# Intuitive understanding of Lagrange point L3

I learned about Lagrange points today and the physics behind L1, L2, L4 and L5 make intuitive sense to me. But I'm stumped by L3.

Wikipedia says "L3 in the Sun–Earth system exists on the opposite side of the Sun, a little outside Earth's orbit but slightly closer to the Sun than Earth is. (This apparent contradiction is because the Sun is also affected by Earth's gravity, and so orbits around the two bodies' barycenter, which is, however, well inside the body of the Sun.) At the L3 point, the combined pull of Earth and Sun again causes the object to orbit with the same period as Earth."

Why doesn't an object at L3 start to drift closer towards the sun? An object at L1 avoids that drift because Earth's gravity counteracts the Sun's pull. But it would seem to me that not only is an object at L3 missing a counteracting force pulling away from the Sun, but perhaps the gravity of the Earth is also pulling it towards the Sun.

• Wouldn't your confusion also apply to L2--that one sits away from both bodies as well. Clearly we can't just put down a static (much less stable) particle to the side of two gravitating bodies, it will fall in because there is no force to counteract the gravitational attraction. I think your confusion might stem from not taking into account the effect of angular momentum. – Surgical Commander Feb 7 '15 at 23:36
• Lagrangian points are points in space in which a small object is provided with the centripetal force required to orbit from two larger bodies. The reason it doesn't 'drift closer towards the sun' is because with Lagrangian points all we are considering is points which it is possible for an object to maintain a stable orbit. – Joshua Lin Feb 7 '15 at 23:37
• It's not clear to me what you're confused about. L3 is a stable orbit around the Sun. So is L1 but L1 can be a lot closer to the Sun because as you point out, the Earth's gravity counteracts some of the Sun's gravity which allows for the stable orbit with a shorter period than would normally be required to orbit at that distance from the Sun. – Brandon Enright Feb 8 '15 at 0:20

The thing about Lagrange points is there is always two ways of viewing the system.

Method 1: The Inertial Frame. Position yourself "above" the Solar System, looking down at the objects moving in ellipses. Here the force of gravity is the only force. Earth doesn't crash into the Sun because the constant inward acceleration merely bends its path into an ellipse. That is, when you take into account both its acceleration and its velocity, the two combine to make it orbit rather than fall directly radially. Similarly, the Lagrange points are moving points such that their velocities and the accelerations on them lead to orbits. Remember that with equal angular velocities, points further from the barycenter have larger linear tangential velocities, and L3 is further from the barycenter than Earth is when it is on that side of the Sun.

Method 2: The Comoving Frame. Now start spinning your camera such that it tracks the motion of Earth. You are anchored to a point directly above the barycenter of the system. Neglecting eccentricity, the Earth and Sun appear at rest in this frame. Gravity still pulls them directly toward one another, but in this non-inertial frame there is a centrifugal force that exactly balances gravity. Similarly, at any of the 5 Lagrange points, the combined gravity of the Earth and Sun on the point is balanced by the centrifugal force that point experiences. At the the middle Lagrange point the two gravitational effects oppose, and there is not much centrifugal force. At the outer two points, the two gravitational effects work together; however, these points are further from the barycenter and so experience a larger centrifugal force.

Everything works out in the end. The error is in either neglecting tangential velocity in the first frame, or neglecting centrifugal force in the second frame.

An object at L3 orbits the Sun through exactly the same mechanisms that the Earth does: it feels the Sun's gravitational attraction, and this is exactly the centripetal force it needs to perform a uniform circular motion about the Sun.

The slight problem, as you point out, is that it also needs to contend with the Earth's gravity, which at L3 also pulls it towards the Sun. This is the reason why L3 is slightly outside the Earth's orbit, which decreases both gravitational forces to the point where they match the Sun's pull on Earth.

It's not hard to do the math for it. Suppose L3 is at radius $r$ from the barycenter, with the Earth at radius $R$. The Sun will therefore orbit the barycenter at radius $(m/M)R$, where $m/M$ is the ratio of the Earth's mass to the Sun; in practice, the barycenter is well inside the Sun. Asking for the gravitational acceleration at L3 to match the Earth's then reads as

$$\frac{M}{(r-\tfrac mM R)^2}+\frac{m}{(r+R)^2}=\frac{M}{(R+\tfrac mM R)^2}.$$ This can be solved for $r$ but it's very messy. One thing you can do instead is to linearize this equation with respect to $m/M$ and $(r-R)/R$, in the expectation that those are small for the Earth-Sun system. Doing that gives a value of

$$r\approx\left(1+\frac{17}{8}\frac mM\right)R,$$

which is indeed slightly bigger than $R$, by about twice $m/M$, which for the Earth-Sun system is only about 6 parts in 106.

• For the first equation, the exact formula (for circular geometry) would have an extra $r/R$ multiplying the RHS, since you really want to set the gravitational acceleration equal to the centripetal acceleration at $r$ rather than $R$, keeping angular velocity constant. Not that this changes the approximation. – user10851 Feb 9 '15 at 18:35
• @ChrisWhite That's a good point. It will probably change the final equation too, though. I'll give it a whirl tomorrow. – Emilio Pisanty Feb 10 '15 at 0:37