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An integral of motion is a function of phase space that is constant along a trajectory. Put $\varphi =-\psi +\mathrm{constant}$, so the stellar acceleration is $\dot{\mathbf{v}}=\nabla \psi$. In a time-independent spherical Newtonian gravitational potential,The second equality uses the radial direction of $\nabla \psi$. Thus both the binding energy and the full specific angular momentum vector are integrals of motion.

The Collisionless Boltzmann equation isFor a stationary galactic distribution function, this says $df/dt=0$ on every characteristic, so $f$ is constant on each stellar orbit. Locally choose a complete set of orbit labels $I_j$, themselves integrals of motion; constancy along the characteristic means $f=F(I_1,\dots ,I_s)$. Conversely, for any such differentiable $F$, the chain rule gives $df/dt={\sum }_j{F}_{,j}{\dot{I}}_j=0$. This proves Jeans theorem in its orbit-invariance form. It does not claim that just two particular integrals classify every orbit in an arbitrary potential. For the specified spherical model, $F(E,L)$ is a rotationally invariant stationary choice.

For the velocity integration, write $v_r=v\cos \eta$, $v_{\perp }=v\sin \eta$, $L=rv\sin \eta$, and set $k=r/r_a$. At fixed position the velocity-dependent factor isUse ${\int }_0^{\infty }{v}^2{e}^{-A{v}^2/(2{\sigma }^2)}dv=\sqrt{\pi /2}{\sigma }^3{A}^{-3/2}$ and integrate over $\chi$. The mass density isSince ${\rho }_c=\rho (0)={\rho }_1{e}^{\psi (0)/{\sigma }^2}$,These integrals use the stated untruncated galactic distribution function over all velocities; an additional escape-energy cutoff would change the result.

In local orthogonal velocity components the same exponent is $-[v_r^2+(1+k^2)(v_{\theta }^2+v_{\varphi }^2)]/(2{\sigma }^2)$. Each component is a centred Gaussian distribution. The one-dimensional identity $\int v^2{e}^{-A{v}^2/(2{\sigma }^2)}dv/\int e^{-A{v}^2/(2{\sigma }^2)}dv={\sigma }^2/A$ yieldsIn particular the velocity-anisotropy parameter is ${\beta }_{\mathrm{an}}=1-({\sigma }_{\theta }^2+{\sigma }_{\varphi }^2)/(2{\sigma }_r^2)=r^2/(r_a^2+r^2)$. This is Gaussian angular-momentum suppression in a stellar distribution.

The radial laws also require self-consistency through the Poisson equation for Newtonian gravity:For $r\ll r_a$, outside a sufficiently small core, the model is almost isotropic and isothermal. Seek its scale-free envelope with $\rho =A/r^2$. Its distribution-density relation gives ${\psi }^{\prime }=-2{\sigma }^2/r$, and the Poisson equation for Newtonian gravity gives $A={\sigma }^2/(2\pi G)$. Thus the intended intermediate singular isothermal sphere behaviour iswhere a core scale is $r_c=\sigma /\sqrt{4\pi G{\rho }_c}$.

The stated $r^{-2}$ law cannot be the actual central limit of a regular finite-density model. For finite ${\rho }_c$ and $\psi (0)$, regularity and the Poisson equation for Newtonian gravity instead giveand $v_c^2=(4\pi G{\rho }_c/3)r^2+O(r^4)$. A broad intermediate $r^{-2}$ region exists only if $r_c\ll r_a$; the radius $r_a$ alone does not ensure it.

For the outer radially anisotropic isothermal halo asymptotics, put $s=\log (r/r_a)$, $y=[\psi (r)-\psi (0)]/{\sigma }^2$ and $\lambda =4\pi G{\rho }_c{r}_a^2/{\sigma }^2$. At $r\gg r_a$ the density relation and Poisson equation becomeThe slow outer balance is $y_s\sim -\lambda e^y$, so $d(e^{-y})/ds\sim \lambda$, giving $e^y\sim 1/(\lambda s)$. Its consistency check is $y\sim -\log (\lambda s)$, for which $y_{ss}/y_s\sim -1/s\to 0$. ThereforeThe reference radius $r_{\ast }$ changes only subleading logarithmic terms. The circular speed falls as $(\log r{)}^{-1/2}$, rather than remaining exactly flat.

