An article I read said we haven’t solved the three body problem and can’t predict the motion of 3 orbiting things in motion (2002 VE “Venus moon”) but we’ve been able to get the telescope to orbit just fine with itself the earth and the sun.. what’s the difference?
In: Physics
We have solved the 3-body problem, *under specific conditions*. Consider a different scenario entirely. I ask you “what will the weather be like in six months’ time?”
You don’t know. The weather is unpredictable, after all. *But*, you know that it’s always gloomy in the great lakes region, so you might say “In Michigan, it’s going to be gloomy”. Even if you can’t predict the weather everywhere, you can predict it in specific places because they are consistent.
Back to the 3-body problem. There are certain specific orbits that are consistent, and these are the lagrange points. As long as the telescope is in the right spot where all of the equations come together nicely, its motion is very predictable.
We can’t predict the orbits over a long period of time. All man made satellites need occasional small corrections to keep it in place. We can predict it’s orbit just fine day to day, but small drift will escalate over time.
> can’t predict the motion of 3 orbiting things
What this means is we can’t say exactly how it will be several orbits ahead, but we can make step by step simulations that are accurate enough to get the orbit almost perfect. The spacecraft will then use its small maneuvering engines to correct its position if it starts to drift.
The 3-body problem is there is no general closed-form solution to where an object will be in the future.
That means there is not an expression you can just input the time in the future and directly get into what position the object will be. I you travel i a straight line at 7 m/s the distance you are along the into at time t is 7*t. So if I calculate where you are in 564 seconds the answer is 7* 564=3948m away from you at time 0.
You can create a formula like that for the orbit of two objects. It will be larger but you can just input the time
For three bodies you need to do a calculation like how will the object move in 1 second, then use the new position and calculate for the next second. The accuracy will depend on the length of steps but also the precision of the number you have in the computation. If the calculation just has 10 decimals you can’t have time steps that only change the 11th decimal. Smaller steps also mean more steps and a longer time to calculate the result.
Even for two bodies in practice, we will not know the exact future position. You need to know the distance exactly and the mass. All real-world measurements have a margin of error.
For the James Webb Space Telescope, we can calculate where it will be with high enough accuracy to know where it should be placed. We can then measure where it is and use the trust to keep it in the required orbit.
Even if we could calculate it all exactly we would not know exactly where it will be. Light and the solar wind result in a force in the spacecraft. How large depends on the exact orientation of the spacecraft. We do not know exactly where it will point and for how long. The exact strength of the solar wind is also not known in advance.
So in practice, all calculated result has a margin of error because we can measure all input variables exactly enough. In addition some parameter are not constant and will change.
The James Webb Space Telescope is located in one of the Sun-Earth [Lagrange points](https://en.wikipedia.org/wiki/Lagrange_point). The Lagrange points are a set of equilibrium solutions of the restricted three-body problem, a version of the three body problem where the mass of the third object is so insignificant that its influence on the two other objects is not considered at all. The restricted three-body problem is mainly a two-body problem, for which there *do* exist analytic solutions.
earth + sun = earth + sun + JWST
…at least for any reasonable or neccessary degree of accuracy.
Three body problems are unsolved, but if one of the three bodies is ***vastly*** less massive than the other two, you can get away with assuming it’s only a two body problem and keep a relatively high degree of accuracy.
The key here is what is meant by “solved”. It’s been proved that there’s no general “closed-form” solution for the three-body problem. This means that, generally, there’s no mathematical formula we can use to give us the future positions of three bodies orbiting each other. But that’s not a problem at all for spacecraft because we can just use computers to simulate their motion to any degree of accuracy. This is a “numerical” solution as opposed to a “closed-form” one.
In practice, the limiting factor on the accuracy of our numerical solutions is in knowing the exact positions and velocities of the objects at some point in time. Small errors in the starting conditions of the simulation lead to larger and larger errors over time.
Anyway, the solar system has many more than three objects orbiting around so, even if we had a closed-form solution to the three-body problem, it wouldn’t give an exact solution. There’s still the problem of not knowing the exact starting conditions and then you have the other planets contributing tiny effects that build up over time. There are also small effects from sunlight and the solar wind. As you get closer to the earth, the fact that it bulges at the equator has a substantial effect and its magnetic field can have effects too.
The five, special-case solutions to the three-body problem discovered by Lagrange are often useful in planning orbits for spacecraft. It’s just like our knowing the two-body problem has solutions like circular and elliptical orbits. The two-body solutions aren’t exactly correct in the real world, because there are many more than two or three bodies, but they’re a close-enough approximation and they’re easy to understand.
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