Editing Lagrange point (section)

==Physical and mathematical details==

[[File:Lagrangian points equipotential.png|thumb|link={{filepath:Lagrangian_points_equipotential.gif}}|Visualization of the relationship between the Lagrange points (red) of a planet (blue) orbiting a star (yellow) counterclockwise, and the [[effective potential]] in the plane containing the orbit (grey rubber-sheet model with purple contours of equal potential).<ref>{{Cite journal|title=The Roche Problem: Some Analytics|first=Zakir F.|last=Seidov|date=1 March 2004|journal=The Astrophysical Journal|volume=603|issue=1|pages=283–284|doi=10.1086/381315|arxiv=astro-ph/0311272|bibcode=2004ApJ...603..283S|s2cid=16724058}}</ref><br />[[:File:Lagrangian points equipotential.gif|Click for animation.]] ]]

Lagrange points are the constant-pattern solutions of the restricted [[three-body problem]]. For example, given two massive bodies in orbits around their common [[barycenter]], there are five positions in space where a third body, of comparatively negligible [[mass]], could be placed so as to maintain its position relative to the two massive bodies. This occurs because the combined gravitational forces of the two massive bodies provide the exact centripetal force required to maintain the [[circular motion]] that matches their orbital motion.

Alternatively, when seen in a [[rotating reference frame]] that matches the [[angular velocity]] of the two co-orbiting bodies, at the Lagrange points the combined [[gravitational field]]s of two massive bodies balance the [[centrifugal force|centrifugal pseudo-force]], allowing the smaller third body to remain stationary (in this frame) with respect to the first two.

==={{L1|nolink=yes}}===
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The location of L<sub>1</sub> is the solution to the following equation, gravitation providing the centripetal force:
<math display="block">\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}</math>
where ''r'' is the distance of the L<sub>1</sub> point from the smaller object, ''R'' is the distance between the two main objects, and ''M''<sub>1</sub> and ''M''<sub>2</sub> are the masses of the large and small object, respectively. The quantity in parentheses on the right is the distance of L<sub>1</sub> from the center of mass. The solution for ''r'' is the only [[Real number|real]] [[Zero of a function|root]] of the following [[quintic function]]

<math display="block">x^5 + (\mu - 3) x^4 + (3 - 2\mu) x^3 - (\mu) x^2 + (2\mu) x - \mu = 0</math>
where <math display="block"> \mu = \frac{M_2}{M_1+M_2} </math>
is the mass fraction of ''M<sub>2</sub>'' and <math display="block"> x = \frac{r}{R} </math>
is the normalized distance. If the mass of the smaller object (''M''<sub>2</sub>) is much smaller than the mass of the larger object (''M''<sub>1</sub>) then {{L1|nolink=yes}} and {{L2|nolink=yes}} are at approximately equal distances ''r'' from the smaller object, equal to the radius of the [[Hill sphere]], given by:
<math display="block">r \approx R \sqrt[3]{\frac{\mu}{3}}</math>

We may also write this as:
<math display="block">\frac{M_2}{r^3}\approx 3\frac{M_1}{R^3}</math>
Since the [[Tidal force|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 L{{sub|1}} or at the L{{sub|2}} point is about three times of that body. We may also write:
<math display="block">\rho_2\left(\frac{d_2}{r}\right)^3\approx 3\rho_1\left(\frac{d_1}{R}\right)^3</math>
where ''ρ''{{sub|1}} and ''ρ''{{sub|2}} are the average densities of the two bodies and ''d''{{sub|1}} and ''d''{{sub|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 ''M''<sub>2</sub> in the absence of ''M''<sub>1</sub>, is that of ''M''<sub>2</sub> around ''M''<sub>1</sub>, divided by {{sqrt|3}} ≈ 1.73:
<math display="block">T_{s,M_2}(r) = \frac{T_{M_2,M_1}(R)}{\sqrt{3}}.</math>