Physically, the model has an almost isotropic central region and, if the scales separate, a flat-rotation isothermal envelope. Beyond $r_a$, suppression of large specific angular momentum leaves predominantly radial orbits: radial velocity dispersion remains $\sigma$, while transverse dispersion falls as $r^{-1}$. The outer density acquires a logarithmic suppression and the galaxy rotation curve declines slowly. The enclosed mass still grows approximately as $r/\log r$, so this untruncated model has infinite total mass.

For a bound Kepler orbit, the semimajor axis is half the long diameter of the ellipse; if $r_p,r_{mathrma}$ are its apsidal distances, $a=(r_p+r_{mathrma})/2$. The orbital eccentricity is $e=(r_{mathrma}-r_p)/(r_{mathrma}+r_p)$, with $0\le e<1$ and semiminor axis $b=a\sqrt{1-e^2}$. The gravitating centre is a focus, displaced by $ae$ from the ellipse's geometric centre.

Choose a reference plane, a positive normal and a fixed reference direction in that plane. The ascending node is where the orbit crosses toward positive normal displacement. The longitude of ascending node $\Omega$ is the reference-plane angle from the fixed direction to that node. The argument of periapsis $\omega$ is the angle from the ascending node to periapsis, measured in the orbital plane in the direction of motion. The orbital inclination $i$ is the angle between the reference normal and the orbital angular-momentum normal, with $0\le i\le \pi$. The diagram separates the ellipse's shape from these orientation angles.

Let $\mu =GM$ and use actions per unit test-particle mass. In the prograde spherical-action chart underlying the stated Hamiltonian, $J_{\varphi }=L_z$ and $J_{\theta }=L-L_z$, so $J_{\theta }+J_{\varphi }=L$. The radial action then satisfies $J_r+L=\sqrt{\mu a}$, consistently with $H_K=-\mu /(2a)$. DefineIf $(w_r,w_{\theta },w_{\varphi })$ are the old angles, a type-two generating function isDifferentiation with respect to the old angles recovers $J_r,J_{\theta },J_{\varphi }$, and differentiation with respect to the new actions gives the canonical transformationIts triangular integer action matrix has determinant one, so it preserves the angle periodicities. In the new coordinates,These are the Delaunay variables:With the usual choices of angle origins, $w_1=\ell$ is mean anomaly, $w_2=\omega$ is argument of periapsis, and $w_3=\Omega$ is longitude of ascending node. The generator $\mathcal{G}=L$ rotates the orbit within its plane, whereas $\mathcal{H}=L_z$ rotates that plane about the reference normal; this explains the angular identification. The fast angle obeys Kepler's equation $\ell =\mathcal{E}-e\sin \mathcal{E}$, where $\mathcal{E}$ is eccentric anomaly.

For retrograde orbits, signed $J_{\varphi }=L_z$ gives $J_{\theta }=L-|L_z|$ and the spherical Hamiltonian uses $J_r+J_{\theta }+|J_{\varphi }|$. The stated plus-sign form is therefore a prograde chart, not a global signed-action formula. The Delaunay variables still use $\mathcal{H}=L_z$; on the retrograde branch the generating function is $(\mathcal{L}-\mathcal{G})w_r+(\mathcal{G}+\mathcal{H})w_{\theta }+\mathcal{H}{w}_{\varphi }$. At $e=0$, periapsis is undefined, and at $i=0$ or $\pi$ the node is undefined: these are angle-chart degeneracies, not physical singularities.

For the perturbed planet orbit, the radial and vertical frequencies mean small oscillations about a circular equatorial orbit of cylindrical radius $R$. Write $r=\sqrt{R^2+z^2}$ and retain the stated quadrupole potential. On the equator,At fixed axial specific angular momentum $L_z$, the radial effective potential is $\varphi (R,0)+L_z^2/(2R^2)$. Its second derivative at circular balance $L_z=R^2{\Omega }_c$ gives the radial epicyclic frequencyReflection symmetry decouples vertical motion to first order. Expanding the potential in $z$ givesso the vertical epicyclic frequency isThe circular orbit is radially stable only for $R^2>3{\alpha }^2$. For $\alpha \ne 0$ in that domain, $\nu >{\Omega }_c>\kappa$. Vertical node crossings repeat at frequency $\nu$ while azimuth advances at ${\Omega }_c$, so the line of nodes has nodal precession rate ${\Omega }_c-\nu <0$. Radial periapsis passages repeat at $\kappa$, giving apsidal precession rate ${\Omega }_c-\kappa >0$. To leading quadrupole order, with $n=\sqrt{\mu /R^3}$,This is oblate-quadrupole epicyclic precession. Here $\varpi$ denotes the nearly equatorial longitude of periapsis, not $\omega$ alone.

