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Let a disk contain fraction $f_d$ of its host mass and let its edge specific angular momentum be $f_j{R}_h{v}_h$. With halo relation $v_h^2=G{M}_h/R_h$ and disk geometry coefficient $\kappa =v_d^2{R}_d/(G{M}_d)$, combining centrifugal balance and the edge angular momentum gives the displayed estimates. The coefficient records the radial mass profile and thickness: enclosed mass does not determine a disc rotation curve. Treating $\kappa =1$ as a monopole estimate must not be mistaken for an exact thin-disk theorem.

Let $M_h,R_h$ describe the halo and let $v=v_{\mathrm{vir}}$. The cosmic baryon fraction is $f_b={\Omega }_b/{\Omega }_m=0.20$, so the specified settled fraction gives $M_d=0.10M_h$. The specific angular momentum constraint at the disk edge isTo infer an actual circular speed one must specify the disk's radial mass distribution: enclosed mass does not determine a disc rotation curve. Introduce a finite geometry coefficient $\kappa$ by $v_d^2=\kappa G{M}_d/R_d$. A rounded, centrally concentrated disk permits a monopole estimate $\kappa \simeq 1$ at its outer edge; it is an explicit approximation, not the spherical shell theorem applied exactly to a razor-thin disk. Ignore the force of the unsettled baryons as well as the dark matter within the disk. Using $v^2=G{M}_h/R_h$, the angular-momentum estimate of a self-gravitating galactic disk givesAn exact disk answer cannot be fixed by total disk mass alone; the dependence on $\kappa$ records that missing input.

For the numerical virial radius of a dark-matter halo, adopt mean density ${\Delta }_c=200$ times the critical density at the formation epoch. This conventional definition givesThe supplied expansion law gives $H(3)=286.49\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$, and hence $R_h=v/(10H)=97.74\mathrm{kpc}$. Taking $\kappa =1$ yieldsThe corresponding virial mass of a dark-matter halo is $M_h\simeq 1.78\times 10^{12}{M}_{\odot }$, and $M_d\simeq 1.78\times 10^{11}{M}_{\odot }$. Other overdensity conventions give $R_d\propto {\Delta }_c^{-1/2}$ at fixed $H,v,\kappa$.

Interpret the wavelength separation as an observed-frame local mean near the redshift in question. Since Lyman-alpha absorption appears at ${\lambda }_{\mathrm{obs}}={\lambda }_{\alpha }(1+z)$, the incidence isAssume one counted absorption system per intercepted disk, no unrelated forest systems or missed absorbers, unity neutral covering fraction, and a locally slowly varying population. Let ${\sigma }_{\mathrm{abs}}$ be the proper interception cross-section and $n_c$ the comoving number density. The proper density is $n_p=n_c(1+z{)}^3$, and the proper line element along the light path is $d{l}_p=cdz/[H(z)(1+z)]$. Thus the absorber incidence and comoving number density relation isFor a thin circular disk, the projected area is $\pi R_d^2|\cos i|$. Isotropically oriented normals have $⟨|\cos i|⟩=1/2$, so the random-orientation absorbing-disk cross-section is ${\sigma }_{\mathrm{abs}}=\pi R_d^2/2$. ConsequentlyIf all disks are taken face-on instead, $n_c\simeq 0.398{\mathrm{Mpc}}^{-3}$. Generally the random-orientation result scales as ${\kappa }^2({\Delta }_c/200)$ and is divided by the neutral covering fraction. The finite mean redshift spacing is sizable, so a precision inference would integrate the incidence over the actual redshift interval rather than identify it with one local value.

There is also a halo abundance mass-budget bound on this formal result. The present mean matter density for these parameters is ${\bar{\rho }}_{m,0}={\Omega }_{m}3H_0^2/(8\pi G)\simeq 3.40\times 10^{10}{M}_{\odot }{\mathrm{Mpc}}^{-3}$. Distinct haloes of this mass cannot have $n_c{M}_h>{\bar{\rho }}_{m,0}$, even if every matter particle belonged to them. Yet the inferred random-orientation population hasEven the face-on estimate exceeds this bound by about twenty-one. The numerical incidence result is conditional; the supplied population assumptions are not cosmologically consistent under this standard virial and compact-disk estimate. A larger neutral-gas absorption radius, a different absorber population, or different physical inputs are needed. For example, random orientations would require an absorbing radius at least about $63\mathrm{kpc}$ merely to reach the all-matter upper bound, much larger than the calculated centrifugal radius. This check does not change the algebraic answer, but prevents treating it as a realizable population.

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.

False. The enclosed-mass formula is a consequence of the spherical shell theorem. In a spherical galaxy, exterior shells exert no radial force and interior mass acts as though concentrated at the centre, giving $v_c^2=GM(<r)/r$.

A razor-thin axisymmetric disc instead hasBoth interior and exterior annuli contribute to that derivative. Its geometry is not determined by $M(<R)$.

A concrete counterexample compares a central point mass $M_0$ plus an exterior spherical shell with the same point mass plus a thin circular ring of mass $m$ and radius $a$. Their enclosed-mass profiles are identical: $M_0$ inside $a$, and $M_0+m$ outside. The interior gravitational force of an exterior thin ring follows by expanding its angularly averaged relative potential:For $R\ll a$, the ring's acceleration is outward, ${\partial }_R{\psi }_{\mathrm{ring}}=GmR/(2a^3)+\cdots$, whereas the spherical shell's acceleration is zero. With $M_0$ large enough to retain circular orbits, the disc-side speed isA narrow smooth annulus gives the same distinction. This proves that enclosed mass does not determine a disc rotation curve, even when the mass profiles of the spherical and disc models agree.