The Hill sphere and the Lagrange points

Builds on: Sections 3.7 and 3.8
Reading time: 35 min
Exercises: 4
Olympiad problems: 1

The James Webb Space Telescope works 1.5 million km beyond Earth, on the side away from the Sun. At that distance, by Kepler's third law, it should take longer than Earth to go round the Sun, and fall behind. It keeps pace with Earth all the same, because Earth's pull adds to the Sun's just enough to make it go round in exactly a year. The point where this happens is one of five Lagrange points of the Sun and Earth. Thousands of asteroids share Jupiter's orbit at two others, 60° ahead of and behind the planet. In this section we find where these points are, and how far out a planet can hold on to a moon.

3.9.1The Hill sphere

Before you readMake a guess first

Could Earth hold a moon on a circular orbit 2 million km away, five times the Moon's distance?

A planet of mass $m$ goes round a star of mass $M$ on a circular orbit of radius $a$ . A moon of the planet feels the planet's pull, the star's tidal stretch ((3.18)), and, seen from the planet, which turns round the star once in each orbit, an outward pull from that turning. The Hill sphere is the region round the planet inside which the planet's pull wins, after George Hill, who studied the Moon's motion in the 1870s.

ResultThe Hill sphere(3.20)

r_{H} = a {(\frac{m}{3 M})}^{1 / 3}

$r_{H}$	the radius of the Hill sphere round the planet
$a$	the radius of the planet's orbit round the star
$m, M$	the masses of the planet and of the star, with $m$ much less than $M$

Moons whose orbits last stay well inside it, within about half of

r_{H}

. The Moon is 0.26 of the way out to the edge of Earth's Hill sphere.

Show the derivation

Seen from the planet, the frame turns at the planet's orbital rate $ω$ , with $ω^{2} = G M / a^{3}$ . A moon at distance $r$ from the planet on the line to the star feels the star's tidal stretch $2 G M r / a^{3}$ outwards ((3.18)). The turning frame adds an outward pull $ω^{2} r = G M r / a^{3}$ beyond what holds the planet itself on its orbit. Together these are $3 G M r / a^{3}$ , and they balance the planet's pull when

\frac{G m}{r^{2}} = \frac{3 G M r}{a^{3}}, r^{3} = \frac{m}{3 M} a^{3} .

For Earth, $m / 3 M = 5.972 \times 10^{24} / (3 \times 1.988 \times 10^{30}) = 1.001 \times 10^{- 6}$ , so $r_{H} = 1.496 \times 10^{8} km \times 0.01000 = 1.50$ million km. This is the distance of the James Webb Space Telescope, for a reason the next part explains.

Quick checkJupiter's Hill sphere

Jupiter has 1/1047.6 of the Sun's mass and orbits at 5.203 au. What is the radius of its Hill sphere?

3.9.2The five Lagrange points

In the frame that turns with two bodies on a circular orbit, there are five points where a third, small body feels no net pull and stays put. These are the Lagrange points, after Joseph-Louis Lagrange, who found them in 1772. A body at one of them goes round the centre of mass of the two in exactly the period of the smaller body, keeping the same place relative to both.

L1 lies between the two bodies and L2 just beyond the smaller one, both close to the edge of its Hill sphere. L3 lies on the far side of the larger body, at almost the same distance from it as the smaller one. L4 and L5 lie on the orbit of the smaller body, 60° ahead of it and 60° behind, each making an equilateral triangle with the two bodies.

Figure 3.8The five Lagrange points

Mass ratio

m / (M + m)

L1 and L2 from m
Stability

The figure shows two bodies, a heavy one M and a lighter one m, in the frame that turns with them, so that both stay still on the page. The lines join points of equal effective potential, the potential energy of a small body in this frame including the outward pull of the turning, and the heavier lines pass through L1, L2 and L3. The crosses mark the five Lagrange points, where the potential is flat. As the mass ratio grows, L1 and L2 move away from m, and L4 and L5 stop being stable once m is more than about a twenty-sixth of the total. The figure can be set to , or .

