2. Sect. 3.6: Closed Orbit Conditions & Stability of Circular Orbits Can still get a LOT more (qualitative & quantitative) info about orbital motion from equivalent 1d (r) problem & orbit eqtn, without specifying V(r). For example, its possible to derive a theorem on the types of attractive central forces which lead to CLOSED ORBITS. (  Bertrand’s Theorem). Also, we will discuss stability of circular orbits. My treatment is similar to Sect. 3.6, but may be slightly different in places. I actually include more details! I get the same results, of course!

3. We’ve seen: For analysis of the RADIAL motion for a “particle” of mass m in a central potential V(r), the centrifugal term V c (r) = [ 2  (2mr 2 )] acts as an additional potential! –Recall: Physically, it comes from the (angular part of) the Kinetic Energy!  Lump V(r) & V c (r) together into an Effective Potential  V´(r)  V(r) + V c (r)  V(r) + [ 2  (2mr 2 )] Circular Orbits

4. Previous discussion: For a given, the orbit is circular if the total energy = min (or max!) value of the effective potential, which occurs at some r (  r 0 ), E  V´(r 0 ) = V(r 0 ) + [ 2  {2m(r 0 ) 2 }] At this value of r, the radial velocity r = 0. [E = (½)mr 2 + [ 2  (2mr 2 )] + V(r) = const] A circular orbit is allowed for ANY attractive potential V(r):  If & only if V´ has a min (or a max!) at r = r 0 (  ρ in my notation, sorry!) Circular orbits are always ALLOWED, but they are not always STABLE! Here, we also examine the stability question.

5. Orbit at r = ρ is circular if the total energy = E  V´(ρ) = V(ρ) + [ 2  (2mρ 2 )] (1) ρ is defined so that V´ is a min or a max at r = ρ.  At r = ρ the “force” coming from the effective potential is zero: f´(ρ) = -(∂V´/∂r)| r = ρ = 0 (2) –For the r motion, the condition for a circular orbit is very much like a general condition for static equilibrium.  At r = ρ the attractive force from V(r) exactly balances the “centrifugal force” from the (repulsive) V c (r) : f(ρ)  -(∂V/∂r)| r = ρ (1) & (2)  f(ρ) = -[ 2  (mρ 3 )] (3) (1) + (3)  Conditions for a circular orbit

6. For a given, whether a circular orbit is stable or unstable depends on whether V´ is a minimum or a maximum at r = ρ. Stable circular orbit at r = ρ Unstable circular orbit at r = ρ –Analogous to conditions for stable & unstable equilibrium in static equilibrium problems!  r = ρ 

7. If V´ = a min at r = ρ, as in fig: If give m an energy slightly above V´(ρ), the orbit will no longer be circular, but will still be bounded (r will oscillate between apsidal values, as for E 3 in figure.)  Stable circular orbit at r = ρ –Analogous to stable equilibrium condition in static problems!  r = ρ

8. If V´ = a max at r = ρ, as in fig: If give m an energy slightly above V´(ρ), the orbit will no longer be circular, & also will now be unbounded (m moves through r = 0 & out to r   ).  Unstable circular orbit at r = ρ –Analogous to unstable equilibrium condition in static problems! r = ρ 

9. Stability of Circular Orbits Stability of circular orbits is determined (naturally!) mathematically by the sign of the 2 nd derivative (curvature) of V´ evaluated at r = ρ: If (∂ 2 V´/∂r 2 )| r = ρ > 0, the orbit is stable. If (∂ 2 V´/∂r 2 )| r = ρ < 0, the orbit is unstable.

10. Summary: A circular orbit at r = ρ exists if r| r = ρ = 0 for all time t. This is possible if (∂V´/∂r)| r = ρ = 0. A stable circular orbit occurs if & only if this effective potential V´(r) has a true minimum (Not a maximum!). All other circular orbits are unstable!  General condition for stability of circular orbits: (∂ 2 V´/∂r 2 )| r = ρ > 0.

