Lagrange points

Potential curves in a system of two bodies (here the Sun and Earth), showing the five points of Lagrange. The arrows point the potential increase direction around points L – approaching or away of them. Against intuition, points L₄ and L₅ They're minimal.

The Lagrange points, also called L points or libration points, are the five positions in an orbital system where a small object, affected only by gravity, can theoretically be stationary with respect to two larger objects, such as an artificial satellite with respect to the Earth and the Moon. The Lagrange points mark the positions where the combined gravitational pull of the two large masses provides the necessary centripetal force to rotate synchronously with the smaller one. They are analogous to geosynchronous orbits that allow an object to be in a "fixed" position in space rather than in an orbit where its relative position is continually changing.

A more precise but technical definition is that the Lagrange points are the stationary solutions of the three-body problem restricted to circular orbits. If, for example, we have two large bodies in a circular orbit around their common center of mass, there are five positions in space where a third body, of negligible mass compared to the other two, can be located and maintain its position. relative to the two large bodies. Viewed from a rotating reference frame that rotates with the same period as the two co-orbital bodies, the gravitational field of two large bodies combined with the centrifugal force balances out at the Lagrange points, allowing the third body to remain stationary with respect to the first two.

History and concepts

In 1772, the Italian-French mathematician Joseph-Louis Lagrange was working on the famous three-body problem when he discovered an interesting peculiarity. Originally, he was trying to discover a way to easily calculate the gravitational interaction of an arbitrary number of bodies in a system. Newtonian mechanics determines that such a system rotates chaotically until either a collision occurs, or one of the bodies is expelled from the system and mechanical equilibrium is achieved. It is very easy to solve the case of two bodies that orbit around the common center of gravity. However, if a third body, or more, is introduced, the mathematical calculations are very complicated, being a situation in which one would have to calculate the sum of all the gravitational interactions on each object at each point along its trajectory..

However, Lagrange wanted to make this easier, and he achieved it by a simple hypothesis: The trajectory of an object is determined by finding a path that minimizes the action over time. This is calculated by subtracting the potential energy from the kinetic energy. Developing this hypothesis, Lagrange reformulated Newton's classical mechanics to give rise to Lagrangian mechanics. With his new way of calculating, Lagrange's work led him to hypothesize a third body of negligible mass orbiting two larger bodies that were already spinning in a quasi-circular orbit. In a reference frame that rotates with the larger bodies, he found five specific fixed points at which the third body, following the orbit of the larger bodies, is subjected to zero force. These points were called Lagrange points in his honor.

In the most general case of elliptical orbits there are no longer stationary points but rather a "area" of Lagrange. The successive Lagrange points, considering circular orbits at every moment, form stationary elliptical orbits, geometrically similar to the orbit of the larger bodies. This is due to Newton's second law (dp/dt=F{displaystyle dmathbf {p} /dt=mathbf {F} }Where P = mv (p is the amount of movement, m mass and v speed). p is an invariant if the strength and position are multiplied by the same factor. A body at a point of Lagrange orbits with the same period as the two large bodies in the circular case, implying, as it happens, that they have the same proportion between gravitational force and radial distance. This fact is independent of the circularity of the orbits and implies that the elliptical orbits described by the Lagrange points are solutions of the movement equation of the third body.

Complications to Kepler's Laws

Both the Earth and the Sun influence each other through their gravitational forces. This means that, although the Sun causes tides on the Earth, this in turn causes disturbances in the movement of the Sun. In fact, both bodies (the Sun-Earth system) move around the point called the center of mass or barycenter, which it is located near the center of the Sun due to the different masses of both bodies and the much greater influence of the Sun due to its mass. In the case of the Sun-Jupiter system, the barycenter is near the solar surface. On the other hand, since the mass of an artificial satellite is negligible with respect to the mentioned bodies, its mass does not have a significant influence on the barycenter of the three.

Kepler's laws describe in a simple way the behavior of two bodies orbiting around each other. The third law says that the square of its orbital period (the time it takes to go around the Sun) is directly proportional to the cube of the average distance from the Sun. For this reason, the increase in radius gives rise to an increase of the orbital period, therefore, two bodies located at different distances from the Sun will never have a synchronized movement.

