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Motion under a central force is planar. In plane polar coordinates, the radial component of the acceleration is $\ddot{r}-r{\dot{\theta }}^2$, soThe conserved specific angular momentum is $h=r^2\dot{\theta }$. Substitution of $\dot{\theta }=h/r^2$ yieldsThe last term is the centrifugal part of the radial effective potential.

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.

The equation has the one-dimensional effective potential per unit massThus the conserved specific energy isThe potential has its minimum $-2mk/a$ at $z=0$ and tends to $0$ from below as $|z|\to \infty$. Therefore $-2mk/a<\mathcal{E}<0$ gives bounded oscillation between two turning points; $\mathcal{E}=0$ gives marginal escape with asymptotic speed zero; and $\mathcal{E}>0$ gives escape with nonzero asymptotic speed. The minimum itself is the equilibrium $z=0$.

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.

Write $f(r)=1-r_{+}^2/r^2$ and let a dot denote differentiation with respect to an affine parameter $\tau$. The time-translation and $z$-translation Killing vector fields, together with rotational symmetry of the unit $S^3$, give the conserved quantitiesDefine $\epsilon =-g_{ab}{\dot{x}}^a{\dot{x}}^b$, so $\epsilon =1,0,-1$ for timelike, null and spacelike geodesics respectively. The normalization equation becomesIt therefore has the effective potential form

$L^2=J^2-(q\beta {)}^2=J^2{\sin }^2\alpha$. Splitting kinetic energy into radial and angular parts therefore gives $E=\frac{1}{2}m{\dot{r}}^2+J^2{\sin }^2\alpha /(2m{r}^2)+V(r)$, proving the stated effective potential.

The radial effective potential is $V_{\mathrm{eff}}=mgr\cot \alpha +5J^2/(14m{r}^2)$. Initially $r=8a$, $J=(7m/5)(8a)v_0$. Comparing the initial energy with $V_{\mathrm{eff}}(2a)$ giveswhich is exactly $v_0^2\ge (4/7)ga\cot \alpha$. Larger speed raises the centrifugal barrier at the neck faster than it raises the initial kinetic energy.

For a cold corotating field line anchored in a circular point-mass orbit, the effective potential has meridional Hessian ${\omega }^2\mathrm{diag}(-3,1)$. Along a straight line inclined by $\alpha$ to the vertical its quadratic curvature is ${\omega }^2(1-4{\sin }^2\alpha )$. Negative curvature gives the strict local launching criterion. Equality is marginal and global escape requires additional information.