Abstract. M 1 r 1 +M 2 r 2 = 0.
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1 Lagrangian Points Dennis Westra July 5, 07 Abstract In this text we shortly address some questions about the Lagrangian points. We show that there are five such points and then consider the stability of two of them. In finding their approximate positions we demonstrate a problem in perturbation theory and give a method to deal with a certain class of problems in approximations. To show stability we consider a powerful method that reduces the problem to calculating eigenvalues of a matrix. The three-body problem is not exactly solvable. If however the mass of the third object is really much smaller than the other two masses, we can ignore the backreaction of the third mass on the other two masses. In this setting the Lagrangian points are discussed: We have a configuration of two masses rotating in circular orbits around their center of mass. The Lagrangian points are those co-rotating points where an infinitesimal mass would experience no force. In these points one could leave a space-ship, without having to burn fuel to stay there. Let M and M be the masses of the two larger objects. Let r and r be the position vectors, where the origin is taken to be the center of mass. We thus have M r +M r = 0. We assume that the masses rotate in circular orbits. The angular velocity ω can be calculated using Kepler s law, or by considering the centripetal force and the gravitational force. In any case, one finds ω = G(M +M r r 3 = GM r r r, where G is Newton s gravitational constant and where the last equality uses M r + M r = 0. In addition to the choice of the origin, we make another choice: We consider a coordinate system that is co-rotating, thus one in which the masses are at fixed positions. Since this is a non-inertial frame, two additional forces show up: the
2 centrifugal force and the Coriolis force. The forces F m on a mass m at position r are now F m = GM m r r 3(r r GM m r r 3(r r +mω r mω ṙ, ( where ω is the angular velocity vector; ω is as above the square of its norm. For an explanation where the Coriolis force and the centrifugal forces come from, one can consult any classical mechanics textbook, or the hand-out For stationary points the Coriolis force is irrelevant, as we want points with zero velocity ṙ. We first consider those points that lie on the line through the masses M and M. We divide this line into three parts: Region is to the right of M, region 3 is between M and M and region is to the left of M see the sketch below, where CM stands for center of mass. We will in general not be able to solve exactly for the positons where F m = 0, but we will find good approximations if M >> M. M CM M Region Region 3 Region Region and region have the same equation for F m = 0, but with M and M interchanged. However, if we assume M << M the approximate equations are different. Region Coordinates on the line through M and M are given as follows: The center of mass ist at x = 0, mass M is at position x = r, mass M is at position x = r. We thus have r r = r +r. Our task is to find a position x such that all forces are balanced at this point, that is F m = 0. Balance of forces in the third region is determined by the following equation GM (x+r + GM (x r = G(M +M (r +r 3 x. This is a quintic equation, so that solving it by hand seems hopeless. We introduce the new parameter α = M M and z = x r and obtain the following equation (z +α(z +α = (+α z(z (z +α. (
3 We now want to use perturbation theory. The perturbation parameter is α. We would like to find a perturbative expansion in α for a solution. The zeroth order problem is obtained by setting α = 0. For α = 0 eqn.( reduces to (z = z 3 (z which is solved for by z =. Naive perturbation theory now would proceed as follows: expand the original equation to first order α and consider a solution of the form z = +λα and put it into the equation, discard all higher order terms α, α 3 and so on and then determine λ. We now show this does not work, why it does not work and how to proceed in this and similar cases. If we use the expansion we find that the function is to first in α given by (+α = α+3α 4α 3... P(z = (z +α(z +α (+α z(z (z +α P(z = (z ( z 3 +αz (+(z 3 +O(α. To find an approximate zero, we write z = +αλ and find 0 = α+o(α. The symbol O(α stands for terms that are at least quadratic in α. The variable λ has dropped out of the equation! The reason for this bad bevavior is that z = is not an ordinary zero of P(z, but a third order zero, indeed P(z = (z 3 (+z+z +O(α. To study this kind of problems, we consider the following problem: Let P(x;ǫ be a polynomial in x that is given as an expansion in ǫ: P(x;ǫ = P 0 (x+ǫp (x+ǫ P (x+... where all the P i (x are polynomials which is not really a necessary requierement yet. We assume P 0 (0 = 0 and want a perturbative expansion for this zero in ǫ. We thus want an expression x(ǫ = ǫx +ǫ x +... such that P(x(ǫ;ǫ = 0 for all orders 3
