Lagrange points L1–L5 — derivation, stability: Visual walkthrough

We build on the setup from Restricted Three-Body Problem and the landscape idea from Effective Potential & Jacobi Constant. The spinning-frame forces come from Coriolis and Centrifugal Forces (Rotating Frames).

Step 1 — The stage: two heavy bodies and a spinning turntable

WHY a turntable? If we stayed still we'd see everything whirling — impossible to find a "parking spot". Freezing the two bodies turns a motion problem into a "find the flat spot" problem. This is the whole trick of a rotating frame.

PICTURE. Below, $M_{1}$ sits left, $M_{2}$ sits right, the barycenter (black dot) is between them. The curved arrow shows the whole picture spinning together at rate $ω$ (Greek "omega", our symbol for how fast the turntable turns, in radians per second).

Figure — Lagrange points L1–L5 — derivation, stability

Step 2 — Choosing rulers so the numbers are clean

WHY $ω = 1$ ? Because Kepler's Third Law links spin to gravity and distance: for the two bodies to circle each other, gravity must supply the centripetal pull, which forces $ω^{2} = \frac{G ( M _{1} + M _{2} )}{D ^{3}} .$ Term by term: $G (M_{1} + M_{2})$ is the total gravitational strength, $D^{3}$ is the cube of separation, and their ratio is the spin-squared. Choosing units where $G (M_{1} + M_{2}) = 1$ and $D = 1$ makes the right side $1$ , so $ω = 1$ . One symbol, gone.

PICTURE. We place $M_{1}$ at $x = - μ$ and $M_{2}$ at $x = + (1 - μ)$ , where $μ$ (Greek "mu") is the fraction of mass in the smaller body: $μ = \frac{M _{2}}{M _{1} + M _{2}} .$ Here $μ$ answers "what share of the total weight lives in $M_{2}$ ?" — a number between $0$ and $\frac{1}{2}$ . The barycenter lands exactly at $x = 0$ .

Recall

Why is $M_{1}$ at $- μ$ and not at $- 1/2$ ? ::: Because the heavier body sits closer to the balance point. The lever law $M_{1} \cdot (arm_{1}) = M_{2} \cdot (arm_{2})$ pulls the barycenter toward the heavy side.

Step 3 — The two forces a still marble feels

WHY only two? The Coriolis force is $- 2 ω \times v$ : it needs a velocity $v$ . A marble at rest has $v = 0$ , so Coriolis is exactly zero. Finding the parking spots needs only gravity + centrifugal. (Coriolis wakes up in Step 8, for stability.)

PICTURE. For a marble at $(x, y)$ we measure two distances: $r_{1} = (x + μ)^{2} + y^{2}, r_{2} = (x - (1 - μ))^{2} + y^{2} .$ Reading it out: $r_{1}$ is the straight-line gap to $M_{1}$ (which sits at $- μ$ ), $r_{2}$ is the gap to $M_{2}$ (at $1 - μ$ ). The magenta and violet arrows are the two gravity pulls; the orange arrow is the centrifugal push.

Step 4 — Packing all forces into one landscape $Ω$

WHY a potential? A force that can be written as "downhill of some height map" is far easier to reason about: instead of balancing vectors, we look for where the ground is level. Both gravity and centrifugal force are of this kind. This is the effective potential.

PICTURE + equation, term by term: $Ω (x, y) = well around M_{1} - \frac{1 - μ}{r _{1}} well around M_{2} - \frac{μ}{r _{2}} centrifugal dome - \frac{1}{2} (x^{2} + y^{2}) .$

$- \frac{1 - μ}{r _{1}}$ : a deep funnel around the big body; $1 - μ = M_{1} / (M_{1} + M_{2})$ is its weight-share, and $1/ r_{1}$ makes it plunge as you approach.
$- \frac{μ}{r _{2}}$ : a shallower funnel around $M_{2}$ (smaller share $μ$ ).
$- \frac{1}{2} (x^{2} + y^{2})$ : an upside-down bowl. Its slope points outward, exactly reproducing the centrifugal push $ω^{2} ρ$ (with $ω = 1$ ). The $\frac{1}{2}$ is there so that differentiating $x^{2}$ gives a clean $x$ .

Flat spot rule: a Lagrange point is where the ground stops tilting in every direction: $\frac{\partial Ω}{\partial x} = 0 and \frac{\partial Ω}{\partial y} = 0.$ The symbol $\partial / \partial x$ ("partial derivative") just asks: if I nudge $x$ a hair and hold $y$ fixed, does the height change? Zero means "no tilt this way".

Step 5 — The sideways-slope equation ( $\partial Ω/ \partial y = 0$ )

WHY start with $y$ ? Because the two big bodies both sit on the $x$ -axis, the $y$ -direction is symmetric and gives the cleanest condition — it will tell us the sum of the gravity strengths must equal the centrifugal strength.

Doing the derivative, term by term. Differentiating each piece of $Ω$ with respect to $y$ : $\frac{\partial Ω}{\partial y} = (\frac{1 - μ}{r _{1}^{3}}) y + (\frac{μ}{r _{2}^{3}}) y - y = 0.$

$\frac{1 - μ}{r _{1}^{3}} y$ : the $y$ -slope of $M_{1}$ 's well. The $r_{1}^{3}$ (cube) appears because differentiating $1/ r_{1}$ where $r_{1} = \dots + y^{2}$ brings down a $y / r_{1}^{3}$ .
$\frac{μ}{r _{2}^{3}} y$ : same story for $M_{2}$ 's well.
$- y$ : the $y$ -slope of the centrifugal dome.

