You may have read about rumors that NASA is considering building a space station at a place called the “Earth-Moon L2 Point.” The “L” is short for “Lagrange,” and this is one of the places in space known as a “Lagrange Point.” Unless you’re familiar with the basics of orbit mechanics, you may be wondering – who the heck is Lagrange, and why does he have points in space? More to the *point* (ha!), why is NASA interested in building a space station there?

To explain what a Lagrange Point is, I’m going to take you through a couple analogies.

Imagine you are standing on the top of a perfectly rounded, symmetric hill.

Right where you are, the ground is flat and level. You don’t feel any forces moving you one way or the other: you are in *equilibrium*. But if you take a step in any direction, the ground begins to slope and a force pulls you out further away from the top of the hill. The magnitude of this force is your weight, times a factor that accounts for the angle of slope:

The force always pulls you *out* from the center of the hill. Let’s call this direction * r*, for “radial.” There’s another direction on the hill, the “circumferential” direction

**– this is a direction that always takes you walking**

*c**around*the hill in a circle. There is no component of force pulling you in this direction.

From physics classes, you are probably familiar with the idea of *potential energy*. Potential energy is a quality associated with points in space, and we express that quality with a single number measured in joules. Where I am sitting, space might have a potential energy of six joules, and where you are standing space might have a potential energy of ten joules. This energy comes from sources like gravity or magnetism. The *difference* in potential energy between two points tells us how much *work* it takes to move something from one point to the other: if I want to visit you, I need to spend four joules of energy. If you visit me, you actually *get* four joules out of the deal, which you can spend on something else (such as moving faster).

Potential energy has a direct connection to force. If you are in a place which has a high potential energy, and nearby is a place with low potential energy, you will feel a force pushing you towards the lower-energy spot. Mathematically, we say that the force is equal to the *gradient* of the potential energy. So, on this hill, the top of the hill has the most potential energy (we’ll call it zero, though) and there is less and less energy as we move off in the +* r* direction. In the

*direction, the potential energy is always the same, depending on your current position in the*

**c***direction. If you go up the hill, in the direction –*

**r***, you will go towards higher potential energy and the force of your weight will work against you. You could imagine making a topographic map of the hill, only instead of the contour representing different heights, they represent different potential energy levels.*

**r**If you were to let yourself go and slide down the hill, your total energy would be about constant. You may recall the definition of *kinetic* energy: *mv*^{2}/2, where *m* is your mass and *v* is your speed. The sum of potential and kinetic energy must stay the same, so as you roll and your *potential *energy drops, your *kinetic* energy (therefore, your speed) will rise. The equation for your total energy will have the pieces from both, though: *E* = *U* + *K* = –*mgr*sin(*theta*) + *mv*^{2}/2. Notice that in this equation we have one term that depends on our *position* in space and on term that depends on our *speed*.

Got the hill down? Great. Now I’m going to stick you someplace else!

Suppose we go and find a merry-go-round, and we convince the operator to clear out all the horses and chariots and stuff so that it’s just a big, flat, rotating disk. Let’s also put a curtain around the outer edge, so that we can’t see outside from within. Then, you go and stand in the very center and we slowly start rotating the disk until it reaches a constant, slow angular velocity.

In the exact middle of the disk, you will notice very little. However, take one step away – in the +* r* direction – and you will begin to feel a

*centrifugal*force: a force that you perceive as pulling you

*outwards*. But thinking about forces in rotating reference frames is hard, and we’ll immediately get into pedantic debates about which forces exist and which don’t. So let’s think about energy again!

The disk is flat, so every point on it will have the same gravitational potential energy – it doesn’t really matter what that energy value actually is, since it’s *differences* in potential energy that are valuable, so let’s call it all zero. As you walk in the +* r* direction, you will have kinetic energy from two sources. First, there is your own motion, which contributes energy

*mv*

^{2}/2. Second, there is the energy you have from spinning in a circle, because your feet are on the disk. If the spin rate

*w*of the disk is slow enough, you might not notice it, but it’s there – and the energy is

*mr*.

