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Central Field Orbits Section 8.5

Central Field Orbits Section 8.5. We had, (for a GENERAL central potential!): Radial velocity r vs. relative coordinate r r(r) = (2 μ) -1 ([E - U(r)] - [  2  (2 μ r 2 )]) ½ Time vs. r t(r) =  (2 μ -1 ) ∫ dr ( [E - U(r)] - [  2  (2 μ r 2 )]) -½

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Central Field Orbits Section 8.5

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  1. Central Field OrbitsSection 8.5 • We had, (for a GENERAL central potential!): Radial velocity r vs. relative coordinate r r(r) = (2μ)-1([E - U(r)] - [2(2μr2)])½ Time vs. r t(r) =  (2μ-1)∫dr([E - U(r)] - [2(2μr2)])-½ Derivative of θ with respect to r (dθ/dr) =  (r-2)(2μ)-½[E - U(r) -{2(2μr2)}]-½ The Orbit θ(r) θ(r) = ∫(/r2)(2μ)-½[E - U(r) - {2(2μr2)}]-½dr

  2. All of these quantities have the common factor: ([E - U(r)] - [2(2μr2)])½(1) • (1) should remind you of 1d where we analyzed the particle motion for various E using an analogous expression & found turning points, oscillations, phase diagrams, etc. • It is similar to conservation of energy in 1d, which gives: (dx/dt) = x(x) =  [(2m-1)(E - U(x))]½ • We now qualitatively analyze the RADIALmotion for a “particle” of mass μ in the presence of a central potential U(r) by methods similar to those used in Sect. 2.6 for 1d motion. • Remember: The ACTUAL motion is 2 d!  We need to superimpose angular the motion (θ) on the results for the radial motion, to get the ORBIT θ(r) or r(θ).

  3. Look at the radial velocity vs. r: r(r) = (2μ)-1([E - U(r)] - [2(2μr2)])½ (a) We could use this to construct the r vs. r phase diagram! See Goldstein’s graduate mechanics text. • Or, conservation of energy: E = (½)μr2 + [2(2μr2)] + U(r) = const (b) • Define:Radial turning points Points where r = 0 (the particle stops moving in the r direction!) • Look at either (a) or (b) & find, at the radial turning points: [E - U(r)] - [2(2μr2)] = 0 (1)

  4. Turning Points [E - U(r)] - [2(2μr2)] = 0 (1) r(r) = (2μ)-1([E - U(r)] - [2(2μr2)])½ (2) • The solutions r to (1) are those r where (2) = 0 Radial Turning Points are called Apsidal Distances • Generally, (1) has 2 roots (the max & min r of the orbit) rmax & rmin  We know that rmax≥ r ≥ rmin The radial motion will beoscillatory between rmax & rmin. • For some combinations of E, U(r), ,(1) has only one root:  In this case, from (2)r = 0, for all t  r = constant  The orbit r(θ) will be CIRCULAR.

  5. Closed & Open Orbits [E - U(r)] - [2(2μr2)] = 0 (1) r(r) = (2μ)-1([E - U(r)] - [2(2μr2)])½ (2) • If the motion for a given U(r) is periodic in r, then  The orbit r(θ) will be closed.  That is, after a finite number of oscillations of r between rmax & rmin, the motion will repeat itself exactly! If the orbit does not exactly close on itself after a finite number of radial oscillations between rmax & rmin, the orbit r(θ) will be open. See figure 

  6. Use the orbit eqtn θ(r) to find the change in θ due to one complete oscillation of r from rmin to rmax & back to rmin: • Angular change = 2  the change in going once rmin to rmax:  Δθ 2∫(/r2)(2μ)-½[E - U(r) - {2(2μr2)}]-½dr (limits rmin r  rmax) Δθ | |

  7. Δθ= 2∫(/r2)(2μ)-½[E - U(r) - {2(2μr2)}]-½dr (rmin r  rmax) • Periodic motion & a closed orbit results if & only if the angular change is a rational fraction of 2π. That is: Δθ 2π(a/b) (a, b integers) periodic, closed orbit. • If the orbit is closed, after b periods, the radius vector the of particle will have made a complete revolutions & the particle will be at its original position. • It can be shown (Prob. 8.35) that if the potential is a power law in r: U(r) = k rn+1a closed NON-CIRCULAR path can occur ONLY forn = -2 (the inverse square law force: gravity, electrostatics; discussed in detail next!) & n = +1 (the 3d isotropic, simple harmonic oscillator of Ch. 3). • Footnote:Some fractional values of n also lead to closed orbits. These aren’t interesting from the PHYSICSviewpoint.

  8. Centrifugal Energy & Effective PotentialSection 8.6 • Consider again the common factor in r, Δθ, θ(r), t(r), etc.: ([E - U(r)] - [2(2μr2)])½ • Qualitatively analyze the RADIAL motion for a “particle” of mass μ in a central potential U(r) by methods similar to those used in Sect. 2.6 for 1d motion. • Remember that the ACTUALmotion is in 2d!  Must superimpose the angular motion (θ) on the radial motion results, to get the ORBIT θ(r) orr(θ).