11. Apply general condition for circular orbit stability: (∂ 2 V´/∂r 2 )| r = ρ > 0 V´(r) = V(r) + [ 2  (2mr 2 )] (∂V´/∂r) = (∂V/∂r) - [ 2  (mr 3 )] = - f(r) - [ 2  (mr 3 )]  (∂ 2 V´/∂r 2 )| r = ρ = -(∂f/∂r)| r = ρ + (3 2 )/(mρ 4 ) > 0 (1) At r = ρ the force balances the centrifugal force: f(ρ) = -[ 2  (mρ 3 )] (2) Combining (1) & (2), the stability condition is: (∂f/∂r)| r = ρ < - 3f(ρ)/ρ (3) Or: (d[ln(f)]/d[ln(r)])| r = ρ < - 3 (4) (3) or (4)  Condition on the Central force f(r) which will give a Stable Circular Orbit at r = ρ.

12.  General condition for stability of a circular orbit of radius ρ with a central force: (∂f/∂r)| r = ρ < - 3f(ρ)/ρ (3) Or: (d[ln(f)]/d[ln(r)])| r = ρ < - 3 (4) –Conditions for Stable Circular Orbit at r = ρ. Suppose, f(r) is an attractive power law force (at least near r = ρ): f(r) = -k r n (k >0) –Using (3), this gives: -knρ n-1 < 3kρ n-1.  A circular orbit is stable (any r = ρ) if n > -3

13.  Stability criterion for circular orbits for power law central force  Stable circular orbits for f(r) = -kr n only exist for n > -3 !  Stable circular orbits for f(r) = -(k/r n ) only exist for n < 3 !  All attractive power law forces f(r) = -kr n with n > -3 can have stable circular orbits (at any r = ρ)  All attractive power law potentials V(r) = -k r n+1 with n > -3 can have stable circular orbits (at any r = ρ) If the other conditions for a circular orbit are satisfied, of course!

14. Related topics: –For “almost” circular orbits: Frequency of radial oscillation about a circular orbit in a general central force field. –Criteria for closed & open orbits. –Some treatment comes from Marion’s text. –Get Eq. (3.45) in Goldstein General Central Force: f(r). Define function g(r): f(r)  - mg(r) = -(∂V/∂r) Lagrangian: L= (½)m(r 2 + r 2 θ 2 ) - V(r) Lagrange’s Eqtn for r: (∂L/∂r) - (d/dt)[(∂L/∂r)]= 0

15.  m(r - rθ 2 ) = -(∂V/∂r) = f(r) = -mg(r) Equivalent to the radial part of Newton’s 2 nd Law (polar coordinates) Dividing by m, this is: r - rθ 2 = -g(r) (1) Angular momentum conservation : = mr 2 θ = const  (1) becomes: r - [( 2 )/(m 2 r 3 )] = -g(r) (2) Suppose the “particle” of mass m is initially in a circular orbit of radius ρ. Suppose, due to some perturbation, the orbit radius is changed from ρ to r = ρ + x, where x << ρ ρ = constant  r = x & (2) becomes:

16.  x - [( 2 )/(m 2 ρ 3 )][1+(x/ρ)] -3 = -g(ρ + x) (3) Since x << ρ, expand the left & right sides of (3) in a Taylor’s series about r = ρ & keep only up through linear terms in x: [1+(x/ρ)] -3  1 - 3(x/ρ) +... g(ρ + x)  g(ρ) + x(dg/dr)| r = ρ +  (3) becomes: x - [( 2 )/(m 2 ρ 3 )][1- 3(x/ρ)]  -[g(ρ) + x(dg/dr)| r = ρ ] (4) Assumption: Initially, a circular orbit at r = ρ (2) evaluated at r = ρ (ρ = constant, r = ρ = 0 in (2)):  g(ρ) = [( 2 )/(m 2 ρ 3 )] > 0 (5)

