Transcript Sect. 3.6

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!
Circular Orbits
• We’ve seen: For analysis of the RADIAL motion for a
“particle” of mass m in a central potential V(r), the
centrifugal 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)]
• 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 ( 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 attractive
potential V(r):  If & only if V´ has a min (or a
max!)
at r = r0 ( ρ in my notation, sorry!)
• Circular orbits are always ALLOWED, but they are
not always STABLE! Here, we also examine the
stability question.
• Orbit at r = ρ is circular if the total energy =
E  V´(ρ) = V(ρ) + [2 (2mρ2)]
(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)
Vc(r) : f(ρ)  -(∂V/∂r)|r = ρ
(1) & (2) 
f(ρ) = -[2 (mρ3)]
(3)
(1) + (3)  Conditions for a circular orbit
• For a given , whether a circular orbit is stable or
unstable depends on whether V´ is a minimum or a
maximum at r = ρ.
r=ρ


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!
• 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.)

r=ρ

Stable circular orbit at r = ρ
– Analogous to stable equilibrium condition in static problems!
• 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  ).
r=ρ


Unstable circular orbit at r = ρ
– Analogous to unstable equilibrium condition in static problems!
Stability of Circular Orbits
• Stability of circular orbits is determined
(naturally!) mathematically by the sign of the
2nd derivative (curvature) of V´ evaluated at
r = ρ:
•
If (∂2V´/∂r2)|r = ρ > 0, the orbit is stable.
•
If (∂2V´/∂r2)|r = ρ < 0, the orbit is unstable.
• 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:
(∂2V´/∂r2)|r = ρ > 0.
• 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 = ρ the 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.
 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 (any r = ρ) if n > -3
 Stability criterion for circular orbits for
power law central force  Stable circular
orbits for f(r) = -krn only exist for n > -3 !
 Stable circular orbits for f(r) = -(k/rn) only exist for n < 3 !
 All attractive power law forces f(r) = -krn
with n > -3 can have stable circular orbits (at
any r = ρ)
 All attractive power law potentials
V(r) = -k rn+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!
• 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

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 ρ to
r = ρ + x , where x << ρ
• ρ = constant  r = x & (2) becomes:
 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)

x + [3(g(ρ)/ρ) + (dg/dr)|r = ρ]x  0
• Rewrite (defining frequency ω0): x + (ω0)2 x = 0
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):
or
x(t) = A exp(iω0t) + B exp(-iω0t)
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.
(6)
 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!
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]
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.
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.
u(θ) = [1/r(θ)] = u0 + a cos(βθ)
• Manipulation gives:
(3.45)
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
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
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)/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?
• 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 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
β2  3 + [ρ/f(ρ)][df/dr]|r = ρ becomes a differential
equation for the force law f(r).
• 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
(A differential 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.
• 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
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!)
 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
• 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 fg(r) varies as r-2.
– That is, fg(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!