The simplicities of Kepler's laws do not hold if multi-body interactions are taken into account instead of two or three, as in the solar system. Even if only a group of three were considered, the Sun, the Earth and an artificial satellite, the predictions become complicated. Thus, a satellite located on the Sun-Earth line and between them should have an orbital period of less than one year, but if it is at a distance of 1.5 million km from Earth, in what will later be called L₁, the attraction of the Earth decreases the attraction of the Sun and its period is the same as that of the Earth. Less distance does not mean less period.

Lagrange points

Diagram that shows the five points of Lagrange in a very different two-body mass system (e.g. the Sun and the Earth). In a system like that, L₄-L₅ They seem to rotate in the same orbit as the second body, although in fact it does slightly further away from the first.

The five Lagrangian points are called and defined as follows:

The L1 point

The point L₁ is between the two large masses M₁ and M₂ on the line that joins them. It is the most intuitive of the Lagrange points, the one in which the opposite attractions of the two larger bodies are balanced.

• Example: an object that orbits the Sun closer than the Earth would have a shorter orbital period than the Earth, but that ignores the gravitational pull effect of the Earth. If the object is directly between the Earth and the Sun, then the effect of the Earth's gravity is to weaken the force that pulls the object toward the Sun and therefore increases the orbital period of the object. The closer the object of the Earth is, the greater this effect. At point L₁the orbital period of the object is exactly the same as the orbital period of the Earth. This point is to be found 1 502 000 km of the earth.

The point L₁ of the Sun-Earth system is ideal for making observations of the Sun. Objects located here are never eclipsed by the Earth or the Moon. The Solar and Heliosphere Observatory (SOHO) space probe is stationed at point L₁, and the Advanced Composition Explorer (ACE) satellite is in a Lissajous orbit around point L_{1 as well} . The L₁ point of the Earth-Moon system allows easy access to lunar and Earth orbit with minimal velocity change, delta-v, and would be ideal for a manned space station located halfway path designed to help transport cargo and personnel to and from the Moon.

The L2 point

Sol-Earth System Diagram, displaying point L₂further away than the lunar orbit.

The point L₂ lies on the line defined by the two large masses M₁ and M₂, and beyond the largest small of the two In it, the gravitational attraction of the two larger bodies compensates for the centrifugal force caused by the smaller one.

• Example: an object that orbits the Sun further than the Earth would have a longer orbital period than the Earth. The additional force of Earth's gravity reduces the orbital period of the object, and precisely point L₂ is that in which the orbital period is equal to that of Earth.

The point L₂ of the Sun-Earth system is a good point for space observatories, because an object around L₂ will maintain the same orientation with respect to the Sun and Earth, and calibration and shielding are easier. The Wilkinson Microwave Anisotropy Probe (WMAP) as well as the Herschel Space Observatory are already in orbit around the L₂ point of the Sun-Earth system. The James Webb Space Telescope is also located at the L₂ point in the Sun-Earth system. The L₂ point in the Earth-Moon system would be a good location for a satellite communications network covering the far side of the Moon.

If M₂ is much smaller than M₁, then L₁ and L₂ are at approximately equal distances r from M₂, equal to the radius of Hill's sphere, given by:

r≈ ≈ RM23M13{displaystyle rapprox R{sqrt[{3}]{frac {M_{2}}}{3M_{1}}}}}}}{3M

where R is the distance between the two bodies.

This distance can be described as that in which the orbital period corresponding to a circular orbit with this distance around M₂ and in the absence of M₁It's time it takes to turn M₂ around M₁, divided by 33≈ ≈ 1,44{displaystyle {sqrt}{3}{3}}approx 1,44}.

Examples:

• Sun and Earth System: 1,500,000 km from Earth
• Earth and Moon System: 61.500 km from the Moon

The L3 point

The point L₃ lies on the line defined by the two large masses M₁ and M₂, and beyond the largest Of the two.

