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For circular orbits in an exponential central potential, a constant positive radius requires $V_{\mathrm{eff}}^{\prime }(r)=0$, orLet $h(r)=r^3{e}^{-r}$. Its derivative is $h^{\prime }(r)=r^2{e}^{-r}(3-r)$: it rises from zero to $h(3)=27/e^3$, then falls back to zero. HenceThe strict lower bound matters: the printed upper bound also allows $l=0$, but then there is no finite-radius circular orbit. Nonzero angular momentum is an implicit hypothesis of the two-orbit claim.

At a circular orbit, use $l^2=r^3{e}^{-r}$ to obtainA small radial displacement $\eta$ satisfies $\ddot{\eta }=-V_{\mathrm{eff}}^{\prime \prime }(r)\eta$. Therefore the inner circular orbit is radially stable at fixed angular momentum, while the outer orbit is unstable. The inner perturbation oscillates with squared frequency $e^{-r_1}(3/r_1-1)$; the outer perturbation has an exponentially growing solution. At the limiting value $l^2=27/e^3$, the radii coalesce at $r=3$ and the linear restoring coefficient vanishes.

For a fixed satellite mass on an adiabatically changing circular orbit, $L_s=M_s\sqrt{G{M}_{\ast }{r}_0}=M_s{r}_0^2\Omega$. The satellite torque is minus the disc torque. HenceThe circular-orbit migration rate from disk torque, in the normalization supplied by the question, isIt is inward for a positive mass density slope and outward for a negative one. The complete-flyby normalization makes its magnitude four times larger, with the same direction. This is a local torque estimate; the hypothesis of a nearly circular slowly migrating orbit is needed when using the derivative of the circular angular momentum.

Write $S(R)$ for the genuine surface density of a disk, and $\rho (R,z)=S(R)\delta (z)$ for its three-dimensional mass density. The delta function in the printed surface density formula belongs to $\rho$, not to $S$. Introduce a reference length $R_{\ast }$ to make logarithms dimensionless. The intended model is the infinite, self-gravitating, scale-free Mestel disk, with no extra source or imposed gravitational field.

The circular speed satisfies $v_c^2=R{\partial }_R\Phi (R,0)$. Thus the flat galaxy rotation curve givesA reflection-symmetric harmonic function with this midplane boundary value isVerify this continuation using the Poisson equation for Newtonian gravity. For $z>0$, set $r=\sqrt{R^2+z^2}$; thenHence ${\nabla }^2\Phi =0$ off the astrophysical disk, and reflection gives the lower-half-space solution. Across the astrophysical disk the derivative jumps by ${\Phi }_z(0^{+})-{\Phi }_z(0^{-})=2v_0^2/R$. The distributional Poisson equation for Newtonian gravity therefore givesAt the origin the enclosed disk mass tends to zero linearly with radius, so there is no additional central point mass. This verifies the Mestel disk potential-density pair. The astrophysical disk has infinite total mass and a logarithmic Newtonian gravitational potential, so it is not an isolated finite-mass model with Newtonian gravitational potential zero at infinity. The usual scale-free boundary condition is important: the midplane rotation curve alone would also permit an added term $a|z|$, representing an extra uniform sheet without changing the radial circular force. That contribution is excluded in the intended Mestel disk model.

Use a mass-weighted planar galactic distribution function, so $F(\mathbf{x},\mathbf{v})d^{2}x{d}^{2}v$ is the stellar mass in a small planar phase space element. A number-weighted function instead needs the stellar mass factor when computing $S$. Assume a steady collisionless stellar system and isotropy in the two in-plane velocity components. Put $u=|\mathbf{v}{|}^2/2$ and write $F=f(\mathbf{x},u)$. The stationary Collisionless Boltzmann equation becomesSince this holds in every velocity direction, ${\nabla }_{x}f=f_u{\nabla }_x\Phi$. In coordinates $(\mathbf{x},E)$ with $E=u+\Phi$, this is exactly ${\nabla }_{x}f{|}_E=0$. Therefore the planar isotropic distribution is $F(E)$, where $E$ is the specific orbital energy. Stationarity is essential; instantaneous isotropy alone would not imply this result.

Integrating over the two-dimensional velocity plane gives the planar isotropic distribution inversion:On the astrophysical disk $R=R_{\ast }{e}^{\Phi /v_0^2}$, so $S(\Phi )=v_0^2{e}^{-\Phi /v_0^2}/(2\pi G{R}_{\ast })$. Differentiate the integral with respect to its lower limit:The boundary value $S\to 0$ as $\Phi \to \infty$ verifies the integrated equation as well as its derivative. Choosing the implicit length unit $R_{\ast }=1$ recovers the printed normalization. Changing the additive energy zero changes this prefactor accordingly.

