Intuition The big pictureTwo big masses (Sun–Earth, Earth–Moon) go round their common centre of mass in nice circles. Now drop a tiny third body (a spacecraft, an asteroid) so light it feels the two masses but never pushes back on them. That's the Circular Restricted Three-Body Problem (CR3BP) .
The magic trick: ride along in a frame that rotates with the two masses . In that rotating frame the two big bodies are frozen in place . Now we hunt for spots where the little body can also sit frozen — the Lagrange points . The characteristic equation is just the eigenvalue equation that tells us whether a body nudged off such a point drifts back (stable) or runs away (unstable).
WHAT: Full 3-body gravity has no closed-form solution. But if m 3 → 0 m_3 \to 0 m 3 → 0 autonomous in a rotating frame — time drops out of the potential.WHY rotating frame: In an inertial frame the two primaries whirl around, so the potential is time-dependent → messy. Rotate with them at their orbital rate ω \omega ω conserved energy-like quantity (Jacobi constant) and fixed equilibria.HOW we pay for it: Rotating frames introduce fictitious forces — centrifugal (pushes outward) and Coriolis (turns moving bodies). These are the price of the frozen picture.
Definition Nondimensional unitsChoose units so that total mass = 1 =1 = 1 = 1 =1 = 1 ω = 1 \omega = 1 ω = 1 G = 1 G=1 G = 1 mass parameter
μ = m 2 m 1 + m 2 , 0 < μ ≤ 1 2 . \mu = \frac{m_2}{m_1+m_2}, \qquad 0<\mu\le \tfrac12. μ = m 1 + m 2 m 2 , 0 < μ ≤ 2 1 . x y xy x y m 1 = 1 − μ m_1=1-\mu m 1 = 1 − μ x = − μ x=-\mu x = − μ m 2 = μ m_2=\mu m 2 = μ x = 1 − μ x=1-\mu x = 1 − μ
Distances to each primary:
r 1 = ( x + μ ) 2 + y 2 + z 2 , r 2 = ( x − 1 + μ ) 2 + y 2 + z 2 . r_1=\sqrt{(x+\mu)^2+y^2+z^2},\qquad r_2=\sqrt{(x-1+\mu)^2+y^2+z^2}. r 1 = ( x + μ ) 2 + y 2 + z 2 , r 2 = ( x − 1 + μ ) 2 + y 2 + z 2 .
In an inertial frame, r ¨ in = − ∇ V grav \ddot{\mathbf r}_{\text{in}} = -\nabla V_{\text{grav}} r ¨ in = − ∇ V grav ω = z ^ \boldsymbol\omega = \hat z ω = z ^ ω = 1 \omega=1 ω = 1 r ¨ in = r ¨ + 2 ω × r ˙ + ω × ( ω × r ) . \ddot{\mathbf r}_{\text{in}} = \ddot{\mathbf r} + 2\,\boldsymbol\omega\times\dot{\mathbf r} + \boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r). r ¨ in = r ¨ + 2 ω × r ˙ + ω × ( ω × r ) .
Why these terms? 2 ω × r ˙ 2\boldsymbol\omega\times\dot{\mathbf r} 2 ω × r ˙ Coriolis term (appears because velocity itself gets "twisted" by rotation); ω × ( ω × r ) \boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r) ω × ( ω × r ) centrifugal term (rotation flings you outward).
