The Circular Restricted Three-Body Problem

Two primaries in a rotating frame, the equations of motion, the Jacobi constant, and the zero-velocity curves that decide where a spacecraft may go

Sriharikota, 2 September 2023

At 11:50 IST a PSLV rocket lifted off from Sriharikota carrying Aditya-L1, India's first solar observatory in space. Its destination was not a planet, a moon or an orbit around the Earth. It was a point in empty space about 1.5 million kilometres away — four times as far as the Moon — on the line from the Earth to the Sun. Think about what that means with the tools of this course. A spacecraft 1.5 million kilometres closer to the Sun than the Earth sits 0.99 AU from it. Kepler's third law gives an orbit at that distance a period of 0.985 years: about 360 days. Yet Aditya-L1 goes round the Sun in exactly one year, keeping station between the Sun and the Earth for as long as its mission lasts. No two-body orbit does that. The Earth's pull, outward towards the Earth, cancels just enough of the Sun's to slow the spacecraft down to the Earth's pace. To describe it, both bodies have to be in the problem at once.

The lesson on spheres of influence drew a sphere of influence around the Earth with radius

a(m/M)^{2/5}

: about 929,000 km. Inside it the Earth is treated as the only attracting body; outside it, the Sun. Aditya-L1's destination lies at about 1,497,000 km — well outside that sphere, yet held in place by the Earth. The patched conic cannot describe it at all.

The lesson on gravity assists and Lagrange points named the five points where this happens and showed the James Webb Space Telescope looping around one of them. This unit derives them, works out which ones a spacecraft can use, and follows a real mission from launch to its first completed orbit around one.

💡The circular restricted three-body problem

Three simplifications make the problem tractable without losing its essence:

Two massive bodies — the primaries — move on circular orbits about their common centre of mass.
The third body is so light that it does not disturb them.
We describe everything in a frame that rotates with the primaries, so they stand still.

Choose units in which the distance between the primaries, their total mass and their angular velocity are all 1. The only parameter left is the mass ratio

\mu = m_2/(m_1 + m_2)

. The big primary sits at

x = -\mu

, the small one at

x = 1 - \mu

, and one unit of time is

1/2\pi

of their orbital period. The first lab sets this up: for the Sun and the Earth (with the Moon's mass added to the Earth's, the standard choice),

\mu = 3.040 \times 10^{-6}

and one time unit is 58.13 days; for the Earth and the Moon,

\mu = 0.01215

and one time unit is 4.342 days.

\ddot x - 2\dot y = \frac{\partial U}{\partial x},\quad \ddot y + 2\dot x = \frac{\partial U}{\partial y},\quad \ddot z = \frac{\partial U}{\partial z},\qquad U = \frac{x^2+y^2}{2} + \frac{1-\mu}{r_1} + \frac{\mu}{r_2}

The 2ẏ and 2ẋ terms are the Coriolis acceleration of the rotating frame (see the lessons on frames of reference and on planetary frames); the (x²+y²)/2 in U is its centrifugal potential. r₁ and r₂ are the distances to the two primaries.

The lesson on planetary frames and mission frame hand-offs introduced this synodic frame as the natural one near the Moon or near the Sun–Earth Lagrange points. Its price is that the equations contain velocity-dependent forces. A spacecraft 'at rest' at a point of this frame is in fact circling the Sun at that point's distance: the lab's spacecraft standing still at Sun–Earth L1 moves round the Sun at 29.486 km/s, slower than the Earth's 29.785 km/s, as it must be to take a full year on a smaller circle.

💡The Jacobi constant

In the inertial frame the spacecraft's energy is not conserved: the primaries move and do work on it. In the rotating frame one combination is conserved — Jacobi's integral, published in 1836:

C = 2U(x, y, z) - v^2

where

v

is the speed in the rotating frame. It is the only conserved quantity the problem has. Heinrich Bruns proved in 1887, and Henri Poincaré in his prize memoir of 1890, that the three-body problem has no further integrals of the kind that solved the two-body problem in closed form. Everything else must be computed.

📐Worked Example: an integrator honesty test

The second lab starts a spacecraft 10,000 km beyond the Moon at 0.4 km/s — a messy, chaotic trajectory — and integrates it for 30 days three ways. Because

C

must stay constant, its drift measures the numerical error:

adaptive DOP853 with tolerance $10^{-12}$ : $|C - C_0| = 1.1 \times 10^{-10}$ ;
fixed-step RK4 with 20,000 steps: $3.4 \times 10^{-5}$ ;
fixed-step RK4 with 2,000 steps: 14 — larger than $C$ itself. That run is not a solution of the three-body problem at all; the close pass to the Moon was taken in steps far too long.

Check the Jacobi constant in every three-body simulation, as you checked energy in two-body ones (in the lesson on orbit propagation).

💡Zero-velocity curves

Because

v^2 \geq 0

, a spacecraft with Jacobi constant

C

can only be where

2U \geq C

. The boundary

2U = C

is the zero-velocity surface, first drawn by George William Hill in his lunar theory of 1878. Large

C

(little energy) confines the spacecraft to a bubble around one primary. As

C

falls, the walls open — first at L1 between the primaries, then at L2 behind the small one, then at L3, and finally the forbidden region shrinks to two islands around L4 and L5. For the Earth–Moon system the third lab finds

C(\text{L1}) = 3.188341

C(\text{L2}) = 3.172160

C(\text{L3}) = 3.012147

and

C(\text{L4}) = C(\text{L5}) = 2.987997

📐Worked Example: can a spacecraft leaving low orbit reach the Moon’s region?

Give a spacecraft at perigee 200 km above the Earth, on the side away from the Moon, a prograde inertial speed

v

, and compute its Jacobi constant:

10.87 km/s: $C = 3.2766$ — every gate closed; it can never get past the Earth–Moon L1 neck;
10.88 km/s: $C = 3.0697$ — L1 and L2 open;
the speed that just opens L1 is 10.8743 km/s, below the two-body escape speed of 11.0086 km/s and below the 10.9156 km/s a two-body transfer needs to reach the Moon's distance.

An open gate is necessary, not sufficient. It says where the spacecraft may go, not that it will — whether a particular trajectory threads the neck has to be found by integration.

For more than sixty years after the publication of the Principia astronomers were puzzled to account for the motion of the lunar perigee, simply because they could not conceive that terms of the second and higher orders, with respect to the disturbing force, produced more than half of it.

– George William Hill, Acta Mathematica (1886)

💡Where this model is good enough — and where it is not

Real primaries are not on circles: the Earth's orbit has eccentricity 0.017, the Moon's 0.055, and the Moon also pulls on everything near the Sun–Earth L1 point. Mission designers therefore use the restricted problem to understand and to find starting solutions, then refine them in a full ephemeris model with the Sun, the Earth, the Moon and the planets at their true positions. That is exactly how the orbits in this unit are used in practice — and why its labs keep comparing with what flew.

↓But there is a problem

Set the velocity to zero in the rotating frame and ask where the forces balance. The answer is five points. Euler found the three on the line through the primaries; Lagrange found two more, each making an equilateral triangle with the primaries. In 1906 an astronomer in Heidelberg found an asteroid sitting exactly where Lagrange had said one of them would be — sixty degrees ahead of Jupiter.