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Sect. 3.8: Motion in Time, Kepler Problem

Sect. 3.8: Motion in Time, Kepler Problem. We’ve seen: Orbital eqtn for r -2 force law is fairly straightforward. Not so, if want r(t) & θ (t).

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Sect. 3.8: Motion in Time, Kepler Problem

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  1. Sect. 3.8: Motion in Time, Kepler Problem • We’ve seen: Orbital eqtn for r-2 force law is fairly straightforward. Not so, if want r(t) & θ(t). • Earlier: Formal solution to Central Force problem. Requires evaluation of 2 integrals, which will give r(t) & θ(t) : (Given V(r) can do them, in principle.) t(r) = ∫dr({2/m}[E - V(r)] - [2(m2r2)])-½ (1) • Limits r0  r, r0determined by initial condition • Invert this to get r(t) & use that in θ(t) (below) θ(t) = (/m)∫(dt/[r2(t)]) + θ0 (2) • Limits 0  t, θ0determined by initial condition • Need 4 integration constants: E, , r0, θ0 • Most cases: (1), (2) can’t be done except numerically

  2. Look at (1) for 1/r2 force law: V(r) = -k/r. t(r) = (m/2)½∫dr[E + (k/r) - {2(m2r2)}]-½ (1´) • Limits r0  r, r0determined by initial condition • For θ(t), instead of applying (2) directly, go back to conservation of angular momentum:  = mr2θ = constant  dθ = (/mr2)dt or dt = (mr2/)dθ (3) • Put orbit eqtn results r(θ) into (3) & integrate: • We had [α/r(θ)] = 1 + e cos(θ - θ´) • With e = [ 1 + {2E2(mk2)}]½ • And 2α = [22(mk)]

  3.  (3) becomes: t(θ) = (3/mk2) ∫dθ[1 + e cos(θ - θ´)]-2 (4) • Limits θ0θ • We had: t(r) = (m/2)½∫dr[E + (k/r) - {2(m2r2)}]-½ (1´) • (1´) & (4) are not difficult integrals. Can express them in terms of elementary functions (tabulated!). • However, they are complicated. Also, inverting to give r(t) and θ(t) is non-trivial. This is especially true if one wants high precision, as is needed for comparison to astronomical observations!

  4. Time: Parabolic Orbit • Even though, of course, we’re primarily interested in elliptic orbits, its instructive to first evaluate (4) for a parabolic orbit: e = 1, E = 0 • From table of planetary properties from earlier: Halley’s Comet, e = 0.967  1  Orbit is  parabolic. Results we are about to get are  valid for it. • In this case, (4) becomes: t(θ) = (3/mk2)∫dθ[1 + cos(θ - θ´)]-2 (4´) • Limits θ0θ • Measure θfrom distance of closest approach (perihelion). That is θ = 0 at r = rmin  θ0 = θ´ = 0

  5. So, we want to evaluate: t(θ) = (3/mk2)∫dθ[1 + cosθ]-2 (5) (Limits 0 θ) • Use trig identity: 1 + cosθ = 2cos2(½θ) So: t(θ) = [(3)/(4mk2)]∫d θsec4 (½θ)(5´) (Limits 0 θ) • Change variables to x = tan(½θ) Gives: t(θ) = [(3)/(2mk2)]∫dx (1+x2)(5´´) (Limits 0  tan (½θ)) t(θ) = [(3)/(2mk2)][tan(½θ)+ (⅓)tan3(½θ)] (6)

  6. t(θ) = [(3)/(2mk2)][tan(½θ)+ (⅓)tan3(½θ)] (6) (-π < θ < π) To understand what the particle is doing at different times, look at orbit eqtn at the same time as (6): [α/r(θ)] = 1 + cosθ(7)  At t  -  (θ = - π), particle approaches from r   At t = 0 (θ = 0), particle is at perihelion r = rmin At t  +  (θ = + π), particle again approaches r   • (6) gives t = t(θ). To invert and get θ = θ(t): • Treat (6) as cubic eqtn in tan(½θ). Solve & compute arctan or tan-1 of result.  θ = θ(t) • To get r(t), substitute resulting θ = θ(t) into (7): [α /r(t)]= [α /r{θ(t)}] = 1 + cos[θ(t)] • Same if do integral & invert to get r(t) (E=0: parabola) t(r) = (m/2)½∫dr[(k/r) - {2(m2r2)}]-½

  7. Time: Elliptic Orbit • Elliptic orbit:[α/r(θ)] = 1 + e cosθ(θ´ = 0) (1) • With e = [1 + {2E2(mk2)}]½ • And 2α = [22(mk)] • Rewrite (2):r = [a(1- e2)]/[1 + e cosθ] (2) a  Semimajor axis  (α)/[1 - e2] = (k)/(2|E|) • Its convenient to define an auxiliary angle: ψ Eccentric Anomaly(elliptic orbits only!)  By definition: r  a(1 - e cosψ) (2´) • (2) & (2´) cosψ = (e + cosθ)/(1 + e cosθ) cosθ = (cos ψ -e)/(1- e cosψ) • ψgoes between 0 & 2π as θ goes between -π&π • Perihelion, rminoccurs at ψ = θ = 0. • Aphelion, rmaxoccurs at ψ = θ = π.