At a fixed radius, the normalized velocity density isIt is a product of centered Gaussian distributions. Differentiating the supplied Gaussian integral with respect to its coefficient gives the second moments, and odd moments vanish. Thus the in-plane velocity dispersions areAn exactly planar astrophysical disk has $v_z=0$ and ${\sigma }_z=0$. The nonzero circular speed is a property of the force field, not a statement that this hot stellar distribution has net rotation.

Reversing every retrograde star folds the azimuthal Gaussian to a half-normal distribution. Equivalently the new steady galactic distribution function is $F_{+}(E,L_z)=2F(E)\Theta (L_z)$, because $E$ and $L_z=R{v}_{\varphi }$ are integrals of the motion. Its density and even velocity moments are unchanged. Its streaming velocity isThis maximally prograde stellar distribution still has radial motion and a spread of azimuthal speeds; it does not place every star on a circular orbit. In particular $⟨v_{\varphi }^2{⟩}_{+}=v_0^2$ while ${\sigma }_{\varphi ,+}^2=v_0^2(1-2/\pi )$. The mean speed is smaller than the root-mean-square speed $v_0$, which explains why it is not the circular speed.

Synchronous radius and its limits. Kepler's third law for a circumplanetary orbit givesHere $G$ is the gravitational constant. A planet-synchronous orbit repeats after one spin period. A planet-stationary orbit additionally requires a circular orbit, zero orbital inclination to the equator and prograde motion. Then an antenna fixed on the ground points continuously at the same satellite; merely matching the orbital period does not give this property.

Write the planetary radius as $R_p=(3M/(4\pi \rho ){)}^{1/3}$. Requiring the semi-major axis to exceed $R_p$ gives $P^2>3\pi /(G\rho )$. The stellar tidal force restricts the circumplanetary orbit to the Hill sphere, of radius $r_H=a(M/(3M_{\star }){)}^{1/3}$. ThusThese are the surface and Hill-radius constraints in the idealized spherical, small-$M/M_{\star }$ model. Long-term prograde stability generally requires a radius appreciably inside the Hill sphere; for nonzero orbital eccentricity, the surface constraint applies to the periapsis, not just to the semi-major axis.

Collision time and the launch population. A phase-mixed isotropic swarm occupies a shell of radial scale $e{a}_s$. Its number density scales as $N_s/(e{a}_s^3)$, its relative speed as $a_s/P$, and its geometric collision cross-section as $r^2$. Consequently the total collision rate scales as $N_s^2{r}^2/(Pe{a}_s^2)$. To give the numerical normalization used here, adopt an effective shell volume $V=4\pi e{a}_s^3$. At a fixed position the velocity directions are uniformly distributed in the tangent plane, so their mean relative speed is $4v_K/\pi =8a_s/P$. For an unordered pair, the collision cross-section is $4\pi r^2$, givingThus the mean interval between collisions is $A/N_s^2$ in this large-population kinetic model. The effective shell width fixes an order-one coefficient: small orbital eccentricity and random planes alone do not specify a unique radial probability density. Phase mixing, negligible gravitational focusing, $r\ll e{a}_s$, and uncorrelated encounters are implicit in this estimate. Exactly identical orbital periods with perfectly fixed phases do not themselves produce a memoryless collision process.

For nearly planet-stationary orbits with $e\ll I\ll 1$, the swarm volume is smaller by a factor of order $I$, while the relative speed is smaller by the same factor, since vertical motion of scale $I{v}_K$ dominates the eccentric motion. The two changes cancel in the rate $n\sigma v$: there is no parametric factor $I$ in the collision time within the same phase-mixed kinetic approximation. Numerical factors and phase correlations can differ. This is not an argument that bringing all satellites into one nearly circular plane makes their phases random.

With $N_s=Rt$, an Inhomogeneous Poisson process has cumulative hazard functionThe survival function of the first collision is $\exp [-H(t)]$. Setting the expected number of collisions to one givesThis is a characteristic first-event population, with probability $1-e^{-1}$ of an earlier event. It is not a median: the median has an additional factor $(\log 2{)}^{1/3}$.

