### Motion in central potentials

Potential energy U = crn

#### Problem:

A mass m moves in a central force field.  The force is  F = f(r)(r/r), where f(r) = -kr and k > 0.  Assume the mass moves at a constant speed in a circular path of radius R.  Calculate the angular velocity of the mass, and show that its energy is E = kR2.

Solution:

• Concepts:
Motion in a central potential, F = f(r)(r/r), f(r) = (-∂U/∂r)
• Reasoning:
f(r) = (-∂U/∂r) = -kr,  U(r) = ½kr2.  We have a central  potential.
The energy of the mass moving at a constant speed v in a circular path of radius R is
E = T + U = ½mv2 + ½kR2.
• Details of the calculation:
For circular motion at constant speed we have |F| = mv2/R = kR.  v = R(k/m)½.
The energy is E = T + U = ½mv2 + ½kR2 = ½mR2(k/m) + ½kR2 = kR2.

#### Problem:

A particle of mass m moves in a central force field such that its potential energy is given by V = krn, where r is the distance from the center of force and k and n are constants.
(a)  Write down the Lagrangian for this system and determine the equations of motion in polar coordinates.
(b)  Show that angular momentum is conserved for the system.
(c)  Find an expression for the total energy of the system that depends only on the radial variable.
(d)  Find the conditions (sign and magnitude of n and k) for a stable circular orbit by investigating the particle at stable equilibrium.

Solution:

• Concept:
Motion in a central potential,  Lagrange's equations, stable orbits
• Reasoning:
We are asked to write down the Lagrangian, find an expression for the total energy, and investigate the stability of orbits.
• Details of the calculation:
(a)  L = T - U = ½m[(dr/dt)2 + r2(dΦ/dt)2] - krn.
The generalized coordinates are r and Φ.  The coordinate Φ is cyclic.
Lagrange's equations yield the equations of motion.
(d/dt)[mr2(dΦ/dt)] = 0.
md2r/dt2 = M2/mr3 - nkrn-1.
(b)  (d/dt)[mr2(dΦ/dt)] = 0.  M = mr2(dΦ/dt) = constant, angular momentum is conserved.
(c)  E = T + U = ½m(dr/dt)2 + M2/(2mr2) + krn = ½m(dr/dt)2 + Ueff(r).
This expression looks like the total energy of a particle moving in one dimension in a potential Ueff(r).
(d)  For a circular orbit at with radius a we need ∂Ueff(r)/∂r|a = 0.
∂Ueff(r)/∂r =  nkrn–1 - M2/(mr3).
nkan – 1 = M2/(ma3),  an+2  =  M2/(mnk),  (n ≠ 0).
For a stable orbit of radius a we need a restoring force.  Let r = a + ρ, ρ << a.
We need md2ρ/dt2 = -kρ = -∂Ueff/∂ρ = -(∂2Ueff/∂ρ2)|ρ=0ρ,
or ∂2Ueff/∂ρ2 > 0 near ρ = 0.
∂Ueff(r)/∂r =  nkrn–1 - M2/(mr3).
∂Ueff(ρ)/∂ρ = nk(a + ρ)n–1 - M2/[m(a + ρ)3]
= nkan-1(1 + ρ/a)n–1 - [M2/(ma3)](1 + ρ/a)-3
= nkan-1(1 + (n-1)ρ/a) - [M2/(ma3)](1 - 3ρ/a).
2Ueff/∂ρ2|ρ=0 = n(n-1)kan-2 + 3M2/(ma4) > 0.
mn(n – 1)kan+2  + 3M2 > 0.
mn(n – 1)k[M2/(mnk)] + 3M2 > 0.
(n – 1)M2 + 3M2 > 0.
(n – 1) + 3 > 0.
n + 2 > 0.
Requirements for stable circular orbit with radius a for potential  V(r) = krn  are:
n > -2 , n ≠ 0,
k = M2/(nman+2).
When n < 0 then k < 0, and when n > 0 then k > 0.

#### Problem:

Two particles with reduced mass μ orbiting each have a potential energy function U = ½kr2, where k > 0 and r is the distance between them.
(a)  Find the equilibrium distance r0 at which the particles can circle each other at a constant distance as a function of the angular momentum M.
(b)  Determine if this is a stable equilibrium distance.
(c)  Assume the particles orbiting at the equilibrium distance r0 are slightly disturbed.  Determine if the disturbed orbits are closed.

