Lagrangian Points Explained

Pretty much everyone understands at least the basic concept of a spacecraft orbiting the Earth.  The Earth's gravity tries to pull the spacecraft towards it, while the motion of the spacecraft perpendicular to the Earth's gravity is just fast enough to keep the spacecraft the same distance from the Earth.  The spacecraft is literally in free-fall towards the Earth, and only the spacecraft's relative motion keeps it from crashing into the planet's surface.

Similarly, the concept of a geosynchronous orbit or geostationary orbit is something most people have heard of, but may not be quite as familiar as just the basic concept of orbit.  The only thing that makes a geosynchronous or geostationary orbit unique is that the spacecraft is orbiting at the same speed as the planet itself does.  To accomplish this, the spacecraft just be at a particular distance from the Earth, because while orbital speeds decrease the further you get from the planet, the Earth's rotation remains constant.  There is a certain distance from the Earth that a spacecraft would have an orbital period of 23 hours, 56 minutes, and 4 seconds (exactly one Earth day), which happens to be 22,236 miles above sea level.  Communication satellites are often placed in geostationary orbit so that ground-based antennae can be pointed at a single point in the sky.

Also, while there is no difference between a geosynchronous and geostationary orbit in popular usage, they are actually slightly different.  Both orbit at the same distance from the Earth, but while a geosynchronous orbit refers to any orbit at this distance, a geostationary orbit only exists at zero inclination, or directly above the equator.  A spacecraft in geostationary orbit will appear in the same point in the sky at all times, while a spacecraft in geosynchronous orbit will move in the sky in an analemma.

The Five Lagrangian Points

An overhead map of the Sun-Earth Lagrangian points, including
gravitational topography (gray lines), "downhill" slopes (blue
arrows), and "uphill" slopes (red arrows). CC Image provided
by Wikipedia.

Something that is not as well known is that there are five orbital locations (known as Lagrangian points L₁ through L₅) in any two-body system (Sun-Earth, Earth-Moon, etc), named after French mathematician and astronomer Joseph-Louis Lagrange, where the spacecraft will not move relative to those two bodies.  These orbits would not be geosynchronous or geostationary, but they would be in balance between the forces from both bodies.  These points are very important to certain space missions, especially for deep-space telescopes or missions to study the Sun.  From the first two of these points, the position relative to the Earth remains constant, so communication between the spacecraft and the Earth are made easier, and the amount of fuel required to keep the spacecraft at that position is minimal since gravitational forces are in equilibrium.

Several space missions have been positioned at one or more of the Sun-Earth Lagrangian points, including the Planck Space Observatory at L₂ , and the Solar and Heliospheric Observatory (SOHO) at L₁.  For a basic overview of all five Lagrangian points, I put together a brief presentation using Prezi (very cool online presentation tool) that covers each point's location and a brief description.  Also, while I use the Sun-Earth system as the example, these points apply equally to any system of two massive bodies, such as the Earth and the moon.  If it is easier for you to visualize, replace the Sun in the presentation with the Earth, and the Earth with the moon.


This simple presentation is probably enough to convey the concept of Lagrangian points.  There is actually quite a bit of math involved in actually describing what is going on here, so if you are interested in delving deeper into the physics and mathematics behind the Lagrangian points, please read on.  If you are one of those people whose head starts to spin upon sight of an equation, please heed the following warning, one which I think I will use for future math-heavy posts:

WARNING: MATH AHEAD

Lagrangian Stability

The L₁ through L₃ points are all inherently unstable.  This means that any slight nudge will cause the spacecraft to start to move away from the L-point, and gravitational forces will continue to pull the spacecraft  further and further away.  This is known as positive feedback, which is actually not usually positive.  Positive feedback means any unbalance in the equation feeds back to make that unbalance worse and worse.

The easiest way of thinking about this is to imagine a ball on top of a hill.  It is completely motionless, at equilibrium, exactly at the top of the hill.  Knock that ball just slightly in any direction, however, and the ball starts to roll down the hill, slowly at first, then faster and faster.  The L₁ , L₂ , and L₃ points can be thought of as gravitational hills (although it is in three dimensions rather than two).  As long as you maintain a spacecraft's position at L₁ , L₂ , or L₃ through periodic adjustments, it will remain in that location.  Were the spacecraft to run out of fuel, it would eventually fall out of this point and fall towards either of the two massive bodies.

The L₄ and L₅ points, while they look like the most complicated and unstable of the L-points, are actually inherently stable, as long as the ratio of the masses of the two bodies is greater than 24.96.  This means that a slight nudge out of the L₄ or L₅ points will not cause the spacecraft to roll down the hill, so to speak, but it will want to fall back towards that point, orbiting the point in a kidney bean shape. This is negative feedback, which opposite of positive feedback, wants to move the spacecraft back towards equilibrium rather than away from it.  The simplest example of this would be the opposite of the gravitational hill, a gravitational valley.  I wish I could say this were true of the L₄ and L₅ points, but it is not.  Their stability is actually due to the Coriolis effect, something I will discuss in more detail later.

