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Use a translating accelerated reference frame; its axes do not rotate. If $\mathbf{R}=(\xi ,\eta ,\zeta )$ is the block's position relative to the table, then its inertial position is $\mathbf{r}=\mathbf{R}+\mathbf{A}\sin \omega t$. Its relative velocity is $\dot{\mathbf{R}}$ and its inertial acceleration is $\ddot{\mathbf{R}}-{\omega }^2\mathbf{A}\sin \omega t$. Let $N$ be the upward normal force. During contact $\zeta =\dot{\zeta }=\ddot{\zeta }=0$, and the linear tangential resistance givesThese are Newton's second law transformed to the table's reference frame; the extra terms arise from the table's acceleration. The last equation gives $N=m(g-A_z{\omega }^2\sin \omega t)$ while contact persists. If contact is lost, both the contact resistance and $N$ vanish, and free flight must be described by the corresponding gravity-only equations.

For $A_z=0$, assume $\lambda >0$ and put $\gamma =\lambda /m$. Seek a periodic particular solution ${\xi }_p=b\sin \omega t+c\cos \omega t$. Substitution into the first differential equation givesSolving yieldsThe homogeneous solutions are a constant and a multiple of $e^{-\gamma t}$. Choose the phase $0<\theta <\pi /2$ with $\tan \theta =\omega /\gamma =m\omega /\lambda$. Then $b=-A_x{\sin }^2\theta$ and $c=-A_x\sin \theta \cos \theta$, soAlso $\eta (t)={\eta }_{\infty }+C_{\eta }{e}^{-\lambda t/m}$. The decaying transients leave the stated steady orbit. Positive damping is essential to this late-time assertion; for $\lambda =0$ an additional constant drift may persist.

No attractive contact force is available, so $N$ cannot be negative. Over a whole oscillation its minimum is $m(g-{\omega }^2|A_z|)$. Thus strictly positive contact loading throughout requires $\boxed{{\omega }^2|A_z|<g}$, or the printed form with $A_z$ taken as a nonnegative vertical amplitude. If ${\omega }^2|A_z|>g$, contact cannot persist: it would require $N<0$ during part of the cycle. At equality, $N$ vanishes only at isolated instants and the ideal constrained motion can still remain in contact. Hence the strict inequality characterizes positive loading; mere nonnegative contact permits the marginal equality case.

Use units $c=1$ and signature $(+---)$. For a massive test particle, write its four-momentum as $p^i=m{u}^i$, where $u^i=d{x}^i/d\tau$, $\tau$ is Minkowski proper time, and $u^i{u}_i=1$. The stated flat-background scalar-force theory of gravity then givesContract with $p_i=m{u}_i$. The two terms cancel:Thus rest mass is conserved and the acceleration law isThere is no mass or composition parameter in this equation. Particles released at the same event with the same velocity consequently have identical trajectories. In a static weak field at small speed, its spatial part becomes $d^2\mathbf{x}/d{t}^2=-\nabla \Phi$, since raising a spatial index introduces a minus sign. The field equation likewise reduces to ${\nabla }^2\Phi =4\pi G\rho$ for nonrelativistic matter, because $T^i{}_i\simeq \rho$ and $\square =-{\nabla }^2$ in the static limit.

Therefore yes: the model agrees with the Eötvös test of universal free fall. In the point-particle/test-body approximation, inertial and passive gravitational masses have the same universal ratio. This is the weak equivalence principle tested by the Eötvös experiment; it is not a claim about strongly self-gravitating bodies or a proof of all aspects of the Einstein equivalence principle.

Take both curvature radii to be one. Coordinate ranges depend on the global identification, which a line element alone cannot fix. For the ordinary two-dimensional de Sitter spacetime,This is the unit hyperboloid $X_0^2-X_1^2-X_2^2=-1$ in ambient signature $(+--)$, parametrized by $X_0=\sinh t$, $X_1=\cosh t\cos \chi$, $X_2=\cosh t\sin \chi$. Unwrapping $\chi$ instead gives its universal cover.