4 in ǫ. Using the first order Taylor expansion P 0 (ǫx = ǫx P (0, we find that the first order equation is ǫx P 0 (0+ǫP (0 = 0. Now we clearly see the structure; this equation can only be solved for x if x = 0 is an ordinary zero of P 0. If it is a double zero, then P 0 (0 = 0 and one cannot solve for x. LetusassumethatP 0 (0 = P 0 (0 = P 0 (0butP 0 (0 0. ThenP 0(x = 6 P 0 (0x3 + higher order terms. This makes clear that x 3 should be of order ǫ, and thus we can expand in powers of ǫ /3, thus x = ǫ /3 a+ǫ /3 b+... We thus consider the problem to find approximate solutions for the zero of P(x;ǫ = x 3 Q(x+ǫR(x+..., where Q(0 0 and the ellipsis contains terms of higher order in ǫ. Putting in x(ǫ = aǫ /3 +bǫ / one finds P(x(ǫ;ǫ = ǫ(a 3 Q(0+R(0+ǫ 4/3 (3a bq(0+a 3 Q (0+aR ( Requiring P(x(ǫ;ǫ = 0 to all orders in ǫ thus results in a = 3 R(0 Q(0. Now we return to our original problem and consider the transformation z = +y, so that the zero lies at y = 0. We then find P(y = y 3 (3+3y+y +α(+α (+y 3 +α (+y( y (+6y+3y +... Identifying Q(y = (3+3y+y and R(y = (+y (+y, we find Q(0 = 3 and R(0 =. Thus we have y(ǫ = ǫ /3 3 ǫ = Retracing back the definitions we find In the literature this point is called L. x = r ( + 3 M 3M. Now let us plug in some numbers. For the system earth-sun we have M = M sun 0 30 kg, M = M earth kg and r km. Thus x r km. That is, L lies at.5 Million km from the earth, away from the sun. Region We let x be the distance between the test mass and the center of mass. Mass M is r left from the center of mass, M is r right from the center of mass. Balance of 4
5 forces in region 3 is then given by the equation ω x = GM (x r + GM (x+r. (3 Again, a general solution is hopeless to find, so we assume M >> M, and thus r << r. We put x = zr, M = ǫm and thus r = ǫr. Together with ω = G(M +M (r +r 3 this turns eqn.(3 into (+ǫ z(z ǫ (z + = (z + +ǫ(z ǫ. (4 For convenience we multiply through by ( + ǫ. Then, to first order in ǫ eqn.(4 reduces to 0 = (z 3 (z + ǫ(z 4 +4z 3 +5z +4z +. (5 The relevant solution for ǫ = 0 is z =. Since z 3 = (z (+z+z we recast eqn.(5 in the form 0 = (z Q(z+ǫR(z. Now z = is an ordinary zero and we thus can savely take z(ǫ = +λǫ and find to first order λ = R(. In our case Q( = and R( = 7. Q( If we now retrace the definitions we find x = r (+ 7 M ; the distance between this point and the mass M is thus ( x r = r + 5 M M whereas the distance between M and M is r +r = r (+ M M. This Lagrangian point, called L 3 in the literature, is thus closer to M than M is to M. For the sun-earth system we find that L 3 is just some 600 km outside the orbit of the earth. M Region 3 Balance of forces is now given by GM (x+r GM (x r = G(M +M (r +r 3 x. (6 We put r +r = a, z = x a, s = r r +r, t = r r +r = s, so that M = s(m +M and M = t(m +M. This reduces eqn.(6 to s(z s t(z +t = z(z +t (z s. 5
6 If s = t = then indeed one finds a solution z = 0, i.e., precisely in between the two masses. If M >> M, then s tends to and t tends to 0. We thus put s = β and t = β and find the following equation for z up to first order in β 0 = (z 3 (+z +z +β(4z 4 6z 3 +4z 4z +3. This equation is solved perturbatively by z = + λβ /3 and one finds λ = 3. 3 Hence we have ( M x = a 3, 3M so that for the system sun-earth we find that this Lagrangian points lies approximately.5 Mio km from the earth in the direction of the sun. In the literature this point is called L. Other Lagrangian points Are there other Lagrangian points? It turns out, yes, but only two more, making thus a total of five. Balance of forces leads to an equation of the form α(r r +β(r r = r, where α and β depend on r through r r and r r. Rearranging one concludes that αr +βr has to be parallel to r. Since r and r lie on the line through M and M asolutionisonlypossibleifmisonthelinethroughm andm,orthecoefficients are chosen such that αr +βr = 0. The center of mass equation M r +M r = 0 gives the ratio between the lengths of r and r and thus αr +βr = 0 if and only if α : β = M : M. Looking at the precise form of α and β, α = GM ω r r, β = GM ω r r, we see that α : β = M : M is only satisfied if r r = r r. Let us consider this case and define s = r r = r r. Balance of forces is then given by M s 3 (r r + M s 3 (r r = M +M r r 3r. It follows straightforwardly that this equation is only satisfied if r r = s. Thus the final two Lagrangian points are those two points in the plane of rotation that make the three masses lie on the vertices of an equilateral triangle. In the literature these are called L 4 and L 5. 6