Rearranging for the rotating-frame acceleration and moving the centrifugal term into an effective potential:
x ¨ − 2 y ˙ = ∂ U ∂ x , y ¨ + 2 x ˙ = ∂ U ∂ y , z ¨ = ∂ U ∂ z \boxed{\ \ddot x - 2\dot y = \frac{\partial U}{\partial x},\qquad \ddot y + 2\dot x = \frac{\partial U}{\partial y},\qquad \ddot z = \frac{\partial U}{\partial z}\ } x ¨ − 2 y ˙ = ∂ x ∂ U , y ¨ + 2 x ˙ = ∂ y ∂ U , z ¨ = ∂ z ∂ U
Equilibrium means ˙ = ¨ = 0 \dot{}=\ddot{}=0 ˙ = ¨ = 0 ∇ U = 0 \nabla U = 0 ∇ U = 0 z = 0 z=0 z = 0 ∂ U ∂ x = x − ( 1 − μ ) ( x + μ ) r 1 3 − μ ( x − 1 + μ ) r 2 3 = 0 , \frac{\partial U}{\partial x}=x-\frac{(1-\mu)(x+\mu)}{r_1^3}-\frac{\mu(x-1+\mu)}{r_2^3}=0, ∂ x ∂ U = x − r 1 3 ( 1 − μ ) ( x + μ ) − r 2 3 μ ( x − 1 + μ ) = 0 , ∂ U ∂ y = y [ 1 − 1 − μ r 1 3 − μ r 2 3 ] = 0. \frac{\partial U}{\partial y}=y\left[1-\frac{1-\mu}{r_1^3}-\frac{\mu}{r_2^3}\right]=0. ∂ y ∂ U = y [ 1 − r 1 3 1 − μ − r 2 3 μ ] = 0.
y = 0 y=0 y = 0 collinear points L 1 , L 2 , L 3 L_1,L_2,L_3 L 1 , L 2 , L 3 Bracket = 0 =0 = 0 y ≠ 0 y\ne0 y = 0 r 1 = r 2 = 1 r_1=r_2=1 r 1 = r 2 = 1 triangular points L 4 , L 5 L_4,L_5 L 4 , L 5
A Lagrange point is a balance point . To know if it's a valley (stable) or a saddle/hill (unstable) we nudge the body by ( ξ , η ) (\xi,\eta) ( ξ , η ) e + λ t e^{+\lambda t} e + λ t e i ω t e^{i\omega t} e iω t characteristic equation gives those λ \lambda λ
Let x = x 0 + ξ x=x_0+\xi x = x 0 + ξ y = y 0 + η y=y_0+\eta y = y 0 + η U U U U x x = ∂ 2 U / ∂ x 2 U_{xx}=\partial^2U/\partial x^2 U xx = ∂ 2 U / ∂ x 2 ξ ¨ − 2 η ˙ = U x x ξ + U x y η , \ddot\xi - 2\dot\eta = U_{xx}\,\xi + U_{xy}\,\eta, ξ ¨ − 2 η ˙ = U xx ξ + U x y η , η ¨ + 2 ξ ˙ = U x y ξ + U y y η . \ddot\eta + 2\dot\xi = U_{xy}\,\xi + U_{yy}\,\eta. η ¨ + 2 ξ ˙ = U x y ξ + U y y η .
Assume ξ , η ∝ e λ t \xi,\eta \propto e^{\lambda t} ξ , η ∝ e λ t { ( λ 2 − U x x ) ξ − ( 2 λ + U x y ) η = 0 ( 2 λ − U x y ) ξ + ( λ 2 − U y y ) η = 0 \begin{cases}(\lambda^2-U_{xx})\xi -(2\lambda+U_{xy})\eta=0\\ (2\lambda-U_{xy})\xi + (\lambda^2-U_{yy})\eta=0\end{cases} { ( λ 2 − U xx ) ξ − ( 2 λ + U x y ) η = 0 ( 2 λ − U x y ) ξ + ( λ 2 − U y y ) η = 0
Non-trivial solution ⇒ determinant = 0 =0 = 0
There U x y = 0 U_{xy}=0 U x y = 0 U x x > 0 U_{xx}>0 U xx > 0 U y y < 0 U_{yy}<0 U y y < 0 U x x U y y − U x y 2 < 0 U_{xx}U_{yy}-U_{xy}^2<0 U xx U y y − U x y 2 < 0 Λ > 0 \Lambda>0 Λ > 0 λ \lambda λ always unstable (saddle-type). That's why halo orbits around L 1 / L 2 L_1/L_2 L 1 / L 2
Here U x x = 3 4 U_{xx}=\tfrac34 U xx = 4 3 U y y = 9 4 U_{yy}=\tfrac94 U y y = 4 9 U x y = ± 3 3 4 ( 1 − 2 μ ) U_{xy}=\pm\tfrac{3\sqrt3}{4}(1-2\mu) U x y = ± 4 3 3 ( 1 − 2 μ ) Λ \Lambda Λ iff the discriminant ≥ 0 \ge0 ≥ 0 1 − 27 μ ( 1 − μ ) ≥ 0 ⇒ μ ≤ μ crit = 1 2 ( 1 − 23 27 ) ≈ 0.03852. 1-27\mu(1-\mu)\ge0 \;\Rightarrow\; \mu \le \mu_{\text{crit}}=\frac{1}{2}\left(1-\sqrt{\tfrac{23}{27}}\right)\approx 0.03852. 1 − 27 μ ( 1 − μ ) ≥ 0 ⇒ μ ≤ μ crit = 2 1 ( 1 − 27 23 ) ≈ 0.03852.
