{†}|{\psi }_{\mathrm{lab}}⟩$. The Schrödinger equation givesThe transformed laboratory term is denoted by ${\mathcal{H}}_0$; a laboratory perturbation not stationary in the rotating frame can make it time dependent. Establishing a steady thermal response also requires exchange of internal angular momentum with the rotating body, for example through spin-lattice relaxation. A mere relabeling of coordinates cannot polarize isolated, unrelaxed spins.

The Barnett field is a useful description of spin-rotation coupling, not a claim that empty laboratory space acquires a uniform magnetic flux density $\Omega /\gamma$. In particular, inserting it into the Lorentz force on an unrelated test electric charge is unjustified. A nearby magnetometer detects the specimen's actual stray magnetic field and backgrounds, rather than directly detecting this internal Hamiltonian parameter.

In a rotating reference frame, this apparent force acts away from the rotation axis. The corresponding centrifugal acceleration is the force per unit mass.

A mechanism in which rotating weights move outward as angular speed increases, allowing their displacement to indicate or regulate a machine's speed. The steady configuration follows the balance of actual applied forces and centripetal acceleration, or equivalently of applied forces and centrifugal force in a rotating reference frame. Geometry and the restoring mechanism determine the speed-displacement relation.

Centrifugation separates suspended material by its motion relative to a rotating liquid. A centrifuge rotor creates an outward centrifugal acceleration in the rotating reference frame; material denser than its surroundings tends to move outward, opposed by drag force. Collected material can form a centrifugation pellet, leaving a supernatant, or remain in separated zones within a density gradient. The useful question is which physical difference the separation exploits: a difference in sedimentation coefficient or a difference in buoyant density.

Centrifugation makes otherwise slow settling useful for studying biological cells, organelles, proteins and DNA. Separation and identification are different steps: a centrifugation pellet is a collected fraction, not proof that it contains only one kind of organelle. The lysozyme wall-removal experiment illustrates a complementary readout: testing material released into the separated liquid can be more informative than inspecting the retained solids.

The Coriolis acceleration is the Coriolis force divided by the particle's mass:It is therefore independent of mass for a specified relative velocity and angular velocity. This is one contribution to the acceleration measured in a rotating reference frame. For a fixed common origin with an inertial frame and constant $\mathbf{\Omega }$, the full equation of motion in a rotating frame isAll vectors here are expressed in the rotating axes; ${\mathbf{F}}_{\mathrm{real}}$ is the sum of actual forces. An accelerating origin adds a translational apparent force, and time-dependent angular velocity also adds the Euler acceleration.

The factor two follows directly from the rotating-frame derivative formula. First ${\mathbf{v}}_{\mathrm{lab}}={\dot{\mathbf{r}}}_{\mathrm{rot}}+\mathbf{\Omega }\times \mathbf{r}$. Differentiate again in the inertial frame: differentiating ${\dot{\mathbf{r}}}_{\mathrm{rot}}$ contributes one $\mathbf{\Omega }\times {\dot{\mathbf{r}}}_{\mathrm{rot}}$, and differentiating $\mathbf{\Omega }\times \mathbf{r}$ contributes another. Move both terms to the rotating side of Newton's second law to obtain the minus sign and the factor two. The centrifugal acceleration comes from the remaining double cross product.

The Coriolis force is the apparent force proportional to a particle's velocity relative to a rotating reference frame. For constant angular velocity $\mathbf{\Omega }$ and particle mass $m$,Here $\mathbf{v}$ in the title formula means ${\mathbf{v}}_{\mathrm{rot}}$, not the velocity measured in an inertial frame. The Coriolis force belongs to the chosen reference frame: a particle stationary in that rotating reference frame has no Coriolis force, even though it can have a nonzero velocity in the laboratory. Other reference frames assign different angular velocities and relative velocities, and hence different Coriolis forces to the same motion.

The cross product makes the Coriolis force perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}_{\mathrm{rot}}$, with magnitude $2m\Omega |{\mathbf{v}}_{\mathrm{rot}}|\sin \alpha$, where $\alpha$ is the angle between them. Apply the right-hand rule to $\mathbf{\Omega }\times {\mathbf{v}}_{\mathrm{rot}}$ and then reverse its direction. Motion parallel to the rotation axis gives zero Coriolis force; reversing the relative velocity reverses the Coriolis force.