For two-dimensional anti-de Sitter spacetime, the embedding $X_0^2+X_1^2-X_2^2=1$ in ambient signature $(++-)$ is parametrized by $X_0=\cosh r\cos t$, $X_1=\cosh r\sin t$, $X_2=\sinh r$. On this hyperboloid $r\in \mathbb{R}$ and $t$ is periodic modulo $2\pi$, creating closed timelike circles. The physically usual universal cover removes that periodicity:Both signs of $r$ are needed for the full two-dimensional spatial line. Taking $r\ge 0$ without an additional construction would omit one of its two ends. The following causal comparison uses the AdS universal cover.

Both models are maximally symmetric Lorentzian manifolds, with constant sectional curvature and no curvature singularities. In the paper's convention their Ricci scalars are respectively $+2$ and $-2$. The de Sitter spatial scale factor $a(t)=\cosh t$ contracts to a nonzero minimum and then expands; $t=0$ is not a big-bang singularity. The AdS metric is static, and ${\partial }_t$ is timelike everywhere because $g_{tt}={\cosh }^{2}r>0$.

The embeddings give a concise classification of all affinely parametrized geodesics. Geodesic acceleration in the ambient space is normal to the hyperboloid, hence proportional to $X$. Differentiating the constraint and setting $X^{\prime 2}=\kappa$, with $\kappa =1,0,-1$ for timelike, null and spacelike curves, givesEvery solution stays in the two-plane spanned by its initial point $P$ and tangent $V$, so every geodesic is a plane section through the ambient origin. For de Sitter, these are $P\cosh s+V\sinh s$ for timelike curves, $P+sV$ for null curves, and $P\cos s+V\sin s$ for spacelike curves. Thus timelike geodesics and null geodesics are complete and nonclosed, while spacelike geodesics are closed circles on the ordinary hyperboloid. These are the geodesics of two-dimensional de Sitter spacetime.

For AdS the roles of the circular and hyperbolic solutions are reversed: timelike curves are $P\cos s+V\sin s$, null curves are $P+sV$, and spacelike curves are $P\cosh s+V\sinh s$. Timelike curves close after proper time $2\pi$ on the hyperboloid, but their lifts on the universal cover continue to ever later time rather than forming closed curves. All three kinds extend for arbitrary affine parameter. This is geodesics of two-dimensional anti-de Sitter spacetime, not an assertion that geodesics reach the conformal boundary in finite physical affine length.

The coordinate first integrals make the behavior concrete. In de Sitter, spatial rotational symmetry gives a conserved $P_{\chi }={\cosh }^{2}td\chi /ds$, and normalization givesConstant-$\chi$ observers are timelike geodesics. Nontrivial null geodesics have affine parameter proportional to $\sinh t$, which diverges at both $t\to \pm \infty$. In AdS the conserved static energy is $E={\cosh }^{2}rdt/ds$, withFor timelike curves, $E\ge 1$ andThey oscillate through the center and never reach spatial infinity; $r=0$ is the $E=1$ geodesic. For null curves, $\lambda =\pm \sinh r/E+\text{constant}$, so both ends are at infinite affine parameter. Spacelike AdS geodesics likewise extend to the two spatial ends.

For the conformal structure, introduce de Sitter conformal timeMultiplying by ${\cos }^2\eta$ gives the conformal cylinder of two-dimensional de Sitter spacetime. The past and future conformal infinities are spacelike circles at $\eta =\mp \pi /2$. Null curves are $\chi =\pm \eta +\text{constant}$ modulo $2\pi$. Although the conformal-time interval is finite, their physical affine parameters and timelike observers' proper times are infinite at its ends. Constant-$\eta$ circles are Cauchy hypersurfaces; global de Sitter is globally hyperbolic.

For AdS putThe conformal strip of two-dimensional anti-de Sitter spacetime has unbounded time and timelike conformal boundaries at $\psi =\pm \pi /2$. Null curves have $t\pm \psi =\text{constant}$. A signal from any finite $r$ reaches the center in coordinate time $|\psi |<\pi /2$, and a null curve crosses the entire conformal strip in time $\pi$. This finite coordinate travel time coexists with infinite physical affine length to the boundary. The covering spacetime has no closed timelike curves, but is not globally hyperbolic: causal curves can arrive from timelike infinity without meeting a proposed initial slice. Field evolution consequently requires boundary conditions as well as initial data. Unwrapping time removes closed timelike curves, not the timelike boundary.