Intuition Feel for the numberμ crit ≈ 0.0385 \mu_{\text{crit}}\approx 0.0385 μ crit ≈ 0.0385 μ ≈ 0.00095 \mu\approx0.00095 μ ≈ 0.00095 Trojan asteroids parked at Jupiter's L 4 , L 5 L_4,L_5 L 4 , L 5
Worked example 1 — Locate the primaries for Earth–Moonm 1 = m_1= m 1 = m 2 = m_2= m 2 = μ = m 2 m 1 + m 2 ≈ 7.35 × 10 22 6.05 × 10 24 ≈ 0.0123 \mu=\dfrac{m_2}{m_1+m_2}\approx\dfrac{7.35\times10^{22}}{6.05\times10^{24}}\approx0.0123 μ = m 1 + m 2 m 2 ≈ 6.05 × 1 0 24 7.35 × 1 0 22 ≈ 0.0123 x = − μ = − 0.0123 x=-\mu=-0.0123 x = − μ = − 0.0123 x = 1 − μ = 0.9877 x=1-\mu=0.9877 x = 1 − μ = 0.9877 Why this step? Because our nondim units force separation = 1 =1 = 1 0 0 0 ( 1 − μ ) ( − μ ) + μ ( 1 − μ ) = 0 (1-\mu)(-\mu)+\mu(1-\mu)=0 ( 1 − μ ) ( − μ ) + μ ( 1 − μ ) = 0
Worked example 2 — Is Earth–MoonL 4 L_4 L 4 27 μ ( 1 − μ ) = 27 ( 0.0123 ) ( 0.9877 ) = 0.328 < 1 27\mu(1-\mu)=27(0.0123)(0.9877)=0.328<1 27 μ ( 1 − μ ) = 27 ( 0.0123 ) ( 0.9877 ) = 0.328 < 1 Why this step? Stability of L 4 / L 5 L_4/L_5 L 4 / L 5 1 − 27 μ ( 1 − μ ) 1-27\mu(1-\mu) 1 − 27 μ ( 1 − μ ) 0.328 < 1 0.328<1 0.328 < 1 linearly stable . (Real dust clumps — "Kordylewski clouds" — are debated evidence.)
Worked example 3 — Sign of the characteristic roots atL 1 L_1 L 1 U x y = 0 U_{xy}=0 U x y = 0 L 1 L_1 L 1 U x x ≈ + 11.3 U_{xx}\approx +11.3 U xx ≈ + 11.3 U y y ≈ − 4.2 U_{yy}\approx -4.2 U y y ≈ − 4.2 = U x x U y y = ( 11.3 ) ( − 4.2 ) ≈ − 47 < 0 =U_{xx}U_{yy}=(11.3)(-4.2)\approx-47<0 = U xx U y y = ( 11.3 ) ( − 4.2 ) ≈ − 47 < 0 Why this step? For Λ 2 + b Λ + c = 0 \Lambda^2+b\Lambda+c=0 Λ 2 + b Λ + c = 0 = c < 0 =c<0 = c < 0 Λ > 0 \Lambda>0 Λ > 0 λ \lambda λ Unstable, as expected.