The precession signs above have a small-eccentricity, small-inclination domain. They are not a universal claim for every inclined Kepler ellipse. To see the distinction with the same Delaunay variables, average the perturbation over an unperturbed orbit. Since $z/r=\sin i\sin (\omega +f)$ and $dt=r^{2}df/\mathcal{G}$, the term linear in $e\cos f$ integrates to zero and the ${\sin }^2(\omega +f)$ average in the $r^{-3}$ integral is $1/2$. ThusDifferentiating at fixed other actions yields the first-order secular ratesTheir nearly equatorial sum reproduces the prograde $\dot{\varpi }$. For a polar orbit, however, the node is fixed and $\dot{\omega }<0$. This supplies a concrete counterexample to extending the requested forward periapsis precession beyond the intended epicyclic regime.

For an equatorial circular orbit in the Paczyński-Wiita potential, radial force balance givesIts specific angular momentum is $j=R^2\Omega =\sqrt{GM}{R}^{3/2}/(R-R_S)$. Perturbing the circular radius at fixed $j$ gives the radial epicyclic frequency squared ${\kappa }^2=R^{-3}d(j^2)/dR$. ThereforeCircular motion is radially stable for $R>3R_S$, marginal at $3R_S$, and unstable between $R_S$ and $3R_S$. Thus the inner edge of the nearly circular thin accretion disk is $R_{\mathrm{in}}=3R_S$, joining a plunging region of a black-hole accretion disk inside the innermost stable circular orbit. This conclusion concerns the circular thin-disk model, not an assertion that all gas disappears inside this radius.

Take the inward mass accretion rate $\dot{M}$ as positive, and write the positive outward viscous torque in an accretion disk asSteady conservation of angular momentum makes the net inward flux $C=\dot{M}j-\mathcal{G}$ independent of radius. Impose a zero-torque inner boundary condition at $3R_S$, appropriate to an approximately stress-free transition to plunging motion. ThenThis is the steady accretion disk with arbitrary rotation law angular-momentum balance.

The local viscous dissipation integrated through the full disk thickness is $D=\nu \Sigma (R{\Omega }^{\prime }{)}^2=-\mathcal{G}{\Omega }^{\prime }/(2\pi R)$. It is the power per unit planar area summed over both radiating faces. Sincesubstitution gives the two-face dissipation of a Paczyński-Wiita disk:In particular $D$ vanishes at the stress-free inner edge and is nonnegative outside it because $j$ increases there. A one-face radiative flux is $D/2$; the total accretion luminosity is therefore $L={\int }_{3R_S}^{\infty }2\pi RDdR$, with no additional factor of two.

For a direct evaluation put $x=R/R_S$. ThenThe first integral is $7/4$, using $(3x-1)/(x-1{)}^3=3/(x-1{)}^2+2/(x-1{)}^3$. For the second,so it equals $2/[\sqrt{3}(3-1)]=1/\sqrt{3}$. HenceThis is the radiative efficiency of black-hole accretion within this pseudo-Newtonian model.

Independently, the specific orbital energy of a circular particle isThus $E(\infty )=0$ and $E(3R_S)=-GM/(8R_S)=-c^2/16$. The energy released per unit accreted mass is exactly $c^2/16$, confirming the integrated accretion luminosity. The Paczyński-Wiita circular orbit reproduces the Schwarzschild inner radius but gives an approximate binding energy; the efficiency derived here is for the stated potential.

Multiplying the tangential equation of motion by $r$ givesTherefore $h=r^2\dot{\theta }$ is constant. It is the signed specific angular momentum; reverse the angular orientation if necessary to take $h>0$ for a nonradial orbit. Set $u=1/r$ and denote differentiation with respect to $\theta$ by a prime. ThenSubstitution into the radial equation of motion and division by $-h^2{u}^2$ give the Binet equationIts general solution is $u=GM/h^2+A\cos \theta +B\sin \theta$. Write $A=(GM/h^2)e\cos {\theta }_0$ and $B=(GM/h^2)e\sin {\theta }_0$, where $e\ge 0$. This produces the Kepler orbitHere $e$ is the orbital eccentricity, and ${\theta }_0$ points toward periapsis when $e>0$. Only portions with $u>0$ describe positive radial distances. The expression assumes $h\ne 0$; a zero-angular momentum trajectory is radial and must be treated separately.