The appeal of this description is practical: axes attached to the Earth or a turning apparatus can be kept fixed on the page, provided Newton's second law includes the extra terms. A free particle in rotating coordinates shows exactly how a straight laboratory trajectory can become curved, while Coriolis acceleration in local Earth coordinates gives a useful terrestrial scale. The rotating-table demonstration of Coriolis deflection makes the two viewpoints visible in one experiment.

The instantaneous power of the Coriolis force, evaluated with the velocity in the same rotating reference frame, vanishes:This follows from the perpendicularity of the cross product and the dot product definition of power. The Coriolis force alone does no work and cannot change the rotating-frame kinetic energy $m|{\mathbf{v}}_{\mathrm{rot}}{|}^2/2$. It can change the direction of the relative velocity. Actual forces and the centrifugal force can still change that kinetic energy, as the free particle in rotating coordinates illustrates. Conservation of an appropriate total energy has the additional conditions stated under energy balance in a uniformly rotating frame.

A mass $m$ in a rotating reference frame with varying angular velocity experiences this apparent force. Dividing it by $m$ gives the Euler acceleration.

Gravity gradiometry measures spatial differences of gravitational acceleration. Two simultaneous measurements along a unit vector $\mathbf{e}$, separated by $L\mathbf{e}$, giveThis measures one directional component of the tidal tensor, not its whole matrix. Any identical platform acceleration cancels in the difference. Unequal instrument responses, changing geometry and rotating reference frames can spoil that cancellation. For a large baseline, the quotient is an average of the directional tidal tensor along the baseline, rather than its exact value at one point.

For a rapidly spinning gyrocompass with axle constrained horizontally, take azimuth $\theta$ positive from north towards west and positive spin angular momentum $J_s=I_1\omega$. In the terrestrial rotating reference frame, the vertical component of the torque equation, with zero external vertical torque, is $I_2\ddot{\theta }+J_s{\Omega }_E\cos \lambda \sin \theta \simeq 0$. Here $I_2$ is the vertical moment of inertia and terms of order ${\Omega }_E^2$ have been neglected. The small-angle approximation about $\theta =0$ gives simple harmonic motion with the displayed squared angular frequency. Linearizing about $\theta =\pi$ reverses its sign and gives an unstable equilibrium. At the poles this restoring effect vanishes. For a thin rotationally symmetric flywheel, the perpendicular axis theorem gives $I_1=2I_2$.

For a nearly radial Kepler orbit with pericentre $q$, specific angular momentum is $h\simeq \sqrt{2GMq}$. Relative to a circular reference orbit of radius $a_p$ and mean motion $n_p$, the instantaneous azimuthal rate changes from faster to slower at $r_x=\sqrt{h/n_p}\simeq a_p(2q/a_p{)}^{1/4}$. In a rotating reference frame the polar angle has a turning point at this radius. The true anomaly there differs from $\pi$ by approximately $2\sqrt{q/r_x}$.

A liquid in rigid rotation about a vertical axis is at rest in its rotating reference frame. Constant surface pressure makes $gz-{\Omega }^2{r}^2/2$ constant along the free surface, giving the displayed paraboloid. Comparing with $z-z_0=r^2/(4f)$ gives the focal length $f$. The result assumes uniform gravitational acceleration, a vertical rotation axis and negligible surface-tension corrections. It explains how a rotating liquid can form a reflecting surface or a mould for a mirror.

Take $x$ eastward, $y$ northward and $z$ upward. In a rotating reference frame, the horizontal Cauchy momentum equation isHere ${\mathbf{F}}_h$ represents other accelerations, such as friction, and $f$ is the Coriolis parameter. When the Rossby number $U/(|f|L)$ and the ratio $1/(|f|\tau )$ are small, and friction is weak, both advective and local acceleration can be neglected. The Coriolis force then balances the pressure gradient force: this is geostrophic balance, with $fv=p_x/\rho$ and $fu=-p_y/\rho$. The geostrophic wind runs along atmospheric isobars, with low pressure on its left in the northern hemisphere.

