How are Lagrange points L4 and L5 formed?

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I have tried reading about them and I have understood how L1,2&3 are made. That was pretty easy to get. But L4 and L5 I am not able to. If anyone could give me a source to read or explain it to me. Thanks.

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4 Answers

Anonymous 0 Comments

They are not formed, they are a result of the geometry/math. If you wiggle those workout ropes, or a slinky on the floor, you will notice for some frequencies of input there are places where the rope/slinky just rotates in place. That is an emergent system mode, not a special formation of rope.

Anonymous 0 Comments

[Hard to do without a diagram.]
Consider a world with two massive bodies rotating around each other.
Look at the world in a rotating frame of reference, so the two bodies are at fixed positions.
Draw a line, put marks for the two bodies (M1, M2) and a third for the center of mass of the system (CM); let’s say that’s close to the left body.

If you do the math, the L4 point is going to be the third corner of an equilateral triangle with the two bodies, so put a little mark there.
At every point in the system, a third body is going to feel three forces: gravitational attraction toward M1, which decreases with increasing distance from M1; gravitational attraction toward M2; and the centrifugal force, which *increases* with increasing distance from the CM.

Now, consider a point to the upper right of L4.
The gravitational force from M1 is a little less, and the centrifugal force is a little greater, so the *net* force, compared to that at L4, is to the upper right.

Next, consider a point to the lower right of L4.
The gravity from M2 is a little more, so the net force, compared to that at L4, is to the lower right.

Similarly, you can work out that the net forces at points to the upper left and lower left of L4 are away from L4.

So *somewhere* in between those four points, there has to be a place where the three forces are in balance, so there’s no net force … and it turns out to be at L4.

Anonymous 0 Comments

So imagine you have an equilateral triangle with a small satellite A, a heavy mass B, and a medium mass C at the three corners. The satellite A is equally far from the heavy and medium masses (by definition of an equilateral traingle), so B and C will pull A towards each of them in proportion to their masses. If the heavy mass is 99x as heavy as the medium mass, it pulls exactly 99x as hard, so if you imagine the line segment BC divided up into 100 equal portions, satellite A’s net force will be directed towards the point at the end of the first portion, exactly one unit away from B and 99 units away from C.

That point is *also* the center of mass of the BC orbital system, for the same reason, that if B is 99x more massive, the center of mass must be 99x closer to B. B and C are both going to be orbiting that point.

So you have a situation where the two massive objects are both going to maintain an orbit around a certain center of mass point, the third satellite is also experiencing a centripetal acceleration directly toward that center of mass point, and the orbit is stable because the object is orbiting as the same rate around the point as the medium mass C is. (If it were an isosceles triangle and A were closer or further away, A would need a greater or lesser orbital velocity than C, and couldn’t maintain the equilateral triangle).

That creates two points where it is conceivable for A to remain in a stable orbit around B thanks to the offsetting pull by C.

Anonymous 0 Comments

I believe they are where the sun and earth are pulling at such an angle and force that the net force is zero.

Takes a bit of trig to work out