The previous post looked at two of the five Lagrange points of the Sun-Earth system. These points, L1 and L2, are located on either side of Earth along a line between the Earth and the Sun. The third Lagrange point, L3, is located along that same line, but on the opposite side of the Sun.
L1, L2, and L3 are unstable, but stable enough on a short time scale to be useful places to position probes. Lagrange points are in the news this week because the James Webb Space Telescope (JWST), launched on Christmas Day, is headed toward L2 at the moment.
The remaining Lagrange points, L4 and L5, are stable. These points are essentially in Earth’s orbit around the Sun, 60° ahead and 60° behind Earth. To put it another way, they’re located where Earth will be in two months and where Earth was two months ago. The points L3, L4, and L5 form an equilateral triangle centered at the Sun.
Lagrange points more generally
Lagrange points are not unique to the Sun and Earth, but also holds for other systems as well. You have two bodies m1 and m2 , such as a star and a planet or a planet and a moon, and a third body, such as the JWST, with mass so much less than the other two that its mass is negligible compared to the other two bodies.
The L1, L2, and L3 points are always unstable, meaning that an object placed there will eventually leave, but the L4 and L5 points are stable, provided one of the bodies is sufficiently less massive than the other. This post will explore just how much less massive.
Mass ratio requirement
Michael Spivak  devotes a section of his physics book to the Trojan asteroids, asteroids that orbit the Sun at the L4 and L5 Lagrange points of a Sun-planet system. Most Trojan asteroids are part of the Sun-Jupiter system, but other planets have Trojan asteroids as well. The Earth has a couple Trojan asteroids of its own.
Spivak shows that in order for L4 and L5 to be stable, the masses of the two objects must satisfy
(m1 – m2) / (m1 + m2) > k
where m1 is the mass of the more massive body, m2 is the mass of the less massive body, and
k = √(23/27).
If we define r to be the ratio of the smaller mass to the larger mass,
r = m2 / m1,
then by dividing by m1 we see that equivalently we must have
(1 – r) / (1 + r) > k.
We run into the function (1 – z)/(1 + z) yet again. As we’ve pointed out before, this function is its own inverse, and so the solution for r is that
r < (1 – k) / (1 + k) = 0.04006…
In other words, the more massive body must be at least 25 times more massive than the smaller body.
The Sun is over 1000 times more massive than Jupiter, so Jupiter’s L4 and L5 Lagrange points with respect to the Sun are stable. The Earth is over 80 times more massive than the Moon, so the L4 and L5 points of the Earth-Moon system are stable as well.
Pluto has only 8 times the mass of its moon Charon, so the L4 and L5 points of the Pluto-Charon system would not be stable.
 Michael Spivak: Physics for Mathematicians: Mechanics I. Addendum 10A.
3 thoughts on “When do two-body systems have stable Lagrange points?”
Are there other stable orbits such as an orbit around the line between the sun and earth with equal pull from each of the two? Would be in similar location as L4 and L5 but in an orbit perpendicular to the line between the sun and earth?
That value of sqrt(23/27) makes my mathematician’s spidey-sense tingle – it looks very precise, i.e. not an approximation, and I don’t think I’ve ever seen the number 23 show up “naturally” in a formula before.
I’ve downloaded this: http://wmap.gsfc.nasa.gov/media/ContentMedia/lagrange.pdf, but have not dug in yet. I’m hoping to stimulate your curiosity and maybe you’ll report back here later. :-)
The 23/27 factor is indeed precise. It all falls out of an eigenvalue calculation, starting with small simple numbers. The 23/27 comes from 1 – 4/27, which looks a little more natural.