17.  x + [3(g(ρ)/ρ) + (dg/dr)| r = ρ ]x  0 Rewrite (defining frequency ω 0 ): x + (ω 0 ) 2 x = 0 (6) with (ω 0 ) 2  [3(g(ρ)/ρ) + (dg/dr)| r = ρ ] (6) is diff eqtn for simple harmonic oscillator, freq. ω 0 ! Solution to (6), for (ω 0 ) 2 > 0 (  ω 0 = real): x(t) = A exp(iω 0 t) + B exp(-iω 0 t) or x(t) = X sin(ω 0 t + δ)  The orbit radius oscillates harmonically about r = ρ  r = ρ is a stable circular orbit! Solution to (6), for (ω 0 ) 2 < 0 (  ω 0 = imaginary): x(t) = C exp(|ω 0 |t) + D exp(-|ω 0 |t)  The orbit radius increases exponentially from r = ρ  r = ρ is an unstable circular orbit.

18.  The condition for oscillation is thus  Condition for stability of a circular orbit. This is: (ω 0 ) 2  [3(g(ρ)/ρ) + (dg/dr)| r = ρ ] > 0 Divide by g(ρ):  Condition for stability of a circular orbit is [(dg/dr)| r = ρ ]/g(ρ) +(3/ρ) > 0 Note that g(r) = -f(r)/m  General condition for stability of circular orbit of radius ρ with a central force: [(df/dr)| r = ρ ]/f(ρ) +(3/ρ) > 0 (same as before) Evaluate this for a power law force f(r) = - kr n, get (3+n)(1/ρ) > 0 or n > - 3, same as before!

19. SUMMARY: General condition for stability of a circular orbit of radius ρ with a central force f(r): [(df/dr)| r = ρ ]/f(ρ) +(3/ρ) > 0 For an orbit which is perturbed slightly away from circular, r = ρ + x, where x << ρ: –If the original circular orbit was stable, there will be harmonic oscillations about r = ρ. That is: x(t) = X sin(ω 0 t + δ) –If the original circular orbit was unstable, the radius will increase exponentially from r = ρ. That is: x(t) = C exp(|ω 0 |t) + D exp(-|ω 0 |t) –In both cases, m(ω 0 ) 2  [3(f(ρ)/(ρ) + (df/dr)| r = ρ ]

20. Example Investigate the stability of circular orbits in a force field described by the potential function: V(r) = -(k/r)e -(r/a)  Screened Coulomb Potential (in E&M)  Yukawa Potential (in nuclear Physics) Using the criteria just discussed, we find stable circular orbits for ρ < ~ 1.62a. See figure.

21. Eq. (3.45) Use a very similar approach to get Eq. (3.45) of Goldstein. This Eq: For small deviations from circular orbit of radius ρ, the orbit has the form: u(θ) = [1/r(θ)] = u 0 + a cos(βθ) where u 0 = [1/ρ]. β, a are to be determined. u = (1/r) undergoes simple harmonic motion about the circular orbit value u 0. The derivation is tedious! Almost like doing the previous calculation over again except for u = 1/r instead of r itself. Also for r(θ) instead of for r(t). Frequency β: From a Taylor’s series expansion of the force law f(r) about the circular orbit radius ρ. Amplitude a : Depends on the deviation of the energy E from its value at the circular orbit of radius radius ρ.

22. u(θ) = [1/r(θ)] = u 0 + a cos(βθ) (3.45) Manipulation gives: β 2  3 + [ρ/f(ρ)][df/dr]| r = ρ = 3 + (d[ln(f)]/d[ln(r)])| r = ρ If β 2 > 0, a cos(βθ) (& thus u( θ )) is oscillatory (harmonic). Corresponds to the stable circular orbit result from before: (df/dr)| r = ρ < - 3f(ρ)/ρ Or: (d[ln(f)]/d[ln(r)])| r = ρ < - 3 If β 2 - 3f(ρ)/ρ Or: (d[ln(f)]/d[ln(r)])| r = ρ > - 3

