Note--figures and the proof for sin(a + b) will be provided in a later version.
## (34b) The L4 and L5 Lagrangian PointsNote: This calculation is intended for users fluent in algebra and fairly familiar with trigonometry. It is longer, more tedious and a bit more intricate than other calculations in "Stargazers." If you plan to study it, a good idea may be to copy it and check it out on paper as you go along.additional station-keeping points of this sort also exist. One of them, L3, is on the Earth-Sun line but one the far side of the Sun, at about the same distance as the Earth. It has no practical use, because at that position, a calculation involving just the Earth and the Sun is a very poor approximation. The pull of other planets can exceed that of the Earth and cannot be ignored.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 Earth-Sun line. At those locations the two-body calculation based on the Earth and the Sun also predicts station-keeping (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
They have an important property (which will not be proved) that
Here we will show that L4 and L5 of the Earth-Moon system are positions of equilibrium
## Tools of the Calculation- We will need Newton's law of gravitation and the fact that the center of the Moon's orbit is only
**approximately**the center of the Earth. The actual center of the orbit is the**center of mass**(or "center of gravity") of the Earth-Moon system (see end of section 11).As shown in section 25, if **m**is the mass of the Moon and**M**that of the Earth, the center of mass is the point that divides the Earth-Moon line by a ratio**m:M**. Say A is the center of the Earth, B the one of the Moon, and**c**is the distance between the two (drawing). Then if D is the center of mass,AD = cm */*(m+M)DB = cM */*(m+M)and it is easy to check that the sum of these distances is c and their ratio is **m/M**. An alternative form of DB (which will be useful) is obtained by dividing numerator and denominator by**M**:DB = c */*(1 + m/M) - From trigonometry we will need the
**"law of sines"**. Suppose we are given a triangle ABC of arbitrary size and shape (drawing). The angles at the three corners will be named A, B and C as well, while the lengths of the sides facing them are denoted**a**,**b**and**c**. Then the law of sines sayssinA */*a = sinB*/*b = sinC*/*cLet us prove it for the angles A and B, by drawing from the third corner C a line perpendicular to the opposite side of the triangle. Let **h**be the length of that line. ThensinA = h */*b b sinA = hsinB = h */*a a sinB = h From that
b sinA = a sinB and dividing both sides by **ab**gives the required result. To prove that the angle C also fulfils the condition, we repeat the calculation with a perpendicular line drawn from A or from B. - We also need a trigonometrical identity for the sine of the sum of two angles. If those angles are denoted by the Greek letters
**a**and**b**sin(a + b) = sina cosb + cosa sinb The proof of this identity is given separately. - Finally, we will need the resolution of vectors (see sec. 14). Suppose a force
**F**acts on an object at some point C, making an angle a with the direction of a given line, marked here with**R**(drawing). Suppose also that we need to resolve**F**into components parallel and perpendicular to**R**. In the triangle CPQ, if CP represents the force**F**, then CQ and QP represent its parallel and perpendicular components. Then sincesin a = QC */*CPcos a = QP we get*/*CPparallel force = CQ = F sina perpen. force = QP = F cosa
## Conditions of EquilibriumTo the diagram drawn earlier to illustrate the center of mass of the Earth-Moon system, we add a spacecraft at some point C, with distancesb 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 (
The question to be answered is:
The calculation is best handled
## Let us collect equations--the ones which the distances and angles must obey.
(1) Note first that the
Denoting the rotational velocity of the Moon by
/ (1 + m/ M) = VT
/ T = v/ R/ T = (V/ c)(1 + m/ M)
The two expressions equal to
(1) v This merely expresses the well-known 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}/ [c/ (1 + m/ M)] = m(V^{2}/ c)(1 + m/ M)
/ c^{2} G is Newton's constant of gravitation, first measured by Henry Cavendish. In a circular orbit. the two must be equal, balancing each other (as in the calculation in section 20):
/ c^{2} = m(V^{2}/ c)(1 + m/ M) m) gives our second equation: / c
(2) GM
(3) Let m' be the mass of the spacecraft. The centrifugal force on it is
^{2}/ RF of the Earth and _{e}F of the Moon. However, only the components of those forces _{m}along the line R are effective in opposing the centrifugal force. Hence
^{2}/R = F_{m} cosb + F_{e} cosa Now by Newton's theory of gravitation _{m} = G m'm/ a^{2}
_{e} = G m'M/ b^{2}
Inserting these in the upper equation and dividing both sides by
(3) v
(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
_{m} sinb = F_{e} sina Gm' leaves
(4) (m/a Collecting all equations once more: -
v
*/*R = (V*/*c)(1 + m*/*M) - GM
*/*c = V^{2}(1 + m*/*M) - v
^{2}*/*R = (Gm*/*a^{2}) cosb + (GM*/*b^{2}) cosa - (m
*/*a^{2}) sinb = (M*/*b^{2}) sina
The quantities appearing here are of 3 types. - ---Some are known
**constants**--**G**,**m**and**M**. They have given values and we do not expect them to change. - --Some are
**distances**--**r**,**a**,**b**and**c**--having to do with the positions of the Earth, Moon and spacecraft in space. The angles (**a**,**b**) depend on those distances too, but we won't need the exact relationships for this. - --And some are
**velocities**, namely**v**and**V**.
Let us get eliminate the velocities, so that the conditions we are left with are purely geometrical, involving only distances and angles.
We already carried out an
From (1), squaring both sides
^{2}/ R^{2} = (V^{2}/ c^{2}) (1 + m/ M) ^{2}
Multiply both sides by (1 + m/ M)
^{2} (c^{2}/ R^{2})/ [1 + m/ M] = V^{2} (1 + m/ M)
But by (2) / c = V^{2} (1 + m/ M)
(5) v
and R
^{2}/ R = (GM/ c^{3}) R (1 + m/ M)
^{2}/R = (Gm/ a^{2}) cosb + (GM/ b^{2}) cosa Therefore (moving a factor of
/ c^{2}) (R/ c) (1 + m/ M) = (Gm/ a^{2}) cosb + (GM/ b^{2}) cosa Dividing everything by
(6) (1
(4) (m
In that case, if the above equations are multiplied by 1) and (/ b^{2}1) all disappear, leaving / c^{2}
(7) (R (8) m sinb = M sina
From here m Substitute this on the right side of (7)
/ c)(1 + m/ M) = sina cosb/ sinb + cosa
sin b (R = sin(a + b) = sin C = sin B
The last equality comes because all three angles (A,B,C) are equal. / [cM/ (M+m)] = sin B/ R That however is no more than the law of sines in the triangle CDB and is therefore obviously fulfilled. Working backwards from here, one can now show that all preceding equations, down to (1) through (4), are satisfied if ABC is equilateral. Therefore C is a point of equilibrium in the rotating frame. If we had As already noted, because L4 and L5 are stable points of equilibrium, they have been proposed for sites of large self-contained "space colonies", an idea developed and advocated by the late Gerald O'Neill. In 1978 Bill Higgins and Barry Gehm even wrote for would-be colonists "The L5 Song " to the tune of "Home on the Range. " Here is its beginning:
## Home on Lagrange-
Oh give me a locus
Where the gravitons focus Where the three-body problem is solved Where the microwaves play Down at 3 degrees K And the cold virus never evolved
CHORUS: |

**Next Stop: #35 To the Planets, to the Stars**

Author and curator: David P. Stern, u5dps@lepvax.gsfc.nasa.gov

Last updated 29 January 1999