7 Stability of L 4 and L 5 The first free Lagrangian points are unstable. This means that if a space-ship is parked there a small deviation from the precise position will let the space-ship wander off from the Lagrangian point. The speed with which this happens differs from point to point, but a typical time-scale is of the order of years. A good discussion on these issues is found in a nice document by Neil Cornish. In fact, we will discuss the stability of L 4 and L 5 by the methods outlined in this document, but with a slightly worse notation perhaps. In order to study stability we slightly perturb the equations of motion around the solution. Since Newton s equations are second order equations, it is convenient to turn them into a system of first-order differential equations. This is done as follows: We first introduce the six-dimensional vector Z = (r,ṙ, we then have Ż = (z 4,z 5,z 6,F,F,F 3. Since the forces depend on positions and velocities the right-hand side is a function of components of Z. In the case at hand a few simplifications are possible; from physical grounds we know that the mass m will always be atracted to the plane of rotation and the Coriolis force does not work in this direction. We may thus simplify and consider motion in the plane of rotation. We thus take Z = (x,y,v x,v y. We have a solution Z 0 = ( a+x, 3 a,0,0 for L 4, say. Of course a = r r is the length of the sides of the equilateral triangle on whose corners the three masses are. The equations of motion are Ż = (v x,v y,a x,a y = G(Z, where G(Z is some vector whose components depend on those of Z, and we consider solutions of the form Z = Z 0 + δz, where δz = (δx,δy,δv x,δv y represents a small deviation from Z 0. Since G(Z 0 = 0 we have d dt δz = G(Z 0 +δz. The trick is now to expand G(Z 0 +δz to first order in δz. The resulting equation will then be of the form d δz = MδZ, (7 dt where M is some matrix. Stability is then guaranteed if all the eigenvalues of M have non-positive real part. Indeed, if λ = a + ib is some eigenvalue for an eigenvector 7
8 W and a > 0, then the vector We (a+ibt solves equation (7. Clearly this solution wanders away from the solution Z 0 on time-scales /a. Below we will perform these calculations for Z 0 = ( a+x, 3 a,0,0. We take M to be at position (x,y = ( M a M +M,0 and M to be at (x,y = ( M a M +M,0. The accelerations are given by and GM GM a x = ((x x +y 3/(x x ((x x +y 3/(x x +ω x+ωv y GM a y = ((x x +y 3/(y y GM ((x x +y 3/(y y +ω y ωv x. We now insert x = a+x +δx, y = 3 a+δy, v x = δv x, v y = δv y, and keep only terms linear in δx, δy, δv x and δv y. A useful expansion is r r 3 = r 0 r 3 3(r 0 r (r r 0 r r 5 +O( r r 0. We however stipulate an even easier way. We write the accelerations as a x = x U(x,y+ωv y, a y = y U(x,y ωv x, with U(x,y = GM + GN + r r r r ω r, which plays the role of the negative of a potential. At the point Z 0 the gradient of U(x,y vanishes. Writing U xx = U(x,y, x U xy = U(x,y, U x y yy = U(x,y evaluated at the point (x,y = ( a+x y, 3 a the matrix M is found to be M = U xx U xy 0 ω. U xy U yy ω 0 Thus we only have to calculate the derivatives of U, evaluate at the solution point, and then calculate the eigenvalues of M. One finds that U xx = 3 4 ω, U xy = 3 4 τ, U yy = 9 4 ω, where τ = G(M M a 3, where we already anticipated that 3 M > M. The equation det(m λ = 0 then becomes λ 4 +ω λ ω4 9 6 τ4, 8
9 whichisquiteremarkable, sincenooddpowersofλappear. Forus, thisisconvenient, as we can now solve exactly by hand. First we remark that 7 6 ω4 9 6 τ4 = 7 4 G M M a 6. Then we calculate the discriminant as D = G a 6 (M +M 5M M, which is smaller than ω 4. We have for λ the expression λ = ω ± D. If D > 0, then λ < 0 since D < ω and we thus have a total of four imaginary eigenvalues λ, which implies stability. If D = 0, then we have two imaginary eigenvalues and thus also stability. If however D < 0, then λ aquires an imaginary part. Thus λ = ρe ±iϕ with π < ϕ < π. Thus λ = ± ρe ±iϕ/ and at least one of them has a positive real part and in this case we thus find an unstable direction. Since the zeros of t 5t + are at t = 5 ( and at t = /t we conclude that L 4 and L 5 are stable if the larger mass M is at least t times as large as the smaller mass M. Indeed, we have D = G M (( M M + 5 a 6 M M which is positive if M M > t or M M < t. Since t < the last inequality cannot be satisfied if M is larger than M, hence we need the first, and we have thus proven stability for large enough M, i.e. M > 5M ( M. For the earth-sun system this is satisfied, and even for the system earth-moon it holds. The stability of the Trojan asteroids in between Mars and Jupiter is explained this way: At the points L 4 and L 5 in the system sun-jupiter little rocks that fly around our solar system accumulate. In these points they can stay for a long time and small perturbations will let them tend back to their position. 9
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