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Conservation Laws and Orbits: Solving the 2-Body Problem

This summary explains the process of solving the 2-body problem with a central force. It discusses the use of conservation laws and provides equations of motion for analyzing orbits in the r-θ plane. The importance of the Lagrangian and conservation of energy is also highlighted.

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Conservation Laws and Orbits: Solving the 2-Body Problem

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  1. Ch. 8: Summary So Far • We’re doing the “2 body”, conservative central forceproblem! • 2 bodies (m1& m2) with a central force directed along the line between them F(r) = -U(r) = -(dU/dr)r • Started with a 6d, 2 body problem. By transforming to CM & relative coordinates (R,r) reduced it to 2, 3d 1 body problems. One (the CM motion, R) is trivial & free particle-like. The 2nd is for a “particle” with mass μ (m1m2)(m1+m2) = reduced mass interacting with F(r). • Lagrangian for relative motion: L  (½)μ|v|2 - U(r) • Angular momentum conservation: • First, constant DIRECTION, effectively reduced the 3d relative coordinate problem (relative motion) from 3d to 2d, motion in a plane ( L)! Then, constant MAGNITUDE (  μr2θ= constant), reduced the 2d problem of motion in a plane from problem with 2 degrees of freedom to a problem with one generalized coordinate = r .

  2. The Lagrangian is: L= (½)μr2 + [2(2μr2)] - U(r) • We also used conservation of angular momentum to prove Kepler’s 2nd Law (areal velocity is constant): The radius vector from the force center sweeps out equal areas in equal times. Valid for any central force, not just gravitation! • The total mechanical energy is conserved since the central force is conservative: E = T + U = constant E = (½)μ(r2 + r2θ2) + U(r) = const • Using   μr2θ=constant: E = (½)μr2 + [2(2μr2)] + U(r) = const

  3. Equations of MotionSection 8.4 E = (½)μr2 + [2(2μr2)] + U(r) = const (1) • Conservation of energy allows us to get solutions to the equations of motion in terms of both r(t) and r(θ) or θ(r)The orbit of the particle(s)! • Equation of motion to get r(t): One degree of freedom  Very similar to 1d problem discussed in earlier chapters! • Solve for r = (dr/dt) :r = ({2/μ}[E - U(r)] - [2(μ2r2)])½ • Note: This gives r(r), the phase diagram for the relative coordinate & velocity. We can qualitatively analyze the (r part of the) motion using it, just as we did in 1d in Ch. 2. • Solve for dt & integrate to get t(r). In principle, then invert this to get r(t).

  4. r = ({2/μ}[E - U(r)] - [2(μ2r2)])½ • Solve for dt & integrate to get t(r). t(r) = ∫dr({2/μ}[E - U(r)] - [2(μ2r2)])-½ • Note the square root in the denominator! • Often, we are much more interested in the path in the r - θ plane: r(θ) or θ(r)The Orbit. • Note the chain rule: (dθ/dr) = (dθ/dt)(dt/dr) = (dθ/dt)(dr/dt) Or: (dθ/dr) = (θ/r) Also,   μr2θ= const  θ = [/(μr2)]  (dθ/dr) =  [/(μr2)] ({2/μ}[E - U(r)] - [2(μ2r2)])-½ Or: (dθ/dr) =  (/r2)(2μ)-½[E - U(r) -{2(2μr2)}]-½ • Integrating this will (in principle) give the orbit θ(r) .

  5. Integrating (dθ/dr) formally gives θ(r) = ∫(/r2)(2μ)-½[E - U(r) - {2(2μr2)}]-½dr • Once the central force is specified, we know U(r) & can, in principle, do the integral & get the orbitθ(r), or, (if this can be inverted!)r(θ). This is quite remarkable! Assuming only a central force law & nothing else: We have reduced the original 6d problem of 2 particles to a 2d problem with only 1 variable (r). The solution can (in principle) be obtained simply by doing the above (1d) integral!

  6. Orbits • The general equation for the orbit (for any Central Potential U(r)) is: (2μ)½θ(r) = ∫ (r-2dr)[D(r)]-1 (1) where D(r)  [E - U(r) - {(2μ)-12r-2}]½ • In general, (1) must be evaluated numerically. This is true even for most power law forces of the form: F(r) = k rn ; U(r) = k rn+1 • For a few integer & fractional values of n, we can express (1) in terms of certain elliptic integrals.