At infinity, the meteorite's incoming asymptote has speed $V$ and perpendicular distance $b$ from the Earth’s centre. The magnitude of its specific angular momentum is the magnitude of $\mathbf{r}\times \mathbf{v}$. On the asymptote, this is $bV$, so conservation of angular momentum gives $\boxed{h=bV}$. Its mechanical energy at infinity is $E=m{V}^2/2$. Comparing with part (ii) givesFor $b>0$ this Kepler orbit is a hyperbola. The smallest $r=1/u$ occurs at periapsis, where $\cos (\theta -{\theta }_0)=1$. ThusThe middle equality follows by substituting $e^2-1=b^2{V}^4/(G^2{M}^2)$. A trajectory entering the Earth satisfies $r_{\min }<R$, equivalentlyTherefore the sufficient collision condition isThis is gravitational focusing: Newtonian gravity increases the collision radius measured at infinity. Equality is the ideal grazing trajectory, which touches the surface; a strict inequality penetrates it. For $b=0$ the incoming motion is radial and collides, with $r_{\min }=0$ as the limiting value of the above expression.

Steady axisymmetric mass conservation is $\nabla \cdot (\rho {\mathbf{u}}_p)=0$, permitting the axisymmetric hydrodynamic mass flux function representationThus stream surfaces are regular level surfaces of $\psi$. Adiabaticity gives $\mathbf{u}\cdot \nabla s=0$, and axisymmetry removes the azimuthal derivative, so the specific entropy is constant along their meridional streamlines. The azimuthal Euler equations for an inviscid fluid has no pressure or gravitational torque:The conserved specific angular momentum is $\ell =R{u}_{\varphi }$. Taking the scalar product of the steady Crocco form of the unsteady Euler equation with $\mathbf{u}$ also givesConsequently, on each connected regular family of stream surfaces,These functions express entropy, angular momentum and the Bernoulli function carried by a stellar wind. Stagnation regions and disconnected branches with identical flux labels require their own matching information; the argument establishes the functions locally where the poloidal flow is nonzero.

To derive the hydrodynamic transfield equation for an axisymmetric wind, decompose the vorticity. The poloidal part comes from the toroidal velocity:The azimuthal part comes from the poloidal velocity:The final divergence is the full axisymmetric cylindrical divergence, including its radial $1/R$ factor.

Since ${\mathbf{u}}_p=\nabla \psi \times {\mathbf{e}}_{\varphi }/(\rho R)$, the meridional part of $\mathbf{\omega }\times \mathbf{u}$ isThe remaining steady Euler equations for an inviscid fluid is thereforeCanceling $\nabla \psi$ on a regular stream surface and inserting ${\omega }_{\varphi }$ givesThe sign of the angular-momentum term follows from ${\mathbf{u}}_p\times {\mathbf{e}}_{\varphi }=-\nabla \psi /(\rho R)$. To determine the wind, this transverse force equation is supplemented by its Bernoulli function relationand the equation of state, with the gravitational potential supplied externally or by the Poisson equation.

Assume an axisymmetric, geometrically thin disc in a fixed Keplerian rotation field, neglecting the disk's self-gravity. Write $v_R$ positive outwards and letThe ring supplies the local specific angular momentum. Its source per unit area iswhere the Dirac delta function is normalized so that $\int 2\pi RsdR=\dot{M}(t)$. Define the positive outward viscous torque in an accretion disk byThe height-integrated conservation of mass and conservation of angular momentum equations areSubtract $h$ times the mass equation. Since $h$ is time-independent, this givesThe last term is zero as a distribution, because the source is supported where $h=h_0$. Since $h^{\prime }=h/(2R)$, substituting the torque givesInserting this into the mass equation proves the Keplerian disk evolution with a ring source:The viscosity may be a prescribed function or a function of the local disk state; the conservation derivation is the same.

