A mathematical aside: I would like to show more explicitly why the form of this instability for L1 through L3 is an exponential, as described

in the article here, but in a simpler, hopefully more approachable way. (Well, it's probably still not very approachable, but we'll see.) As prerequisite, I'll assume some knowledge of calculus (specifically, we'll need to do a few simple derivatives and integrals). But you don't really need to know how to do them, and can just skim over the actual calculations and still get some intuition.

First, we'll use the fact that the effective potential at these three Lagrange points is a saddle. It "bends upward" in one direction, and downward in another. Bending upward means it is like climbing up a hill. A displacement in that direction will result in a restoring force that brings the particle back toward the saddle point, just like a ball rolling back and forth in a bowl. But displacements in the other direction will result in a force that pulls it away, like a ball rolling downhill, faster and faster. This is the source of the instability.

Let's derive the equation of motion along that direction, for which the shape of the potential looks like the top of a hill. As mentioned back in the discussion of using potentials to analyze the motion of Mercury, a useful analytic technique is that the shape of a curve around any maximum or minimum can be approximated as a parabola, for displacements that are not too large from that location. So for the shape of the potential near L1, L2, or L3, along this axis of instability, we can treat the potential as a parabola opening downward.

In physics, the potential determines the force on a particle placed there. The steeper the potential, the greater the force. So let's calculate the force, and hence the acceleration, of a particle near the top of this hill. If we can calculate the acceleration, we can calculate its motion!

So, let the potential near these points along the axis bending downward be a negative parabola:

where k is just some constant describing how fast the parabola opens. The force is the negative gradient (negative slope) of the potential, and the acceleration is force per mass:

where kappa is just a new constant that we've mashed the factors of 2, m, and k into.

Acceleration is the 2nd time derivative of position, so let's write it out as

This is a 2nd order differential equation, and a pretty simple one. Actually it's exactly like Hooke's law for a mass on a spring, except instead of the force being attractive, it's repulsive (because there's no minus sign). Right away that should hint that this is bad for the particle and it will fly off faster and faster. Even if the solution wasn't already spoiled, it might not be hard to guess.

*"What kind of function has a 2nd derivative proportional to itself?"* But, let's solve it as if we didn't already know. First we can use an integration trick: multiply both sides by dx/dt dt, and use the subtitution u = dx/dt. Then du = d

^{2}x/dt

^{2} dt, so that on the left side we have the integral of udu. On the right side, the dt's cancel and we have integral of kappa xdx. Both are easy integrals:

which evaluate to:

Waving my hands vigorously, I argue we can drop this integration constant of c at the end, on the grounds that we're considering the velocity (dx/dt) to be very small when the displacement (x) from the Lagrange point is very small. In other words, by choice of boundary conditions, we can say c is zero.

Then, multiplying both sides of the equation by 2 and taking the square root,

We've now turned it into a 1st order separable differential equation! We love separable differential equations. Separate it by moving the x to the left side and the dt to the right side. Then we can integrate again:

So the natural logarithm of the position is proportional to time, plus some new integration constant.

How to get rid of the natural log? Exponentiate both sides!

And there it is! For a particle released close to the L1 - L3 Lagrange points, whose potentials are shaped like saddles, we have shown that in this direction of downward sloping potential, the instability leads to

*exponential* growth of any small perturbation from those points.

Now in this derivation, I've excluded a really important fact: in addition to the potential (which accounts for the gravitational + centrifugal forces), there is the Coriolis force. Ignoring it leads to the exponential instability that we see here. What's interesting is that when you do include the Coriolis term, you still get the exponential instability for L1 through L3, but it ends up stabilizing motion around L4 and L5. I don't think this is intuitively obvious at all, and to show it required the more complicated analysis seen in the above article on Lagrange point stability. That analysis also accounted for the motion in 2 dimensions, whereas here we limited to 1D, which still captures the equation of motion that we were after.