^{2}w^{2}Your total energy, therefore, is *U* + *K* = *mr ^{2}w^{2}* +

*mv*

^{2}/2. Now–

Huh. Wait a second. That equation has a term that depends on your *position* in space and a term that depends on your *speed*. The piece that depends on speed is exactly what we had on the hill, too!

The piece that depends on position doesn’t quite look the same as it did on the hill. However, it has one very similar property: when you move out in +* r*, the value of that term changes. And, therefore, the magnitude of your speed

*v*must change to keep the total energy constant! We can debate about whether centrifugal or centripetal forces are real, but

*effectively*, the equation for your total energy behaves in the same kind of way on the spinning disk as it does on a hill.

*Effectively*, your entire kinetic energy trades back and forth between the “translational”

*/2 part and the “rotational”*

*mv*^{2}*mr*part, just as on the hill it traded between

^{2}w^{2}*K*and

*U*.

So let’s call *mr ^{2}w^{2}* your

*“effective potential energy”*on the disk! It behaves just like any other kind of potential energy – gravitational, magnetic, chemical, whatever – would, because it is energy that depends

*only on your position in space*, even though it’s

*actually*kinetic energy. We could even make a contour map of the effective potential energy.

Okay, then. Lagrange points, right?

Imagine the Earth and the Moon, sitting in space near each other. Don’t worry about orbital motions yet – just pretend that the two are fixed. Each body has a gravitational field, which we can visualize by a contour map of potential energy levels: far away from both the Earth and Moon, an object would fall generally inwards toward them, with potential energy decreasing as it goes in. The closer an object gets to either body, the stronger the gravitational pull, so the contour lines must be spaced closer together. And somewhere in the middle, the gravitational force of the Earth and Moon will balance each other exactly, so there is a level spot in the potential energy map.

This isn’t the whole story about bodies in space, though, because the Earth and the Moon aren’t fixed. They orbit around each other. An object we place near the Earth and Moon will also orbit around them. And because of that orbital motion, the energy of the object must include a component from rotation – which we can incorporate into the *effective* potential energy map around the Earth and Moon. Picture sitting in a spaceship somewhere “above” the Earth-Moon system that rotates at the same rate as the Moon orbits the Earth, such that from your perspective the Earth and Moon appear fixed in space. Then the effective potential energy map must have a component accounting for that rotation, just like on the merry-go-round. The map will look something like this:

Notice that there are five places on the map where the “topography” is locally *flat* – meaning that there is no net force acting on an object there. Between the Moon’s gravity, the Earth’s gravity, and the objects’ own orbital rotation, objects in those locations are at *equilibrium**!*

These are the Lagrange points, and this is what makes them special: place a satellite at a Lagrange point, and it will stay there.

The reason why these points are attractive places to put a space station is because it’s easier to get to Lagrange points from the Earth’s surface than it is to go all the way to the Moon – and vice-versa. In terms of our effective potential energy map, we have to cross fewer contour lines to get from the Earth to, say, L2 than we do to get to the Lunar surface. Every time we want to cross a contour line, we gave to make our spaceship gain or lose kinetic energy, and that means firing the engine – so crossing fewer contour lines directly corresponds to using less propellant or power.

If NASA located a space station at L2, then it could launch crews to the station with a smaller rocket than it would need to put the same crew on the Moon. NASA could also launch exploration vehicles and extra fuel to the station, so that the crew could eventually shuttle from the Earth to the station, and then take the station-to-Moon express from that point, at their leisure.

So: The reason why a station at L2 is exciting is not that L2 is an especially exciting place, but that the station would be part of a larger space exploration *architecture*. Not just flags and footprints, but more stations and vehicles and astronauts!

Great post—but a colleague pointed out to me that doesn’t have units of energy. Is that a typo? Shouldn’t it be $momega^2 r^2$?

Quite right! That’ll teach me to have my notes sitting someplace other than where I’m typing. I was mixing up the energy in terms of omega with the energy in terms of v. E = Iw^2/2, and I goes as mr^2. Thanks for the correction!