Common mistake Steel-manned trapsTrap A: "Coriolis force does work and changes energy." Feels right because it's a real term in the equations. Fix: Coriolis is ∝ ω × r ˙ \propto \boldsymbol\omega\times\dot{\mathbf r} ∝ ω × r ˙ F ⋅ r ˙ = 0 \mathbf F\cdot\dot{\mathbf r}=0 F ⋅ r ˙ = 0 C J C_J C J
Trap B: "L 4 / L 5 L_4/L_5 L 4 / L 5 U U U Feels right — L 4 , L 5 L_4,L_5 L 4 , L 5 maxima of U U U Fix: Ordinary intuition ("balls roll off hilltops") ignores the Coriolis term , which steers the runaway motion into a stable orbit around the point — as long as μ < 0.0385 \mu<0.0385 μ < 0.0385
Trap C: "Restricted means the third body is small in size." Fix: It means small in mass — it exerts negligible gravity on the primaries. Size is irrelevant.
Trap D: Forgetting the "4 4 4 λ 4 + ( 4 − U x x − U y y ) λ 2 + … \lambda^4+(4-U_{xx}-U_{yy})\lambda^2+\dots λ 4 + ( 4 − U xx − U y y ) λ 2 + … Fix: the 4 4 4 2 η ˙ , 2 ξ ˙ 2\dot\eta,2\dot\xi 2 η ˙ , 2 ξ ˙
Recall Feynman: explain to a 12-year-oldImagine a merry-go-round with two heavy kids sitting still on it (because you're spinning along with them). Somewhere on the floor there are special spots where a marble would just sit without rolling — the gravity-pull, the spinning-fling, all cancel. Some spots are like the bottom of a bowl: push the marble and it comes back. Others are like the top of a hill: push it and it rolls away forever. The "characteristic equation" is our little math machine: we give it the shape of the ground around a spot, and it tells us "bowl" (safe) or "hill" (runaway). Surprise: even a hilltop spot can be safe here, because the spinning keeps curving the marble back — that's the secret of L 4 L_4 L 4 L 5 L_5 L 5
"Collinear = Cliff, Triangular = Trampoline (if Tiny)."
L 1 L 2 L 3 L_1L_2L_3 L 1 L 2 L 3 L 4 L 5 L_4L_5 L 4 L 5 tiny (μ < 0.0385 \mu<0.0385 μ < 0.0385 "4 comes from the 2-plus-2 of Coriolis."
What does "restricted" mean in CR3BP? The third body has negligible mass , so it doesn't perturb the two primaries.
Why use a rotating frame in CR3BP? The two primaries become stationary and the potential becomes time-independent, giving a conserved Jacobi constant and fixed equilibria.
Define the mass parameter μ \mu μ μ = m 2 / ( m 1 + m 2 ) \mu=m_2/(m_1+m_2) μ = m 2 / ( m 1 + m 2 ) , the mass fraction of the smaller primary (
0 < μ ≤ 1 / 2 0<\mu\le1/2 0 < μ ≤ 1/2 ).
Write the effective potential U U U U = 1 2 ( x 2 + y 2 ) + 1 − μ r 1 + μ r 2 U=\tfrac12(x^2+y^2)+\dfrac{1-\mu}{r_1}+\dfrac{\mu}{r_2} U = 2 1 ( x 2 + y 2 ) + r 1 1 − μ + r 2 μ .
What is the Jacobi constant? C J = 2 U − ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) C_J=2U-(\dot x^2+\dot y^2+\dot z^2) C J = 2 U − ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) ; the sole integral of motion in CR3BP.
Why is the Jacobi constant conserved despite Coriolis force? Coriolis force is perpendicular to velocity, so it does no work.
State the (planar) characteristic equation. λ 4 + ( 4 − U x x − U y y ) λ 2 + ( U x x U y y − U x y 2 ) = 0 \lambda^4+(4-U_{xx}-U_{yy})\lambda^2+(U_{xx}U_{yy}-U_{xy}^2)=0 λ 4 + ( 4 − U xx − U y y ) λ 2 + ( U xx U y y − U x y 2 ) = 0 .
