The Newtonian Three-Body Problem

The movement of two bodies under the influence of mutual gravitation is amenable to a thoroughly detailed solution. It is natural enough, then, to consider the case of three or more bodies' movement under their mutual gravitational attraction. However, even the three-body problem has been found to be intractable in general; it is, indeed, one of the classic examples of an intractable problem seemingly only slightly more complex than a problem whose solution is entirely straight-forward. All the same, there are some tractable special cases. In this page I intend to explore some of these.

Stable points near one body in orbit about another

In the two-body problem, it proves to be possible to identify some positions, near the mutually orbiting bodies, at which a much smaller body's position will remain fixed in relation to the orbiting bodies. There are five positions, known as the Lagrange points, at which a third body will remain stably in an orbit controlled by two others; when one of these two is sufficiently smaller than the other, two of these positions - a sixth of a turn ahead of and behind this secondary body in its orbit about the primary - are actually stable; and two more are meta-stable (that is, it only takes a little effort to keep a body near to the meta-stable point - NASA's SOHO station occupies one of these for an ideal vantage point on the Sun).

Consider, then, a pair of bodies in mutual elliptical orbit with small eccentricity (i.e. almost in mutual circular orbit). Deploy a uniformly rotating frame of reference centred on their mutual centre of gravity with period of rotation equal to the mutual orbital period. In this frame, we can describe the motion of our original two bodies in terms of centrifugal force balancing gravity's centripetal force, with a Corriolis effect perturbing their remaining residual motion. We can likewise describe a pseudo-gravitational potential, comprised of a difference between a potential for the centrifugal force and a sum of the two bodies' gravitational potentials. The third body's motion will then be controlled by that potential, though perturbed by Coriolis effects similar to those affecting the mutually orbiting bodies.

eh - ignore most of the following. convert to cocentric's formalism, possibly with the help of a Lagrangian ... Consider a very large compact (i.e. functionally point-like) body of mass M being orbitted by a smaller compact body of mass m, at constant radius R with angular velocity w = (G*M/R)**.5 / R We can use G*M = w*w*R*R*R to eliminate G from the system. Work in the rotating frame centred on M with x axis through m, y axis so chosen that w is positive and z axis chosen right-handedly relative to these; denote positions by their [x,y,z] list of co-ordinates. Thus M is at [0,0,0], m at [R,0,0]. We want to find stable stationary (in the chosen frame of reference) points of the gravitational field; i.e. where a body tiny compared even to m would remain if placed there. First, solve for the stationary points, then examine each for its stability. Introduce a vector W = [0,0,w] in the direction of the rotation axis so that a body following trajectory (: r = [x,y,z] ←t :) in our given co-ordinates has acceleration a = r'' +2*W^r' +W^(W^r) which we can equate to the force on it, per unit mass, which is the sum of gravitational attractions (per unit mass) due to M and m; a = -w*w*(q(r) +q(r-[R,0,0])*m/M) where q = (: r * R**3 / |r|**3 ←r :) using |r| to denote the usual length of r. If r has non-zero z component, then q(r) +q(r-[R,0,0])*m/M will have a non-zero z component. However, for a stationary point (with r' and r'' negligible) a = W^(W^r), which has no z component; thus we must have zero z component. Cancelling the factor of -w*w, we are thus left with r = q(r) +q(r-[R,0,0])*m/M Define [u,v] = [x/R, y/R] so that, when r = [R*u,R*v,0], q(r) = R*[u,v,0] / (u*u +v*v)**3/2 to re-express the equation as [u,v] * (1 - 1 / (u*u+v*v)**1.5) = [u-1,v] *m/M / (u*u -2*u +1 +v*v)**1.5 The case v = 0 yields u - 1/u/u = m/M/(u*u -2*u +1) or u*u*m/M = (u*u -2*u +1) * (u*u*u -1) = u**5 -2*u**4 +u**3 -u*u +2*u -1 so introduce f(u) = u**5 -2*u**4 +u**3 -(1+m/M)*u*u +2*u -1 f'(u) = 5*u**4 -8*u**3 +3*u**2 -2*(1+m/M)*u +2 Both f(1) and f'(1) are of order m/M, i.e. tiny, so let's consider u = 1+e for some e (presumed small, though e might not be as tiny as m/M) and suppose some such e gives f(1+e) = 0; then 0 = (1+e)**5 -2*(1+e)**4 +(1+e)**3 -(1+m/M)*(1+2*e+e*e) +2+2*e -1 = 1 -2 +1 -1 -m/M +2 -1 +e*(5 -8 +3 -2 -2*m/M +2) +e*e*(10 -12 +3 -1 -m/M) +e*e*e*(10 -8 +1) +e*e*e*e*(5 -2) +e**5 = (e**3 -m/M)*(1+e)**2 +e**4 so we can expect a root with e just less than the cube root of m/M. Let c be the cube root of m/M and e = c+a, yielding: 0 = (3*c*c*a +3*c*a*a +a*a*a)*(1 +2*c +2*a +c*c +2*c*a +a*a) +c*m/M +4*a*m/M +6*a*a*c*c +4*a*a*a*c +a**4 = (3*c*c*a +3*c*a*a +a*a*a)*(1 +2*c +2*a +c*c +2*c*a +a*a) +c*m/M +4*a*m/M +6*a*a*c*c +4*a*a*a*c +a**4 For Earth orbitting Sun: m/M = 3.003e-6 whose cube root is 14.428e-3.

Sun, Planet, Moon

The Sun is a third of a million times as massive as the Earth, which is about 81 times as massive as the Moon. The Earth is nearly 400 times as far from the Sun as from the Moon. The Sun is about a thousand times as massive as Jupiter; again, it is much closer to its moons than to the Sun. The huge variations in scale make it possible to work out an approximately correct answer - by ignoring much smaller quantities in the context of much larger ones - and then refining it by treating the ignored quantities as small perturbations.

Written by Eddy.
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