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The rotating-frame effective potential of the circular restricted three-body problem has negative singular wells at both primaries. Its three Collinear Lagrange points are planar saddles, and its two Triangular Lagrange points are planar local maxima for nonzero primary masses. These geometric labels do not replace a dynamical stability analysis including the Coriolis force.

Let $M=M_1+M_2$ and $\alpha =M_2/M$, with $0<\alpha <1$. In the centre of mass frame the two primary distances are $r_1=\alpha R$ and $r_2=(1-\alpha )R$. Applying Newton's law of universal gravitation and circular centripetal acceleration to the first primary gives $M_1{\omega }^2\alpha R=G{M}_1{M}_2/R^2$, and henceThe same result follows from the second primary or from the relative two-body equation of motion.

A rotating-frame equilibrium has zero rotating-frame velocity, so its Coriolis force vanishes. Above the orbital plane both primaries have downward gravitational components; below it both have upward components. The centrifugal force has no $z$ component. More explicitly,which vanishes only at $z=0$ for finite nonsingular points. Thus all equilibria lie in the orbital plane.

The equation of motion in a rotating frame for the dwarf isOn the orbital plane the centrifugal contribution is $m{\omega }^2\mathbf{r}$. Thus the rotating-frame effective potential is $E=U-m{\omega }^2(x^2+y^2)/2$; equilibria are exactly its stationary points. The Coriolis force affects motion about those equilibria but not their positions. This is the circular restricted three-body problem, with the dwarf treated as a test mass.

The printed energy unit has the wrong dimensions. The correct total energy unit is $mGM/R$, or $GM/R$ for specific energy; $GM/R^2$ is an acceleration. Dividing $E$ by $mGM/R$ and measuring lengths in units of $R$ gives the dimensionless effective potentialThe primary coordinates are $(-\alpha ,0)$ and $(1-\alpha ,0)$. Restricting to $y=0$ gives $F(x)=\mathcal{E}(x,0)$ andThe absolute values are essential; dropping them would reverse the attraction on the left of a primary. On each of the intervals separated by the two primaries, $F^{\prime }$ decreases strictly from $+\infty$ to $-\infty$. There is exactly one Collinear Lagrange point on each interval, a maximum along the $x$ direction.

For the inner Lagrange point, write $x=1-\alpha -\delta$ with $\delta >0$. For the L2 Lagrange point, write $x=1-\alpha +\delta$. The respective equations areBalancing the leading terms gives $\delta =(\alpha /3{)}^{1/3}+O({\alpha }^{2/3})$. Therefore the small-mass-ratio collinear Lagrange points obeyThe omitted $-\alpha$ displacement of the secondary is smaller than the stated error. The leading distance from the secondary is the dimensionless Hill radius.

For the L3 Lagrange point put $x=-1+c\alpha +O({\alpha }^2)$. It lies to the left of both primaries, soExpanding the two positive inverse-square factors gives $(1-\alpha )/(x+\alpha {)}^2=1+(2c+1)\alpha +O({\alpha }^2)$ and $\alpha /(x-1+\alpha {)}^2=\alpha /4+O({\alpha }^2)$. Thus $F^{\prime }=-(3c+5/4)\alpha +O({\alpha }^2)$, soAll three have $y=z=0$. The small-$\alpha$ equalities in the paper are asymptotic expressions, not exact coordinates for finite mass ratio.

For the remaining Lagrange points, introduce plane polar coordinates about the first primary: $x=-\alpha +s\cos \theta$, $y=s\sin \theta$. Its distance to the second primary is $d=(s^2+1-2s\cos \theta {)}^{1/2}$, andThe angular stationary condition is ${\partial }_{\theta }\mathcal{E}=\alpha s\sin \theta (d^{-3}-1)=0$. For an off-axis point $s>0$ and $\sin \theta \ne 0$, so $d=1$. The radial stationary condition becomesHence $s=1$, and $d=1$ implies $\cos \theta =1/2$. The Triangular Lagrange points are exactlyBoth triangles are equilateral. The collinear and off-axis calculations together exhaust the stationary points.