Which population collides next? Immediately after the first disruption let $S=N_{s0}-2$ and $F=2x^{-3}$. The latter follows from mass conservation for equal-density spherical pieces. Using the same unordered-pair counting and geometric collision cross-sections as above, define $b=(1+x{)}^2/2$; thenHere $S^2,F^2$ stand for the usual large-population approximations to $S(S-1),F(F-1)$. The factor two distinguishing identical and different species is essential. The probability that the next collision is fragment–fragment isIt is the most likely type if $S<4/[x(1+x{)}^2]$ and $S<2/x^2$. It has probability greater than one half ifFor $x\ll 1$, a strongly fragment-dominated next event therefore requires $Sx\ll 4$; the fragment–satellite comparison is more restrictive than the satellite–satellite comparison. If $S=0$, the next event must be fragment–fragment, provided fragments remain.

Population equations and the normalization discrepancy. Every satellite–satellite event produces $2x^{-3}$ fragments and destroys two satellites. Every satellite–fragment event produces a net $x^{-3}-1$ fragments and destroys one satellite; every fragment–fragment event destroys two fragments. Consistent collision counting therefore givesFor $x\ll 1$, $b\simeq 1/2$ and $x^{-3}-1\simeq x^{-3}$. The last two terms in $A{\dot{N}}_f$ are then twice those in the printed equation. This is a genuine factor-of-two inconsistency: equal-size fragment pairs have a collision cross-section smaller by $x^2$, so their event rate must be $x^2{N}_f^2/A$ if the satellite event rate is $N_s^2/A$; destroying both fragments necessarily gives the sink $-2x^2{N}_f^2/A$.

If the printed equation is taken as a prescribed approximate rate model instead, its implicit event rates are $N_s^2/A$, $N_f{N}_s/(4A)$ and $x^2{N}_f^2/(2A)$. Under precisely that mixed normalization, its corresponding satellite equation isThe printed source term also neglects the consumed fragment in a satellite–fragment event, a legitimate relative $O(x^3)$ approximation. That approximation does not repair the pair-counting discrepancy.

The ensuing cascade. This is a two-size fragmentation cascade. First use the printed approximate model, dropping satellite–satellite events as requested. Put $S=N_s$, $F=N_f$, $k=4x^2$ and take $x<1/2$. ThenIntegrating this linear differential equation, with $F_0=2x^{-3}$ and $S_0=N_{s0}-2>0$, givesInitially the collisional cascade grows if $S_0>8x^2$, with approximate early exponential growth time $4A{x}^3/S_0$ when the satellite population is nearly fixed. The fragment population reaches its maximum atAfterwards satellites are depleted and fragment–fragment losses dominate. For $S_0>0$ the continuum solution has $S\to 0$, $F\to 0$, withWhen no satellites remain initially, $F(t)=F_0/[1+x^2{F}_{0}t/A]$ directly. Small integer populations eventually invalidate these deterministic differential equations.

The consistently counted model has, to leading order in $x$, exactly the same curve $F(S)$ and peak, but both retained time derivatives are twice as large. Its growth time is $2A{x}^3/S_0$, and $F(t)\sim A/(2x^{2}t)$. Keeping $(1+x{)}^2$ and the consumed fragment replaces $k$ by $4x^2/(1+x{)}^2$ and $x^{-3}$ in the curve by $x^{-3}-1$. This explicitly separates the physical collision bookkeeping from the printed normalization while giving the evolution under both conventions.

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.

A phase-space distribution function describes the mass in a small cell of positions and velocities: $F(\mathbf{x},\mathbf{v},t)d^{3}x{d}^{3}v$ is that mass. Its velocity integral is the spatial mass density, and $F/\rho$ at a fixed position is a normalized velocity probability density. A number-normalized galactic distribution function differs by the stellar mass convention.

Take $a>0$ and $v_0>0$. For the cusped logarithmic spherical galaxy model, differentiate the gravitational potential and use the spherical Poisson equation for Newtonian gravity:ThusA circular orbit requires $v_c^2/r={\Phi }^{\prime }$. Its galaxy rotation curve is thereforeIt rises as $v_0\sqrt{r/a}$ at small radius and approaches $v_0$ from below at large radius.

The outer flat galaxy rotation curve is qualitatively useful for galaxies. However, the density has an inner $1/r$ cusp and an outer $1/r^2$ tail, while $M(r)\sim v_0^{2}r/G$ diverges. The model therefore requires an outer truncation for a finite galaxy, and does not describe a central constant-density core. Its gravitational potential cannot be normalized to zero at infinity.

For a star, the specific energy is $E=v^2/2+\Phi$ and the specific angular momentum is $\mathbf{L}=\mathbf{x}\times \mathbf{v}$. Since the gravitational potential is stationary and spherical,Thus both the specific energy and the entire angular-momentum vector are conserved.