The contrast in observer event horizons follows directly from these null curves. For the complete de Sitter observer $\chi =0$, let $d(\chi ,0)\in [0,\pi ]$ be shortest angular distance. An event can send a signal to this observer before its future endpoint precisely whenEquality is its future observer event horizon. It can receive a signal emitted by the observer after its past endpoint precisely when $d(\chi ,0)<\eta +\pi /2$, whose equality is the past horizon. In the observer's fundamental domain these horizons are null linesTheir intersection encloses the observer's static patch of de Sitter spacetime, the diamond $d(\chi ,0)<\pi /2-|\eta |$. They are observer-dependent cosmological horizons, not curvature singularities or Cauchy horizons. Every inertial observer has the corresponding horizons by de Sitter symmetry.

For the complete AdS static observer at $r=0$, every event at finite $r$ can send a signal to the observer, and can receive one, because time is unbounded and the coordinate distance $|\psi |$ is finite. Thus there is no corresponding global static-observer event horizon. The lapse never vanishes in these coordinates. Restricted accelerated-observer patches can have observer horizons, but those are distinct from the global static model. On the original periodic-time AdS hyperboloid, the more serious issue is closed timelike curves.

The figure identifies the de Sitter spatial edges and displays only a finite time window of the AdS covering strip, whose time continues indefinitely. Solid red curves are null geodesics; the oscillating AdS curve is a timelike geodesic. The shaded de Sitter diamond is the static observer's two-way communication region, not the entire spacetime.

The concise comparison is de Sitter has spacelike conformal infinities and cosmological observer horizons; covering AdS has timelike conformal boundaries and no horizon for an eternal global static observer. Both are constant-curvature and geodesically complete, but their geodesic recurrence, global topology, causal boundaries and initial-value properties differ.

Use the equal-time points to estimate velocity and acceleration with centered finite differences:At the endpoints use one-sided formulas of the desired order. A differentiated path and its time parametrization can instead supply these quantities. Let $g_k$ denote the gravitational acceleration, directed downwards. In the accelerating camera, the effective downward vector is $g_k-a_k$, so the locally felt upward vector is $U_k=a_k-g_k$. For a stationary camera this gives $-g_k$, fixing the sign.

The apparent-gravity camera frame is obtained by orthogonal projection and a cross product:When $W_k\ne 0$, $X_k,Z_k$ are perpendicular unit vectors, and $Y_k$ completes a right-handed orthonormal basis: $X_k\times Y_k=Z_k$. Also $U_k=(U_k\cdot X_k)X_k+\Vert W_k\Vert Z_k$, so the felt-up vector lies in the $XZ$ plane, with its transverse component pointing towards positive $Z$.

The stated exclusion of $a_k=g_k$ ensures $U_k\ne 0$, but it does not ensure $W_k\ne 0$. If felt-up is parallel to travel, the roll angle is undetermined rather than the construction being impossible. Every transverse orthonormal basis then has the required plane property. For continuity, project the previous $Z$ into $X_k^{\perp }$ and normalize it, using $Y_k=Z_k\times X_k$. If this projection is also too small, choose a coordinate unit vector least aligned with $X_k$ and project that vector instead. Switch to this transported choice near the parallel case to avoid numerical amplification. A stopped vehicle likewise needs a chosen limiting travel tangent vector, since its zero velocity itself has no direction. None of this requires a Frenet frame, which can fail where curvature vanishes.

Take upward velocity as positive, and consider the closed material system consisting of the rocket and the fuel it ejects during a short interval $dt$. Initially its mass and momentum are $m$ and $mv$. Let $\delta m>0$ be the ejected mass, so the rocket's mass change is $dm=-\delta m$. At the end its velocity is $v+dv$, while the ejected gas has inertial velocity $v-u$ to first order; its downward relative velocity need not be downward in the inertial frame.

The final total momentum, to first order, isNeglect air resistance and take the gravitational acceleration $g$ constant. The external gravitational impulse on this material system is $-mgdt$ to first order. Newton's second law therefore gives $mdv-u\delta m=-mgdt$, orThe positive thrust is $-u\dot{m}$ because fuel loss makes $\dot{m}<0$. Applying $d(mv)/dt$ to the rocket alone without accounting for the escaping momentum would give the wrong equation.