To track signs in the steady problem, define the inward mass fluxThen $0=F_M^{\prime }/(2\pi R)+s$. Away from the source $F_M$ is constant, with $F_M={\dot{M}}_{\mathrm{in}}$ on the inner side and $F_M=-{\dot{M}}_{\mathrm{out}}$ on the outer side, where both loss rates are positive. Integrating through the source gives $F_M(R_0^{+})-F_M(R_0^{-})=-\dot{M}$. Therefore

The outward angular momentum flux is $F_J=\mathcal{G}-F_{M}h$. The steady angular-momentum equation gives $F_J^{\prime }=2\pi Rs{h}_0$. At the two torque-free boundaries its values are $F_J(R_{\mathrm{in}})=-{\dot{M}}_{\mathrm{in}}{h}_{\mathrm{in}}$ and $F_J(R_{\mathrm{out}})={\dot{M}}_{\mathrm{out}}{h}_{\mathrm{out}}$. Integrating across the disk therefore yieldsEliminate ${\dot{M}}_{\mathrm{out}}=\dot{M}-{\dot{M}}_{\mathrm{in}}$ and use $h\propto R^{1/2}$. The two-boundary mass split of a ring-fed disk isThe complementary outward fraction is $(\sqrt{R_0}-\sqrt{R_{\mathrm{in}}})/(\sqrt{R_{\mathrm{out}}}-\sqrt{R_{\mathrm{in}}})$. Both fractions lie between zero and one for a source strictly between the boundaries.

As a local check, integrating the steady mass flux from each zero-torque boundary givesContinuity of this quantity at $R_0$ gives the same split. The split depends on boundary and source angular momenta, rather than on the detailed viscosity law.

Interpret the model comparison at the exam's 2008 epoch. A simulation's failure depends on its treatment of baryons, resolution, and adopted stellar feedback; it is not a theorem that the Lambda-CDM model cannot produce the observation under any physical implementation. The following were important tensions for then-standard realizations, with different implications for monolithic galaxy formation.

Old, alpha-enhanced massive spheroids and downsizing. The most massive quiescent elliptical galaxies tend to contain older stellar populations and elevated alpha element abundances relative to iron, indicating that much of their star formation occurred early and rapidly. Prompt core-collapse enrichment followed by shutdown before substantial delayed iron enrichment accounts for the abundance pattern. Models that spread a massive galaxy's star formation over many late progenitor-merger episodes can make its stars too young or insufficiently alpha enhanced. A rapid early dissipative collapse naturally gives a short formation interval and is therefore more immediately consistent with this particular observation. However, downsizing in galaxy formation is about stellar formation, not necessarily final assembly: late dry galaxy mergers can assemble already old alpha-enhanced stars. Early quenching and an appropriate chemical-enrichment model can consequently reconcile hierarchical assembly with the fossil record.

Thin, high-angular-momentum stellar discs. The existence of extended thin galactic disks, including systems with small galactic bulges, challenged simulations that produced compact central stellar concentrations or overly thick discs. Premature gas cooling into small progenitors, transfer of specific angular momentum to the halo, and repeated dynamical perturbations could cause the galactic angular-momentum catastrophe; satellite encounters also increase galactic disk heating. A comparatively quiet dissipative collapse with approximate conservation of baryonic specific angular momentum more readily produces an extended rotating disc, and avoiding subsequent strong perturbations helps retain thinness. Nevertheless, rapid collapse alone does not explain a disc's prolonged star formation and gas replenishment. Hierarchical histories with quieter late accretion, better feedback, delayed cooling, and dissipative disc regrowth can reduce these tensions; a problem in early implementations is not an unavoidable property of the cosmological framework.

Cores inferred in low-mass galaxies. Resolved inner galaxy rotation curves in a number of low-surface-brightness and dwarf systems favoured shallower central mass densities than the cusps of collisionless cold-dark-matter simulations. For a roughly constant-density dark-matter core, $M(<r)\propto r^3$ and $V_c\propto r$; an inner Navarro--Frenk--White profile has $\rho \propto r^{-1}$, hence $M(<r)\propto r^2$ and $V_c\propto r^{1/2}$. The differing slopes explain the cusp--core problem. Beam smearing, inclination, and noncircular motions complicate inference, but do not make the issue merely a normalization mismatch. A simple early dissipative collapse does not automatically solve it: baryonic contraction can deepen the central concentration. Potential fluctuations from gas loss and stellar feedback can instead redistribute dark matter and modify a cusp. Thus this observation tests the joint halo-and-baryon model rather than preferentially vindicating the old collapse picture.

Rapid formation of the stars and hierarchical assembly of the final galaxy are compatible histories. The comparison must keep those clocks separate, and the historical modelling tensions should not be presented as universal failures of all later simulations.

