Sect. 3.6: Closed Orbit Conditions & Stability of Circular Orbits

2. Sect. 3.6: Closed Orbit Conditions & Stability of Circular Orbits • Can still get aLOTmore (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. Circular Orbits • We’ve seen: For analysis of the RADIALmotion for a “particle” of mass m in a central potential V(r), thecentrifugal term Vc(r) = [2 (2mr2)] acts as an additional potential! • Recall:Physically, it comes from the (angular part of) the Kinetic Energy!  Lump V(r) &Vc(r) together into an Effective Potential V´(r)  V(r) + Vc(r)  V(r) + [2(2mr2)]

4. Previous discussion: For a given ,the orbit is circularif the total energy = min (or max!) value of the effective potential, which occurs at some r ( r0), E  V´(r0) = V(r0)+ [2 {2m(r0)2}] At this value of r, the radial velocity r = 0. [E = (½)mr2 + [2(2mr2)] + V(r) = const] • A circular orbit is allowed for ANY attractivepotentialV(r): If & only if V´ has a min (or a max!) at r = r0( ρin my notation, sorry!) • Circular orbits are always ALLOWED, but they are not always STABLE! Here, we also examine the stability question.

5. Orbit atr = ρis circular if the total energy = E  V´(ρ) = V(ρ)+ [2 (2mρ2)](1) • ρis 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) Vc(r) : f(ρ) -(∂V/∂r)|r = ρ (1) & (2) f(ρ)= -[2 (mρ3)](3) (1) + (3)  Conditions for a circular orbit

6. For a given , whether a circular orbit isstableor unstable depends on whether V´ is a minimum or a maximum at r = ρ. Stable circular orbit at r = ρUnstable circular orbit at r = ρ • Analogous to conditions for stable & unstable equilibrium in static equilibrium problems! r = ρ   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 E3 in figure.)  Stable circular orbit at r = ρ • Analogous to stable equilibrium condition in static problems!  r = ρ

8. If V´ = a max atr = 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 2nd derivative (curvature) ofV´ evaluated at r = ρ: • If (∂2V´/∂r2)|r = ρ > 0, the orbit is stable. • If (∂2V´/∂r2)|r = ρ < 0, the orbit is unstable.

10. Summary: A circular orbit at r = ρexists if r|r = ρ = 0 for all time t. • This is possible if (∂V´/∂r)|r = ρ = 0. • Astablecircular 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: (∂2V´/∂r2)|r = ρ > 0.

11. Apply general condition for circular orbit stability: (∂2V´/∂r2)| r = ρ > 0 V´(r) = V(r)+ [2 (2mr2)] (∂V´/∂r) = (∂V/∂r) - [2 (mr3)] = - f(r) - [2 (mr3)]  (∂2V´/∂r2)|r = ρ = -(∂f/∂r)|r = ρ + (32)/(mρ4) > 0 (1) • At r = ρthe force balances the centrifugal force:f(ρ)= -[2 (mρ3)](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 rn(k >0) • Using (3), this gives: -knρn-1 < 3kρn-1 .  A circular orbit is stable(anyr = ρ)if n > -3

13.  Stability criterion for circular orbits for power law central force Stable circular orbits for f(r) = -krnonly exist for n > -3 !  Stable circular orbits for f(r) = -(k/rn) only exist for n < 3 !  All attractive power law forcesf(r) = -krnwith n > -3 can have stable circular orbits(at anyr = ρ)  All attractive power law potentials V(r) = -k rn+1with n > -3 can have stable circular orbits(at anyr = 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(r2 + r2θ2) - V(r) • Lagrange’s Eqtn for r: (∂L/∂r) - (d/dt)[(∂L/∂r)]= 0

15.  m(r - rθ2) = -(∂V/∂r) = f(r) = -mg(r) Equivalent to the radial part of Newton’s 2nd Law(polar coordinates) • Dividing by m, this is: r - rθ2 = -g(r) (1) • Angular momentum conservation: = mr2θ = const  (1) becomes: r - [(2)/(m2r3 )] = -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 ρto r = ρ + x , where x << ρ • ρ = constant  r = x & (2) becomes:

16.  x - [(2)/(m2ρ3 )][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)/(m2ρ3)][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)/(m2ρ3 )] > 0 (5)