For Jeans theorem, follow a stellar trajectory in phase space. The Collisionless Boltzmann equation is preciselyIn a steady system ${\partial }_{t}F=0$, so the value of the galactic distribution function is constant along every trajectory: it is itself an integral of motion. Conversely, if $I_1,\dots ,I_m$ are time-independent integrals of motion, then $F=h(I_1,\dots ,I_m)$ satisfies $dF/dt=\sum h_{,a}{\dot{I}}_a=0$. Locally, a complete set of invariant orbit labels expresses any steady galactic distribution function this way. This is the weak form of Jeans theorem; a stronger assertion about a small set of isolating integrals needs additional orbital regularity or mixing assumptions. In particular, the theorem does not say that every steady galaxy must have a galactic distribution function depending on only $E$ and $L$.

For the proposed constant-anisotropy distribution function, write $v_T=(v_{\theta }^2+v_{\varphi }^2{)}^{1/2}$ and $L=r{v}_T$. Introduce a polar angle $\psi$ in the tangential velocity plane. Then $d^{3}v=d{v}_r{v}_{T}d{v}_{T}d\psi$, so the apparent $1/L$ singularity cancels against the velocity Jacobian. This inverse-angular-momentum velocity integration makes the density finite for every $r>0$.

Let $t=e^{-\Phi /v_0^2}=a/(a+r)$, $c_1=1$, $c_2=2$, and ${\alpha }_k=k/(2v_0^2)$. Each exponential component contributesHere the two Gaussian integrals and the angular integral have product ${\pi }^2/{\alpha }_k$. Consequently,The galactic distribution function is nonnegative, gives the required density, and depends on conserved $E,L$; by Jeans theorem it is a self-consistent steady solution.

Differentiating the Gaussian integral with respect to ${\alpha }_k$ shows that each component has $⟨v_r^2{⟩}_k=1/(2{\alpha }_k)=v_0^2/k$, and the same value for $⟨v_T^2{⟩}_k$. Angular averaging gives $⟨v_{\theta }^2{⟩}_k=⟨v_{\varphi }^2{⟩}_k=v_0^2/(2k)$. Weighting these by ${\rho }_1,{\rho }_2$ yields the velocity dispersionsThe radial numerator includes $2r$, as in the original PDF. All mean velocities vanish by parity. The same parity, or the zero angular average of $\sin \psi \cos \psi$, gives $⟨v_r{v}_{\theta }⟩=⟨v_r{v}_{\varphi }⟩=⟨v_{\theta }{v}_{\varphi }⟩=0$.

The velocity-anisotropy parameter isThe constant-anisotropy distribution function therefore favors radial motion at every radius: the radial variance is twice either individual tangential variance. It still has a substantial population with nonzero angular momentum, rather than exclusively radial orbits. The $L^{-1}$ factor enhances low-angular-momentum stars. The radial velocity dispersion approaches $3v_0^2/4$ at the centre and $v_0^2$ far out; each individual tangential variance is half as large. There is no mean rotation, despite the well-defined galaxy rotation curve for hypothetical circular orbits.

In a stationary axisymmetric gravitational potential, conservation of angular momentum gives $L_z=R^2\dot{\varphi }$. Eliminating $\dot{\varphi }$ introduces the effective potentialFor an equatorial circular orbit, radial balance is ${\partial }_R{\Phi }_{\mathrm{eff}}(R_c,0)=0$, givingVertical balance also requires ${\Phi }_{,z}(R_c,0)=0$.

For epicyclic motion, assume a twice differentiable stationary gravitational potential symmetric under $z\mapsto -z$, small displacements $|x|,|z|\ll R_c$, and a stable circular orbit. Choose the epicyclic guiding center using the conserved $L_z$. Reflection symmetry makes ${\Phi }_{,Rz}(R_c,0)=0$, eliminating radial-vertical coupling at first order. Expanding the equations about $(R_c,0)$ givesThe derivatives are evaluated at the guiding centre, and stability requires ${\kappa }^2,{\nu }^2>0$. Without midplane symmetry, a mixed Hessian term can couple the two oscillations. The radial epicyclic frequency and vertical epicyclic frequency are the frequencies of these independent linear oscillations.