  8. Back to time for the elliptic orbit: t(r) = (m/2)½∫dr[E + (k/r) - {2(m2r2)}]-½ (3) • Limits r0  r, r0  rmin = perihelion distance • Rewrite (3) using eccentricity e [1 + {2E2(mk2)}]½ a  Semimajor axis  (α)/[1 - e2] = (k)/(2|E|), α [2(mk)]  (3) becomes: t(r) = -(m/2k)½∫rdr[r - (r2/2a) – (½)a(1-e2) ]-½ (3´) By definition: r  a(1 - e cosψ) (2´) Change integration variables from r to ψ  (3´)is: t(ψ) = (ma3/k)½∫dψ(1 - e cosψ) (4) Limits 0  ψ Given t(ψ), combine with (2´) to get t(r). Invert to get r(t).

  9. t(ψ) = (ma3/k)½∫dψ(1 - e cosψ) (4) Limits 0  ψ r  a(1 - e cosψ) or cosψ = (1-r/a)/e (2´) ψ = cos-1[(1-r/a)/e] (excludes e = 0!) • Convenient to defineω (k/ma3)½ • Will show ω frequency of revolution in orbit.  ωt(ψ) = ∫dψ(1 - e cosψ) (Limits 0  ψ) Integrates easily toωt(ψ) = ψ - e sinψ (4´) Note that: sin ψ = [1 - cos2ψ]½ • Combining, (4´) becomes: ωt(r) = cos-1[(1-r/a)/e] + [1 - {(1-r/a)/e}2]½(4´´) • Inverting this to get r(t) can only be done numerically!

  10. Time: Orbit Period • Back briefly ωt(ψ) = ∫dψ(1 - e cosψ) ω (k/ma3)½ • If integrate over full range, ψ= 0 to 2π  t  τ = period of orbit. • This gives, τ = 2π(m/k)½a3/2  2π/ω τ2 = [(4π2m)/(k)] a3 Same as earlier, of course! Kepler’s 3rd Law! • Clearly, ω frequency of revolution in orbit: ω 2π/τ

  11. Elliptic Orbit • Back to general problem: ωt(ψ) = ψ - e sinψ (4´) ω (k/ma3)½(4´) Kepler’s Equation • Recall also: r  a(1 - e cosψ)  [a(1- e2)]/[1 + e cosθ] • Terminology(left over from medieval astronomy): ψ “eccentric anomaly”. Medieval astronomers expected angular motion of planets to be constant (indep of time). That is, they expected circular orbits (r =a & e = 0 above).  Deviations from a circle were termed “anomalous”! For similar reasons θ “true anomaly”. Still use these terms today. From earlier table, eccentricities e forMOST planets are very small! Except for Mercury (e = 0.2056) & Pluto (e = 0.2484) all planet’s have e < 0.1. Several have e < 0.05. Earth (e = 0.017), Venus (e = 0.007), Neptune (e = 0.010)  Orbits are  circles & “anomalies” are small!

  12. Summary: Motion in time for elliptic orbits: ωt(ψ) = ψ - e sinψ, ω (k/ma3)½Kepler’s Equation • Also: r  a(1 - e cosψ)  [a(1- e2)]/[1 + e cosθ] • Combining (e  0): ωt(r) = cos-1[(1-r/a)/e] + [1 - {(1-r/a)/e}2]½ • Inverting this to get r(t) can only be done numerically! • Comparing 2 eqtns for r gives: cosψ = (e + cosθ)/(1 + e cosθ) orcosθ = (cosψ -e)/(1-e cosψ) Using some trig identities, this converts to: tan(½θ) = [(1-e)/(1+e)]½tan(½ψ) • Use this to get θ once ψ is known.

  13. tan(½θ) = [(1-e)/(1+e)]½tan(½ψ) Use this to getθonce ψis known. • Solving this & Kepler’s Equation ωt(ψ) = ψ - e sinψ, ω (k/ma3)½ to get ψ(t), θ(t) & r(t): A classic problem, first posed by Kepler. Many famous mathematicians worked on it, including Newton. To study motion of bodies in solar system & to understand observations on such bodies one needs this solution to very high accuracy! • Goldstein: “ The need to solve Kepler’s equation to accuracies of a second of arc over the whole range of eccentricities fathered many developments in numerical mathematics in the eighteenth and nineteenth centuries.” • More than 100 methods of solution have been developed! Some are in problems for the chapter! (# 2,3,25,27)

  14. Sect. 3.9: Laplace-Runge-Lenz Vector • Conserved Quantities(1st integrals of motion)in the Central Force Problem(& so in Kepler r-2 force problem): Total mechanical energy: E = (½)m(r2 + r2θ2) + V(r) = const = (½)mr2 + [2(2mr2)] + V(r) Total angular momentum: L = r  p = const(magnitude & direction!)  3 components or 2 components + magnitude: Constant magnitude    pθ mr2θ= const. • Can show: There is also another conserved vector quantity  Laplace-Runge-Lenz Vector, A

