Lagrange points: balance in the sky

A spacecraft further from the Sun than Earth should circle it more slowly and fall behind. Yet the James Webb Space Telescope, 1.5 million km beyond Earth, keeps pace with us all year round. It sits at a Lagrange point, one of five places where the pull of two bodies and the motion of going round balance out.

Five parking spots in the sky, and only two of them are free of charge.

Look

Imagine a landscape where height means “how much energy it takes to be here” for something that turns around the Sun with Earth:

• Two deep wells at the Sun and at Earth.
• L1 and L2: two mountain passes (saddles) on either side of Earth, about 1.5 million km away. Balanced, but only just: nudge a spacecraft and it slides off. Missions there make small corrections every few weeks.
• L3: another saddle on the far side of the Sun. Nothing useful goes there, because the Sun blocks it from us.
• L4 and L5: two gentle hilltops 60 degrees ahead of and behind Earth. You would expect things to roll off, but the Coriolis effect of the rotation turns the slide into a loop. They are stable, and nature fills them: Jupiter’s L4 and L5 hold more than ten thousand Trojan asteroids.

Who lives there: SOHO has watched the Sun from L1 since 1996; it sees solar storms before they reach us. Webb works around L2, where the Sun, Earth and Moon all stay on one side, behind its sunshield, keeping the telescope at about −230 °C.

Understand

In the frame that turns with Earth, a small object feels the Sun’s gravity, Earth’s gravity and the outward centrifugal effect of rotation. Where all three cancel, it can stay still in that frame, that is, orbit the Sun in exactly one year.

For L1 and L2 the distance from the smaller body is roughly the Hill radius:

r ≈ a · ∛(m / 3M)

For the Sun and Earth: 149.6 million km × ∛(3.04 × 10⁻⁶ / 3) ≈ 1.5 million km. Solving the balance exactly gives 1.49 million km for L1 and 1.51 million km for L2.

Master

In the circular restricted three-body problem, with distances in units of the separation and μ = m/(M+m), the effective potential is

Ω(x, y) = −(1−μ)/r₁ − μ/r₂ − ½(x² + y²)

The collinear points solve ∂Ω/∂x = 0 on the x-axis (a quintic, solved numerically here); L4 and L5 form equilateral triangles with the two bodies. Linear stability analysis shows L1-L3 are always unstable, and L4/L5 are stable when μ < μ_Routh = ½(1 − √(23/27)) ≈ 0.0385. Spacecraft at L1 and L2 fly halo or Lissajous orbits around the point rather than sitting on it.