Now use $y \neq = 0$ (we want an off-axis point). Divide the whole equation by $y$ : $\frac{1 - μ}{r _{1}^{3}} + \frac{μ}{r _{2}^{3}} = 1.$ Read it: the two gravity strengths, weighted by mass-share and by $1/ r^{3}$ , must add up to exactly $1$ (the centrifugal strength). Remember this box — call it Equation (Y).

Step 6 — The along-axis slope forces $r_{1} = r_{2}$

WHY equal distances? Intuitively: the $y$ -condition already fixed the total gravity strength. The $x$ -condition then balances the left–right components, and the only way to keep the barycenter's lever honest is for both bodies to sit at the same distance from the marble.

The $x$ -derivative, term by term: $\frac{\partial Ω}{\partial x} = \frac{1 - μ}{r _{1}^{3}} (x + μ) + \frac{μ}{r _{2}^{3}} (x - (1 - μ)) - x = 0.$

$(x + μ)$ : how far east of $M_{1}$ we are.
$(x - (1 - μ))$ : how far east of $M_{2}$ we are.

Substitute Equation (Y) to eliminate $x$ . Replacing the " $- x$ " using $x = x \cdot 1 = x (\frac{1 - μ}{r _{1}^{3}} + \frac{μ}{r _{2}^{3}})$ and simplifying, every $x$ cancels and you are left with: $(1 - μ) μ (\frac{1}{r _{1}^{3}} - \frac{1}{r _{2}^{3}}) = 0.$ Since $μ \neq = 0$ and $1 - μ \neq = 0$ (two real bodies), the only escape is: $r_{1} = r_{2} .$

PICTURE. The marble is now equidistant from both bodies — it lives on the perpendicular bisector of the $M_{1} M_{2}$ line (the dashed vertical line at $x = \frac{1}{2} - μ$ ).

Step 7 — Snapping the triangle shut: $r_{1} = r_{2} = D = 1$

Plug $r_{1} = r_{2} = r$ into (Y): $\frac{1 - μ}{r ^{3}} + \frac{μ}{r ^{3}} = \frac{( 1 - μ ) + μ}{r ^{3}} = \frac{1}{r ^{3}} = 1 ⟹ r^{3} = 1 ⟹ r = 1 = D .$ Every symbol earns its place: the numerator collapses because mass-shares sum to $1$ , and $r^{3} = 1$ forces $r = 1$ , which is exactly the separation $D$ .

WHY this is the equilateral triangle. Both big bodies are a distance $D$ apart, and the marble is a distance $D$ from each. Three points, all pairwise distances equal to $D$ — that is the definition of an equilateral triangle.

PICTURE + coordinates. Solving "distance $1$ from $M_{1}$ at $- μ$ and from $M_{2}$ at $1 - μ$ ": $x = \frac{1}{2} - μ, y = \pm \frac{3}{2} .$

$x = \frac{1}{2} - μ$ : midway between the bodies (the bisector from Step 6).
$y = + \frac{3}{2}$ is L4 (leads $M_{2}$ by $6 0^{\circ}$ ); $y = - \frac{3}{2}$ is L5 (trails by $6 0^{\circ}$ ). The $3 /2$ is just the height of a unit equilateral triangle.

Step 8 — The twist: L4/L5 are hilltops, yet stable

WHY they still hold. The moment the marble starts to slide, it is moving — and now the sleeping Coriolis force wakes up. Coriolis, $- 2 ω \times v$ , bends the drifting marble sideways, curling its escape into a tiny loop around the hilltop. The marble endlessly circles the peak instead of leaving.

The catch (a condition on masses). Coriolis only wins if the spin is fast enough relative to how steep the hill is — i.e. if $M_{2}$ is small enough. Linearising the motion gives the characteristic equation $λ^{4} + λ^{2} + \frac{27}{4} μ (1 - μ) = 0,$ and bounded (oscillating) motion requires its discriminant $\geq 0$ : $27 μ (1 - μ) \leq 1 ⟺ μ < μ_{c} = \frac{1}{2} (1 - \frac{23}{27}) \approx 0.0385,$ equivalently $M_{1} / M_{2} > \frac{25 + 621}{2} \approx 24.96$ .

PICTURE. Left: the hilltop with a marble spiralling (Coriolis-trapped). Right: for a too-heavy $M_{2}$ the loop unwinds and the marble escapes.

The one-picture summary

The whole derivation on one canvas: two bodies frozen on a turntable → build the height map $Ω$ → "no north tilt" gives Equation (Y) → "no east tilt" forces $r_{1} = r_{2}$ → together they snap the triangle shut at $r_{1} = r_{2} = D$ → Coriolis decides whether the hilltop traps or releases.

Recall Feynman retelling — the walkthrough in plain words

We hopped onto a merry-go-round spinning with the Sun and Jupiter, so both of them sit still for us. A crumb placed on the ride feels two tugs (gravity from each) plus one outward fling (centrifugal) — and while it sits still, the sideways Coriolis nudge is asleep. We drew a landscape of hills and valleys whose slope is that total tug, so a parking spot is just a flat patch of ground. Asking "is the ground level north–south?" told us the two gravity tugs must add up to the fling. Asking "level east–west?" told us the crumb must be equally far from both bodies. Put those together and the distances both lock to exactly the Sun–Jupiter gap — three points all the same distance apart, a perfect triangle. Those tips are L4 and L5, one $6 0^{\circ}$ ahead and one $6 0^{\circ}$ behind Jupiter. The funny part: they're hilltops, not valleys, so a crumb should roll off — but the instant it moves, the sleeping Coriolis nudge curls it into a little loop around the peak, trapping it forever, as long as the second body isn't too heavy. Jupiter is featherweight next to the Sun, so its Trojan asteroids have parked there since the Solar System was born.

Hill Sphere · Halo Orbits & Station-Keeping (JWST, SOHO) · Roche Limit · Kepler's Third Law · Restricted Three-Body Problem