in the plane containing the orbit (grey rubber-sheet model with purple contours of equal potential).
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 fields of two massive bodies balance the
centrifugal pseudo-force, allowing the smaller third body to remain stationary (in this frame) with respect to the first two. ====== The location of L1 is the solution to the following equation, gravitation providing the centripetal force: \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} 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. The solution for
r is the only
real root of the following
quintic function x^5 + (\mu - 3) x^4 + (3 - 2\mu) x^3 - (\mu) x^2 + (2\mu) x - \mu = 0 where \mu = \frac{M_2}{M_1+M_2} is the mass fraction of
M2 and x = \frac{r}{R} is the normalized distance. If the mass of the smaller object (
M2) is much smaller than the mass of the larger object (
M1) then and are at approximately equal distances
r from the smaller object, equal to the radius of the
Hill sphere, given by: r \approx R \sqrt[3]{\frac{\mu}{3}} We may also write this as: \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 L or at the L point is about three times of that body. We may also write: \rho_2\left(\frac{d_2}{r}\right)^3\approx 3\rho_1\left(\frac{d_1}{R}\right)^3 where
ρ and
ρ are the average densities of the two bodies and
d and
d 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 ≈ 1.73: T_{s,M_2}(r) = \frac{T_{M_2,M_1}(R)}{\sqrt{3}}. ====== –
Earth system The location of L2 is the solution to the following equation, gravitation providing the centripetal force: \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. The corresponding quintic equation is x^5 + x^4 (3 - \mu) + x^3 (3 - 2\mu) - x^2 (\mu) - x (2\mu) - \mu = 0 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: r \approx R \sqrt[3]{\frac{\mu}{3}} The same remarks about tidal influence and apparent size apply as for the L point. For example, the angular radius of the Sun as viewed from L2 is arcsin() ≈ 0.264°, whereas that of the Earth is arcsin() ≈ 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.
L3 The location of L3 is the solution to the following equation, gravitation providing the centripetal force: \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 being defined such that the distance of L3 from the center of the larger object is
R −
r. If the mass of the smaller object (
M2) is much smaller than the mass of the larger object (
M1), then: r\approx R(1-\tfrac{5}{12}\mu).3 to the center of mass, here we are showing the distance between L3 and the orbit of the smaller object --> Thus the distance from L3 to the larger object is less than the separation of the two objects (although the distance between L3 and the barycentre is greater than the distance between the smaller object and the barycentre).
and The reason these points are in balance is that at and 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.
Radial acceleration The radial acceleration
a of an object in orbit at a point along the line passing through both bodies is given by: 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} 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. ==Stability==