Central Field Orbits Section 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

(dx/dt) = x(x) =  [(2m-1)(E - U(x))]½ 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 RADIAL motion 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).

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)

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 be oscillatory 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.

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 

 Δθ  2∫(/r2)(2μ)-½[E - U(r) - {2(2μr2)}]-½dr 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 | |

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+1 a closed NON-CIRCULAR path can occur ONLY for n = -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 PHYSICS viewpoint.

Centrifugal Energy & Effective Potential Section 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 ACTUAL motion is in 2d!  Must superimpose the angular motion (θ) on the radial motion results, to get the ORBIT θ(r) or r(θ).

Consider: ([E - U(r)] - [2(2μr2)])½ Recall the physics of the term [2(2μr2)]! From previous discussion (conservation of angular momentum;  = μr2θ):  [2(2μr2)]  (½)μr2θ2  The angular part of the kinetic energy of 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

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”

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 not an 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 an Inertial Frame is directed INWARD TOWARDS THE FORCE CENTER  Centripetal Force [F = μrθ2 = μrω2 = μ(v2/r)]

Effective Potential V(r)  U(r) + [2(2μr2)] 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)]

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.

F(r) = - kr-2  U(r) = - kr-1 = - (k/r) 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)]

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

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 (E1 in figure): This case corresponds to unbounded radial motion. μ moves 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

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 (E2 in figure): This case corresponds to oscillatory radial motion. μ moves 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

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) (E3 in the figure): This case corresponds to no radial motion at all because r = 0 for all time. That is, the orbit radius is constant (r = r3 in the figure). That is, μ moves in a Circular Orbit! Note: E < Vmin is Not Allowed! (r would be imaginary!) r = r3  E = E3  Vmin 