Lagrange point

The L₁ point lies on the line defined between the two large masses M₁ and M₂. It is the point where the gravitational attraction of M₂ and that of M₁ combine to produce an equilibrium. An object that orbits the Sun more closely than Earth would normally have a shorter orbital period than Earth, but that ignores the effect of Earth's own gravitational pull. If the object is directly between Earth and the Sun, then Earth's gravity counteracts some of the Sun's pull on the object, and therefore increases the orbital period of the object. The closer to Earth the object is, the greater this effect is. At the L₁ point, the orbital period of the object becomes exactly equal to Earth's orbital period. L₁ is about 1.5 million kilometers from Earth, or 0.01 au.[1]

The L₂ point lies on the line through the two large masses, beyond the smaller of the two. Here, the gravitational forces of the two large masses balance the centrifugal effect on a body at L₂. On the opposite side of Earth from the Sun, the orbital period of an object would normally be greater than that of Earth. The extra pull of Earth's gravity decreases the orbital period of the object, and at the L₂ point that orbital period becomes equal to Earth's. Like L₁, L₂ is about 1.5 million kilometers or 0.01 au from Earth. An example of a spacecraft at L₂ is the James Webb Space Telescope, designed to operate near the Earth–Sun L₂.[7] Earlier examples include the Wilkinson Microwave Anisotropy Probe and its successor, Planck.

The L₃ point lies on the line defined by the two large masses, beyond the larger of the two. Within the Sun–Earth system, the L₃ point exists on the opposite side of the Sun, a little outside Earth's orbit and slightly closer to the center of the Sun than Earth is. This placement occurs because the Sun is also affected by Earth's gravity and so orbits around the two bodies' barycenter, which is well inside the body of the Sun. An object at Earth's distance from the Sun would have an orbital period of one year if only the Sun's gravity is considered. But an object on the opposite side of the Sun from Earth and directly in line with both "feels" Earth's gravity adding slightly to the Sun's and therefore must orbit a little farther from the barycenter of Earth and Sun in order to have the same 1-year period. It is at the L₃ point that the combined pull of Earth and Sun causes the object to orbit with the same period as Earth, in effect orbiting an Earth+Sun mass with the Earth-Sun barycenter at one focus of its orbit.

Gravitational accelerations at L₄

The L₄ and L₅ points lie at the third vertices 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 60° ahead of (L₄) or behind (L₅) the smaller mass with regard to its orbit around the larger mass.

The triangular points (L₄ and L₅) are stable equilibria, provided that the ratio of M₁/M₂ is greater than 24.96.[note 1] This is the case for the Sun–Earth system, the Sun–Jupiter system, and, by a smaller margin, the Earth–Moon system. When a body at these points is perturbed, it moves away from the point, but the factor opposite of that which is increased or decreased by the perturbation (either gravity or angular momentum-induced speed) will also increase or decrease, bending the object's path into a stable, kidney bean-shaped orbit around the point (as seen in the corotating frame of reference).[8]

The points L₁, L₂, and L₃ are positions of unstable equilibrium. Any object orbiting at L₁, L₂, or L₃ will tend to fall out of orbit; it is therefore rare to find natural objects there, and spacecraft inhabiting these areas must employ station keeping in order to maintain their position.

The location of L₁ is the solution to the following equation, gravitation providing the centripetal force:

$${\displaystyle {\frac {M_{1}}{(R-r)^{2}}}-{\frac {M_{2}}{r^{2}}}=\left({\frac {M_{1}R}{M_{1}+M_{2}}}-r\right){\frac {M_{1}+M_{2}}{R^{3}}}}$$

where r is the distance of the L₁ point from the smaller object, R is the distance between the two main objects, and M₁ and M₂ are the masses of the large and small object, respectively. The quantity in parentheses on the right is the distance of L₁ from the center of mass. Solving this for r involves solving a quintic function, but if the mass of the smaller object (M₂) is much smaller than the mass of the larger object (M₁) then L₁ and L₂ are at approximately equal distances r from the smaller object, equal to the radius of the Hill sphere, given by:

$${\displaystyle r\approx R{\sqrt[{3}]{\frac {M_{2}}{3M_{1}}}}}$$

We may also write this as:

