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The spherical shell theorem yields $v_c^2=GM(<r)/r$ only in spherical geometry. A razor-thin axisymmetric disc instead requires $v_c^2=R{\partial }_R\Phi (R,0)$, including forces from exterior annuli. The interior gravitational force of an exterior thin ring gives a concrete counterexample with identical enclosed mass but different circular speeds.

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.