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For an exponential galactic disk, the first nonspherical term in the Legendre expansion of thin-disk gravity is positive and proportional to $R^{-3}$ in $v_c^2$. The missing enclosed mass and exterior-ring terms are exponentially small at large radius. Hence the galaxy rotation curve approaches the point-mass Keplerian curve from above.

A Mestel disk is a thin axisymmetric astrophysical disk with surface density inversely proportional to radius. Its galaxy rotation curve is exactly flat, $v_c=v_M$, and $v_c^2=GM(R)/R$ despite its nonspherical geometry. In the Legendre expansion of thin-disk gravity, all positive-degree interior and exterior terms cancel. Its total mass is infinite, so only potential differences, rather than a potential zero at infinity, are defined.

Put $x=r/R_d$, $f(x)=1-(1+x)e^{-x}$ and $f_t=f(5)=1-6e^{-5}$. Direct integration of the exponential surface density givesBeyond the disk, $M_d(<r)=M_{\star }$.

A razor-thin disk is not a spherical mass distribution. Its exact exterior force does not become the point-mass force immediately at $r=R_t$. For all $r>R_t$, an exact expression isThis follows by differentiating the Newtonian potential of each mass element. At distant radii its leading behavior is Keplerian:The first disk-shape correction from the Legendre expansion of thin-disk gravity is $V_d^2=(G{M}_{\star }/r)[1+3⟨s^2⟩/(4r^2)+\cdots ]$, with positive mass-weighted $⟨s^2⟩$. This supplies a concrete counterexample to treating $G{M}_{\star }/r$ as exact for a flattened disk merely because the observation point is outside its edge.

If the outer dark matter is approximately spherical, an exactly flat speed requiresIn the halo-dominated leading approximation, this is $M_h(<r)\simeq V_{\infty }^{2}r/G$ and ${\rho }_h\simeq V_{\infty }^2/(4\pi G{r}^2)$. Such a flat curve cannot continue to arbitrarily large radius without giving infinite total mass.

Interpret the new force as attractive. With outward position vector $\mathbf{r}$, this requires a minus sign in front of the source's printed expression; the literal positive sign and positive $a_0$ would repel a test particle. The radial force magnitude is $G{M}_d(r)m/(a_0{r}^{\alpha -1})$, so circular balance gives $V_c^2=G{M}_d(r)r^{2-\alpha }/a_0$. Constant exterior $M_d$ therefore requiresThis $a_0$ is not an acceleration parameter. Matching the outer speed sets $a_0=G{M}_{\star }/V_{\infty }^2$. The entire rotation curve defined by this enclosed-mass toy law isSince $f(x)=x^2/2+O(x^3)$, it rises linearly from zero at small radius and flattens outside. The enclosed-mass prescription defines this modified model; it is not a derivation of a general modified field equation for arbitrary flattened matter.

The central obstruction to an outer halo mass offset requires an explicit distinction. Extending the full distant-outer expression $V_{\infty }^{2}r/G-M_{\star }$ to zero gives $M_{\mathrm{DM}}(0)=-M_{\star }$. It becomes negative for $r<G{M}_{\star }/V_{\infty }^2$ and cannot be a physical halo. Its formal total squared speed, $V_d^2+V_{\infty }^2-G{M}_{\star }/r$, even becomes negative near the centre. Thus there is no physical entire rotation curve under that literal extrapolation.

If instead one extends only the positive halo-dominated leading term, $M_h=V_{\infty }^{2}r/G$, the entire Newtonian model isFor a full calculation of the disk contribution, the elliptic-integral circular-speed kernel of a thin disk giveswhere $K$ is the complete elliptic integral of the first kind and $E$ the complete elliptic integral of the second kind. The principal value is needed inside the disk; outside it the integral is ordinary. The disk term vanishes at the centre, so this halo extrapolation has $V_{\mathrm{N}}(0^{+})=V_{\infty }$, develops a stellar contribution at intermediate radii and approaches $V_{\infty }$ from above. A discontinuous, exactly zero-thickness truncation has an artificial logarithmic force singularity right at the edge; a smooth edge or finite thickness removes it. The figure omits the single singular edge point.

Observed galaxy rotation curves generally rise from the central region rather than beginning with a nonzero constant. The modified toy curve therefore captures that qualitative behavior better than this particular singular isothermal halo extrapolation, without establishing that its force law is correct. If dark matter is the explanation, the inner halo cannot universally retain $M_h\propto r$, ${\rho }_h\propto r^{-2}$ down to zero. For a spherical density ${\rho }_h\propto r^{-{\gamma }_h}$, $V_h\propto r^{1-{\gamma }_h/2}$ tends to zero when ${\gamma }_h<2$; a constant-density core gives a linear rise. A shallower cusp can also tend to zero. Stellar mass-to-light ratio, baryonic structure and disk–halo degeneracy prevent this qualitative argument alone from proving that every halo has a constant-density core.

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.

For an exponential galactic disk, write $\Sigma (s)={\Sigma }_0{e}^{-s/R_d}$, so $M_{\infty }=2\pi {\Sigma }_0{R}_d^2$ and $M(R)=M_{\infty }[1-(1+R/R_d)e^{-R/R_d}]$. At large $R$, the missing mass and exterior-ring terms are exponentially small. The leading nonspherical interior term is $k=1$, with ${\alpha }_2=\pi /4$ and $I_2(\infty )=6{\Sigma }_0{R}_d^4$. Consequently the exponential-disk Keplerian asymptotic isThe positive leading correction shows that the rotation curve approaches the Keplerian limit from above. The finite-order large-radius expansion is sufficient here; an infinite moment expansion need not converge for a disk extending to infinity.

A useful special example is the Mestel disk, with surface density $\Sigma (s)=C/s$ for $s>0$, $C>0$. It has $M(R)=2\pi CR$. For every positive even $n$,All the nonspherical corrections cancel, givingThus the Mestel disk has a perfectly flat galaxy rotation curve.

This example has infinite total mass and a singular central surface density. The absolute gravitational potential cannot be set to zero at infinity, but the radial force exists as a limit of disks with increasing outer cutoff. Potential differences are $\Phi (R)-\Phi (R_0)=v_M^2\log (R/R_0)$. The earlier potential integral is therefore interpreted up to a radius-independent divergent constant for this example; the force calculation remains valid. Truncating the Mestel disk gives a more physical finite system but changes the exact flat curve near its edges.