• Example: Point L₃ in the Sun-Earth system is on the opposite side of the Sun, a little closer to the Sun than the Earth itself. This apparent contradiction is explained because the Sun is also affected by the Earth's gravity, and so it revolves around the common mass center or varnish which, however, is within the Sun. In L₃ the combined gravitational force of the Earth and the Sun makes the object orbit with the same period as the Earth. Point L₃ in the Sol-Earth system was a popular place used to place a "Contra-Earth" in science fiction books or in comics; although direct observation by probes and satellites later demonstrated their absence. In reality, L₃ in the Sol-Earth system is very unstable, as the gravitational forces of the other planets can overcome that of the Earth (Venus, for example, passes to 0.3 AU of L₃ every twenty months).

The points L4 and L5

Gravitational actions in L₄.

The point L₄ and the point L₅ are at the vertices of equilateral triangles whose common base is the line joining the two masses, such that the point L₄ precedes the small body by an angle of 60º seen from the large mass, while L₅ turns behind the small body, although with a greater radius than this, with a delay of 60º seen in turn from the large body. These points, as well as the smaller body of mass M₂, do not revolve around the large body, but rather around the centroid of both bodies marked b in the figure. The large body also rotates about b with radius r₁

The radio r{displaystyle r} common orbit to points L₄ and L₅ can be deduced from the figure by geometric reasoning:

Bearing in mind that the radios of the orbits of the large bodies r1{displaystyle r_{1}} and r2{displaystyle r_{2}} they are in reverse relation to their masses: r1r2=M2M1=γ γ {displaystyle {frac {r_{1}}{r_{2}}}}}{{frac {M_{2}}{M_{1}}}}}=gamma }, solves the triangle formed by L₄, b and the lower body mass center; resulting in the relationship r2=r12+r1r2+r22{displaystyle r^{2}=r_{1}{2}+r_{1}r_{2}r_{2}r_{2}{2}{2}}{2}}.

Demonstration

Geometric scale for the calculation of the rotation radius of the points L4 and L5 We will use the attached figure that outlines the geometric situation of the previous image. Here, P and Q are the respective mass centers of the larger and smaller bodies, while B is the center of gravity of the system around which the three objects rotate. Since the triangles GBQ and FPB are equilateral, the quadrilateral LGBF is parallelogram and therefore LG too. r1. We then apply law of cosenos in the triangle LBQ with respect to the angle LQB=60° to get: LB2=BQ2+LQ2− − 2BQLQ# 60 {displaystyle LB^{2}=BQ^{2}+LQ^{2}-2,BQcos 60^{circ}}}} which corresponds to: r2=r22+(r1+r2)2− − 2r2(r1+r2)# 60 {displaystyle r^{2}=r_{2}^{2}+(r_{1})^{2}-2r_{2}(r_{1}+r_{2})cos 60^{circ}}}}} r2=r22+r12+2r1r2+r22− − (2r1r2+2r22)/2{displaystyle r^{2}=r_{2}^{2}+r_{1}^{2+}2r_{1}r_{2}+r_{2}{2}{2}-(2r_{1}r_{2} +2r_{2}{2}{2})/2} since # 60 =12{displaystyle cos 60^{circ }={frac {1}{2}{2}}. Making simplification throws the desired result: r2=r12+r1r2+r22{displaystyle r^{2}=r_{1}{2}+r_{1}r_{2}r_{2}r_{2}{2}{2}}{2}}.

Expressing the outcome on the basis of γ γ {displaystyle gamma } results:

r=r2⋅ ⋅ γ γ 2+γ γ +1{displaystyle r=r_{2}cdot {sqrt {gamma ^{2+}gamma +1}}}}}

Demonstration

Post γ γ =r1r2{displaystyle gamma ={frac {r_{1}}{r_{2}}}, then r1=γ γ r2{displaystyle r_{1}=gamma r_{2},}. We make the replacement r2=r12+r1r2+r22{displaystyle r^{2}=r_{1}{2}+r_{1}r_{2}r_{2}r_{2}{2}{2}}{2}}. as a result r2=(γ γ r2)2+(γ γ r2)r2+r22{displaystyle r^{2}=(gamma r_{2})^{2}+(gamma r_{2})r_{2}+r_{2}{2}{2}}}}}. The right side has a common factor r22{displaystyle r_{2}{2}{2}}} and so r2=r22(γ γ 2+γ γ +1){displaystyle r^{2}=r_{2}{2}(gamma ^{2}+gamma +1)}. Finally, the square root is applied on both sides of the expression to conclude r=r2γ γ 2+γ γ +1{displaystyle r=r_{2}{sqrt {gamma ^{2}+gamma +1}}}}}{sqrt {gamma ^{2}.