Where does the "4" in the characteristic equation come from? From the Coriolis coupling terms
2 η ˙ 2\dot\eta 2 η ˙ and
2 ξ ˙ 2\dot\xi 2 ξ ˙ (each factor of 2, squared in the determinant).
Condition for stability of the roots? Both
Λ = λ 2 \Lambda=\lambda^2 Λ = λ 2 real and negative ⇒
λ = ± i ω \lambda=\pm i\omega λ = ± iω (pure oscillation).
Why are L 1 , L 2 , L 3 L_1,L_2,L_3 L 1 , L 2 , L 3 U x x U y y − U x y 2 < 0 U_{xx}U_{yy}-U_{xy}^2<0 U xx U y y − U x y 2 < 0 , so one
Λ > 0 \Lambda>0 Λ > 0 , giving a real positive
λ \lambda λ (saddle).
Stability condition for L 4 , L 5 L_4,L_5 L 4 , L 5 1 − 27 μ ( 1 − μ ) ≥ 0 1-27\mu(1-\mu)\ge0 1 − 27 μ ( 1 − μ ) ≥ 0 , i.e.
μ ≤ μ crit ≈ 0.03852 \mu\le\mu_{\text{crit}}\approx0.03852 μ ≤ μ crit ≈ 0.03852 .
Give a real physical example of stable L 4 / L 5 L_4/L_5 L 4 / L 5 Jupiter's Trojan asteroids (
μ Sun-Jup ≈ 0.00095 \mu_{\text{Sun-Jup}}\approx0.00095 μ Sun-Jup ≈ 0.00095 ).
How do collinear points differ physically from triangular ones? Collinear lie on the primary–primary line (unstable saddles); triangular form equilateral triangles with the primaries (
r 1 = r 2 = 1 r_1=r_2=1 r 1 = r 2 = 1 ).
Lagrange points — the five equilibria this note stabilisesJacobi constant & zero-velocity curves — the conserved energy and forbidden regionsRotating reference frames — Coriolis & centrifugal forces Eigenvalues & linear stability analysis — where the characteristic equation comes from generallyHalo orbits & station-keeping — practical use of unstable L 1 / L 2 L_1/L_2 L 1 / L 2 Trojan asteroids — natural bodies at L 4 / L 5 L_4/L_5 L 4 / L 5 Two-body problem & Kepler orbits — the μ → 0 \mu\to0 μ → 0
Assume m3 to 0 and circular primaries
Fixed primaries and equilibria
omega cross omega cross r
Rotating-frame equations of motion
Intuition Hinglish mein samjhoDekho, CR3BP ka funda simple hai: do heavy bodies (jaise Earth aur Moon) apne common centre of mass ke around circle mein ghoom rahe hain, aur ek bahut halka teesra body (spacecraft) unki gravity feel karta hai par unpe koi asar nahi daalta — isiliye "restricted". Ab trick yeh hai ki hum ek rotating frame mein baith jaate hain jo in dono ke saath ghoomta hai. Is frame mein dono bade bodies ek jagah freeze ho jaate hain, aur problem time-independent ban jaati hai. Iska inaam milta hai ek conserved quantity — Jacobi constant C J = 2 U − v 2 C_J = 2U - v^2 C J = 2 U − v 2
Rotating frame mein do naye "fake" forces aate hain: centrifugal (bahar ki taraf phenkta hai) aur Coriolis (chalti hui cheez ko mod deta hai). Effective potential U = 1 2 ( x 2 + y 2 ) + 1 − μ r 1 + μ r 2 U=\tfrac12(x^2+y^2)+\frac{1-\mu}{r_1}+\frac{\mu}{r_2} U = 2 1 ( x 2 + y 2 ) + r 1 1 − μ + r 2 μ ∇ U = 0 \nabla U=0 ∇ U = 0