Write $\mu =G{M}_{\odot }$, and interpret the specified radial force field as acceleration, or force per unit test-particle mass. Since it is a central force, the specific angular momentum $\mathbf{h}=\mathbf{r}\times \mathbf{v}$ is conserved: $\dot{\mathbf{h}}=\mathbf{r}\times \mathbf{F}=0$. For $h=|\mathbf{h}|\ne 0$, motion is confined to the fixed plane perpendicular to $\mathbf{h}$. Choose the orientation of the polar angle so that $h=r^2\dot{\theta }>0$.

To derive the Binet equation here, let $u=1/r$ and differentiate with respect to $\theta$. Then $\dot{r}=-h{u}^{\prime }$ and $\ddot{r}-r{\dot{\theta }}^2=-h^2{u}^2(u^{\prime \prime }+u)$. Equating this to the radial acceleration givesPut $K=h^2+C$ and $\alpha =\sqrt{K/h^2}$. For $K>0$, the solution isThe radial potential energy per unit mass is $V(r)=-\mu /r+C/(2r^2)$, so the conserved specific orbital energy isSubstitute the displayed $u$ and $\dot{r}=-h{u}^{\prime }$ to obtain $\mathcal{E}={\mu }^2(e^2-1)/(2K)$. Therefore defineFor regular bound motion, $K>0$ and $-{\mu }^2/(2K)\le \mathcal{E}<0$, giving $a>0$ and $0\le e<1$. Since $K/\mu =a(1-e^2)$, choose the angular origin at a pericentre, so ${\theta }_0=0$, and obtainThe turning radii are $r_{\mathrm{p}}=a(1-e)$ and $r_{\mathrm{a}}=a(1+e)$. Thus $a$ is the arithmetic mean of the turning radii; when the orbit precesses it should not be interpreted as the semimajor axis of a fixed ellipse. If $K\le 0$, the centrifugal barrier needed for a regular two-turning-point bound orbit is absent, and this bound-orbit parametrization does not apply.

For $\alpha =1$, one has $C=0$ and the orbit is a closed Kepler orbit: an ellipse of eccentricity $e$ with the Sun at a focus. In addition to $\mathcal{E}$ and the three components of $\mathbf{h}$, the Laplace-Runge-Lenz vectoris conserved. Indeed,so $\dot{\mathbf{A}}=0$. It points toward the fixed pericentre. The constraints $\mathbf{A}\cdot \mathbf{h}=0$ and $A^2={\mu }^2+2\mathcal{E}{h}^2$ leave five generically independent isolating integrals. One may takechoosing a component $A_j$ whose differential is independent of the other four on the local region under consideration. The extra integral fixes the apsidal direction in the already fixed orbital plane.

For $\alpha \ne 1$, successive pericentres are separated by $2\pi /\alpha$ in polar angle. The generic bound orbit is a rosette in that same plane. Its perihelion advance per radial cycle isFor $\alpha <1$ this is prograde; for $\alpha >1$ it is retrograde. If $\alpha$ is rational, the orbit closes after a finite number of radial cycles; if it is irrational and $0<e<1$, its successive radial phases densely fill the accessible annulus. A generic independent set of single-valued isolating integrals isThe apsidal direction is no longer fixed. For irrational frequency ratio, any continuous integral of motion must be constant on the whole two-dimensional orbital torus, because one orbit is dense there; there is no further independent continuous integral specifying a fixed apsidal direction. Exceptional closed resonant orbits do not provide a fifth global integral on an open region of phase space. Circular and radial orbits are degenerate cases of these generic counts.

The plotted precession is enlarged for visibility. Both examples have $e=0.5$ and the same turning radii; the second has $\alpha =\sqrt{0.9}$.

For the precession rate, it is useful to derive the radial period rather than assume an unperturbed angular period. Set $r=a(1-e\cos \chi )$ in the radial energy equation. On a complete radial cycle,Hence $n=2\pi /P_r=\sqrt{\mu /a^3}$. Also $K=\mu a(1-e^2)$ and $h^2=K-C$, so ${\alpha }^{-1}=\sqrt{1-C/K}$. For $|C|\ll K$, inverse-cube perturbation of Kepler apsidal motion givesThere is a dimensional issue with the printed parameter: $C$ has units $L^4/T^2$ and $\mu$ has units $L^3/T^2$, so $\eta =aC/\mu$ has units $L^2$. Retaining that definition literally, the requested expression isA dimensionless perturbation coefficient would instead be $\tilde{\eta }=C/(\mu a)=\eta /a^2$. In that convention the rate is $-n\tilde{\eta }/[2(1-e^2)]$. These two conventions must not be conflated.