For constant $u$ and positive mass, divide by $m$ and integrate from release at $t=0$:Thus the gravity-corrected rocket equation is equivalent to

Use $D=(d/dt{)}_S$, $D^{\prime }=(d/dt{)}_{S^{\prime }}$, and ${\mathbf{v}}^{\prime }=D^{\prime }\mathbf{r}$. By the rotating-frame derivative formula, $D\mathbf{r}={\mathbf{v}}^{\prime }+\mathbf{\omega }\times \mathbf{r}$. Apply the product rule for the cross product and the same rotating-frame derivative formula again:We have $D{\mathbf{v}}^{\prime }=D^{\prime }{\mathbf{v}}^{\prime }+\mathbf{\omega }\times {\mathbf{v}}^{\prime }$ and $D\mathbf{\omega }=D^{\prime }\mathbf{\omega }$, since $\mathbf{\omega }\times \mathbf{\omega }=0$. ThereforeRearranging this identity into an equation of motion in a rotating frame produces the apparent Coriolis acceleration, Euler acceleration and centrifugal acceleration, with signs opposite to their corresponding terms on the right of this inertial acceleration identity. No translational term is needed because the two origins coincide.

Let $\Omega$ be the Earth's angular velocity magnitude. Use the local east, north and upward unit vectors ${\mathbf{e}}_E,{\mathbf{e}}_N,{\mathbf{e}}_U$, with ${\mathbf{e}}_E\times {\mathbf{e}}_N={\mathbf{e}}_U$. At latitude $\lambda$, the Earth's angular velocity and the train's ground-relative velocity areThus the east component of the Coriolis force isFor a north–south track, the train has no eastward acceleration relative to Earth. Its path curvature lies in the local north–vertical plane. The centrifugal acceleration and Newtonian gravity also lie in that plane, and the Earth's constant angular velocity gives no Euler acceleration. Hence the sideways force from the track must cancel the eastward Coriolis force:Its magnitude is $2m\Omega V\sin \lambda$, and its direction is west in the Northern hemisphere. The eastward tendency is the apparent Coriolis force; the requested track force has the opposite direction.

During ejection, the remaining mass is $m(t)=m_o+m_w-Qt$, and the water is exhausted at $t_c=m_w/Q$. Consider the momentum of the octopus and the small portion of water expelled during $dt$. Initially it is $mu$. At the end of this interval, the remaining body has mass $m-Qdt$ and velocity $u+du$, while the expelled mass $Qdt$ has laboratory velocity $u-V$, to first order. Hence the total final momentum isThe external impulse from quadratic drag is $-\alpha u^{2}dt$. Equating this to the change in total momentum, and dividing by $dt$, proves the equation of motionThe outgoing water's momentum flux is essential: applying $d(mu)/dt$ to the remaining body alone would omit it.

For dimensional analysis, the independent physical inputs are $m_o,m_w,Q,V,\alpha$, with dimensionsThe scales $m_o$, $m_o/Q$ and $V{m}_o/Q$ fix units of mass, time and length. In these units, the only remaining independent dimensionless variables areFor example, with $s=Qt/m_o$ and $w=u/V$, the equation of motion becomes $(1+\lambda -s)dw/ds=1-\mu w^2$, with $w(0)=0$ and ejection ending at $s=\lambda$. Thus the dimensionless terminal value depends only on $\lambda ,\mu$, proving $u_c=Vf(\lambda ,\mu )$.

For its explicit value, put $m_0=m_o+m_w$ and $U=\sqrt{QV/\alpha }$. The initially stationary solution stays below $U$, since the acceleration vanishes at $U$ and is positive below it. Separation of variables givesThe two integrals yieldThis is constant mass-loss propulsion with quadratic drag. At $t_c$, the remaining mass is $m_o$, givingThus $f(\lambda ,\mu )={\mu }^{-1/2}\tanh (\sqrt{\mu }\log (1+\lambda ))$, verifying the required dimensional form. The limit as drag tends to zero is $V\log (1+\lambda )$, the ordinary rocket equation, while the positive-drag answer is below the terminal velocity $U$.