$${\displaystyle {\frac {M_{2}}{r^{3}}}\approx 3{\frac {M_{1}}{R^{3}}}}$$

Since the tidal effect of a body is proportional to its mass divided by the distance cubed, this means that the tidal effect of the smaller body at the L₁ or at the L₂ point is about three times of that body. We may also write:

$${\displaystyle \rho _{2}(d_{2}/r)^{3}\approx 3\rho _{1}(d_{1}/R)^{3}}$$

where ρ₁ and ρ₂ are the average densities of the two bodies and ${\displaystyle d_{1}}$ and ${\displaystyle d_{2}}$ are their diameters. The ratio of diameter to distance gives the angle subtended by the body, showing that viewed from these two Lagrange points, the apparent sizes of the two bodies will be similar, especially if the density of the smaller one is about thrice that of the larger, as in the case of the earth and the sun.

This distance can be described as being such that the orbital period, corresponding to a circular orbit with this distance as radius around M₂ in the absence of M₁, is that of M₂ around M₁, divided by √3 ≈ 1.73:

$${\displaystyle T_{s,M_{2}}(r)={\frac {T_{M_{2},M_{1}}(R)}{\sqrt {3}}}.}$$

The Lagrangian L₂ point for the Sun–Earth system

The location of L₂ is the solution to the following equation, gravitation providing the centripetal force:

$${\displaystyle {\frac {M_{1}}{(R+r)^{2}}}+{\frac {M_{2}}{r^{2}}}=\left({\frac {M_{1}}{M_{1}+M_{2}}}(R+r)\right){\frac {M_{1}+M_{2}}{R^{3}}}}$$

with parameters defined as for the L₁ case. Again, if the mass of the smaller object (M₂) is much smaller than the mass of the larger object (M₁) then L₂ is at approximately the radius of the Hill sphere, given by:

$${\displaystyle r\approx R{\sqrt[{3}]{\frac {M_{2}}{3M_{1}}}}}$$

The same remarks about tidal influence and apparent size apply as for the L₁ point. For example, the angular radius of the sun as viewed from L₂ is arcsin(695.5×10³/151.1×10⁶) ≈ 0.264°, whereas that of the earth is arcsin(6371/1.5×10⁶) ≈ 0.242°. Looking toward the sun from L₂ one sees an annular eclipse. It is necessary for a spacecraft, like Gaia, to follow a Lissajous orbit or a halo orbit around L₂ in order for its solar panels to get full sun.

The location of L₃ is the solution to the following equation, gravitation providing the centripetal force:

$${\displaystyle {\frac {M_{1}}{\left(R-r\right)^{2}}}+{\frac {M_{2}}{\left(2R-r\right)^{2}}}=\left({\frac {M_{2}}{M_{1}+M_{2}}}R+R-r\right){\frac {M_{1}+M_{2}}{R^{3}}}}$$

with parameters M₁, M₂, and R defined as for the L₁ and L₂ cases, and r now indicates the distance of L₃ from the position of the smaller object, if it were rotated 180 degrees about the larger object, while positive r implying L₃ is closer to the larger object than the smaller object. If the mass of the smaller object (M₂) is much smaller than the mass of the larger object (M₁), then:[18]

$${\displaystyle r\approx R{\frac {7M_{2}}{12M_{1}}}}$$

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 the system (indeed, the third body needs to have negligible mass). The general triangular configuration was discovered by Lagrange working on the three-body problem.

Net radial acceleration of a point orbiting along the Earth–Moon line

The radial acceleration a of an object in orbit at a point along the line passing through both bodies is given by:

$${\displaystyle a=-{\frac {GM_{1}}{r^{2}}}\operatorname {sgn}(r)+{\frac {GM_{2}}{(R-r)^{2}}}\operatorname {sgn}(R-r)+{\frac {G{\bigl (}(M_{1}+M_{2})r-M_{2}R{\bigr )}}{R^{3}}}}$$

where r is the distance from the large body M₁, R is the distance between the two main objects, and sgn(x) is the sign function of x. The terms in this function represent respectively: force from M₁; force from M₂; and centripetal force. The points L₃, L₁, L₂ occur where the acceleration is zero — see chart at right. Positive acceleration is acceleration towards the right of the chart and negative acceleration is towards the left; that is why acceleration has opposite signs on opposite sides of the gravity wells.