Positive Versus Negative Feedback

I described the difference between positive and negative feedback as the difference between a hill nad a valley.  To try and put it in a perspective more related to the L-points, imagine L₁ , with the Sun on one side and the Earth on the other.  Both bodies are pulling on the spacecraft with a force equal to 

where F is the force of gravity, G is the gravitational constant, m₁ is the mass of the Sun (or Earth), m₂ is the mass of the spacecraft (negligible compared to the Sun or Earth), and r is the distance between the Sun (or Earth) and the spacecraft.  Since this equation is the same for both the Sun and Earth pulling on the spacecraft, the equilibrium point is where the equations are equal,

where R is the distance from the Sun to the Earth, and r is the distance from the Earth to the spacecraft (and R-r is the distance from the Sun to the spacecraft).  Being that the mass of the Sun is much greater than the mass of the Earth, R-r would have to be much greater than r for the equations to become equal.  As you can see (or at least understand intuitively), if the spacecraft moves from the L₁ point in either direction, the force from the body it is moving towards becomes greater, while the force from the body it is moving away from becomes less.  This is positive feedback because the natural system pulls the spacecraft away from the equilibrium rather than towards it.

A similar situation that would be an example of negative feedback would be an object (representing the spacecraft), being suspended by two springs on opposite sides of it.  Like the example of the spacecraft at L₁ , the forces pulling on it from either side are equal and the spacecraft remains stationary.  Unlike the spacecraft at L₁ , if the object were to be moved away from equilibrium, the natuaral system will want to pull the object back towards equilibrium.  This is because springs pull in relation to how far they are stretched.  The farther you stretch the spring, the stronger the force gets, and the less you stretch it, the weaker the force gets.  So if you try to move the object away from equilibrium, the spring the object is moving away from pulls back harder, while the spring the object is moving towards pulls back less.

L₁ and L₂ Points

For the above example of the spacecraft at the L₁ point, I ignored the effects of centripetal forces due to the rotation of the spacecraft for the sake of simplicity.  I treated it as a statics problem (forces on non-moving objects), when it is really a dynamics problem (forces of moving objects).  Including in the centripetal forces gets a little messy due to rotation about the center of mass of the two-body system (the mass of the Earth moves the center of mass of the system from the center of the Sun to somewhere inside the Sun but off-center), so I won't go into the whole thing.  Since the mass of the Sun is so much greater than that of the Earth, however, the equation for both the L₁ and L₂ points is just:

where r is the distance from the Earth to L₁ and L₂ and R is the distance from the Sun to the Earth.

The only difference between L₁ and L₂ are the direction of the forces.  At L₁, the Sun is pulling in one direction (left in the presentation above), and the Earth and centripetal forces are pulling in the opposite direction (right in the presentation).  At L₂, the Sun still pulls left, and the centripetal force pulls right, but now that you are on the opposite side of the Earth, the Earth also pulls left with the Sun.

L₃ Point

The L₃ point is the other unstable Lagrangian point.  It is also pretty easy to visualize, as it is essentially in orbit around the Sun in almost the same orbit as that of Earth, just on the opposite side of the Sun.  The only difference is that the mass of the Sun-Earth system combined is slightly greater, and the center of mass of the system is slightly offset from the center of the sun.  The end result is an orbit that is slightly farther from the Sun than Earth, but keeps tracking exactly opposite the Sun from Earth. Simplified for Earth's negligible mass compared to that of the Sun, the equation for the location of the L₃ point is

where r is the distance from the Earth's orbit to L₃ and R is the distance from the Sun to the Earth.

The L₃ point is less useful to space missions because it would put any spacecraft out of direct contact to the Earth (the Sun would always be in between them).  It was theorized that a small planet or "Counter-Earth" could exist at this point, however direct observation has concluded that there is not one.

L₄ and L₅ Points

The L₄ and L₅ points are not intuitively at equilibrium, but they are actually the only ones that are inherently stable.  The reason these points are at equilibrium is because they are the only two points within the system that are equal distance from both large bodies.  The resultant gravitational force of the two bodies runs directly through the center of mass of the two-body system, also known as the barycenter.  This is also true of any point along a line drawn equidistant from both the Sun and Earth, however the orbital velocity required changes depending on distance from the center of mass.  The only two points along this line that have orbital velocities equal to that of Earth are 60^o and -60^o from the sun (see the above presentation for a graphical example).

For the reason why the L₄ and L₅ points are stable, I actually had to do some digging.  I was not able to intuitively determine what was causing a spacecraft at L₄ or L₅ to not just fall down the gravitational hill so to speak.  After searching the internet a bit, I determined the reason for the stability was because of the Coriolis effect, the deflection of an object in a rotating frame of reference.  This is sort of like the idea of trying to hit a moving target.  Say you were trying to throw a ball from the North Pole to the equator (work with me here).  If your target were directly "below" you at the equator, you would actually have to throw the ball somewhat to your left to account for the Earth's rotation.  That point directly "below" you won't be in the same spot when the ball arrives at the equator.  It would have rotated a bit, requiring the change in direction.  This is the same effect that causes hurricanes and cyclones to spin in opposite directions in the northern and southern hemisphere.

The same effect occurs at L₄ and L₅ (and actually elsewhere too, but the directions of the forces end up balancing at these points).  The end result is that a spacecraft not quite at L₄ or L₅ would orbit around the center of either L-point in a kidney bean orbit.  To help better visualize, below is an video demonstrating the orbits around L₄ (plus it has some very exciting music to go along with it.