  15. Newton’s 2nd Law for a central force: (dp/dt) = p = f(r)(r/r) (1) • Cross product of p with angular momentum L: p  L = p  (r  p) = p  [r  (mr)] (1)  p  L = [mf(r)/r][r  (r  r)] Use a  (b  c) = b(ac) - c(ab)  p  L = [mf(r)/r][r(rr) - r2r] Note that rr = rr. L = constp  L = d(p  L)/dt Combining these gives: d(p  L)/dt = - [mf(r)r2][(r/r) - (rr/r2)] Or: d(p  L)/dt - [mf(r)r2][d(r/r)/dt] (2) • Valid for a general central force!

  16. General central force: d(p  L)/dt= - [mf(r)r2][d(r/r)/dt] (2) • Look at (2) in case of r-2 force: f(r) = -(k/r2)  - mf(r)r2 = mk • For r-2 forces, (2) becomes: d(p  L)/dt= [d(mkr/r)/dt] Or: d[(p  L) - (mkr/r)]/dt] = 0 (2´) • Define Laplace-Runge-Lenz Vector A A  (p  L) - (mkr/r) (3) (2´)  (dA/dt) = 0 or A = constant (conserved!)

  17. Summary: For r-2 Central Forces (Kepler problem) the Laplace-Runge-Lenz Vector A  (p  L) - (mkr/r) (3) is conserved(a constant, a 1st integral of the motion). • Question: That A is conserved is all well and good, but PHYSICALLYwhat is A? • What follows is more of aGEOMETRIC interpretation than a PHYSICAL interpretation. • By relating A to elliptic orbit geometry, perhaps the physics in it can be inferred.

  18. A  (p  L) - (mkr/r) Definition  AL = 0 (L is  to p  L;r is  L = r  p)  A = fixed (direction & magnitude) in the orbit plane.

  19. A  (p  L) - (mkr/r) AL = 0 A = fixed in orbit plane θ angle between r & fixed A direction  Ar = Ar cos θ = r(p  L) - mkr Identity:r(p  L) = L(r  p) = LL  2  Ar cosθ =2 - mkr Or: (1/r) = (mk/2) [1 + (A/mk)cosθ] (1) Identify θ as orbital angle: A is in perihelion direction (θ = 0, A || rmin) see diagram. (1)  Another way to derive that, for the Kepler Problem, orbit eqtn is a conic section!

  20. The Laplace-Runge-Lenz Vector A = (p  L) - (mkr/r) (1/r) = (mk/2) [1 + (A/mk) cosθ] (1) (1)  The orbit eqtn is a conic section. Also, A is in the perihelion direction. • Earlier, we wrote: (α/r) = 1 + e cosθ (2) α [2(mk)]; e  [ 1 + {2E2(mk2)}]½ • Comparison of (1) & (2) gives relation between A and the eccentricity e (& thus between A, energy E, & angular momentum ): A  mke = mk[ 1 + {2E2(mk2)}]½

  21. Physical interpretation of Laplace-Runge-Lenz Vector, A = (p  L) - (mkr/r) • Direction of A is the same as the perihelion direction: (A || rmin). • Magnitude of A: A  mke = mk[ 1 + {2E2(mk2)}]½ (3) • For the Kepler problem, we’ve found7 conserved quantities: 3 components of vector angular momentum, L 3 components of Laplace-Runge-Lenz Vector, A 1 scalar energy E

  22. A  mke = mk[ 1 + {2E2(mk2)}]½ (3) • 7 conserved quantities: • 3 components of L, 3 components of A, energy E • Recall original problem: 2 masses, 3 dimensions  6 degrees of freedom  6 independent constants of motion.  The 7 quantities aren’t independent. Reln between them is (3):  Reducing the number of independent ones to 6. • All 7 also are functions of r & p which describe the orbit in space. None relate to the initial conditions of the orbit (r(t=0)). Mathematically, one const of motion must contain such initial condition info. There must be a const (say time when r = rmin) indep of the 7 listed above  The 6 consts resulting from using (3) on the 7 cannot all be indep either! There must be one more reln between them. This is supplied by orthogonality of A & L: AL = 0

  23.  For Kepler (r-2 force) problem, we have 5 indep consts of motion containing orbital info (+ one containing initial condition info). Usually choose these as: 3 components of angular momentum L, energy E, and magnitude of Laplace-Runge-Lenz Vector, A • Question:Is there a similar conserved quantity to A for the general central force problem (or for specific central forces which are not r-2 forces)? • Answer: Yes, sometimes, but these usually have no simple interpretation physically. • Can show: such a quantity existsonly for force laws which lead to closed orbits.  Also: the existence of such a quantity is another means to show that the orbit is closed. Bertrand’s theorem: Happens for power laws forces only for f  r-2 & f  r (Hooke’s “Law”).

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