- 13 Aug '10 01:44Why are the lagrangian points always +/- 60 degrees away from the planet?

I was trying to visualize that effect and was thinking if you have a huge planet like Jupiter and it has it's lagrangians at +/- 60, why would a much smaller Earth have the same thing? It would seem to me Earth's gravity well would equal the sun's gravity at a much closer distance to Earth than that of Jupiter. How does that work? - 14 Aug '10 13:49

I know that, but I read the lagrangians are all +/-60 degrees from each orbit, of course moving with the planets but do you see the conundrum here? If Jupiter has them at +/- 60 degrees and is many times larger than Earth, why wouldn't our lagrangians with respect to the sun be say, +/- 20 degrees or something since the Earth's mass is so much less and we are closer to the sun so its gravity is so much stronger than at Jupiter distance, it would seem reasonable to assume the place where the fields are equal would be closer to Earth.*Originally posted by flexmore***These points are not stationary; they spin around the combined centre of mass just as the two masses do.**

With disbalanced pairings the combined centre is shifted just as much as the gravitational tug is shifted. - 14 Aug '10 16:03

Two of the L-points are always 60 degrees behind and 60 degrees in font. The objects mass has nothing to do with it.*Originally posted by sonhouse***I know that, but I read the lagrangians are all +/-60 degrees from each orbit, of course moving with the planets but do you see the conundrum here? If Jupiter has them at +/- 60 degrees and is many times larger than Earth, why wouldn't our lagrangians with respect to the sun be say, +/- 20 degrees or something since the Earth's mass is so much less and we a ...[text shortened]... would seem reasonable to assume the place where the fields are equal would be closer to Earth.**

An L-point is a small well in the space, where things can stick. But if an small object is going into the well, but have too much velocity, it climbs up again at the other side and leave the L-point. Jupiter mass is big enough to trap objects, therefore it has its Trojans.

Why always exactly 60 degrees? I suppose of geometric reasons. - 14 Aug '10 19:27

I don't get that one. Lagrangians are due to gravitational interactions between a major mass and its orbiter, some planet, or like Earth and Luna. If you did an analysis of gravitational strength you will find the place where gravity cancels due to Earth V the sun, it is a lot closer to Earth than the same thing at Jupiter. Why should the lagrangians be only at 60 degrees ahead and behind the orbit? BTW, they have found a few in Earth's lagrang too.*Originally posted by FabianFnas***Two of the L-points are always 60 degrees behind and 60 degrees in font. The objects mass has nothing to do with it.**

An L-point is a small well in the space, where things can stick. But if an small object is going into the well, but have too much velocity, it climbs up again at the other side and leave the L-point. Jupiter mass is big enough to trap ob ...[text shortened]... therefore it has its Trojans.

Why always exactly 60 degrees? I suppose of geometric reasons. - 14 Aug '10 20:03

That's geometry only. 60 degrees in front and behind the object.*Originally posted by sonhouse***I don't get that one. Lagrangians are due to gravitational interactions between a major mass and its orbiter, some planet, or like Earth and Luna. If you did an analysis of gravitational strength you will find the place where gravity cancels due to Earth V the sun, it is a lot closer to Earth than the same thing at Jupiter. Why should the lagrangians be only at 60 degrees ahead and behind the orbit? BTW, they have found a few in Earth's lagrang too.** - 14 Aug '10 23:27

It seems counter-intuitive that mass has nothing to do with it. That would mean that a 1 kg pebble would have a corresponding lagrangian at +/- 60 degrees? That does not make sense on the surface of it.*Originally posted by FabianFnas***That's geometry only. 60 degrees in front and behind the object.** - 15 Aug '10 06:40

If it is a very small object, like an asteroid in orbit aroudn the sun, would you rather have the point only degrees or a fraction of a degree of the object?*Originally posted by sonhouse***It seems counter-intuitive that mass has nothing to do with it. That would mean that a 1 kg pebble would have a corresponding lagrangian at +/- 60 degrees? That does not make sense on the surface of it.**

Or if it was a giant star orbiting a black hole, would you rather have the point big as 80 degrees in front and behind?

When I look at a drawing of where the points cold be found, it looks just fine with 60 degrees. Of gemoetrical reasons. We're talking about L4 and L5. L1 and L2 seems to be variable in position dependant of the mass of the oject. L3 is also fixed in position.

http://en.wikipedia.org/wiki/Lagrangian_point - 15 Aug '10 15:27It is explained on Wikipedia here:

http://en.wikipedia.org/wiki/Lagrangian_point#L4_and_L5

I don't claim to understand it all not having done much physics of that nature in a very long time.

