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Use rotating barycentric coordinates $\mathbf{R}=(x,y,z)$, rotating velocity $\mathbf{w}$, and $\mathbf{\Omega }=\hat{\mathbf{z}}$. Multiplying the three equations by the corresponding velocity components and adding cancels the Coriolis force terms:Thus the Jacobi constant is $C=2U-|\mathbf{w}{|}^2$. The inertial barycentric velocity is $\mathbf{V}=\mathbf{w}+\mathbf{\Omega }\times \mathbf{R}$. If $H_z=(\mathbf{R}\times \mathbf{V}{)}_z$ and $E_b=|\mathbf{V}{|}^2/2-{\mu }_1/r_1-{\mu }_2/r_2$, thenThe inertial energy $E_b$ and inertial angular momentum $H_z$ need not separately be constant. Their combination is constant because the two gravitational sources rotate steadily at unit angular speed.

For the Tisserand parameter take ${\mu }_2\ll {\mu }_1\simeq 1$, and compare the osculating orbital elements well outside a close encounter, where ${\mu }_2/r_2$ and barycentre-to-primary corrections are negligible. ThenThis requires a circular secondary orbit, the restricted three-body problem approximation, and no dissipative force or other perturber. $T$ is conserved to leading order between encounter episodes; the osculating $T$ during a close encounter need not be constant. The exact invariant is $C$.

For clarity the exact two-body energy relation is $a={\mu }_1/(2{\mu }_1/r_1-v_1^2)$, rather than the printed formula without the numerator ${\mu }_1$. At the leading order ${\mu }_1=1$ used for $T$, the two agree; retaining finite ${\mu }_1$ in just part of that formula is inconsistent.

At a close encounter put the secondary radius and speed equal to one at leading order, and denote the particle's incoming relative speed outside the secondary's strong-deflection region by $u$. Conservation of relative speed in the short gravitational encounter givesAn incoming orbit with $a=1$ has $|{\mathbf{v}}_1|=1$ at $r_1=1$ by the vis-viva equation; escape there requires an outgoing speed exceeding $\sqrt{2}$. The triangle inequality therefore gives the necessary single-encounter escape velocity boundEquality only gives a marginal parabolic limit. This argument deliberately gives a weak necessary bound, not a sufficient scattering criterion.

The planet-encounter relative velocity is related to the Tisserand parameter byHere $v_{1,\mathrm{tangential}}=h_z$ at the encounter radius, and the formula also applies to inclined passages at the planet's orbital plane. For the intended prograde coplanar incoming orbit, $I=0$ and $a=1$, so $u^2=2-2\sqrt{1-e^2}$. Applying the preceding necessary bound givesCoplanar alone also allows $I=\pi$. For that retrograde case $u^2=2+2\sqrt{1-e^2}$, and even a circular incoming orbit can be scattered onto an escaping orbit; the orbital eccentricity restriction therefore presumes prograde motion. Also, retaining the outgoing relative-velocity geometry gives the sharper necessary condition $|{\mathbf{v}}_{1,\mathrm{out}}|\le 1+u$, hence $u>\sqrt{2}-1$ and $e>\sqrt{4\sqrt{2}-5}/2$ for the same prograde incoming orbit. This stronger bound is consistent with, and implies, the weaker bound above.

For the Tisserand periapsis bound, first writeA bound orbit capable of another close encounter must intersect the secondary's circular radius, so $q\le 1\le Q$. For fixed $q<1$, the largest possible $T$ occurs at $I=0$. The remaining expression decreases with $Q\ge 1$, sinceThus every such orbit obeysOn $0<q<1$, $g(q)$ increases monotonically from $2$ to $3$. For $2<T<3$ the least accessible pericentre is the unique root of $T=g(q)$: the limiting orbit is prograde and coplanar, with apocentre just at the planet. Introduce $s=\sqrt{3-T}$. Its tangential speed at this apocentre is $1-s$, and its vis-viva equation gives the particularly well-behaved answerEquivalently, setting $w=\sqrt{2q/(1+q)}$ gives $T=2-w^2+2w$ and $w=1-s$, with $q=w^2/(2-w^2)$. Rationalizing the previous result yieldsThe apparent singularity at $T=\sqrt{8}$ is removable; the first form gives $q_{\min }=(\sqrt{2}-1)/2$ there. The limits are $q_{\min }\to 0$ as $T\downarrow 2$ and $q_{\min }\to 1$ as $T\uparrow 3$. The same lower envelope also excludes a smaller pericentre on an escaping trajectory when $2<T<3$. To see this without introducing an apocentre, use the planet-frame velocity sphere: $u=\sqrt{3-T}<1$ forces the tangential component at an encounter to be at least $w=1-u>0$, so total specific angular momentum obeys $h\ge w$. At pericentre the stellar energy obeys $h^2=2q+2E{q}^2$, while $E=h_z-T/2\le h-T/2$. ThusIf $q<q_{\min }=w^2/(2-w^2)$, then $q^2<w\le h$, and the left side is increasing with $h$. Hence $w^2\le 2q+(2w-T)q^2$. But $2w-T=w^2-2$, and the right side is smaller than $w^2$ for $0\le q<q_{\min }$, a contradiction. This establishes the barrier for all Kepler trajectories that reach the encounter radius, including unbound ones.

For $0<T\le 2$, bound encounter-crossing orbits can approach $q=0$ (take $Q\to 2/T$), so the positive root extrapolated from the displayed expression is not a universal barrier. For $T>3$ there is no trajectory reaching a close encounter in this approximation, since $u^2=3-T$ would be negative. Multiple encounters can approach the derived lower envelope by changing energy, angular momentum and inclination while preserving the Tisserand parameter; the invariant alone does not guarantee a particular encounter history reaches it.