This radio, as seen in the figure is generally greater than radio r2{displaystyle r_{2}} of the small body because 1,}" xmlns="http://www.w3.org/1998/Math/MathML">γ γ 2+γ γ +1▪1{displaystyle gamma ^{2}+gamma +1pur1,}1,}" aria-hidden="true" class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/e9473183f42113e26a3c67008b9e8103efa4d946" style="vertical-align: -0.838ex; width:15.088ex; height:3.176ex;"/> and therefore the radical that multiplies also has a value greater than one.

The true precession angle of L₄, that is to say the angle that forms L₄with the small body seen from the turning center b, can also be calculated with geometric procedures, obtaining: α α =arctan (3(1+γ γ 1− − γ γ )){displaystyle alpha =arctan left({sqrt {3}}}left({frac {1+gamma }{1-gamma }}right)}}}}.

Demonstration

Precession angle calculation. Taking the geometric scheme, we trace the height of the triangle LBQ Which goes by L. As noted above, γ γ =r1r2{displaystyle gamma ={frac {r_{1}}{r_{2}}} for what r1=γ γ r2{displaystyle r_{1}=gamma r_{2},} and therefore r1+r2=γ γ r2+r2=r2(γ γ +1){displaystyle r_{1}+r_{2}=gamma r_{2}+r_{2}=r_{2}(gamma +1),}. The trigonometric relations of the rectangle triangle LBQ imply LT=(r1+r2)without 60 =r2(γ γ +1)32{displaystyle LT=(r_{1}+r_{2})sin 60^{circ }=r_{2}(gamma +1){frac {sqrt {3}}{2}}}}{2}}}}, TQ=(r1+r2)# 60 =r2(γ γ +1)⋅ ⋅ 12{displaystyle TQ=(r_{1}+r_{2})cos 60^{circ }=r_{2}(gamma +1)cdot {frac {1}{2}}}}}}}. Now in the triangle rectangle LBT: So... α α =LTBT=LTr2− − TQ{displaystyle tan alpha ={frac {LT}{BT}}}}={frac {LT}{r_{2}-TQ}}}}}. And replacing the expressions found results in So... α α =r2(γ γ +1)32r2− − r2(γ γ +1)⋅ ⋅ 12{displaystyle tan alpha ={frac {r_{2}(gamma +1){frac {sqrt {3}}{2}}}}{r_{2}-r_{2}(gamma +1)cdot {frac {1}{2}}}}}}}}}}{gamma So... α α =r2(γ γ +1)32r2(1− − γ γ +12){displaystyle tan alpha ={frac {r_{2}(gamma +1){frac {sqrt {3}{2}}}}}{r_{2}left(1-{frac {gamma +1}{2}}{2}}}}}}}}}}}}{ So... α α =r2(γ γ +1)32r2(1− − γ γ 2){displaystyle tan alpha ={frac {r_{2}(gamma +1){frac {sqrt {3}}{2}}}}}{r_{2}left({frac {1-gamma }{2}}}{2}}}}}}}}}{ and after canceling terms in the numerator and denominator, you get So... α α =3(1+γ γ 1− − γ γ ){displaystyle tan alpha ={sqrt {3}}left({frac {1+gamma }{1-gamma }}}right)}.

Examples:

• For the Earth-Luna system we have.

