Why are the Trojan libration points equidistant and not mass weighted?

The locations of the collinear Lagrange points L1, L2, L3 are mass weighted, so for example Sun-Earth-L1 is only 1% of the distance to the Sun from Earth. But L4 and L5 are one AU from both the Sun and from Earth. Shouldn't Earth's gravity be negligible that far away? Are L4 and L5 another kind of phenomena than L1, L2, L3?

• Maybe because L4, L5 are in a stable free fall orbit, like Earth is. And then mass doesn't matter? But being on a straight line between Earth and the Sun, then mass does matter alot. – LocalFluff Aug 30 '15 at 14:05
• They are mass-weighed relative to the Sun - in exactly the same mass-distance relation as Earth is. – SF. Dec 22 '15 at 2:16

From Wikipedia (emphasis mine)

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 three-body system, this resultant force is exactly that required to keep the smaller body at the Lagrange point in orbital equilibrium with the other two larger bodies of system. (Indeed, the third body need not have negligible mass.) The general triangular configuration was discovered by Lagrange in work on the three-body problem.

Also from the article: • I somewhere miss you there, Organic Marble. Barycenter is very close to Earth' center of gravity. ESL4 is obviously too far away from Earth or our gravity to be important over there, compared to that of the Sun. Is it the orbital movements relative to Earth and Sun that evens it out? – LocalFluff Aug 29 '15 at 15:11
• Sun/Earth barycenter would be close to the Sun, no? – Organic Marble Aug 29 '15 at 15:12
• "Same distance". So why isn't 0.5 AU a Lagrange point linearly between Earth and Sun? – LocalFluff Aug 29 '15 at 15:15
• I'm not following why you think "the force is the same", nothing I have posted implies that. The point is that the ratio of the forces and masses is the same. Maybe you aren't familiar with the triangle of forces concept? easycalculation.com/trigonometry/… – Organic Marble Aug 29 '15 at 16:20
• I think I get it now, finally. – LocalFluff Dec 21 '15 at 12:19

The brief answer is that the L1 and L2 points are different in the sense that the gravity of Earth exerts a far more significant force on objects at L1 and L2. Earth's gravity is about 10,000 times stronger at L1/L2 than L4/L5 and 40,000 times stronger than at L3.

Objects at L3, L4 and L5 are orbiting the sun at very nearly but not quite exactly 1AU. They naturally orbit in about 365.25 days, would continue to do so if the Earth was zapped out of existence. Earth's gravity isn't exerting much force on them.

In contrast, the heliocentric distance of L1 is about 0.99AU and L2 about 1.01AU. Normally an object at 0.99AU would orbit the sun in about 363.4 days and an object at 1.01AU would orbit the sun in about 367.1 days. But at L1 Earth's gravity partially counteracts the Sun's gravity - by about 3% - sufficient to allow an object to orbit in 365.25 days despite being closer to the Sun. And at L2 the Earth's gravity adds to the Sun's - again about 3% - sufficient to allow an object to orbit in 365.25 days despite being further from the Sun. Here's a neat webpage on the ESA site with animations and detailed descriptions

Okay so what's the difference? Any object at very nearly 1AU and not too close to Earth will maintain its position relative to Earth relatively well for a time span measured in decades, although over time the cumulative effects of Earth's gravity will perturb the orbit. L3 is more stable and L4 and L5 are stable infinitely for idealized 3 body.

In contrast any object with a semi-major axis around the Sun of 0.99AU or 1.01AU will normally very quickly change its position relative to Earth, being at least 2 days ahead or behind the Earth within a single year. The exception is L1 and L2 where the gravity of the Sun and Earth combine such that an object with a suitable velocity and minor station-keeping can follow an orbit synchronized with Earth's. In that sense, the L1 and L2 points are pretty special and Earth's gravity plays a much larger role in their stability.

• Wouldn't any object anywhere in Earth' orbit have a stable heliocentric orbit? If you by station keeping mean to compensate for influence from Jupiter and Venus passing by, I still don't see why anything special should happen when the Sun and Earth distances are equal. And for every planet and single moon regardless of masses. Actually I think Jupiter's gravity is more influential at the Trojans 1 AU from Earth, than is that of Earth. 317 times more massive, only 4½ to 6½ times further away. It's not as if Earth has a 2 AU wide Hill sphere, right? – LocalFluff Aug 30 '15 at 3:56
• @LocalFluff actually at all locations within Earth's orbit except L4 and L5, Earth's gravity itself will gradually perturb the object, until it crashes into Earth/Luna, or is flung out of orbit by Earth. The L points are defined in terms of the 3 body problem so this is true independently of the influence of 4th bodies. In reality it is true that other bodies may have a larger perturbing effect, but that is beside the point to the theory of Lagrange points. – Blake Walsh Aug 30 '15 at 8:39
• This answer is incorrect. Look at L2 -- From L2 both the orbiting body and central body are pulling in the same direction. So obviously gravity from the two bodies aren't canceling out. The L points are where gravity from the two bodies and centrifugal force cancel out. – HopDavid Dec 21 '15 at 16:46
• @HopDavid what I intend to mean is that in the absence of Earth's gravity the object at that distance from the sun would need to take a faster (or slower) orbit around the Sun in order to maintain a circular orbit. – Blake Walsh Dec 21 '15 at 17:45
• There it is again: "L1 and L2 where the gravity of the Sun and Earth balance perfectly…" At L1 sun's gravity is about triple earth's gravity. At L2 both the sun's gravity and earth's gravity pull in the same direction. So they reinforce each other rather than cancel each other out. – HopDavid Dec 21 '15 at 23:56

It's a tug of war between three accelerations: Central body gravity, orbiting body gravity, and centrifugal acceleration (inertia in a rotating frame).