For the complete uniform disc, the surface density is $\sigma =M/(\pi a^2)$. An annular strip of radius $r$ and thickness $dr$ has mass $dm=2\pi \sigma rdr$, so its contribution to the axial moment of inertia is $r^{2}dm$. HenceThis derives the moment of inertia of a uniform disc rather than assuming it.

Removing a concentric disc of radius $a/2$ at the same density removes mass $M/4$ and inertia $\frac{1}{2}(M/4)(a/2{)}^2=M{a}^2/32$. The remaining annulus therefore has mass $m_A=3M/4$ andThe moment of inertia of a uniform annulus gives the same value directly.

Let $A_c$ be the downhill centre-of-mass acceleration and let $F$ denote the magnitude of the uphill static friction. Translation, rotation about the centre, and rolling without slipping giveThus $F=I_A{A}_c/a^2=(5/8)m_A{A}_c$. Eliminating $F$ gives the rolling acceleration with rotational inertiaSubstitution gives the contact forcesThe normal force balances the perpendicular component of gravity because the centre remains at a constant distance from the plane. Although the annulus accelerates downhill, friction must act uphill to supply the torque that increases its rolling angular speed. For a plane with $0\le \alpha <\pi /2$, the assumed no-slip motion requires a coefficient of static friction satisfying ${\mu }_s\ge F/N=(5/13)\tan \alpha$. With that condition, contact friction does no work because the instantaneous contact point is stationary; it redistributes the gravitational energy between translational and rotational kinetic energy.

Take upward as positive and let $dm<0$ be the change in rocket mass during $dt$. The exhaust emitted has mass $-dm$ and, to first order, upward velocity $v-U$. The linear momentum of the rocket and just-emitted gas changes byThe external gravitational impulse is $-mgdt$ to first order. Applying the momentum balance to this material system, rather than applying a fixed-mass formula to the rocket alone, gives the rocket equationThe term $-U\dot{m}$ is the upward thrust.

For the specified mass loss and exhaust law, $\dot{m}=-\alpha$ andThus the condition for strictly positive upward acceleration at launch isAt equality the initial thrust just balances the weight; under this particular exhaust law the acceleration becomes positive immediately afterwards. If the initial thrust is smaller than the weight and the rocket initially rests on a launch surface, a nonzero supporting force persists until a later lift-off, so the free-flight equation does not describe its earlier supported motion.

For immediate free flight, integrate the rocket equation with $v(0)=0$:This applies while $0\le t<m_0/\alpha$ and the nonrelativistic approximation remains valid. The formal divergence as the model's remaining mass approaches zero is not a prediction within that approximation.

For dimensional analysis, write the mass, length and time dimensions as $\mathsf{M},\mathsf{L},\mathsf{T}$. Then $[m_0]=[m]=\mathsf{M}$, $[\alpha ]=\mathsf{M}{\mathsf{T}}^{-1}$, $[t]=\mathsf{T}$, $[U_0]=[U]=[v]=\mathsf{L}{\mathsf{T}}^{-1}$, and $[g]=\mathsf{L}{\mathsf{T}}^{-2}$. Both $m_0$ and $\alpha t$ have mass dimension, so $\alpha t/(m_0-\alpha t)$ is dimensionless; both terms in the boxed velocity have dimension $\mathsf{L}{\mathsf{T}}^{-1}$. The launch inequality compares two forces, each of dimension $\mathsf{ML}{\mathsf{T}}^{-2}$.

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.

Let $M=M_1+M_2$ and $\alpha =M_2/M$, with $0<\alpha <1$. In the centre of mass frame the two primary distances are $r_1=\alpha R$ and $r_2=(1-\alpha )R$. Applying Newton's law of universal gravitation and circular centripetal acceleration to the first primary gives $M_1{\omega }^2\alpha R=G{M}_1{M}_2/R^2$, and henceThe same result follows from the second primary or from the relative two-body equation of motion.

A rotating-frame equilibrium has zero rotating-frame velocity, so its Coriolis force vanishes. Above the orbital plane both primaries have downward gravitational components; below it both have upward components. The centrifugal force has no $z$ component. More explicitly,which vanishes only at $z=0$ for finite nonsingular points. Thus all equilibria lie in the orbital plane.