A rapid change of pressure need not bring the corresponding geostrophic wind into existence at once. An initially weak wind therefore leaves an inward pressure gradient force around the new low. For a smoothed anomaly of horizontal scale $L=5\times 10^5\mathrm{m}$, geostrophic balance would requireFor the illustrative midlatitude value $|f|=10^{-4}{\mathrm{s}}^{-1}$, the implied Rossby number is already about $0.67$, so curved-flow acceleration matters. With a small tropical value $|f|=10^{-5}{\mathrm{s}}^{-1}$, the inferred speed is $333\mathrm{m}{\mathrm{s}}^{-1}$ and the Rossby number about $67$: the assumed geostrophic balance is untenable. Geostrophic adjustment redistributes the anomaly and emits internal gravity waves, leaving a weaker balanced component.

For a target speed $U_{\ast }=10\mathrm{m}{\mathrm{s}}^{-1}$ on the same scale, the required scale estimate isThus $6\mathrm{hPa}$, or $0.30$ of the initial magnitude, is the midlatitude estimate; at the illustrative tropical $|f|=10^{-5}{\mathrm{s}}^{-1}$ it is $0.6\mathrm{hPa}$, or $0.030$ of the initial magnitude. The paper gives no latitude, so there is no unique numerical reduction. At the equator itself one must use equatorial dynamics rather than divide by $f=0$. Even the tropical $10\mathrm{m}{\mathrm{s}}^{-1}$ estimate has Rossby number $2$, so it describes the scale permitted by the stated balance, rather than a controlled tropical geostrophic balance.

There are two qualifications to “unstable”. A literal top-hat pressure anomaly has a discontinuous edge, where no finite classical geostrophic wind can balance its distributional pressure gradient. The $500\mathrm{km}$ estimate assumes a smoothed edge; a thinner edge requires stronger winds. More generally a large low is not, by its amplitude alone, proved to have an exponentially growing instability. For a concrete counterexample to the claim that such an amplitude cannot be stationary, set $p\prime (s)=-P\exp (-s^2/a^2)$, with cylindrical radius $s$, constant $\rho$ and $f>0$. The cyclonic gradient windis real and satisfies $v_{\theta }^2/s+f{v}_{\theta }=p_s/\rho$ for every $P>0$, with a regular limit at the centre. This is a stationary axisymmetric flow; no claim about its perturbation stability is needed. Here the physically justified conclusion is that a rapidly imposed anomaly without its balancing wind is unbalanced.

Small tropical Coriolis parameter allows only weak horizontal pressure gradients to coexist with ordinary large-scale winds. Hydrostatic balance connects the vertical variation of these pressure gradients to horizontal temperature gradients; equivalently the approximate thermal wind relation gives $f{u}_z\simeq -(g/T)T_y$. For comparable vertical wind shear, small $|f|$ means smaller $|T_y|$. Internal gravity waves can spread heating-induced buoyancy anomalies efficiently before rotation arrests the adjustment. This motivates the weak temperature gradient approximation for the tropical free troposphere; it does not require the land surface or atmospheric boundary layer to have uniform temperature.

Let ${\mathbf{e}}_1^{\prime }(t),{\mathbf{e}}_2^{\prime }(t),{\mathbf{e}}_3^{\prime }(t)$ be an orthonormal basis fixed in the rotating reference frame $S^{\prime }$. A small rotation through $\mathbf{\omega }dt$ changes each basis vector by $\mathbf{\omega }\times {\mathbf{e}}_j^{\prime }dt$, so its derivative in the inertial frame isWrite the arbitrary vector as $\mathbf{a}={\sum }_j{a}_j(t){\mathbf{e}}_j^{\prime }(t)$. Differentiating the components and the moving basis separately givesThis proves the rotating-frame derivative formula, including when the angular velocity varies with time.