  7. (2μ)½θ(r) = ∫ (r-2dr)[D(r)]-1 (1) D(r)  [E - U(r) - {(2μ)-12r-2}]½ • It can be proven (see the graduate text by Goldstein), that only forn = 1, -2, and -3 can (1) be integrated to give “simple” functions (like trigonometric functions). n = 1, F(r) = kr. The isotropic, 3d simple harmonic oscillator. n = -2, F(r) = kr-2. Inverse square law force: Gravitation, Coulomb. Treated in detail this chapter! • Other cases: Homework Problems!

  8. Recap:We’ve solved the problem for r(t) & for the orbit θ(r) or r(θ) using conservation laws exclusively: • We combined conservation of angular momentum with conservation of energy into a single result which gives the orbit θ(r) in terms of a SINGLE INTEGRAL! • Before doing this in detail for the r-2 force, it’s useful to first take another (equivalent, of course!) approach which will result in a differential equation for r(θ) • Back to the Lagrangian for the relative coordinate(before using conservation of angular momentum): L  (½)μ|v|2 - U(r) Or: L = (½)μ(r2 + r2θ2) - U(r)

  9. L = (½)μ(r2 + r2θ2) - U(r) • The Lagrange equation forr is: (∂L/∂r) - (d/dt)(∂L/∂r) = 0 (∂L/∂r) = μrθ2 - (dU/dr) (d/dt)(∂L/∂r) = μr; - (dU/dr) = F(r)  Differential Equation of Motionfor a “particle”, mass μsubject to the Central Force F(r): μ(r- rθ2) = F(r)(2) (2) is just Newton’s 2nd Law in plane polar coordinates!

  10. The Differential Equation of Motion: μ(r- rθ2) = F(r)(2) • Its most convenient to solve (2) by first making a change of variables: Let u  (1/r) = r-1  (du/dr) = - (1/r2) = - r-2 ( -u2) • We are more interested in the orbit θ(r) or r(θ) than in r(t). Manipulation (with repeated use of the chain rule!): (du/dθ) = (du/dr)(dr/dθ) = - r-2(dr/dθ) = - r-2(dr/dt)(dt/dθ) = - r-2(dr/dt)(dθ/dt) = - r-2(r/θ)  (du/dθ) = - r-2(r/θ) (3) • Conservation of angular momentum gives:   μr2θ= const  θ = [/(μr2)] (4)

  11. Combining (3) & (4):  (du/dθ) = -(μ/)r (5) • Similar manipulationfor: (d2u/dθ2) = (d/dθ)[-(μ/)r] = -(μ/)(dt/dθ)(dr/dt) = -(μ/)(r/θ) Again substituting θ = [/(μr2)]: (d2u/dθ2) = - (μ2/2)(r2 r) (6) • Solving (6) for r (& using u2 = r-2) :  r = - (2/μ2) u2(d2u/dθ2) (7)   μr2θ= const; θ = [/(μr2)] (but u = r-1)  rθ2 = (2/μ2) u3 (8) • The Equation of Motion is: μ(r - rθ2) = F(r)(2) • Putting (7) & (8) into (2) gives (on simplifying!): (d2u/dθ2) + u = - (μ/2)u-2F(u-1) (9)

  12. DIFFERENTIAL EQUATION for the ORBIT: (d2u/dθ2) + u = - (μ/2)u-2F(u-1) (9) With u = r-1. Alternatively, we could write: (d2[r-1]/dθ2) + r-1 = - (μ/2)r2F(r) (9) • Note! Because of the right hand side, (9) & (9) are nonlinear differential equations in general! The EXCEPTION is when F(r)  r-2 (Inverse Square Law), where the right side = const! • (9) or (9) could, in principle, be used to solve for the orbit r(θ) or θ(r) given the force law F(r). • The result, of course, will be the same as if the previous integral version of θ(r) is evaluated. Can show that the integral form for θ(r) is the solution to (9) & (9) .

  13. DIFFERENTIAL EQUATION for the ORBIT: (d2u/dθ2) + u = - (μ/2)u-2F(u-1) (9) With u = r-1. Alternatively, we could write: (d2[r-1]/dθ2) + r-1 = - (μ/2)r2F(r) (9) • Usually, rather than solve (9) or (9) for orbit, given F(r), it’s usually more convenient to use the integral formulation. • However, where (9) or (9) is the most useful is for THE INVERSE PROBLEM: Given a known orbit r(θ) or θ(r), determine the force law F(r) !!

  14. Examples 8.1, 8.2, 8.3 • 8.1: Find the force law for a central force field that allows a particle to move in a logarithmic spiral orbit given by r = k eαθ, where k and α are constants. • 8.2: Find r(t) and θ(t) for the same case as in Example 8.1 • 8.3: What is the total energy of the orbit of the previous two examples?

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