The equation of motion in a rotating frame for the dwarf isOn the orbital plane the centrifugal contribution is $m{\omega }^2\mathbf{r}$. Thus the rotating-frame effective potential is $E=U-m{\omega }^2(x^2+y^2)/2$; equilibria are exactly its stationary points. The Coriolis force affects motion about those equilibria but not their positions. This is the circular restricted three-body problem, with the dwarf treated as a test mass.

The printed energy unit has the wrong dimensions. The correct total energy unit is $mGM/R$, or $GM/R$ for specific energy; $GM/R^2$ is an acceleration. Dividing $E$ by $mGM/R$ and measuring lengths in units of $R$ gives the dimensionless effective potentialThe primary coordinates are $(-\alpha ,0)$ and $(1-\alpha ,0)$. Restricting to $y=0$ gives $F(x)=\mathcal{E}(x,0)$ andThe absolute values are essential; dropping them would reverse the attraction on the left of a primary. On each of the intervals separated by the two primaries, $F^{\prime }$ decreases strictly from $+\infty$ to $-\infty$. There is exactly one Collinear Lagrange point on each interval, a maximum along the $x$ direction.

For the inner Lagrange point, write $x=1-\alpha -\delta$ with $\delta >0$. For the L2 Lagrange point, write $x=1-\alpha +\delta$. The respective equations areBalancing the leading terms gives $\delta =(\alpha /3{)}^{1/3}+O({\alpha }^{2/3})$. Therefore the small-mass-ratio collinear Lagrange points obeyThe omitted $-\alpha$ displacement of the secondary is smaller than the stated error. The leading distance from the secondary is the dimensionless Hill radius.

For the L3 Lagrange point put $x=-1+c\alpha +O({\alpha }^2)$. It lies to the left of both primaries, soExpanding the two positive inverse-square factors gives $(1-\alpha )/(x+\alpha {)}^2=1+(2c+1)\alpha +O({\alpha }^2)$ and $\alpha /(x-1+\alpha {)}^2=\alpha /4+O({\alpha }^2)$. Thus $F^{\prime }=-(3c+5/4)\alpha +O({\alpha }^2)$, soAll three have $y=z=0$. The small-$\alpha$ equalities in the paper are asymptotic expressions, not exact coordinates for finite mass ratio.

For the remaining Lagrange points, introduce plane polar coordinates about the first primary: $x=-\alpha +s\cos \theta$, $y=s\sin \theta$. Its distance to the second primary is $d=(s^2+1-2s\cos \theta {)}^{1/2}$, andThe angular stationary condition is ${\partial }_{\theta }\mathcal{E}=\alpha s\sin \theta (d^{-3}-1)=0$. For an off-axis point $s>0$ and $\sin \theta \ne 0$, so $d=1$. The radial stationary condition becomesHence $s=1$, and $d=1$ implies $\cos \theta =1/2$. The Triangular Lagrange points are exactlyBoth triangles are equilateral. The collinear and off-axis calculations together exhaust the stationary points.

For the requested sketch, $\mathcal{E}$ tends to $-\infty$ at either primary and at large in-plane radius. Each Collinear Lagrange point is a saddle of the planar effective potential: its $x$ curvature is negative and its $y$ curvature is positive. To see the latter, let $d_1=|x+\alpha |$, $d_2=|x-1+\alpha |$ and $D=(1-\alpha )/d_1^3+\alpha /d_2^3$. At an inner point both distances are below one, so $D>1$. At an exterior stationary point, use barycentric positions $a_1=-\alpha$, $a_2=1-\alpha$ and weights $1-\alpha ,\alpha$: the equation implies $x(D-1)=\sum w_i{a}_i/|x-a_i{|}^3$. To the right, the weighted sum is positive because the positive primary is closer; to the left it is negative because the negative primary is closer. Thus $D>1$ there as well, and ${\mathcal{E}}_{yy}=D-1>0$.

At either Triangular Lagrange point, the planar Hessian isIts trace is $-3$ and determinant $27\alpha (1-\alpha )/4>0$, so these are local maxima. Their height is $\mathcal{E}=-3/2+\alpha (1-\alpha )/2$. The sketch below uses a finite mass ratio to separate the five points clearly; its singular wells are clipped only for display. These curvature labels concern the effective potential, not a stability test that ignores the Coriolis force.