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Sect. 3.12: The Three Body Problem

Sect. 3.12: The Three Body Problem. So far: We’ve done the 2 Body Problem . Central forces only. Eqtns of motion are integrable. Can write as integrals. For most forces, must evaluate numerically. Solvable in closed form for r -2 central forces.

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Sect. 3.12: The Three Body Problem

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  1. Sect. 3.12: The Three Body Problem • So far: We’ve done the 2 Body Problem. • Central forces only. • Eqtns of motion are integrable. • Can write as integrals. For most forces, must evaluate numerically. • Solvable in closed form for r-2 central forces. • Solvable in closed form for some power law forces:V(r) = k rn+1 • Now: Add a 3rd mass 3 Body Problem • Even for r-2 central forces (3 body Kepler Problem), there is no known (closed form!) solution. • That is, cannot integrate the eqtns of motion. Cannot even easily express in terms of integrals that could be evaluated numerically! • Here: Set up the problem & discuss some methods of solution in some simple cases. Some approximations. Some qualitative discussion.

  2. Newtonian 3 body problem: Masses, m1, m2, m3at positions r1, r2, r3 position vectors in the Center of Mass system. • Each of the 3 masses is attracted to the 2 others with the gravitational force given by Newton’s Law of Gravitation. The eqtn of motion for each is given by Newton’s 2nd Law & equating mass  acceleration to the gravitational force. • For example, eqtn of motion for m1: m1r1 = -[G m1m2(r1 - r2)]/|r1 - r2|3-[G m1m3(r1 - r3)]/|r1 - r3|3  r1 = -[Gm2(r1 - r2)]/|r1 - r2|3 -[Gm3(r1 - r3)]/|r1 - r3|3 • Similar, of course, for m2, m3

  3. Define relative position vectors:si  rj - rk • i, j, k = 1, 2, 3 &cyclic permutations. See figure. • xi’s in figure should be ri’s • From the figure, s1 + s2 + s3 = 0 • Algebrato rewrite eqtns of motion in terms of si’s

  4. In this notation, the eqtns of motion are symmetrical: si = -mG[si/(si)3] + miG (i = 1, 2, 3) G  G[s1/(s1)3+ s2/(s2)3 + s3/(s3)3] m  m1 + m2 + m3 • These 3 eqtns cannot be solved in general. Can, however, be solved in some simple special cases.

  5. Euler’s Solution • Euler’s solution(Euler’s special case):Assume: m2 always lies on the straight line between m1 & m3.  r1, r2, r3, s1, s2, s3 & Gare all collinear with each other. • Figure: Euler’s negative energy (bound orbit) solution in special case where mass ratios are: m1: m2 : m3 = 1:2:3 • Results: Masses move on confocal ellipses with the same period τ. During each τ, they pass through the perihelion (on the mutual axis of ellipses & close together) & aphelion (on same axis & far apart).

  6. Lagrange’s Solution • Special Case:G  G[s1/(s1)3+ s2/(s2)3 + s3/(s3)3] = 0  Eqtns of motion decouple to 3 independent eqtns of form:si = -mG[si/(si)3] Geometrically, this decoupling occurs when the3 masses are on vertices of an equilateral triangle. As motion continues, eqtns remain uncoupled. Masses stay on equilateral triangle forever. Triangle’s orientation changes with time to make this possible. Figure:Lagrange’s elliptic solution with mass ratios:m1: m2 : m3 = 1:2:3 Solution:Each mass moves in the same elliptical orbit (in the same plane, with the same focal point & period).

  7. Other special cases: • Asymptotic solutions: • For total E > 0, the 3 masses can move away from each other. Or one can escape, leaving the other 2 behind in an elliptic orbit. • For total E < 0, one mass can escape, leaving the other 2 behind in an elliptic orbit, or the 3 can have mutually bound orbits.

  8. “Restricted” 3 body problem:2 masses are large and in bound (almost) elliptic orbits with each other & the third mass is very small compared to the other 2. This small mass merely slightly perturbs the orbits of the 2 large masses. Examples: • Spacecraft in orbit between the Earth & the Moon. First Approximation: Assume the Earth & the Moon are in unperturbed orbits. The spacecraft interacts with them through the gravitational force. Use perturbation theory to treat effect of this on the orbits. • Perturbation of the Sun on the Moon’s orbit around the Earth • Others

  9. “Restricted” 3 body problem: • Spacecraft interacting with both Earth & Moon: Complicated by the distribution of gravitational PE near Earth-Moon system. Close to Earth, the net force is towards Earth. Close to the Moon, the net force is towards the Moon. Equipotential surfaces (curves of const PE) are closed curves around Earth (m1) & Moon (m2). Figure:

  10. Equipotential surfaces around Earth (m1) & Moon (m2). At some point  The “Lagrange point”L2, along line between E & M attraction of the spacecraft to E is = & opposite to its attraction to M  Net force = 0 at L2. Or: L2 = point of local PE minimum. Mathematically, L2 =“saddlepoint” because PE = min there along axis direction but decreases further in directions  E-M axis. 2 other “Lagrange points” L1 & L3 : Also “saddlepoints” along E-M axis. L1 & L3 : Transition points between orbits around E & orbits around M. 2 more at L4 & L5 : Local minima in PE. Spacecraft near L4 & L5 are attracted to them & can orbit elliptically around them!

  11. To understand this further, consider the Kepler solution to orbits for 2 massive bodies (m1, m2) in the CM frame. Ask: Are there locations where a small “test mass” m (<< m1, m2) will remain at rest (relative to m1, m2)? • Simple special case of this: m1& m2 are in CIRCULAR orbits about the CM. Lagrangian for m is then: L= (½)m(r2 + r2θ2) - V(r,θ,t) V(r,θ,t)  t & θ dependent potential m sees due to m1 & m2. Circular orbits  Radius vector r(NOT the same r as in the Lagrangian above!) between m1& m2 is const in length & rotates with const angular frequencyω(in the inertial frame). Transform to a coordinate system rotating at ω.

  12. m1& m2 inCIRCULARorbits about the CM. Lagrangian for m: L= (½)m(r2 + r2θ2) - V(r,θ,t) • Transform to frame rotating at ω, the angular rotation speed of m1& m2 in their circular orbits: Define: θ´ θ - ωt, ρ distance from the CM to m. Solve using cylindrical coordinates: ρ,θ,z (θ angle the line joining the 2 masses makes with ρ) NOTE: There are several typos in the text!!!!  L= (½)m[ρ2 + ρ2{(θ´)2- ω2} + z2] - V´(ρ,θ,z,t) Some manipulation gives: (There are several typos in the text!!!!) L= (½)m[ρ2 + ρ2(θ´)2 + z2] + (½)mρ2ω2 - V´(ρ,θ,z,t) (½)mρ2ω2 centrifugal energy, V´ gives the Coriolis effect

  13. L= (½)m[ρ2 + ρ2(θ´)2 + z2] + (½)mρ2ω2 - V´(ρ,θ,z,t)Procedure: Get Lagrange Eqtns. Look for solutions where ρ = θ = z = 0. Lagrange showed that there are only five of these: The Lagrange points: L1, L2,L3, L4,L5: Figure: • Look at stability: L2: Not stable against displacements along line between masses. For Sun- Earth system L2  location of 1990’s Solar and Heliospheric Observatory (SOHO). SOHO orbits L2 in plane  E-S axis. Still orbiting & transmitting data.

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