
Index
28.Spaceflight  29. Spacecraft (and 5 more)  30.To Space by Cannon? 30a.Project HARP 31.Nuclear Spaceflight? 32. Solar Sails 32a. Early Warning of Solar Shocks 33. Ion Rockets 34.Orbits in Space 34a. L1 Lagrangian pt. 34b. L4/L5 Points (1) 34c.L4/L5 Points (2) 35. Gravity Assist 36. Pelton Turbine Afterword 
A different derivationshorter, more elegant, more general but using vectors and working in a rotating frame of referencecan be found in section (34c). The other two Lagrangian points, L4 and L5, are on the Earth's orbit, with the lines linking them to the Sun making 60° angles with the EarthSun line. At those locations the twobody calculation based on the Earth and the Sun also predicts stationkeeping (that is, equilibrium in a frame of reference rotating with the Earth). Again, however, L4 and L5 are so distant that for a realistic calculation of the motion of a spacecraft near them, the pull of other planets must be included. However, the EarthMoon system also has its L4 and L5 points, and these have received some attention as possible sites for observatories and for self contained "space colonies." They have an important property (which will not be proved) that they are stable. In contrast, equilibrium at the L1 and L2 points is unstable, like that of a marble perched atop a bowling ball. 
If positioned exactly on the top, the marble will stay in place, but the slightest push will make it move further and still further from equilibrium, until it falls off. By contrast, the equilibrium at L4 or L5 are like that of a marble at the bottom of a spherical bowl: given a slight push, it rolls back again. Thus the spacecraft at L4 or L5 do not tend to wander off, unlike those at L1 and L2 which require small onboard rockets to nudge them back into place from time to time. Here we will show that L4 and L5 of the EarthMoon system are positions of equilibrium in a frame of reference rotating with the Moon, assuming that the Moon's orbit is circular. Noncircular orbits and the question of stability are beyond the scope of this discussion.
Tools of the Calculation
Conditions of EquilibriumTo the diagram drawn earlier to illustrate the center of mass of the EarthMoon system, we add a spacecraft at some point C, with distances b from the Earth, a from the Moon and R from the center of mass D. As in the derivation of the law of sines, we name (A,B,C) the angles at the corner points marked with those letter, and (a,b,c) will be the lengths of the sides facing the corners (A,B,C).We furthermore label as (α,β) the two parts into which R divides the angle C. Check all these out before continuing. The question to be answered is: Under what conditions does the satellite at C maintain a fixed position relative to the Earth and Moon? The calculation is best handled in the frame rotating with the Moon. In that frame, if a satellite at point C is in equilibrium, it will always keep the same distance from the Moon and from Earth. The center of rotation is the point Deven the Earth rotates around itand if the spacecraft at C is in equilibrium, all three bodies have the same orbital period T. If C is motionless in the rotating frame, there exists no Coriolis force (it only acts on objects moving in that frame), but the spacecraft will sense a centrifugal force, as will the Moon and the Earth.
Let us collect equationsthe ones which the distances and angles must obey.
(1) Note first that the radius of rotation R of the spacecraft will Denoting the rotational velocity of the Moon by V and that of the spacecraft by v, since distance = velocity x time
The two expressions equal to 2π/ T must also be equal to each other, hence (1) v/R = (V/c)(1 + m/M) This merely expresses the wellknown observation that if two objects share a rotation, the one more distant from the axis rotates faster, and their velocities are proportional to their distances from the axis.
(2) The centrifugal force on the Moon is
(2) GM/c = V^{2} (1 + m/M)
(3) Let m' be the mass of the spacecraft. The centrifugal force on it is
Now by Newton's theory of gravitation
Inserting these in the upper equation and dividing both sides by m' gives the 3rd equation: (3) v^{2}/R = (Gm/a^{2}) cosβ ;+ (GM/b^{2}) cosα
(4) Finally, the forces pulling the spacecraft in directions perpendicular to R must cancel. Otherwise, the spacecraft would be pulled by the stronger of the two and would not stay at C, that is, would no longer be in equilibrium. That requires
(4) (m/a^{2}) sinβ = (M/b^{2}) sinα
Collecting all equations once more:
The quantities appearing here are of 3 types.
We already carried out an elimination earlier. We had two equations which involved the orbital period T, each was used to express 2π/T, and by setting those two expressions equal to each other, we obtained a single expression which did not contain T (we always "give up one equation" in an elimination processstart with two, end with one). The plan then is as follows. We will eliminate V between (1) and (2), leaving an equation involving only v. Then we will eliminate v between it and (3), winding up with an equation not involving velocitiesplus (4), which also contains neither v nor V. From (1), squaring both sides
Multiply both sides by c^{2}and divide them by (1 + m/M)
But by (2) (5) v^{2} (c^{2}/R^{2}) /[1 + m/M] = GM/c and V has just been eliminated. Now multiply both sides by (1 + m/M), divide them by c^{2} and multiply them by R
Therefore (moving a factor of 1/c to R)
Dividing everything by GM gives one of the equations we are left with, while the other one is (4): (6) (1/c^{2}) (R/c) (1 + m/M) = (1/a^{2})(m/M) cosβ+ (1/b^{2}) cos α
(4) (m/a^{2}) sinβ = (M/b^{2}) sinα
Let R_{1} = BD be the distance from the Moon to the center of gravity (or center of mass) point D, which stays at rest in the EarthMoon system (see section #25); it is just a little less than the EarthMoon distance AB = c. As noted in the drawing So (6) becomes Substituting from (4) and cancelling a factor a^{2} along the way = (1/b^{2}sinβ) [sinα cosβ + cosα sinβ] = (1/b^{2}sinβ) sin(α + β) = (1/b^{2}sinβ) sin C (9) By the law of sines in the triangle BCD But from the law of sines in the triangle ABC An important fact of this calculation is that neither m nor M appear in the final result. We can therefore revise our notation, making M the mass of the Moon and m the mass of the Earth. (The point D in the diagram would be shifted, but it is inaccurate anyway, actually located below the Earth's surface). In the revised scheme, b stands for the distance from the Moon to the spacecraft, originally designated "a". The calculation now shows that the spacecraftMoon distance also equals the EarthMoon distance c. It follows that ABC is an equilateral triangle.
As already noted, because L4 and L5 are stable points of equilibrium, they have been proposed for sites of large selfcontained "space colonies", an idea developed and advocated by the late Gerald O'Neill. In 1978 Bill Higgins and Barry Gehm even wrote for wouldbe colonists "The L5 Song " to the tune of "Home on the Range. " Here is its beginning:
Home on Lagrange
Where the gravitons focus Where the threebody problem is solved Where the microwaves play Down at 3 degrees K And the cold virus never evolved
CHORUS:
Exploring Further:About space colonies at Lagrangian points:
About the L4 and L5 points and about asteroids locked into the neighborhoods of L4 and L5 of the SunJupiter system: "When Trojans and Greeks Collide" by I. Vorobyov, Quantum, SeptemberOctober 1999, p. 1619. That article contains an alternative proof of the equilibrium of motion at L4 and L5, more general (no limitation on the masses) but using a rotating frame of reference and twodimensional vectors. The calculation can be found in section (34c) of this web site. 
Optional: #34c Another derivation of the L4 and L5 Points (see paragraph above). Next Stop: #35 To the Planets, to the Stars Timeline Glossary Back to the Master List
Author and Curator: Dr. David P. Stern 
Curators: Robert Candey, Alex Young, Tamara Kovalick
NASA Privacy, Security, Notices