  9. Consider: ([E - U(r)] - [2(2μr2)])½ • Recall the physicsof the term[2(2μr2)]! From previous discussion (conservation of angular momentum; = μr2θ):  [2(2μr2)]  (½)μr2θ2  The angular part of the kinetic energyof mass μ. • When it is written in the form [2(2μr2)], this contribution to the energy depends only on r. Because of this, when analyzing the r part of the motion, we can treat this as an additional term in the potential energy. It is often convenient to call it another potential energy term THE “CENTRIFUGAL” POTENTIAL ENERGY

  10. Centrifugal Potential [2(2μr2)] “Centrifugal” Potential EnergyUc(r) • As just discussed, this is really the angular part of the Kinetic Energy!  Consider the “Force” associated with Uc(r): Fc(r)  - (∂Uc/∂r) = [2(μr3)] Or, using  = μr2θ: Fc(r) = [2(μr3)]= μrθ2 “Centrifugal Force”

  11. Centrifugal Force Fc(r) = [2(μr3)] “Centrifugal Force” • Fc(r) = A fictitious “force” arising due to fact that the reference frame of the relative coordinate r (of the “particle” of “mass” μ) is notan inertial frame! • Its not a force in the Newtonian sense! (It doesn’t come from any interaction of the “mass” μ with its environment!) It’s a part of the “μa” (right!) of Newton’s 2nd Law, rewritten to appear on the “F” (left) side. For more discussion, see Ch. 10. • “Centrifugal Force” is an unfortunate terminology! Its confusing to elementary (& also sometimes to advanced!) physics students. Direction of Fc: Outward from the force center! • I always tell students that there is no such thing as centrifugal force! • For a particle moving in a circular arc, the actual, physical force in anInertial Frameis directedINWARD TOWARDS THE FORCE CENTER  Centripetal Force[F = μrθ2 = μrω2 = μ(v2/r)]

  12. Effective Potential • Consider again: [E - U(r) - {2(2μr2)}]½ • For both qualitative & quantitative analysis of the RADIAL motion for the “particle” of mass μ in the central potential U(r), Uc(r) = [2(2μr2)] acts as an additional potential & we can treat it as such! • Recall that physically, it comes from the Kinetic Energy of the particle!  Its convenient to lump U(r) &Uc(r) together into an Effective Potential V(r)  U(r) + Uc(r) V(r)  U(r) + [2(2μr2)]

  13. Effective Potential V(r)  U(r) + [2(2μr2)] • Consider now: r  [E - V(r)]½(1) • Given U(r), we can use (1) to qualitatively (& quantitatively) analyze the RADIAL motion for the “particle”. Get (radial) turning points, (radial) oscillations, etc. • Very similar to 1d where we analyzed particle motion for various E using an analogous expression.

  14. Consider now, an important special case (in the mathematical sense) which is the MOST important case in the physics sense! • Consider the Inverse Square Law central force: F(r) = - kr-2 U(r) = - kr-1 = - (k/r) • For convenience, take U(r)  0 Gravitational Force:k = GmM Coulomb Force: (SI Units): k = (q1q2)/(40) • The Effective Potential for this case is: V(r)  -(k/r) + [2(2μr2)]

  15. Effective Potential, Inverse Square Law force. (see figure): V(r) = -(k/r) + [2(2μr2)]

  16. Lets now qualitatively analyze the radial motion using V(r) for the Inverse Square Law force (figure). E = (½)μr2 + V(r)  E - V(r) = (½)μr2  r = 0 at turning points (where E = V(r)) • Consider E ≥ 0 (E1in figure): This case corresponds to unbounded radial motion. μmoves in from r = towards the force center (at r = 0), “hits” the barrier at (turning point) r1( r = 0), and is “reflected” back to r = . E = E1   (½)μr2    r = r1

  17. Qualitative analysis of the radial motion using V(r) for the Inverse Square Law force (figure). E = (½)μr2 + V(r)  E - V(r) = (½)μr2  r = 0 at turning points (where E = V(r)) • Consider Vmin < E < 0 (E2in figure): This case corresponds to oscillatory radial motion.μmoves back & forth between (turning points) r2 & r4 ( r = 0). These turning points are the Apsidal Distances of the orbit (the rmax& rmin from before). r = r2  r = r4  E = E2    (½)μr2 Vmin

  18. Qualitative analysis of the radial motion using V(r) for the Inverse Square Law force (figure). E = (½)μr2 + V(r)  E - V(r) = (½)μr2  r = 0 at turning points (where E = V(r)) • Consider E = Vmin = -(μk2)/(22) (E3in the figure): This case corresponds to no radial motion at all becauser = 0 for all time. That is, the orbit radius is constant (r = r3 in the figure). That is, μmoves in a Circular Orbit! • Note: E < Vmin is Not Allowed! (r would be imaginary!) r = r3  E = E3   Vmin

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