1 / 24

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).

garren
Télécharger la présentation

Sect. 3.8: Motion in Time, Kepler Problem

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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”).

More Related