ResultThe positions of the Lagrange points(3.21)

\begin{aligned} d_{L1} \approx d_{L2} & \approx a {(\frac{m}{3 M})}^{1 / 3} \\ L4, L5 & : 60 ° ahead of and behind m \end{aligned}

$d_{L1}, d_{L2}$	the distances of L1 and L2 from the smaller body, on either side of it along the line to the larger body
$a$	the distance between the two bodies
$m, M$	the smaller and the larger mass

The first line is the Hill radius, and it is close only when

m

is much less than

M

. For Earth and the Moon, the exact distances of L1 and L2 from the Moon are 0.1509 and 0.1678 of the Earth–Moon distance, against 0.1601 from the Hill radius. L4 and L5 are each the same distance

a

from both bodies.

Spacecraft use these points because they stay in a fixed place relative to Earth and the Sun. The SOHO observatory watches the Sun from the Sun–Earth L1, 1.49 million km towards the Sun, and the James Webb Space Telescope works near L2, 1.50 million km the other way, where Earth, the Moon and the Sun are always on one side of its sunshield. The Trojan asteroids gather round Jupiter's L4 and L5.

*3.9.3The stability of L4 and L5Optional extension

L1, L2 and L3 are like the top of a saddle. A body that drifts a little away from one of them keeps drifting, so a spacecraft there must fire its engines from time to time to stay near. L4 and L5 are the tops of hills of the effective potential, which might seem even less stable. A body that starts to slide off, however, picks up speed in the turning frame, and the Coriolis effect, the sideways push that a turning frame gives any body moving in it, bends its path round the point. If the larger mass is more than about 25 times the smaller, so that $m / (M + m)$ is less than 0.0385, this keeps the body circling near L4 or L5 for good. The Sun and Jupiter, and Earth and the Moon, both pass this test easily, which is why Jupiter has thousands of Trojan asteroids. Even Earth has a few, such as 2010 TK7 near the Sun–Earth L4.

Common errorsSection 3.9

The 3 in the Hill radius comes from the star's tidal stretch, which gives 2, and the turning of the frame, which gives 1 more. Leaving out the turning gives $(m / 2 M)^{1 / 3}$ , which is too large.
L1 and L2 are close to the Hill radius but not at it. L1 is a little nearer the smaller body and L2 a little further, and the approximation gets worse as the mass ratio grows.
L4 and L5 are on the orbit of the smaller body, at the same distance from both bodies, 60° ahead and behind. They are not at right angles to the line between the bodies.
The Lagrange points are fixed only in the frame that turns with the two bodies. Seen from outside, a body at any of them goes round the centre of mass of the two once in each orbit of the smaller body.

ExercisesSection 3.94 questions

3.9.1
What is the radius of Earth's Hill sphere? Take Earth's mass as $5.972 \times 10^{24}$ kg, the Sun's as $1.988 \times 10^{30}$ kg and Earth's distance from the Sun as $1.496 \times 10^{8}$ km.
million km
3.9.2
The Moon orbits 384 400 km from Earth. What fraction of the radius of Earth's Hill sphere is this?
3.9.3
Where are the Trojan asteroids of Jupiter found?
3.9.4
Which of the Earth–Moon Lagrange points are stable?

Olympiad problemsPast IOAA problems, restated1 problem

2019A relay satellite beyond the Moon20 points

Adapted from IOAA 2019 (Keszthely), theory problem 9. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

On 3 January 2019 the Chinese spacecraft Chang'e-4 landed on the far side of the Moon. Earth is always below the horizon there, so a relay satellite, Queqiao, was placed in an orbit round the Earth–Moon L2 point, beyond the far side of the Moon. Find the height $h$ of L2 above the Moon's surface. Take the Moon's orbit as a circle of radius $R = 384 400$ km, the Moon's mass as 0.0123 of Earth's and its radius as 1737 km, and ignore the Sun and planets. You may use $1 / (1 + x)^{2} \approx 1 - 2 x$ for small $x$ .

I solved it without the solution

AnswerThe height is about 59 700 km above the Moon's surface. Solving the equation exactly, without the approximation, puts L2 at 0.1678 of the Earth–Moon distance, 64 500 km from the Moon's centre and 62 800 km above its surface, so the approximation is about 5 per cent short.