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The Lagrange point lying between the components of a circular restricted three-body problem. The critical effective potential through this saddle defines their Roche lobes; gas passing through it can initiate Roche-lobe overflow.

The collinear Lagrange point beyond the smaller primary, away from the larger one, is $L_2$. For a small mass ratio it lies approximately one Hill radius from the secondary and has a growing linear mode.

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.

The normalization of the circular restricted three-body problem implies ${\mu }_1+{\mu }_2=1$, by Kepler's third law. The two bodies have barycentric positions $(-{\mu }_2,0)$ and $({\mu }_1,0)$. ConsequentlyDifferentiate the effective potential directly:Since $x={\mu }_1(x+{\mu }_2)+{\mu }_2(x-{\mu }_1)$ and $y=({\mu }_1+{\mu }_2)y$, these becomeOn $y=0$, centrifugal acceleration is the straight line $a_{\mathrm{cf}}=x$. Each gravitational term points towards its source, diverges at that source, and tends to zero far away. Let $f(x)=U_x(x,0)$; away from the sources,Thus the net acceleration is strictly increasing on each of the three intervals separated by the bodies. On the left interval $f$ runs from $-\infty$ to $+\infty$; on the middle interval it runs from $-\infty$ just right of $M_1$ to $+\infty$ just left of $M_2$; on the right interval it runs from $-\infty$ to $+\infty$. The intermediate value theorem and strict monotonicity give exactly three collinear Lagrange points, $L_3,L_1,L_2$ in increasing $x$ order. The sketch plots gravity and centrifugal acceleration separately; their intersections with $a_{\mathrm{grav}}=-x$ locate the equilibria.

Write $x_{L2}={\mu }_1+\delta$, so $r_{2,L2}=\delta$ and $r_{1,L2}=1+\delta$. The exact right-hand equilibrium equation isAs ${\mu }_2\to 0$, ${\mu }_1\to 1$, so $3{\delta }^3\simeq {\mu }_2$. Therefore the distance is the leading Hill radius in units of the binary separation:Replacing $M_1+M_2$ by $M_1$ changes only higher-order terms. In dimensional units the right-hand side is multiplied by the binary separation.

Use the displacement vector $\mathbf{q}=(X,Y,\dot{X},\dot{Y}{)}^T$, where $X=x-x_{L2}$ and $Y=y$. The absolute coordinate $x$ cannot be the first entry of a homogeneous linear system about $L_2$; it must be translated. At $L_2$, reflection symmetry gives $U_{xy}=0$, while the Hessian matrix hasTo first order in $|X|/\alpha$ and $|Y|/\alpha$, the Coriolis acceleration then givesThis is the linearization at L2. The expression for $B$ is exact at the true equilibrium; in the small-secondary limit, ${\mu }_2/{\delta }^3\simeq 3$ and ${\mu }_1/(1+\delta {)}^3\simeq 1$, so $B\simeq 4$.

For an eigenmode proportional to $e^{\lambda t}$, eliminate the velocity coordinates. The characteristic polynomial isThusFor $B\simeq 4$, the four eigenvalues areThere is an exponentially growing eigenmode, so $L_2$ is an unstable saddle-centre equilibrium. A generic displacement has an unstable component whose distance grows as $e^{\sqrt{1+2\sqrt{7}}t}$. The L2 escape e-folding time isThis is a local exponential timescale, not the time to reach a prescribed distance from an unspecified initial offset. Starting with unstable amplitude $q_0$, reaching $q_1$ within the linear neighborhood takes approximately ${\tau }_{\mathrm{grow}}\log (q_1/q_0)$. Initial data exactly on the centre-stable subspace need not grow forward at this linear order; the existence of the growing eigenmode already proves instability.

Equating the satellite's mean density inside $r_t$ with the host's mean density inside orbital radius $R$ givesThus a measured tidal radius and independently estimated satellite mass infer $M(<R)\simeq m(R/r_t{)}^3$. Measurements from satellites over a range of $R$ trace the host's enclosed-mass profile and hence its gravitational potential. Globular clusters are more often tidally limited: their stellar extent can approach the Jacobi boundary, whereas dwarf galaxies commonly occupy extended dark-matter haloes and their observed stars may substantially underfill it. Orbital eccentricity, mass loss, and nonequilibrium structure complicate this inference.

Let the separation vector point from the host to the satellite. Subtracting the two Newton's second law equations givesThe relative orbit is therefore a one-body Kepler orbit with gravitational parameter $G(M+m)$. A circular orbit of separation $R_0$ hasThe centre-of-mass condition gives $x_m=M{R}_0/(M+m)$ for the satellite's distance from the barycentre.

Consider the inner collinear equilibrium a distance $r_J$ toward the host from the satellite. Taking the positive axis from the host toward the satellite, differentiation of the gravitational plus centrifugal effective potential givesSince $r_J\ll R_0$,The constant host-gravity term cancels ${\Omega }^2{x}_m=GM/R_0^2$. The remaining equation isFor a low-mass satellite, the term proportional to $m{r}_J$ is negligible compared with $3M{r}_J$, and the Jacobi tidal radius isThe outer collinear point gives the same leading result.

Stars that cross the two nearby Lagrange points are no longer bound to the satellite. Small energy and angular-momentum offsets place them on slightly different host orbits, producing one leading and one trailing tidal tail. Differential orbital frequency stretches these streams around the host, while epicyclic motion can create density clumps. The tail's position and velocity structure retain information about the satellite orbit and host potential.

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.

The Roche lobe of star 1 is the volume around it bounded by the critical closed equipotential passing through the inner Lagrange point $L_1$. Material on a lower potential surface remains confined to star 1; at the critical surface a path opens toward star 2.

Let $\mathbf{g}=-\nabla \Psi$ and take $d\mathbf{S}$ outward. For a nearly spherical interior surface, the divergence theorem givesThe self-gravity term contributes $-4\pi G{M}_1$. The companion lies outside $S$, so its potential is harmonic inside and contributes zero net flux. For the centrifugal term, ${\nabla }^2(-{\Omega }^2{s}^2/2)=-2{\Omega }^2$, so its acceleration has divergence $2{\Omega }^2$ and contributes $2{\Omega }^{2}V$. Using $V=4\pi r^3/3$ and $S=4\pi r^2$,The sign is the outward-normal component; the dominant self-gravity is inward and therefore negative.

In barycentric coordinates with unit primary separation, a small secondary mass fraction $\alpha$ gives distances $(\alpha /3{)}^{1/3}$ from that secondary to the two neighboring Lagrange points at leading order. This is the dimensionless Hill radius. The opposite exterior L3 Lagrange point has coordinate $-1-5\alpha /12+O({\alpha }^2)$.