Distance Earth-Luna: d = r₁ + r₂ = 3,844·10⁸ m

Earth Mass: M₁ = 5,974·10²⁴ kg

Moon Mass: M₂ = 7.35·10²² kg

Value of γ = M₂/M₁ = 12,30·10^-3

So, like:

γ γ =r1r2{displaystyle gamma ={frac {r_{1}}{r_{2}}},

d=r1+r2=r2γ γ +r2=r2(γ γ +1){displaystyle d=r_{1}+r_{2}=r_{2}gamma +r_{2}=r_{2}(gamma +1),}

you have:

r2=d1+γ γ =3,7972⋅ ⋅ 108m{displaystyle r_{2}={frac {d}{1+gamma }}=3,7972cdot 10^{8}m}.

r1=d− − r2=4,6719⋅ ⋅ 106m{displaystyle r_{1}=d-r_{2}=4,6719cdot 10^{6}m}

With this data and the previous formula is evaluated:

r=r12+r1r2+r22=3,8208⋅ ⋅ 108m{displaystyle r={sqrt {r_{1}{2}+r_{1r_{2}+r_{2}{2}}{2}}}}=3,8208cdot 10^{8}m}

For α, using the other formula, you have: α = 60,6067o

I mean, the common orbit of L₄ and L₅ exceeds that of the Moon in 2360 km and those points form with it angles of 60o 18' 22" regarding the baricentro b system

• Yes M₁ = M₂, case of the symmetrical double stars, the gamma parameter is made equal to one.

In these conditions the two masses occupy a common orbit, the α angle increases to 90o and the radius of the L orbit₄ and L₅ equals the radius of the common orbit of the stars multiplied by the root of 3. This radio coincides with the height of the equilateral triangle whose base coincides with the distance between the stars.

The reason these points are in equilibrium is that point L₄ and point L₅ are the same distance from the two masses. Therefore, the gravitational forces of the two bodies are in the same relationship as their respective masses, and the resulting force acts through the system's barycenter; furthermore, the triangle geometry means that the resulting acceleration is at the distance from the center of gravity in the same proportion as for the two larger bodies. And since the centroid is the center of mass and the center of rotation of the system, this resultant force is exactly what is required to keep a body at the Lagrange point in equilibrium with the rest of the system.

L₄ and L₅ are sometimes called "triangular Lagrange points" or "Trojan points". The name "Trojan points" comes from the Trojan asteroids in the Sun-Jupiter system, named after characters from Homer's Iliad—the legendary Trojan War. The asteroids at point L₄, preceding Jupiter, are the “Greek camp”, the “Greeks”, while those at point L₅ are the “Trojan camp”. ». The names are taken from characters from the Iliad.

Examples:

• Points L₄ and L₅ The Sun-Earth system contains only interplanetary dust and the Earth's trojan asteroid 2010 TK7.
• Points L₄ and L₅ the Earth-Luna system whose location has been calculated earlier, contain interplanetary dust, the so-called Kordylewski clouds.
• Points L₄ and L₅ of the Sun-Jupiter system are occupied by the Trojan asteroids.
• Neptune has Trojan Kuiper belt objects on its L points₄ and L₅.
• The moon of Saturn Tetis has two smaller satellites on its L points₄ and L₅, by name Telesto and Calipso, respectively.
• The moon of Saturn Dione has smaller moons, Helena and Pollux, at its points L₄ and L₅respectively.
• The big impact hypothesis suggests that an object (Theia) was formed in L₄ o L₅ and crashed against Earth by entering into unstable orbit, thus giving rise to the Moon.

Stability

The first three Lagrange points are technically stable only in the plane perpendicular to the line between the two bodies. This can be seen most easily by considering the point L₁. A test mass displaced perpendicular to the center line would feel a force pulling it toward the equilibrium point. This is so because the lateral components of gravity of the two masses add to produce this force, while the along-axis components cancel out. However, if an object located at point L₁ were pulled towards one of the masses, the gravitational attraction it feels for that mass would be greater, and it would be pulled towards it (the model is very similar to that of the tidal force).

Although the points L₁, L₂ and L₃ are nominally unstable, it turns out that it is possible to find stable periodic orbits around them points, at least in the restricted three-body problem. These perfectly periodic orbits, called "halo" orbits, do not exist in a dynamical n-body system like the solar system. However, quasi-periodic Lissajous orbits do exist, and they are the orbits that have been used in all space missions to libration points. Although the orbits are not perfectly stable, a relatively modest effort keeps it in the Lissajous orbit for a long period. It is also useful in the case of the L₁ point of the Sun-Earth system to put the spacecraft in a large amplitude Lissajous orbit (100,000–200,000 km) instead of parking it at the point of the libration, because this keeps the spacecraft out of the direct Sun-Earth line and thus reduces solar interference to Earth-spacecraft communications.