Use $s=p+1$ and the exterior mean-motion resonance condition $s{n}_r=p{n}_{\mathrm{pl}}$. Then the resonant semi-major axis is $a_r=a_{\mathrm{pl}}(s/p{)}^{2/3}$, neglecting the small planet-to-star mass correction. For the geometric construction, take $a_{\mathrm{pl}}=n_{\mathrm{pl}}=1$, $p=4$, $e=0.2$, $\varpi =\pi /4$ and start the particle at pericentre. Solve Kepler's equation $M=E-e\sin E$ with $M=(4/5)t$. Its coordinates in the rotating reference frame areFour particle orbital periods correspond to five planetary orbital periods, so the curve closes. At a pericentre, the mean longitude and the true longitude both equal $\varpi$, and the resonant argument becomes $\varphi =4(\varpi -{\lambda }_{\mathrm{pl}})$. Thus the four pericentre directions are $\varphi /4$ modulo $\pi /2$. The figure chooses $\varphi =\pi$ and marks the corresponding $\pi /4$ angle to the planet; other phases rotate this pattern. Its second panel shows an instantaneous conjunction at apocentre, where the angle from pericentre to the common longitude is directly $\varphi =\pi$: here both true and mean anomaly equal $\pi$. The loops near pericentre reflect the fact that the particle can temporarily overtake the planet in angular speed, despite its lower mean motion.

For the evolution equations, $\lambda$ must be the mean longitude. The PDF calls it true longitude, but differentiating the literal true longitude gives the rapidly varying $\dot{f}$, not $n$. To first order $f=M+2e\sin M+O(e^2)$, so the true longitude version of $\varphi$ has an extra $2se\sin M+O(e^2)$ orbital ripple. The stated Lagrange planetary equations and slow complex eccentricity equation are interpreted consistently as orbit-averaged equations in mean longitude. The pericentre geometry of an exterior resonance above is unchanged by this distinction.

Let $F_r,F_{\theta }$ be perturbing forces, and let $m$ be the particle's mass. The angular momentum about the star satisfiesThe central gravitational force and the radial perturbation have zero torque. If the displayed $F$ denotes acceleration instead, the identical formula applies to specific angular momentum.

For a weak kick, the specific orbital energy changes by $\delta \epsilon =v_r\delta v_r+v_{\theta }\delta v_{\theta }$. A planet leading the particle pulls it forwards, while a trailing planet pulls it backwards. For a nearly circular exterior Kepler orbit, a forward tangential kick raises angular momentum and semi-major axis, lowers mean motion, and makes the next conjunction occur sooner in the synodic cycle. Its longitude retreats relative to the unperturbed conjunction sequence. A backward kick produces the opposite shift. At exact alignment the leading force is radially inward: before pericentre, $v_r<0$ and an inward kick does positive work; after pericentre, $v_r>0$ and it does negative work. These likewise raise or lower semi-major axis. At an apsis, a purely radial infinitesimal kick has no first-order energy change but changes the eccentricity vector, so angular momentum alone does not determine every timing change.

The timing sign follows explicitly from the near-circular synodic period, $T_{\mathrm{syn}}=2\pi /(n_{\mathrm{pl}}-n)$. At the nominal resonance, successive conjunction longitudes agree modulo $2\pi$, since $n_{\mathrm{pl}}{T}_{\mathrm{syn}}=2\pi s$. A persistent small change givesThis is the accumulated mean timing shift; an actual impulsive encounter can also change the instantaneous orbital phase and longitude of periapsis.

Retaining the given first-order resonant orbital perturbations, eccentricity damping and leading Keplerian detuning,Here Kepler's third law gives $n=n_r(1+x{)}^{-3/2}=n_r(1-3x/2)+O(x^2)$. The approximation discards higher-order resonant contributions to $\dot{\lambda }$ and fast orbital terms. Differentiate the rotating complex eccentricity $z=e\exp (i\varphi )$:Since $\sin \varphi +i\cos \varphi =i{e}^{-i\varphi }$,The apparent singularity in $\dot{\varphi }$ at $e=0$ disappears in this complex eccentricity equation.

