Lagrange points solved one of physics' biggest problems
space.com
space.com
At each of the five points, there’s a net gravitational force towards the sun (to be more precise: towards the common barycenter). This is easiest to see at L2 and L3 where the gravitational forces pull in the same direction.
There are now two ways to look at this combined gravitational force:
1. In an inertial reference frame (the one we imagine when we say the Earth revolves around the Sun), this combined gravitational force acts as a centripetal force. At the Lagrange points, it has precisely the correct value to keep the object in an orbit with period 1 year.
2. In the reference frame where Sun and Earth don’t move (that is, the reference frame that rotates at 1/year around the barycenter), every point experiences an apparent centrifugal force. At the Lagrange point, this centrifugal force cancels with the gravitational forces to keep the object stationary in this reference frame.
The math is completely the same (x=y vs. x-y=0). I personally just find the first view to be more intuitive.
You could say that the forces cancel out, if you consider pseudo-forces like centrifugal force or Coriolis force as real forces. However, that view sometimes lead to very misleading notions like 'there is no gravity in space'. It also makes explanation of other phenomena like tidal forces needlessly complicated. For this reason alone, I discourage people from analyzing in non-inertial frames. Inertial frames make life simpler and clearer by removing caveats on Newton's second law. You no longer need to invent imaginary forces.
A classic example is a person inside a falling elevator. From their own perspective, they're floating with no external forces acting upon them (until, of course, the elevator crashes into the ground). Only when the elevator is held stationary (from the POV of an external reference frame) does the occupant feel an acceleration force upward pushing back against the gravitational well they're standing in.
We can say the sun is at rest. Here the satellite accelerates towards the sun; we say that is due to a force called gravity. This is nice and simple.
We can instead say the satellite is at rest. In this frame there is no net force on the satellite. But now the sun accelerates, as do the distant stars; why? And how to explain second-order gravitational effects like tides? It's awkward to work in and leads to bad intuition.
To extend it to your freefall elevator example: everything inside the elevator obeys Newton's First Law so we are justified in treating it as inertial. Extend the frame to the surface of the earth and we have a problem: we either have to explain what force causes the earth to accelerate towards the elevator, or admit we have a non-inertial frame. Simplest to just say the earth is at rest and the elevator is accelerating due to gravity.
However centrifugal force is not all bad, it has a really cool interpretation which yields a powerful insight. Consider a satellite in an elliptical orbit. As it reduces its radial distance to the sun, it must pump energy into its tangential velocity, to conserve angular momentum. So its tangential kinetic energy is a function of its position. Hey, that sounds like a potential!
You can just add this "tangential kinetic potential" to the gravitational potential, yielding a bathtub shaped "effective potential." Now we're in 1D, since we're identifying angular effects with gravity. When the radial distance is large, the satellite will accelerate towards the sun (radial velocity decreasing), as it falls down the gravitational potential well. Eventually it begins to climb the "tangential kinetic potential" hill, and starts accelerating away from the sun (radial velocity increasing). In this 1D world, the sun is attractive at long range and repulsive at short range; the turning points are the apahelion and perihelion of the orbit.
What's so cool about this is that we've reduced a 2D kinematics problem to a 1D problem (radial distance varying over time) by absorbing the tangential kinetic energy into the gravitational potential energy. This connects planetary orbits to more familiar problems like the 1D harmonic oscillator. The cost of doing so was the appearance of this "tangential kinetic potential;" the derivative of a potential is a force and that's centrifugal force.
Now let's see why there is no gravity from the elevator's frame of reference. Remember that by Newton's law of gravitation, gravitational force is GMm/r^2. If you plug in the values, you won't get a zero. This means that the person inside the elevator is definitely facing gravitational 'force'. (I'm assuming classical mechanics where gravity is a force. Relativity considers it as another pseudo force. But that view won't affect our current discussion)
Consider the non-inertial frame inside the elevator first. In order for the person to be experiencing no force, all forces on him must be balanced and cancelling out. We know of only gravitational force that is acting downwards. So to cancel it out, we need to invent an equal and opposite force on the person. That force is called 'rectilinear acceleration force' and its value is given by ma (upwards), where m is the mass of the person and a is the elevator's downwards acceleration. So the net force on the person is mg-ma. Since elevator is in free fall, a=g and so the force on the person is zero.
You can see the mess here. You have to invent a new force that doesn't exist and have to be careful about its direction too. If the frame is rotating, then more pseudo forces have to be invented - the Coriolis force, Centrifugal force and Euler's force. All these 4 forces don't exist in reality, and are there purely to account for non-inertial frames.
Now let's see how the forces play out when considered from the inertial frame of the external observer. Let's consider if the lift is stationary first. Obviously the person inside the elevator is feeling gravitational force that is equal to his weight. In reality, he is not actually sensing gravity acting on him - he is instead sensing the 'reaction force'. The reaction force is the force that the floor of the elevator is applying on him to counteract his weight. In other words, the reaction force from the floor actually cancels his weight - so he remains stationary instead of accelerating towards the ground.
Now, unlike pseudo forces, reaction forces are real forces. In this case, it's provided by the electromagnetic repulsion between the molecules on the floor and in the person. This reaction forces don't just act on the person's foot. It acts throughout his body. For example, his internal organs don't fly off under gravity because the reaction force provided by the connective tissues hold them in place, counteracting gravity.
Finally what happens when the elevator is falling? The elevator is in free fall, so its acceleration is g towards ground. The person is also falling with the same acceleration g. This means that there is no reaction force acting on him to counteract the gravity. In other words, the person is no longer supported by the floor - that support was the reaction force. Remember that I said that the reaction force was the force that the person was sensing as gravity. Now that reaction force has disappeared. So the person in effect feels that he is weightless, even though gravity is still acting on him.