==={{L2|nolink=yes}}===
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[[File:L2 rendering.jpg|thumb|right|upright=1.35|The Lagrangian L<sub>2</sub> point for the [[Sun]]–[[Earth]] system]]
The location of L<sub>2</sub> is the solution to the following equation, gravitation providing the centripetal force:
<math display="block">\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}</math>
with parameters defined as for the L<sub>1</sub> case. The corresponding quintic equation is
<math display="block">x^5 + x^4 (3 - \mu) + x^3 (3 - 2\mu) - x^2 (\mu) - x (2\mu) - \mu = 0</math>

Again, if the mass of the smaller object (''M''<sub>2</sub>) is much smaller than the mass of the larger object (''M''<sub>1</sub>) then L<sub>2</sub> is at approximately the radius of the [[Hill sphere]], given by:
<math display="block">r \approx R \sqrt[3]{\frac{\mu}{3}}</math>

The same remarks about tidal influence and apparent size apply as for the L{{sub|1}} point. For example, the angular radius of the Sun as viewed from L<sub>2</sub> is arcsin({{sfrac|{{val|695.5e3}}|{{val|151.1e6}}}})&nbsp;≈&nbsp;0.264°, whereas that of the Earth is arcsin({{sfrac|6371|{{val|1.5e6}}}})&nbsp;≈&nbsp;0.242°. Looking toward the Sun from L<sub>2</sub> one sees an [[annular eclipse]]. It is necessary for a spacecraft, like [[Gaia (spacecraft)|Gaia]], to follow a [[Lissajous orbit]] or a [[halo orbit]] around L<sub>2</sub> in order for its solar panels to get full sun.

===L<sub>3</sub>===
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The location of L<sub>3</sub> is the solution to the following equation, gravitation providing the centripetal force:
<math display="block">\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}</math>
with parameters ''M''<sub>1</sub>, ''M''<sub>2</sub>, and ''R'' defined as for the L<sub>1</sub> and L<sub>2</sub> cases, and ''r'' being defined such that the distance of L<sub>3</sub> from the center of the larger object is ''R''&nbsp;−&nbsp;''r''. If the mass of the smaller object (''M''<sub>2</sub>) is much smaller than the mass of the larger object (''M''<sub>1</sub>), then:<ref>{{Cite web|url=https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec18.pdf |title=Widnall, Lecture L18 - Exploring the Neighborhood: the Restricted Three-Body Problem}}</ref>

<math display="block">r\approx R\tfrac{7}{12}\mu.</math><!-- in the source the factor that appears in the equation is 5/12, but that is the distance from L3 to the center of mass, here we are showing the distance between L3 and the orbit of the smaller object -->

Thus the distance from L<sub>3</sub> to the larger object is less than the separation of the two objects (although the distance between L<sub>3</sub> and the barycentre is greater than the distance between the smaller object and the barycentre).

==={{L4|nolink=yes}} and {{L5|nolink=yes}}===
{{Further|Trojan (celestial body)}}

The reason these points are in balance is that at {{L4|nolink=yes}} and {{L5|nolink=yes}} 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 [[Barycentric coordinates (astronomy)|barycenter]] of the system.  Additionally, the geometry of the triangle ensures that the [[Parallelogram law|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 [[Dynamic equilibrium|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]].

===Radial acceleration===
[[File:Radial acceleration Earth-Moon Lagrangian.svg|thumb|upright=1.5|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:
<math display="block">a = -\frac{G M_1}{r^2}\sgn(r)+\frac{G M_2}{(R-r)^2}\sgn(R-r)+\frac{G\bigl((M_1+M_2) r-M_2 R\bigr)}{R^3}</math>
where ''r'' is the distance from the large body ''M''<sub>1</sub>, ''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 ''M''<sub>1</sub>; force from ''M''<sub>2</sub>; and centripetal force. The points L<sub>3</sub>, L<sub>1</sub>, L<sub>2</sub> 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.