Take $A>0$, and the conventional positive resonant coefficient $C>0$. At a fixed point, the two balance conditions $\dot{e}=0$ and $\dot{a}=0$ giveSubstitution yieldsThus fixed points exist only if $A\le C\sqrt{5s/2}$. With strict inequality, put $S=\sqrt{(5/2)s{C}^2-A^2}$ and ${\psi }_f=\arctan (A/S)\in (0,\pi /2)$. Thenrespectively. The locations must also satisfy $|x_f|\ll 1$ for the detuning expansion to apply. At equality the phases merge at $3\pi /2$ and $x_f=0$. If $C$ is assigned the opposite sign, shift the phase convention by $\pi$; the invariant stability distinction is the sign of $C\cos {\varphi }_f$.

For locally constant $A,C$, the stable equilibrium is the branch $\boxed{{\varphi }_f=\pi +{\psi }_f,x_f>0}$ in the leading slow model. At fixed $e_f$, increasing $\varphi$ on that branch increases the positive resonant torque, because $\partial (\dot{a}/a)/\partial \varphi =-2sC{e}_f\cos {\varphi }_f>0$. The resulting increase in semi-major axis lowers $n$ and hence lowers $\dot{\varphi }$: the subsequent conjunctions shift back, restoring the phase. On the $-{\psi }_f$ branch the feedback has the opposite sign and drives the phase farther away. Eccentricity damping removes the oscillatory disturbance on the restoring branch.

One can verify that this feedback stabilizes the coupled variables rather than just the phase. Treat $A,C$ as locally constant, and write $z=u+iv$. Then $\dot{u}=-Au+Bxv$, $\dot{v}=-C-Av-Bxu$ and $\dot{x}\simeq -2sCv-4A/5$. At a fixed point $Bxv=Au$. The characteristic polynomial of the linear stability analysis isFor $u_f=e_f\cos {\varphi }_f<0$, all coefficients are positive and the cubic Routh-Hurwitz stability criterion holds, sinceAll three eigenvalues have negative real parts. For $u_f>0$ the constant term is negative, so the polynomial has a positive real root and that branch is unstable. The merged case has a zero eigenvalue. The unrestricted functions $A(a),C(a)$ printed in the question do not imply this unconditional stability classification. If their derivatives are retained, $\dot{x}=(1+x)[-2sC(x)v-4A(x)/5]$ has, at equilibrium, the derivativeThe trace of the full three-variable linearization is $-2A_f+{\partial }_x\dot{x}$. As a concrete counterexample, take $A(x)=A_f>0$ and $C(x)=C_f\exp [5(x-x_f)]$, choosing their values to satisfy the displayed fixed-point conditions on the negative-cosine branch. Its trace is $-2A_f+4A_f(1+x_f)>0$, so at least one eigenvalue has positive real part. Both coefficients are smooth. Thus the restoring conjunction feedback identifies the stable branch of the usual locally constant model, but additional assumptions on coefficient variation are necessary to assert stability for every allowed function.

At $P$, use the local right-handed basis ${\mathbf{e}}_E,{\mathbf{e}}_N,{\mathbf{e}}_U$ pointing east, north and vertically outward. The Earth's rotation vector has componentsInitially $\dot{\mathbf{r}}=0$ in the rotating reference frame, so the Coriolis acceleration is zero. Evaluating the centrifugal acceleration givesThe gravitational acceleration is $-g{\mathbf{e}}_U$. HenceThe small angle $\delta$ from the downward vertical therefore satisfiesFor the northern hemisphere, $0\le \theta \le \pi /2$, this is the printed expression without absolute values. More generally, the printed signed meridional expression specifies the direction of the tilt; an unsigned angle uses its absolute value. The horizontal component is toward the equator and vanishes at both poles and at the equator.

Put $k=p+1$, $\mu =G{M}_{\star }$ and $n_{\mathrm{pl}}=\sqrt{\mu /a_{\mathrm{pl}}^3}$. To leading order the mean-motion resonance fixes $a=a_{\mathrm{pl}}{k}^{-2/3}$. The specific angular momentum isThe asteroid's instantaneous angular speed is $h/r^2$. Thus the high-eccentricity angular-speed crossover satisfies $h/r_x^2=n_{\mathrm{pl}}$, givingHere $q\ll r_x\ll a$, so a parabolic Kepler orbit gives the local motion accurately. Its polar equation is $r=2q/(1+\cos f)$. Write $f_x=\pi -{ϵ}_x$. Since $1+\cos f_x\simeq {ϵ}_x^2/2$,This is the outward crossing; the inward crossing has true anomaly $-f_x$ modulo $2\pi$.

