Lagrange points L1–L5 — derivation, stability: Question bank

Before we begin, a one-line reminder of the cast so no symbol is unearned:

True or false — justify

L1, L2, L3, L4, L5 all lie in the same orbital plane as the two big bodies.

True — the effective potential

Ω

is built only from in-plane distances and the centrifugal term about the spin axis, so all five equilibria live in that plane; there is no out-of-plane parking spot.

There are exactly five Lagrange points for any two-body system, regardless of the masses.

True — the collinear condition always gives 3 real roots and the triangular solution

r_{1} = r_{2} = 1

always exists; the count is fixed at five, only their stability depends on

μ

L4 and L5 are stable only for certain mass ratios.

True — they are stable only when

M_{1} / M_{2} > 24.96

(equivalently

μ < 0.0385

); below that threshold the Coriolis force can no longer trap a drifting particle.

The centrifugal force is a "real" force that a fixed observer in space would also measure.

False — it is an apparent force that appears only because we chose a rotating frame; a non-rotating observer sees just gravity plus the body accelerating in a circle.

Because L4/L5 sit at maxima of

Ω

, a particle placed there must always roll away.

False — a static hilltop picture would say so, but the velocity-dependent Coriolis force (invisible in

Ω

) curls the drift into a small loop, trapping the particle when the mass ratio is small enough.

The Coriolis force is needed to locate the Lagrange points.

False — Coriolis

= - 2 ω \times v

vanishes for a particle at rest, and Lagrange points are defined as rest positions; only gravity + centrifugal set the locations. Coriolis enters solely in the stability analysis.

If you doubled both masses but kept their separation fixed, the five points would move.

False — the geometry depends only on the ratio

μ

, not the absolute masses; doubling both leaves

μ

unchanged, so the points stay put (though

ω

and orbital period change).

A spacecraft sitting exactly at L2 needs zero fuel to stay there forever.

False — L2 is a saddle point, so any tiny perturbation (radiation pressure, solar wind, numerical drift) grows exponentially; real missions like JWST fire small station-keeping burns.

Spot the error

"L1 is halfway between the two bodies by symmetry."

Wrong — L1 is pulled toward the smaller body

M_{2}

; for Sun–Earth it sits only ~1.5 million km from Earth, about 1% of the way, not 50%.

"L4 and L5 are stable because they are the minima of the effective potential."

Wrong — they are actually maxima of

Ω

; their stability comes entirely from the Coriolis deflection, which the static potential landscape cannot show.

"The equilateral-triangle result

r_{1} = r_{2} = D

for L4/L5 only works when

M_{1} = M_{2}

Wrong — it holds for every mass ratio, because

(1 - μ) + μ = 1

makes

r_{1} = r_{2} = 1

satisfy both equilibrium equations regardless of

μ

; only the apex's x-shift

\frac{1}{2} - μ

depends on the masses.

"Since the Coriolis force does no work, it cannot affect stability."

Wrong — it changes nothing about energy but it changes the direction of motion, bending an outward drift into a closed loop; that geometric steering is exactly what stabilizes L4/L5.

"Trojan asteroids sit exactly at the geometric L4/L5 points."

Wrong — because L4/L5 are potential maxima, real Trojans librate (swing) in wide tadpole-shaped paths around the points; they orbit the Lagrange point, they don't rest on it.

"We can only find L1, L2, L3 with a formula because the quintic has no closed form; L4/L5 also need numerics."

Wrong — L4/L5 have the exact closed form

x = \frac{1}{2} - μ, y = \pm \frac{3}{2}

; it is only the collinear points whose quintic requires numerical solution.

"In the rotating frame the total apparent force on a resting particle at a Lagrange point is nonzero but centripetal."

Wrong — the defining condition is

\nablaΩ = 0

, i.e. the net apparent force is exactly zero; that is why the particle stays at rest.

Why questions

Why do we bother going into a rotating frame at all?