This is the advantage of inertial frames. All Newton's laws can be applied in terms of real and concrete forces with well known causes. You don't need to invent anything new and then justify their existence.
[1] Solar System Dynamics, Murray and Dermott, Cambridge University Press.
People don't choose adversity, they only like to dispense it.
Simulation is fun, it's a safe way of exploring danger.
However the hazards up on hills are more typically those of exposure, altitude and isolation, not incoming gravitationally-propelled objects.
"We don't."
>Although the L4 and L5 points are found at the top of a "hill", as in the effective potential contour plot above, they are nonetheless stable. The reason for the stability is a second-order effect: as a body moves away from the exact Lagrange position, Coriolis acceleration (which depends on the velocity of an orbiting object and cannot be modeled as a contour map)[20] curves the trajectory into a path around (rather than away from) the point.[20][22] Because the source of stability is the Coriolis force, the resulting orbits can be stable, but generally are not planar, but "three-dimensional": they lie on a warped surface intersecting the ecliptic plane. The kidney-shaped orbits typically shown nested around L4 and L5 are the projections of the orbits on a plane (e.g. the ecliptic) and not the full 3-D orbits.
> Webb will need to conduct occasional small thruster burns for "station keeping" and "momentum management" to retain its proper location and orientation in space.
> should have enough propellant to allow support of science operations in orbit for significantly more than a 10-year science lifetime
https://www.space.com/james-webb-space-telescope-fuel-lifeti...
There was a project and/or company that was looking into building a spacecraft that could make its way to geostationary satellites about to retire, latch onto them (by putting a docking rod into their exhaust for example) and act as its new engine, to extend their lifespans. Which is neat, and even though they may be 20-30 years old, they're still antennas that can transfer data around.
* escape from the solar system altogether? That requires a big change in velocity, about half as much again as what it took to get to its current orbit.
* chuck it into the sun? needs more change than escaping the solar system
* or just a different orbit to the current one with less chance of being inconvenient? then you may as well leave it where it is - there are already a bunch of Earth Trojan asteroids around the Lagrange points which pose a similar likelihood of becoming inconvenient and being much bigger pose a greater hazard.
In this case "outside of the Earth-moon system" is what I mean. As you say, other changes such as out of solar system or into sun will require massive changes in velocity.
In most cases "out of their original Low Earth orbit" is a good definition, but unusually, JWST isn't in LEO to start with. Most satellites can be easily or no-cost de-orbited (chuck it into the earth's atmosphere, allow the orbit to decay until same) from LEO, but with the JWST, I would guess that that would also be expensive.
In any case, it's not perfectly stable. Occasional perturbations are bigger than the energy needed to leave. So, things put there don't stay long without help.
Total pop-science neophyte here. I found it very very interesting to hear about the work required of the “flight dynamics” team to keep the telescope in its orbit.
Wondering how this project stacks up against other man-made orbiting bodies in this regard. Is there a set-it-and-forget it to Webb range of complexity story here?
[https://eyes.nasa.gov/dsn/dsn.html Deep Space Network]
It is used also in quantum field theory due to its treatment of time, unlike Hamilton.
That is very strange. But it works.
Also, the principle of action is so odd (sort of Feynman Diagram but in classical world) you really just have to believe it instead of thinking too hard on it.
Not sure if that's what you mean, but if you're happy to assume that the massive bodies (e.g. Earth and Sun) are spherical, then you can exactly replace them with point masses, without approximation, so there's no multiplicity in Lagrange points coming from that.
Of course, orbits are not circular, and there are external forces too - other planets' gravity, (tiny) orbital drag, solar wind pressure etc. so in practice it's all a bit different anyway.
I mean, every earth coordinate (e.g. New York, Tokyo, etc.) has it‘s lagrange point, right?
You can see it better in this contour plot:
https://upload.wikimedia.org/wikipedia/commons/thumb/e/ee/La...
Does anyone else find it fascinating how quickly things get complicated? As the article mentions, the three body problem is the classic example. For two bodies it's easy. Add just one more and there's now no general solution and you get a bunch of edge cases like Lagrange points.
I too thought I understood the general principle of Lagrange points (including stability) but then JWST comes along and starts orbiting Earth's L2.
For example, look at the shape of the electron orbitals in a hydrogen atom [0]. These are not shapes one could reproduce using just gravity.
[0]: https://upload.wikimedia.org/wikipedia/commons/e/e7/Hydrogen...
Given that there's no known analytical closed-form solution to the unrestricted 3-body problem, let alone the arbitrary n-body problem, I think there are good reasons to be skeptical that such a generalization is possible.
Not to mention the idea of a Lagrange Point arguably becomes iffy in such situations. For example, what might a Lagrange Point be for the figure-8 3-body orbit [0]? And how would you define Lagrange points for a chaotic system?
> and that too could apply to anything orbital, atomic including?
Not really. As I said, the name is pretty much the only common thing between atomic orbitals and orbits in space. The equations used to describe the motions of celestial bodies are completely inapplicable to the motions of subatomic particles.
As a result it doesn't really matter if a spacecraft is exactly at the Lagrange point or 100 km away from it, so there's no shortage of space.
Also, this might be obvious, but it usually wouldn't make sense to park a spacecraft exactly at a Lagrange point because the spacecraft had to be moving in the first place in order to reach the Lagrange point. It would be a waste of fuel to slow down, so an orbit would typically be used instead.
https://static.scientificamerican.com/sciam/assets/Image/202...