Set $\Omega =L_z/R_c^2$. Along the family of equatorial circular orbits, ${\Phi }_{,R}=R{\Omega }^2$. Differentiating this relation and adding the centrifugal contribution givesTo interpret the common frequency range, let $q=d\log v_c/d\log R$. Since $v_c=R\Omega$, one has ${\kappa }^2=2(1+q){\Omega }^2$. A Keplerian disk has $q=-1/2$ and $\kappa =\Omega$, a flat galaxy rotation curve has $q=0$ and $\kappa =\sqrt{2}\Omega$, and solid-body rotation has $q=1$ and $\kappa =2\Omega$. Typical galactic rotation curves lie between these slopes. Thus $\Omega \lesssim \kappa \lesssim 2\Omega$ is a useful galactic range, not a theorem for every possible axisymmetric potential. As a precise sufficient example, epicyclic frequency bounds for monotone spherical density follow from ${\kappa }^2={\Omega }^2+4\pi G\rho$ and $0\le \rho \le 3M/(4\pi R^3)$.

Use a Cartesian frame rotating with the epicyclic guiding center, with $x$ pointing radially outwards and $y=R_c(\varphi -\Omega t)$ in the direction of rotation. To first order, conservation of angular momentum givesThe radial harmonic oscillator solution and its azimuthal integral areA constant in $y$ merely changes the azimuthal origin of the guiding centre. The epicyclic ellipse obeys $x^2/X^2+y^2/Y^2=1$. In the usual frequency range it is elongated azimuthally, and the star travels clockwise when $x$ points right and $y$ up: its small motion relative to the prograde guiding centre is retrograde.

For the Oort constants, subtract the defining expressions to obtain $\Omega =A-B$ and insert $R{\Omega }^{\prime }=-2A$ into the radial epicyclic frequency formula:The solar-neighbourhood values give $\Omega =26.5\mathrm{km}{\mathrm{s}}^{-1}\mathrm{kp}{\mathrm{c}}^{-1}$, $\kappa =\sqrt{1272}\mathrm{km}{\mathrm{s}}^{-1}\mathrm{kp}{\mathrm{c}}^{-1}$, andThe solar epicyclic ellipse is therefore about $1.49$ times longer azimuthally than radially.

A complete radial oscillation takes $T_r=2\pi /\kappa$. The oscillatory part of $\dot{\varphi }$ has zero average over this interval, so the epicyclic azimuthal advance isThis describes the advance between consecutive radial turning points of the same type; it need not be a full $2\pi$ revolution. For the Sun,The azimuthal advance estimate uses the same linear epicyclic motion approximation as the axis ratio.

For a spherical system, use the spherical shell theorem with shell mass $dm=4\pi \rho (s)s^{2}ds$. Inner shells contribute $-Gdm/r$ to the gravitational potential, while an outer shell contributes its constant interior potential $-Gdm/s$. With $\Phi (\infty )=0$, assuming the required integrals converge,On differentiation, the two terms containing $\rho (r)r$ cancel. Hence ${\Phi }^{\prime }=GM(r)/r^2$, where $M(r)=4\pi {\int }_0^r\rho (s)s^{2}ds$. Radial balance for a circular orbit gives $v_c^2=r{\Phi }^{\prime }=GM(r)/r$.

For a razor-thin axisymmetric astrophysical disk, the element of mass is $dm=\Sigma (s)sdsd{\varphi }^{\prime }$, where $\Sigma$ is the surface density. Superposing Newtonian gravitational potentials therefore givesAxisymmetry permits $\varphi =0$. Unlike the spherical case, an exterior annulus generally exerts a radial force: enclosed mass does not determine a disc rotation curve.

For the Legendre expansion of thin-disk gravity, split the radial integral at $s=R$. In the inner part the kernel expands in $s/R$, while in the outer part it expands in $R/s$. The angular integrals of odd Legendre polynomials vanish, and those of even degree $n$ equal $2{\alpha }_n$, whereIntroduce $I_n(R)={\int }_0^R\Sigma (s)s^{n+1}ds$ and $J_n(R)={\int }_R^{\infty }\Sigma (s)s^{-n}ds$. The gravitational potential becomesWhen differentiating each bracket, the moving-limit terms cancel: $R^{-(n+1)}{I}_n^{\prime }=\Sigma (R)$ and $R^n{J}_n^{\prime }=-\Sigma (R)$. Since $v_c^2=R{\Phi }^{\prime }$, this givesThe zeroth term has ${\alpha }_0=\pi$ and $M(R)=2\pi I_0$. Separating it provesThe inner correction is inward, while the exterior correction is outward. For a smooth surface density, the original potential singularity at coincident points is integrable. Its radial force is understood through a symmetric Cauchy principal value or a vanishing-thickness regularization; the paired interior and exterior terms above retain the cancellation at $s=R$. This avoids treating the two singular local force contributions separately.