Because in it the two big bodies are frozen, so "equilibrium relative to them" becomes a genuine rest condition; in a fixed frame the whole configuration spins and nothing is ever at rest.

Why does the centrifugal potential appear as

- \frac{1}{2} ω^{2} (x^{2} + y^{2})

with a minus sign?

So that its (negative) gradient points outward,

+ ω^{2} ρ \overset{ρ}{^}

, reproducing the outward centrifugal force; the sign is chosen to make the derivative give the correct push.

Why is the "3" in the Hill-radius formula

r \approx D (μ /3)^{1/3}

so specific?

It comes from adding the tidal-stretch factor of 2 to the centrifugal factor of 1 when the collinear equation is expanded near the small body; those two effects together set the balance length scale.

Why do L1 and L2 sit almost the same distance from Earth despite being on opposite sides of it?

Both are governed by the same local balance of Earth's gravity against the combined tidal + centrifugal gradient, which to leading order is symmetric in

\pm r

, giving nearly equal distances

\sim D (μ /3)^{1/3}

Why can Trojan asteroids survive at L4/L5 for billions of years while telescopes at L1/L2 need constant correction?

L4/L5 (with a small mass ratio) are genuinely stable — Coriolis returns any drift — while L1/L2 are saddle points where perturbations grow, so anything there must be actively held in place.

Why does the stability threshold depend on the mass ratio rather than the individual masses?

The linearized equations of motion reduce to a characteristic equation depending only on the product

μ (1 - μ)

; the absolute masses scale out along with the length and time units.

Why is L3 the least useful of the five for real missions?

L3 sits far on the opposite side of

M_{1}

, permanently hidden behind the larger body, and is an unstable saddle — hard to reach, hard to hold, and out of sight for communication.

Edge cases

What happens to the triangular points as

μ \to 0

(test body negligible, one mass dominates)?

The apex

x = \frac{1}{2} - μ \to \frac{1}{2}

and the triangle becomes equilateral about a nearly central primary; stability is guaranteed since

μ ≪ 0.0385

What happens at the exact threshold

μ = μ_{c} \approx 0.0385

The discriminant

1 - 27 μ (1 - μ)

hits zero, the two oscillation frequencies merge, and stability is marginal — perturbations neither clearly grow nor decay, so the point is borderline.

For equal masses

M_{1} = M_{2}

(so

μ = \frac{1}{2}

), are L4/L5 stable?

No —

μ = \frac{1}{2}

far exceeds

μ_{c} \approx 0.0385

, so the Coriolis force is too weak to trap the drift and the triangular points become unstable.

What is

ω

in the limit where the two bodies are very far apart (

D

large)?

By Kepler's third law

ω^{2} = G (M_{1} + M_{2}) / D^{3}

, so

ω \to 0

; the system spins ever more slowly, and the centrifugal term shrinks accordingly, though the five points still exist.

If the test mass were not negligible (comparable to

M_{2}

), would these five points still be exact equilibria?

No — the restricted-three-body assumption (

m

affects nothing) is essential; a heavy third body would perturb the primaries' orbit and the clean five-point structure would break down.

What does the effective potential look like at the location of a big body (

r_{1} \to 0

r_{2} \to 0

The gravity term

- \frac{1 - μ}{r _{1}}

- \frac{μ}{r _{2}}

diverges to

- \infty

, forming an infinitely deep well; Lagrange points sit at the saddles and ridges between these wells, never at the bottom.

Recall One-line summary you should be able to recite

Five points, always; three collinear saddles (unstable, need fuel), two triangular maxima (stable only if $M_{1} / M_{2} > 24.96$ , saved by Coriolis); Coriolis sets stability not location.

Related: Restricted Three-Body Problem · Effective Potential & Jacobi Constant · Coriolis and Centrifugal Forces (Rotating Frames) · Hill Sphere · Trojan Asteroids · Kepler's Third Law · Halo Orbits & Station-Keeping (JWST, SOHO) · Roche Limit