17. x + [3(g(ρ)/ρ) + (dg/dr)|r = ρ]x  0 • Rewrite (defining frequency ω0):x + (ω0)2 x = 0(6) with (ω0)2  [3(g(ρ)/ρ) + (dg/dr)|r = ρ] • (6) is diff eqtn for simple harmonic oscillator, freq. ω0! • Solution to (6), for (ω0)2> 0 ( ω0 = real): x(t) = A exp(iω0t) + B exp(-iω0t) or x(t) = X sin(ω0t + δ)  The orbit radius oscillates harmonically about r = ρ  r = ρ is a stable circular orbit! • Solution to (6), for (ω0)2< 0 ( ω0 = imaginary): x(t) = C exp(|ω0|t) + D exp(-|ω0|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: (ω0)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) = - krn, 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(ω0t + δ) • If the original circular orbit was unstable, the radius will increase exponentially from r = ρ. That is: x(t) = C exp(|ω0|t) + D exp(-|ω0|t) • In both cases, m(ω0)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)] = u0 + a cos(βθ) where u0 = [1/ρ]. β, a are to be determined. u = (1/r) undergoes simple harmonic motion about the circular orbit value u0. 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)] = u0 + a cos(βθ) (3.45) • Manipulation gives: β2 3 + [ρ/f(ρ)][df/dr]|r = ρ = 3 + (d[ln(f)]/d[ln(r)])|r = ρ • If β2> 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 β2< 0, a cos(βθ)  a cosh(βθ)(& thus u(θ)) is an exponentially increasing function of θ. Corresponds to the unstable circular orbit result from before: (df/dr)|r = ρ > - 3f(ρ)/ρ Or: (d[ln(f)]/d[ln(r)])|r = ρ > - 3

23. Open & Closed Circular Orbits u(θ) = [1/r(θ)] = u0 + a cos(βθ) (3.45) β2 3 + [ρ/f(ρ)][df/dr]|r = ρ = 3 + (d[ln(f)]/d[ln(r)])|r = ρ • Consider the stable circular orbit case, so β2> 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

24. Closed, Almost Circular Orbits • Consider an almost circular orbit: u(θ) = [1/r(θ)] = u0 + a cos(βθ) (3.45) β2 3 + [ρ/f(ρ)][df/dr]|r = ρ = 3 + (d[ln(f)]/d[ln(r)])|r = ρ • Stable initial circular orbit, so β2> 0, and (df/dr)|r = ρ < - 3f(ρ)/ρOr: (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 toclosedalmost 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 β2 3 + [ρ/f(ρ)][df/dr]|r = ρ& should thus beρ dependent), if the orbit is closed, βMUSTbe the same rational number for all possibleρ. That isβ = constant, independent of the radius ρ of the original circular orbit!.See text for further discussion.  By this reasoning, (β = constant), the expression β2 3 + [ρ/f(ρ)][df/dr]|r = ρbecomes a differential equation for the force law f(r).

26. Under the specific conditions just described, we have β2 3 + [ρ/f(ρ)][df/dr]|r = ρ β2 3 + [r/f(r)][df/dr] = const (Adifferential equation for the force law f(r)!) • Rewrite this as: (d[ln(f)]/d[ln(r)]) = β2- 3 • Integrating this gives a force law: f(r) = -(k/rα), with α 3 - β2  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/rα), with α 3 - β2 • Examples: β = 1  f(r) = -(k/r2) (Inverse r squared law!) β =  2  f(r) = -kr (Isotropic harmonic oscillator!) β = q/p (q, p integers)  f(r) = -(k/rα), 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/rα), with α 3 - β2 still give closed orbits? • Answer: Keep additional terms in Taylor’s series expansion (to compute β2) & solve orbit equation. • Solved by J. Bertrand (1873). Proved that in such cases,the orbits are closed ONLY for: β =  1  f(r) = -(k/r2) (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 areOBSERVEDto 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 !  Usingonly celestial observationsPLUS Bertrand’s theorem, one can conclude that the gravitational force fg(r) varies as r-2. • That is, fg(r)  -kr-2. • Observations + Bertrand’s Theorem REQUIREthe 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!