The satellite ACE in an orbit around Sun–Earth L₁

Gaia (yellow) and James Webb Space Telescope (blue) orbits around Sun–Earth L₂

Sun–Earth L₁ is suited for making observations of the Sun–Earth system. Objects here are never shadowed by Earth or the Moon and, if observing Earth, always view the sunlit hemisphere. The first mission of this type was the 1978 International Sun Earth Explorer 3 (ISEE-3) mission used as an interplanetary early warning storm monitor for solar disturbances.[23] Since June 2015, DSCOVR has orbited the L₁ point. Conversely it is also useful for space-based solar telescopes, because it provides an uninterrupted view of the Sun and any space weather (including the solar wind and coronal mass ejections) reaches L₁ up to an hour before Earth. Solar and heliospheric missions currently located around L₁ include the Solar and Heliospheric Observatory, Wind, and the Advanced Composition Explorer. Planned missions include the Interstellar Mapping and Acceleration Probe (IMAP) and the NEO Surveyor.

Sun–Earth L₂ is a good spot for space-based observatories. Because an object around L₂ will maintain the same relative position with respect to the Sun and Earth, shielding and calibration are much simpler. It is, however, slightly beyond the reach of Earth's umbra,[24] so solar radiation is not completely blocked at L₂. Spacecraft generally orbit around L₂, avoiding partial eclipses of the Sun to maintain a constant temperature. From locations near L₂, the Sun, Earth and Moon are relatively close together in the sky; this means that a large sunshade with the telescope on the dark-side can allow the telescope to cool passively to around 50 K – this is especially helpful for infrared astronomy and observations of the cosmic microwave background. The James Webb Space Telescope was positioned in a halo orbit about L₂ on January 24, 2022.

Sun–Earth L₁ and L₂ are saddle points and exponentially unstable with time constant of roughly 23 days. Satellites at these points will wander off in a few months unless course corrections are made.[8]

Sun–Earth L₃ was a popular place to put a "Counter-Earth" in pulp science fiction and comic books, despite the fact that the existence of a planetary body in this location had been understood as an impossibility once orbital mechanics and the perturbations of planets upon each other's orbits came to be understood, long before the Space Age; the influence of an Earth-sized body on other planets would not have gone undetected, nor would the fact that the foci of Earth's orbital ellipse would not have been in their expected places, due to the mass of the counter-Earth. The Sun–Earth L₃, however, is a weak saddle point and exponentially unstable with time constant of roughly 150 years.[8] Moreover, it could not contain a natural object, large or small, for very long because the gravitational forces of the other planets are stronger than that of Earth (for example, Venus comes within 0.3 AU of this L₃ every 20 months).

A spacecraft orbiting near Sun–Earth L₃ would be able to closely monitor the evolution of active sunspot regions before they rotate into a geoeffective position, so that a seven-day early warning could be issued by the NOAA Space Weather Prediction Center. Moreover, a satellite near Sun–Earth L₃ would provide very important observations not only for Earth forecasts, but also for deep space support (Mars predictions and for crewed mission to near-Earth asteroids). In 2010, spacecraft transfer trajectories to Sun–Earth L₃ were studied and several designs were considered.[25]

Earth–Moon L₁ allows comparatively easy access to Lunar and Earth orbits with minimal change in velocity and this has as an advantage to position a habitable space station intended to help transport cargo and personnel to the Moon and back.

Earth–Moon L₂ has been used for a communications satellite covering the Moon's far side, for example, Queqiao, launched in 2018,[26] and would be "an ideal location" for a propellant depot as part of the proposed depot-based space transportation architecture.[27]

Scientists at the B612 Foundation were[28] planning to use Venus's L₃ point to position their planned Sentinel telescope, which aimed to look back towards Earth's orbit and compile a catalogue of near-Earth asteroids.[29]

In 2017, the idea of positioning a magnetic dipole shield at the Sun–Mars L₁ point for use as an artificial magnetosphere for Mars was discussed at a NASA conference.[30] The idea is that this would protect the planet's atmosphere from the Sun's radiation and solar winds.