Centrifugal acceleration is $\omega^2*r$ where $\omega$ is how fast the system's spinning and r is distance from barycenter. In the case of L1, centrifugal force and orbiting body gravity are on the same team. These two exactly cancel central body gravity. In this case all accelerations are parallel or anti-parallel.

At L4 and L5 none of these forces are parallel. Centrifugal pushes away from barycenter. Central body's gravity pulls towards center of central body. Orbiting body's gravity pulls towards center of orbiting body.

The way to add vector forces is to put the foot of one vector on the head of the other vector. The sum is a vector from from the first vector's foot to the 2nd vector's head. Adding the two gravities we get a vector that exactly cancels centrifugal acceleration.

If we shrink the orbiting body's mass, barycenter moves closer to center of larger body. Central body's gravity becomes more dominant and the orbiting body gravity vector gets smaller. But the two gravities still sum to cancel the $\omega^2r$ which points from barycenter to L4.

• So, in a Sun-Earth system, a lower mass Earth would pull lesser on the L-point object. But then also the barycenter would be nearer the Sun. So those two even out exactly, so that L4, L5 always describe equilateral triangles? (It's as if some Greek mathematician created the universe) – LocalFluff Dec 21 '15 at 18:06

Let's start by looking at the equations of motion for R3BP.
$\ddot x -2\Omega\dot y-\Omega^2x = -\dfrac{\mu_1}{r_1^3}(x+\pi_2r_{12})-\dfrac{-\mu_2}{r_2^3}(x-\pi_1r_{12})$
$\ddot y+2\Omega\dot x-\Omega^2y = -\dfrac{\mu_1}{r_1^3}y-\dfrac{\mu_2}{r_2^3}y$
$\ddot z = -\dfrac{\mu_1}{r_1^3}z-\dfrac{\mu_2}{r_2^3}z$
Where $\pi_1 = \dfrac{m_1}{m_1+m_2}$ and $\pi_2 = 1-\pi_1$ are the mass ratio, $\Omega$ is the angular velocity of reference frame wtih respect to inertial frame; $r_1$ and $r_2$ are distances of third body located at $(x,y,z)$ from $m_1$ and $m_2$ respectively. And the center of coordinate axes lies at the barycenter.

Your solution for $L_4$ and $L_5$ (trojan points) comes from equilibrium condition thus $\dot x= \dot y = \dot z = 0$ and $\ddot x = \ddot y= \ddot z=0$. Which on solving gives
$x = \dfrac{r_{12}}{2}-\pi_2 r_{12}; \hspace{10pt} y = \pm\dfrac{\sqrt{3}}{2}r_{12}; \hspace{10pt} z = 0$
The remaining 3 points solution requires moving one step further by imposing the constrains $y=0$ and $z=0$ and solving the following polynomial for given mas ratio to obtain $\gamma$
$\dfrac{1-\pi_2}{|\gamma+\pi_2|^3}(\gamma+\pi_2)+\dfrac{\pi_2}{|\gamma+\pi_2-1|^3}(\gamma+\pi_2-1)-\gamma = 0$
giving 3 unique solutions; where $\gamma = \dfrac{x}{r_{12}}$

If you look at $L_4$ and $L_5$ the $y$ coordinate the general solution tells you that they lie along the vertex of an equilateral triangle with other two vertices lying at two masses. This is because the function $\pi_2$ is generally $<<1$ so $x$ becomes essentially $r_{12}/2$. Thus they are function of majorly $r_{12}$.
This unique geometric position enables the force acting on $3^{rd}$ body to pass through the barycenter, which essentially makes it motionless wrt to one of the bodies as seen from other. For better understanding you consider two bodies of comparable masses, then your barycenter position $\rightarrow r_{12}/2$ and your mass ratio $\pi_2 \rightarrow 1/2$, so your $x$ becomes zero. Based on vector diagram it is easier to see that net force will then pass through the barycenter in such case. So all points are equilibrium points and none of them have something special attached to them.

On effect of Earth gravitation at $L_4$ and $L_5$, using Newton's Law we can show that on a unit mass at those points $\dfrac{F_e}{F_s} = 0.000003003$. This small yet significant force arising due to Earth changes the application of resultant force on unit mass to act on barycenter instead of pointing directly at the focus (the Sun center).