Lagrange Points and Trojan Orbits 2
Insights for a complementary explanation of Lagrange points from inertial reference frames
I want to try to address this. As Watsisname said; establishing a non-inertial co-rotating frame of reference has many advantages in terms of understanding the actual situation but it also has some disadvantages (non-inertial forces as Coriolis and centrifugal have to be invoked and even then are difficult or impossible to represent in a potential dwell map). Inertial frames of reference are great because of one thing: they are what we are used to picture in our minds frequently and they are the ones where physical laws are coherent without any extra tweaking needed (no fictitious forces needed). The explanation from the point of view of an inertial observer should be equivalent to Watsisname’s but it should be given in other terms. I also feel that the actual inertial perspective on the subject should be extremely complex to wrap our heads around without doing many alienating calculations.
I haven’t been able to come with a simple discourse for the stability of L4 and L5 nor for Trojan orbits, but I think the fact that Lagrange points exist as “dynamical” equilibrium points is feasible within an inertial framework, so here it goes:
Since we are not co-rotating is useful to remember that Lagrange points are not fixed in space and move themselves. Also very important to remember is Kepler’s third law, which here will be expressed by the fact that objects farther from the Sun have lower orbital speeds (the angular speed will be dependent as the -3/2 power of the distance to the Sun). We place ourselves at rest relative to the Sun-Earth barycenter.
So. Lets say we want our probe to orbit the Sun but always been at the same distance from Earth without any additional effort, profiting from the gravitational landscape of the Solar System to accomplish that. If the Earth was a massless object then any point in Earth’s orbit would be a suitable place to put our probe since it would travel just as Earth does but leading the march or lagging always the same constant amount. But the Earth has mass and some influence so your probe would be pulled back or ahead by the planet and it would be unable to maintain its heliocentric Earth-like orbit. This is interesting because Watsisname did the reverse, he started with a huge mass and then slowly went to the small Earth mass to show how nearly the entire orbit of Earth was covered with a low-inclination plain potential field. If we went a step further, to a massless Earth, we indeed would have equilibrium points all across Earth’s orbit.
So the L1 point is the easiest to understand. But first you have to abandon the misconception that the gravitational tug of the Sun equals that of Earth in that point. This is not at all about that (in fact as we saw in previous posts, even the Moon is farther from Earth than the equal-force point). So no, L1 is not the place where the Earth screens the Sun’s gravity and you can move freely as if there was no Sun nor Earth until they move and lose that particular vantage point, no! L1 is located closer to the Sun than Earth so an object at that distance in a circular heliocentric orbit should get ahead of the Earth since it moves faster in its orbit, but the Earth here is creating his magic; having the Earth right in the opposite direction of the Sun what your probe feels is that the Sun-Earth system is equivalent to a Sun without Earth but this Sun having a little less gravitational attraction. So the Earth is orbiting the Sun but our probe orbits as if it orbited a lighter Sun, this means that even if it is closer to it our probe can still move as slow as Earth due to the lesser influence of this lighter Sun. So the the probe moves at the same angular velocity around the Sun as the Earth because it feels the Sun less and it is closer (both effects balance out) so the L1 – Sun – Earth configuration is maintained constant. Bare in mind that the vector sum of the gravitational force of Earth and that of the Sun on an object placed here at L1 will have the same direction as to point to the Sun-Earth barycenter (so in fact L1 is orbiting the Sun-Earth barycenter also).
L2 obeys a similar interaction. Instead of the tug of Earth screening a small part of the Sun’s influence we get the Earth adding to the pull of the Sun (since L2 the Sun and the Earth are both in the same direction). A probe in L2 would, in principle, lag behind Earth because of Kepler’s third law since it is farther from the Sun, but at the same time the Sun “feels stronger” (thanks to the extra pull of Earth) so you end up going faster than expected. Both effects balance out in the L2 point. And since you move at the same rate as the Earth here (same period or same angular velocity) you will always maintain the same configuration with respect to the Sun and to Earth and that in turn means that you will maintain this behaviour and the L2 point wouldn’t go away from you (you will move with it). Again, as in L1, bare in mind that the vector sum of the gravitational force of Earth and that of the Sun on an object placed here (at L2) will have the same direction as to point to the Sun-Earth barycenter.
L3 is similar to L2 but symmetrical to the Sun. At the opposite side of Earth’s orbit you notice the pull of the Sun as if Earth nearly didn’t existed (you are ~2 AU from it) but Earth still contributes to make you feel the combined pulls of the Earth and the Sun in the same direction as if the Sun had gained some mass and was a little stronger than the Sun we feel from Earth. This means that you can orbit this “more massive Sun” quicker if you are in Earth’s orbit but if you get a little farther from the Sun than Earth’s orbit you would reach a state of “dynamical equilibrium” again were you are moving at the same speed as Earth does. This is interesting because we often see pictures of L3 been located in Earth’s orbital path but in fact it is outside of it, like L2. The difference with L2 is that for you to beat the angular speed increase of the strong gravitational pull of the Sun and Earth you (now in L3) just need to get a little farther but since in L2 the Earth is quite close the influence that has to be overcome is greater and thus you move considerably farther from Earth’s orbit than in the case of L3 to archive the same balance. Again, as in L1 and L2, bare in mind that the vector sum of the gravitational force of Earth and that of the Sun on an object placed here (at L3) will have the same direction as to point to the Sun-Earth barycenter.