Another interesting and useful property of collinear equilibrium points and their associated Lissajous orbits is that they serve as gateways to control the chaotic trajectories of an interplanetary transport network.

In contrast to the instability of collinear points, triangular points (L₄ and L₅) have a stable equilibrium (see attractor), provided that the ratio of the masses M₁/M₂ is > 24.96. This is the case for the Sun/Earth and Earth/Moon systems, although by a smaller margin in the latter case. When a body at these points is disturbed and moves out of the point, a Coriolis effect acts that returns it to the point.

Solar System Values

This table shows values for L₁, L₂, and L₃ within the solar system. The calculations assume that the two bodies orbit in a perfect circle with separation equal to the semi-major axis (SEM) and there are no other bodies nearby. Distances are measured from the center of mass of the largest body with L3 showing a negative location. The percentage columns show how the distances compare to the semimajor axis. For example: For the Moon, L1 is located 326400 km from the center of the Earth, which is 84.9% of the Earth-Moon distance or 15.1% in front of the Moon; L2 is located 448900 km from the center of the Earth, which is 116.8% of the Earth-Moon distance or 16.8% beyond the Moon; and L3 is located -381,600 km from the center of the Earth, which is 99.3% of the Earth-Moon distance or 0.7084% ahead of the "negative" of the moon. The value of L3 percent has been increased by 100.

Space missions at libration points

Libration point orbits have unique characteristics that make them a good choice for locating some types of missions. NASA has sent several spacecraft to points L₁ and L₂ in the Sun-Earth system:

The L5 Society is a precursor to the National Space Society, and promoted the possibility of establishing a colony at points around the L₄ or L₅ system of Earth Moon (see space colonization and colonization of Lagrange points).

Natural examples

In the Sun-Jupiter system there are several thousand asteroids, called Trojan asteroids, which are in orbits around the Sun, at points L₄ or L₅ of the Sun-Jupiter system. Other bodies can be found at the same points in the Sun-Saturn, Sun-Mars, Sun-Neptune, Jupiter - Jovian satellites, and Saturn - satellites of Saturn systems. 2010 TK7 is a Trojan from the Sun-Earth system, at point L₄. Dust clouds surrounding the points L₄ and L₅ were discovered in the 1950s. These dust clouds were called Kordylewski clouds, and even weaker the gegenschein, it is also present at the point L₄ and L₅ of the Earth-Moon system.

Saturn's moon Tethys has two smaller moons at its L₄ and L₅ points called Telesto and Calypso. Saturn's moon Dione also has two coorbital Lagrangian satellites, Helena at its point L₄ and Pollux at L₅. The moons oscillate around Lagrange points, and Polydeuces has the largest deviations, moving up to 32 degrees away from the L₅ point of the Saturn-Dione system. Tethys and Dione are hundreds of times bigger than their "escorts"; (see the moon articles for exact dimensions; in many cases the masses are not known), and Saturn is much more massive, making the system very stable.

Other coorbital examples

Earth has a companion (3753) Cruithne that has an orbit similar to Earth's. He is not a true Trojan. Rather, it occupies one of two regular solar orbits, one slightly smaller and faster than Earth's and the other slightly larger and slower, periodically alternating as it approaches Earth. As the asteroid approaches Earth, inside Earth's orbit, it picks up orbital energy from Earth and moves into a larger, higher energy orbit. Then the Earth catches up with the asteroid, which is in a larger and therefore slower orbit. Now it is the Earth that takes energy and makes the asteroid fall into a smaller, faster orbit and in the future it will be the asteroid that catches the Earth to start the cycle again. This has no noticeable impact on the length of the year, because Earth is more than 20 billion times more massive than 3753 Cruithne.

Saturn's moons Epimetheus and Janus have a similar relationship, although they are of similar masses and actually swap orbits with each other periodically (Janus is about four times as massive, but it's enough for its orbit to be disturbed). Another similar configuration known as orbital resonance means that bodies tend to have periods that are in simple relationships to larger ones due to their interaction.