23. u(θ) = [1/r(θ)] = u 0 + a cos(βθ) (3.45) β 2  3 + [ρ/f(ρ)][df/dr]| r = ρ = 3 + (d[ln(f)]/d[ln(r)])| r = ρ Consider the stable circular orbit case, so β 2 > 0. As the radius vector r sweeps around the plane, u goes through β cycles of oscillation. See fig. If β = q/p with, q, p integers, (so β is a rational number) then after q revs of the radius vector, the orbit retraces itself.  The orbit is closed Open & Closed Circular Orbits

24. Closed, Almost Circular Orbits Consider an almost circular orbit: u(θ) = [1/r(θ)] = u 0 + a cos(βθ) (3.45) β 2  3 + [ρ/f(ρ)][df/dr]| r = ρ = 3 + (d[ln(f)]/d[ln(r)])| r = ρ Stable initial circular orbit, so β 2 > 0, and (df/dr)| r = ρ < - 3f(ρ)/ρ Or: (d[ln(f)]/d[ln(r)])| r = ρ < - 3 At each value of r = ρ for which this stability criterion is met, can, by definition, get a stable circular orbit. Question: What are the conditions on the force law f(r) which will lead to closed almost circular orbits?

25. Goldstein’s reasoning: If the circular orbit is stable & the almost circular orbit is closed, β = q/p (= rational number), where q, p are integers. He argues that (even though β 2  3 + [ρ/f(ρ)][df/dr]| r = ρ & should thus be ρ dependent), if the orbit is closed, β MUST be the same rational number for all possible ρ. That is β = constant, independent of the radius ρ of the original circular orbit!. See text for further discussion.  By this reasoning, (β = constant), the expression β 2  3 + [ρ/f(ρ)][df/dr]| r = ρ becomes a differential equation for the force law f(r).

26. Under the specific conditions just described, we have β 2  3 + [ρ/f(ρ)][df/dr]| r = ρ  β 2  3 + [r/f(r)][df/dr] = const (A differential equation for the force law f(r)!) Rewrite this as: (d[ln(f)]/d[ln(r)]) = β 2 - 3 Integrating this gives a force law: f(r) = -(k/r α ), with α  3 - β 2  All force laws of this form (with β a rational number) lead to closed, stable, almost circular orbits.

27. We’ve shown that closed, stable, almost circular orbits result from all force laws of the form (β a rational number) f(r) = -(k/r α ), with α  3 - β 2 Examples: β =  1  f(r) = -(k/r 2 ) (Inverse r squared law!) β =  2  f(r) = -kr (Isotropic harmonic oscillator!) β = q/p (q, p integers)  f(r) = -(k/r α ), with α = 3 - (q/p) 2

28. Bertrand’s Theorem If initial conditions are such that the perturbed circular orbit is not close to those required for circular orbit (the orbit is not “almost” circular!), will the same type of force law ( a rational number) f(r) = -(k/r α ), with α  3 - β 2 still give closed orbits? –Answer: Keep additional terms in Taylor’s series expansion (to compute β 2 ) & solve orbit equation. Solved by J. Bertrand (1873). Proved that in such cases, the orbits are closed ONLY for: β =  1  f(r) = -(k/r 2 ) (Inverse r squared law!) β =  2  f(r) = -kr (Hooke’s “Law”: Isotropic harmonic oscillator!)

29.  Bertrand’s Theorem: The only central forces that result in bound, closed orbits for all particles are the inverse-square law and Hooke’s “law”. A very important result! For example, bound celestial objects (planets, stars, etc.) all are OBSERVED to have orbits that are  closed. –Deviations are from perturbations due to other bodies

30. Ruling out the (unphysical at large r) Hooke’s “law” force, this  The force (gravitational) holding the objects in their orbits varies as r -2 !  Using only celestial observations PLUS Bertrand’s theorem, one can conclude that the gravitational force f g (r) varies as r -2. –That is, f g (r)  -kr -2. –Observations + Bertrand’s Theorem REQUIRE the gravitational force to have the r dependence given by Newton’s Law of Gravitation! –The observed character of the orbits (closed) fixes the form of the force law!