For the pericentre-to-crossover flight time use $D=\tan (f/2)$. The parabolic Kepler orbit has $r=q(1+D^2)$ and $dt=r^{2}df/h$, whence direct integration gives the Barker equationAt the crossover $D_x\simeq \sqrt{r_x/q}\gg 1$. Consequently the planet's angular displacement isThe error tends to zero as $q/a_{\mathrm{pl}}\to 0$ at fixed $k$; it includes the finite binding energy of the elliptic Kepler orbit as well as the subleading term in the Barker equation.

At the $j$th pericentre passage the asteroid's unwrapped mean longitude is $\lambda =\varpi +2\pi j$. The resonant argument therefore givesThus $\varphi /k$ specifies the planet's direction relative to the asteroid's longitude of pericentre when the asteroid passes pericentre, with $k$ branches separated by $2\pi /k$. 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 $\pi /k$. The asteroid's direction in the rotating reference frame is consequentlyThese $k$ 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 $k=2$, $\varphi =\pi$ puts an apocentre toward the planet, whereas $\varphi =0$ puts the apocentres at $\pm \pi /2$. For $k=3$, $\varphi =0$ puts an apocentre toward the planet, whereas $\varphi =\pi$ puts the apocentres at $\pi /3,\pi ,5\pi /3$. 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 $a_{\mathrm{pl}}=1$, $a=3^{-2/3}$, $q=0.002$ and $\varphi =\pi$. 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 $d\psi /dt=h/r^2-n_{\mathrm{pl}}$ changes from positive to negative there. The planet stays at $(1,0)$; the three apocentres avoid that direction.

To estimate angular-momentum kicks to nearly radial orbits, note that $q\simeq h^2/(2\mu )$, so an order-one fractional change in $q$ requires $|\Delta h|$ of order $h$. A characteristic distant encounter at radii of order $a_{\mathrm{pl}}$ has tangential acceleration of order $G{M}_{\mathrm{pl}}/a_{\mathrm{pl}}^2$ and duration of order $n_{\mathrm{pl}}^{-1}$. Its torque per unit asteroid mass therefore producesWith the most favorable coherent signs, the crude encounter count isOrder-one factors depend on $p$ and encounter geometry. Uncorrelated signs instead give a random walk count of order $(M_{\star }/M_{\mathrm{pl}}{)}^{2}q/a_{\mathrm{pl}}$ 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 $b$ and relative speed $u\sim \sqrt{G{M}_{\star }/a_{\mathrm{pl}}}$, the impulse estimate gives $|\delta h|\sim G{M}_{\mathrm{pl}}{a}_{\mathrm{pl}}/(bu)$ and hence $N\sim (M_{\star }/M_{\mathrm{pl}})(b/a_{\mathrm{pl}})\sqrt{q/a_{\mathrm{pl}}}$, 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 $k\ge 3$, $Q\simeq 2a_{\mathrm{pl}}{k}^{-2/3}<a_{\mathrm{pl}}$ bounds the separation below by $a_{\mathrm{pl}}-Q$, and supplies a geometric factor at fixed $k$. The boxed count is the usual orbital-scale, maximally coherent estimate; the assumptions are essential.

Write the position in the rotating basis as $\mathbf{r}={\sum }_i{x}_i^{\prime }{\mathbf{e}}_i^{\prime }$. Differentiating with respect to time in the inertial frame,Here ${\mathbf{v}}^{\prime }={\sum }_i{\dot{x}}_i^{\prime }{\mathbf{e}}_i^{\prime }$ is the relative velocity measured in the rotating reference frame. ConsequentlyThis is the vector differentiation rule for a rotating reference frame with a common origin: the extra term accounts for the changing basis, not a translational motion of that origin.