L4 and L5 points
Since the gravitational field of the Sun-Earth system is mirror-symmetric to the Sun-Earth line then both L4 and L5 points have the exact same underlying explanation. Remember that L1, L2 and L3 all were orbiting the Sun-Earth barycenter because the net force (the result of adding Earth’s tug and the Sun’s tug) always pointed there? Well this fact was simple to grasp because in the 3 cases we had only vectors in the Sun-Earth line (so the net result would always lie in the same line as the Sun’s and Earth’s pull). But what if we could have a net force still pointed to the Sun-Earth barycenter but with the Earth’s pull and the Sun’s not been parallel from our position? Well, that’s exactly the L4 and L5 points. Here you will experience the Sun pulling in one direction strongly and the Earth pulling in an other (not parallel to that of the Sun) more weakly but enough to have your net force dragging you to the Sun-Earth barycenter instead of just the Sun. Since this geometrical situation can be archived in many places we have to also include the restriction that you maintain the same orbital speed as Earth (if not you would feel the net force pointing to the Sun-Earth barycenter just for a brief moment and then you would be pulled slowly to a different direction as the distance to the Sun or to the Earth would vary). With both restrictions (having your net force pointing to the barycenter and moving at the same orbital period as Earth as to maintain that net force pointing to the barycenter) you end up with only two places in the entire system, and those are the L4 and L5 positions.
This is difficult to address in an inertial frame but let’s try. To understand if L1 is a stable or unstable equilibrium point we are going to imagine our probe placed in 4 different places (all very close to L1). Let’s take a slightly higher orbit from L1 (this means that we are in a Sun – system barycenter – L1 – our probe – Earth – L2 configuration). In this situation we have a greater pull from Earth so our net force will be pointing the same way (to the barycenter) but with less intensity (since here the Earth is now screening more of the Sun’s influence). Been in a higher orbit alone would mean we are moving slower due to Kepler’s 3th law but also we move slower because of the “lighter Sun” we have now. Because of that our probe would lag behind Earth. As this situation continues to evolve the net force would start to have a component pointing outside of the barycenter of the system (as the angle between the pull from the Sun and the pull from Earth gets lower than 180º). This in principle should be nice because it should impart some tangential motion on our probe so that it reaches L1 again, but sadly we are in a to high orbit for this to work. Our probe will gain a little acceleration but would end of still losing the race to Earth. But! this could happen if instead of moving to a higher orbit from L1 we maintain the same orbit as L1 but place our probe behind L1 as it moves. Then the orbit would have a smaller circumference than in the previous case and this new tangential pull would be enough to make our probe reach L1. Once the probe reached L1 it would probably have to much speed to maintain that position so it would come ahead of L1 and the reverse would hold. The probe would be oscillating like in a valley in this case. Lastly let’s think about the Sun – system barycenter – our probe – L1 – Earth – L2 configuration (our probe is in a closer orbit than L1). In this situation we are faster than Earth since the pull of Earth is not enough (since we are farther from it) to overcome the effect of a smaller quicker orbit around the Sun. The probe would surpass Earth and it would end pulling us backwards, but not enough strongly. As the tangential acceleration performs its magic the probe would slow down and lower the orbit even farther from L1 making the probe lose its place of near equilibrium.
What does all of this means? It means that any displacement from L1 performed in the Sun-Earth line would end up been amplified and any deviation in the perpendicular direction of that line will be corrected and the equilibrium restored. So here we have for L1 a saddle point as was expected (stable equilibrium in one direction but unstable in other direction).
L2 and L3 stability
Same thing as before but here the situation is more simple. Any small deviation from L2 in the Sun-Earth line can be explained in the same way for higher or lower orbits from L2; we are now considering a “heavier Sun” (due to Earth’s contributing to the net pull in the same direction as the Sun). If we are in a lower orbit (with respect to L2) the probe will move faster and will overcome the Earth, but then a tangential force would try to slow down the probe because Earth would be a little behind it. If it slows down the orbit shrinks again so a positive feedback loop is generated where the probe ends been separated more and more from L2. For a higher orbit the the probe would lag behind Earth so a tangential pull would appear that would speed up the probe throwing it to an even higher orbit and escaping L2 more and more. So L2 is an unstable equilibrium point for any direction you choose to displace your probe from it. L3 behaves exactly in the same manner due to the same reasoning but since it is farther away from Earth the instability is less abrupt and we still have a lot of margin to move the spacecraft around L3 without loosing control so quickly (this is a peak with a less inclined slope).
L4 and L5 stability:
Here is where I get lost. The geometrical configuration here is way more complex than in the other cases since it ceases to be a 1-dimensional problem and becomes 2D. I don’t know exactly hot to picture this in my mind but any small deviation from L4 has to be diminished since as Watsisname explained this is a stable equilibrium. My guess is that this problem is equivalent to the problem where we only have the Sun and the probe, but this Sun changes mass slightly and harmonically and wobbles from one place to the other in such a way that it pulls the probe as if it was Earth. An animation should make this more visually clear. Let’s say, for example, that we are ahead of L4 (still behind Earth). In this situation Earth’s pull will increase in magnitude and the direction of the net force will change slightly (from the barycenter towards the Sun). The speed then would increase so the probe would archive a higher orbit and slow down, but this somehow would happen in such a way that L4 would surpass our probe because of that slowdown and we would end up behind L4 quickly, where the reverse would hold. We would be somehow oscillating around L4. How is this the case for displacements from L4 in any direction I don’t really know but sure an animated diagram with the forces involved would clarify this a lot (maybe one day I try to do it).