As far as I can tell, the satellite is orbiting the center of mass of the whole system, but is also being pulled towards both large objects in proportion to their masses. But the centre of mass of the whole system is also between the two masses in proportion to their masses and somehow these two proportions cancel each other out so that it doesn't matter what the actual masses are. - 15 Aug '10 17:27

Well, suppose you have two stars, exactly the mass of our sun. They are in orbit around each other at some distance, say Jupiter's distance from the sun or something like that, say 1 billion Kilometers.*Originally posted by FabianFnas***If it is a very small object, like an asteroid in orbit aroudn the sun, would you rather have the point only degrees or a fraction of a degree of the object?**

Or if it was a giant star orbiting a black hole, would you rather have the point big as 80 degrees in front and behind?

When I look at a drawing of where the points cold be found, it looks just f ...[text shortened]... ss of the oject. L3 is also fixed in position.

http://en.wikipedia.org/wiki/Lagrangian_point

With equal masses, the dead center of zero gravity from those two objects would be dead center on a line directly between the centers of each star. If you calculate the +/- 60 degree points around one, say we have a situation where they are not mutually orbiting but one is stationary (not true in real life but for the sake of argument)

The circumference would be 6.28 billion Km (2 PI times 2r, r=1E9 km) and 1/6th of that is 1.04666 E9 km. That is the curved distance but gravity only goes by the actual distance, so it looks like you are describing an equilateral triangle. I think I am starting to get it now. With 1 r as the separation, (the distance between the stars) we make an equal distance to meet with the same r 60 degrees away, which describes an equilateral triangle, that would be the point where the star's gravity would not be canceled but intersected. I guess that angle is what makes for a stable orbit of Trojans there. So I guess the next step is analyzing the same thing for different masses but I don't see how something with a lot less mass like Sol V jupiter, how the same intersection point would make for a stable orbit. Maybe it works out that a Trojan can only be in mass up to the mass of the lesser body, that is to say if you replaced Jupiter with Neptune, a much smaller body, and put Jupiter at one of the Trojan points, it would soon overwhelm the gentle guiding force there. That's my take on it. A Trojan can only be stable up to a certain mass related to the primary orbiting body. - 16 Aug '10 05:50

No, you aren't. You are completely ignoring the fact that they are all rotating about each other, or in a more simplified view, the small mass at the Lagrangian is orbiting the center of gravity of the whole system. What we are looking for is not forces that 'cancel out', but forces that maintain the mass at the Lagrangian.*Originally posted by sonhouse***I think I am starting to get it now.** - 16 Aug '10 17:45

I didn't say the forces canceled out, they couldn't because the angle is not right, the cancelation point would be on a line between the to objects dead center if they had equal mass and closer to the smaller of the two masses. I know it's a guiding force, the natural perturbation forces being smaller than the force keeping on that orbital path.*Originally posted by twhitehead***No, you aren't. You are completely ignoring the fact that they are all rotating about each other, or in a more simplified view, the small mass at the Lagrangian is orbiting the center of gravity of the whole system. What we are looking for is not forces that 'cancel out', but forces that maintain the mass at the Lagrangian.** - 17 Aug '10 01:16

equilateral triangle = 60 degrees.*Originally posted by twhitehead***It is explained on Wikipedia here:**

http://en.wikipedia.org/wiki/Lagrangian_point#L4_and_L5

I don't claim to understand it all not having done much physics of that nature in a very long time.

As far as I can tell, the satellite is orbiting the center of mass of the whole system, but is also being pulled towards both large objects in proportion to ...[text shortened]... se two proportions cancel each other out so that it doesn't matter what the actual masses are.

http://en.wikipedia.org/wiki/Lagrangian_point#L4_and_L5

The L4 and L5 points lie at the third corners of the two equilateral triangles in the plane of orbit whose common base is the line between the centers of the two masses, such that the point lies behind (L5) or ahead of (L4) the smaller mass with regard to its orbit around the larger mass.

The reason these points are in balance is that, at L4 and L5, the distances to the two masses are equal. Accordingly, the gravitational forces from the two massive bodies are in the same ratio as the masses of the two bodies, and so the resultant force acts through the barycenter of the system; additionally, the geometry of the triangle ensures that the resultant acceleration is to the distance from the barycenter in the same ratio as for the two massive bodies. The barycenter being both the center of mass and center of rotation of the system, this resultant force is exactly that required to keep a body at the Lagrange point in orbital equilibrium with the rest of the system. (Indeed, the third body need not have negligible mass). The general triangular configuration was discovered by Lagrange in work on the 3-body problem.

L4 and L5 are sometimes called triangular Lagrange points or Trojan points. The name Trojan points comes from the Trojan asteroids at the Sun–Jupiter L4 and L5 points, which themselves are named after characters from Homer's Iliad (the legendary siege of Troy). Asteroids at the L4 point, which leads Jupiter, are referred to as the "Greek camp", while at the L5 point they are referred to as the "Trojan camp". These asteroids are (largely) named after characters from the respective sides of the Trojan War.