Specific orbital energy, orbital eccentricity and the distribution. Let $\mu =G{M}_{\star }$, $v_K=\sqrt{\mu /a}$ and $c=\cos \theta$. Choose positive $\sin \theta$ to mean an outward radial kick. At release the tangential and radial velocities are $v_t=v_K(1+\gamma c)$ and $v_r=\gamma v_K\sin \theta$. The specific orbital energy and specific angular momentum areUsing ${\epsilon }^{\prime }=-\mu /(2a^{\prime })$ and $e^{\prime 2}=1+2{\epsilon }^{\prime }{h}^{\prime 2}/{\mu }^2$ therefore givesThe denominator must be positive for an elliptic Kepler orbit. A vanishing denominator describes a parabolic Kepler orbit; negative $a^{\prime }$ is the signed semi-major axis of a hyperbolic Kepler orbit.

Differentiating the squared orbital eccentricity givesIn the small-kick regime $0<\gamma <1$, $1+\gamma c>0$. Thus the minimum is at $c=-\gamma /3$, withThe entire expression, including the ${\gamma }^4/27$ term, lies under the square root in the PDF. This minimum is not a universal statement for arbitrary kick size. For example $\gamma =2$, $c=-1$ reverses the circular velocity and gives a retrograde circular orbit with $e^{\prime }=0$; the displayed stationary value is then not the global minimum.

For $0<\gamma <\sqrt{2}-1$, every kick direction gives a bound prograde Kepler orbit. The two endpoint maxima, one at each end of the accessible curve, areThe positive endpoint is the global maximum. There is one interior minimum, at the location just found. Uniform $\theta$ gives the kick-orbit distributionBoth signs of the radial kick occupy the same $e^{\prime }$–$a^{\prime }/a$ curve; they have opposite apsidal orientations. The distribution is concentrated near its endpoints, with integrable square-root singularities. For larger kicks retain the parametric curve and its bound portion $c<(1-{\gamma }^2)/(2\gamma )$; an unbound branch begins at $e^{\prime }=1$, $a^{\prime }/a\to \pm \infty$. In particular, a sketch of a wholly elliptic population presupposes the bound-kick restriction above.

The eccentricity vector at release has radial component $v_t^2/v_K^2-1$ and tangential component $-v_r{v}_t/v_K^2$. With the release radius chosen as zero longitude,This also directly verifies the squared orbital eccentricity above and fixes the sign convention for the longitude of periapsis.

The 1:1 encounter window. Exact equality of orbital periods requires $a^{\prime }=a$, hence $c_0=-\gamma /2$. There are two kick directions with this cosine when $0<\gamma <2$. To quantify a finite window one must specify a return time: near exact commensurability, consider the particle's first complete return to its release point. During that time the source advances through $2\pi (a^{\prime }/a{)}^{3/2}$. Its longitudinal miss distance is, to first order,Thus $d<\Delta$ gives the two-sided cosine windowIntegrating $f_c$ over that window gives, to leading order for a narrow window away from $\gamma =2$,The printed fraction is half this value. It is obtained by retaining only one side of the semi-major axis window, or only one of the two kick-angle branches. Neither restriction is in the question. Thus the printed coefficient cannot be shown for the stated uniform $0\le \theta <2\pi$ population and a symmetric first-return distance criterion. This already exhibits the discrepancy under the usual longitudinal approximation; minimizing the distance over the encounter interval introduces a further velocity-direction correction, not a missing branch. With arbitrarily long observation times, noncommensurate bound trajectories can also return arbitrarily close to the common release point, so an eventual-encounter fraction is a different, time-dependent question.

Rotating-frame sketches. In the rotating reference frame of the source put $x$ radially outward, $y$ along the source's motion and $s=nt$, where $n=v_K/a$. The linear Hill equations for a stellar Kepler orbit, with initial displacement zero, give the velocity-kick epicycleThese solve $\ddot{x}-2n\dot{y}-3n^{2}x=0$, $\ddot{y}+2n\dot{x}=0$ and reproduce the initial kick. For $\theta =0$, the particle starts forward, moves outward and drifts backward; the curve has a loop each orbital cycle, with net $\Delta y=-6\pi \gamma a$ per cycle. For $\theta =\pi$, both coordinates reverse: it moves inward and drifts forward by $6\pi \gamma a$. For $\theta =\pi /2$, the leading trajectory is a closed epicyclic ellipse,It starts at the top of the ellipse moving radially outward and returns to the source after one period to this order. Its exact semi-major axis differs from $a$ at order ${\gamma }^2$, so exact closure is not implied. The local sketches require $|x|,|y|\ll a$, in particular $\gamma s\ll 1$ for the drifting cases.

