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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.