For the requested sketch, $\mathcal{E}$ tends to $-\infty$ at either primary and at large in-plane radius. Each Collinear Lagrange point is a saddle of the planar effective potential: its $x$ curvature is negative and its $y$ curvature is positive. To see the latter, let $d_1=|x+\alpha |$, $d_2=|x-1+\alpha |$ and $D=(1-\alpha )/d_1^3+\alpha /d_2^3$. At an inner point both distances are below one, so $D>1$. At an exterior stationary point, use barycentric positions $a_1=-\alpha$, $a_2=1-\alpha$ and weights $1-\alpha ,\alpha$: the equation implies $x(D-1)=\sum w_i{a}_i/|x-a_i{|}^3$. To the right, the weighted sum is positive because the positive primary is closer; to the left it is negative because the negative primary is closer. Thus $D>1$ there as well, and ${\mathcal{E}}_{yy}=D-1>0$.

At either Triangular Lagrange point, the planar Hessian isIts trace is $-3$ and determinant $27\alpha (1-\alpha )/4>0$, so these are local maxima. Their height is $\mathcal{E}=-3/2+\alpha (1-\alpha )/2$. The sketch below uses a finite mass ratio to separate the five points clearly; its singular wells are clipped only for display. These curvature labels concern the effective potential, not a stability test that ignores the Coriolis force.

At a Triangular Lagrange point, $y\ne 0$, so $A=1$. The $x$ equation then givesSubstitution into $A=1$ and ${\mu }_1+{\mu }_2=1$ yieldsIntersecting these two circles givesAs ${\beta }_1$ rises from zero to one, $s$ falls from one to zero. The two points move along the unit circle about $M_2$, from the classical equilateral positions toward $M_1$, where they coalesce when the attraction of $M_1$ is fully cancelled.

The two halo centres orbit their centre of mass with separation $R$. Their relative coordinate obeys $\ddot{\mathbf{R}}=-G(M_1+M_2)\mathbf{R}/R^3$, hence circular motion requiresReflection symmetry about the orbital plane makes the vertical force point back toward $z=0$, while the centrifugal force has no vertical component. An equilibrium away from that plane is therefore impossible.

In the uniformly rotating frame, an equilibrium is a stationary point of the gravitational plus centrifugal effective potentialSet $R=1$, divide by $G(M_1+M_2)/R$, and write $\alpha =M_2/(M_1+M_2)$. The primary and secondary lie at $x=-\alpha$ and $x=1-\alpha$, so on their line

For the two roots near the secondary, put $x=1-\alpha \mp d$ in $F^{\prime }(x)=0$. Dominant balance gives $3d\simeq \alpha /d^2$, so $d=(\alpha /3{)}^{1/3}$. Expanding the root beyond the primary directly in powers of $\alpha$ givesto the requested orders. The other equilibria are the two Triangular Lagrange points

The distance from $H_2$ to either nearby collinear point is the Hill radius. Since $\alpha \simeq M_2/M_1$,Inside this tidal radius, the subhalo's gravity dominates the host's differential gravitational field; outside it, material can escape through the neighborhoods of $L_1$ and $L_2$. For an extended spherical host, $M_1$ is replaced by enclosed mass and the coefficient becomes $3-d\log M_1/d\log R$, giving the Jacobi tidal radius. An extended subhalo requires the bound mass inside $r_t$ to be found self-consistently. On an eccentric orbit there is no time-independent rotating potential or exact tidal boundary; stripping is strongest near pericentre and the instantaneous radius varies around the orbit.

Because the dark component is more extended, tidal stripping first sends dark matter through both $L_1$ and $L_2$, producing leading and trailing dark-matter tidal tails. The compact stellar component is stripped more deeply and also produces a leading and a trailing stellar tail. The two constituents therefore give four tails distinguished by composition, with the dark tails broader and more extended.

At a Lagrange point, the test particle is stationary in the rotating reference frame, so $F=G=0$. On the $x$ axis, $y=0$ makes $G=0$ automatically. The singularities at the two masses split the axis into three intervals, and the balance between gravity and centrifugal acceleration gives one root of $F=0$ in each interval. These are the three Collinear Lagrange points $L_1,L_2,L_3$.

For an off-axis equilibrium, $y\ne 0$. Put $A_i=1-r_i^{-3}$. ThenwhileHence $A_1=A_2=0$ and therefore $r_1=r_2=1$. The two intersections of unit circles centered at $M_1$ and $M_2$ form equilateral triangles with the binary, giving the Triangular Lagrange pointsTogether with the three collinear points, these are the five equilibria of the circular restricted three-body problem.