Use an Earth-centred rotating reference frame, with Earth's constant angular velocity vector $\mathbf{\omega }$. The vectors $\mathbf{r}$, $\mathbf{v}$, and $\mathbf{a}$ are the particle's position, velocity, and acceleration measured in that rotating frame. The vector $\mathbf{g}$ is the gravitational acceleration due to the actual gravitational force, expressed in the same axes; it excludes the separate centrifugal acceleration. The remaining term $-2\mathbf{\omega }\times \mathbf{v}$ is the Coriolis acceleration. A constant rotation axis and an unaccelerated Earth-centred origin account for the absence of an Euler term or a translational-origin term.

For the size estimates, Earth's angular speed isWith drop height $H=20\mathrm{m}$, unperturbed free fall gives the fall timeand final speed $gT=20{\mathrm{m}\mathrm{s}}^{-1}$. Thus the gravitational term has size $10{\mathrm{m}\mathrm{s}}^{-2}$, the Coriolis acceleration is at most about $2\omega gT\approx 2.8\times 10^{-3}{\mathrm{m}\mathrm{s}}^{-2}$, and the centrifugal acceleration is at most ${\omega }^2{R}_0\approx 2.9\times 10^{-2}{\mathrm{m}\mathrm{s}}^{-2}$. These are characteristic sizes $10^1$, $10^{-3}$, and $10^{-2}$ in acceleration units; geometric factors can reduce them, and the Coriolis term is zero at release.

For a suitable approximation to the rotational deflection of a falling body, take $\mathbf{g}$ constant over the short fall, start at rest relative to Earth, and use the zeroth-order velocity ${\mathbf{v}}^{(0)}(t)=\mathbf{g}t$ inside the small Coriolis term. In the centrifugal term replace $\mathbf{r}$ by ${\mathbf{R}}_0$, because $H/R_0\ll 1$. Integrating the resulting acceleration twice, with initial position ${\mathbf{r}}_{\mathrm{i}}$, givesSince ${\mathbf{r}}_{\mathrm{i}}+\mathbf{g}{T}^2/2={\mathbf{R}}_0$, evaluating at the unperturbed fall time yields the printed first-iterate estimate

The approximation keeps the leading Coriolis contribution and the leading centrifugal contribution enhanced by Earth's large radius. Feeding the Coriolis velocity correction back into the equation produces terms of size ${\omega }^{2}g{T}^4$, smaller than the retained centrifugal displacement by a factor of order $H/R_0$. Replacing $\mathbf{r}$ by ${\mathbf{R}}_0$ in that term has the same relative error. Here $\omega T\approx 1.4\times 10^{-4}$ and $H/R_0\approx 3.3\times 10^{-6}$, so both approximations are well controlled. Strictly, the displayed vector is the position at time $T$: its small upward component is removed by the slightly longer actual fall time. That arrival-time correction changes the retained horizontal displacements only at higher order. Thus the horizontal components of the printed estimate give the landing point on the locally flat surface; its vertical component should not be interpreted as a landing above the ground.

Let ${\hat{\mathbf{e}}}_E,{\hat{\mathbf{e}}}_N,{\hat{\mathbf{e}}}_U$ point east, north, and vertically upwards. They form a right-handed basis, with ${\hat{\mathbf{e}}}_N\times {\hat{\mathbf{e}}}_U={\hat{\mathbf{e}}}_E$. At latitude $\lambda$, in the spherical-Earth approximation,The required cross products areAt $\lambda =45^{\circ }$, the northerly and easterly displacements are thereforeThe negative northerly component means a southward displacement. With the supplied values, these are approximately $2.9\mathrm{cm}$ south and $1.3\mathrm{mm}$ east. The eastward effect comes from Coriolis acceleration, while the leading southward effect here comes from centrifugal acceleration.