Solution:

• Concept:
Motion in a central potential
• Reasoning:
For motion in a central potential energy and angular momentum are conserved.  These two equations are enough to answer the questions.
• Details of the calculation:
(a)  M = μr2(dΦ/dt) = constant, angular momentum is conserved.
E = T + U = ½μ(dr/dt)2 + M2/(2μr2) + ½kr2 = ½μ(dr/dt)2 + Ueff(r).
This expression looks like the total energy of a particle moving in one dimension in a potential Ueff(r).
(b)  For a circular orbit at with radius r0 we need ∂Ueff(r)/∂r|r0 = 0.
∂Ueff(r)/∂r =  kr - M2/(μr3).
kr0 = M2/(μr03),  r04  =  M2/(μk).
(b)  For a stable orbit of radius a we need a restoring force.  Let r = a + ρ, ρ << a.
We need μd2ρ/dt2 = -½kρ = -∂Ueff/∂ρ = -(∂2Ueff/∂ρ2)|ρ=0ρ.
We need ∂2Ueff/∂ρ2 > 0 near ρ = 0, or ∂2Ueff/∂r2 > 0 near r = r0.
2Ueff/∂r2|r0 = k + 3M2/(μr04)  = 4k > 0.
Since k > 0, this is a position of stable equilibrium.
(c)  μd2ρ/dt2 = -4kρ for small displacements from equilibrium.
The angular frequency of small oscillations about r0 is ωosc = 2(k/μ)½.
The angular frequency of the orbit with radius r0 is ωorbit = M/(μr02) = r02(μk)½/(μr02) = (k/μ)½.
ωosc = 2ωorbit.
The relative distance between the particles goes through exactly two cycles for each full orbit.  The orbits are closed.

#### Problem:

A particle of mass m moves under the action of a central force whose potential energy function is U(r) = kr4,  k > 0.
(a)  For what energy and angular momentum will the orbit be a circle of radius a about the origin?  What is the period of this circular motion?
(b)  If the particle is slightly disturbed from this circular motion, what will be the period of small radial oscillations about r = a?

Solution:

• Concepts:
A particle moving in a central potential
U = U(r),  F = f(r)(r/r),  f(r) = -∂U/∂r.
In a central potential the energy E and the angular momentum M are conserved.
E = ½m(dr/dt)2 + M2/(2mr2) + U(r) = ½m(dr/dt)2 + Ueff(r),
with Ueff(r) = M2/(2mr2) + U(r).
md2r/dt2 = -∂Ueff(r)/∂r.
• Reasoning:
A potential energy function U(r) is given.  The particle moves under the action of a central force.
• Details of the calculation:
(a)  For a circular orbit with radius a we need r = a, dr/dt = 0, d2r/dt2 = 0.
For a circular orbit at with radius a we therefore need ∂Ueff(r)/∂r|a = 0.
∂Ueff(r)/∂r =  4kr3 - M2/(mr3).
4ka3 = M2/(ma3),  a6  =  M2/(4mk),  M = 2a3(km)½.
E = Ueff(a) = ka4 + M2/(2ma2) = ka4 + 4mka6/(2ma2) = 3ka4.
The period for the circular motion is τ = 2π/ω, ω = dΦ/dt = M/ma2 = 2a(k/m)½.
τ = (π/a)(m/k)½.
(b)  For a stable orbit we need a restoring force.  Let r = a + ρ, r >> ρ.
We need md2ρ/dt2 = -αρ.
md2ρ/dt2 = -∂Ueff(ρ)/∂ρ = -(∂Ueff/∂ρ)|ρ=0 - (∂2Ueff/∂ρ2)|ρ=0ρ (series expansion).
We need ∂2Ueff/∂ρ2 > 0 at ρ = 0, or ∂2Ueff/∂r2 > 0 at r = a.
∂Ueff(r)/∂r =  4kr3 - M2/(mr3).
2Ueff(r)/∂r2 =  12kr2 + 3M2/(mr4).
2Ueff/∂ρ2|r=a = 12ka2 + 3M2/(ma4) = 12ka2 + 3kma6/(ma4) = 24 ka2 > 0.
The orbit with radius a is stable, α = 24 ka2.  The period of small radial oscillation about a is
τ = 2π/ωr,  ωr = (α/m)½ = (24 ka2/m)½.  τ = (π/a)(m/(6k))½.

Potential energy U proportional to 1/rn

#### Problem:

A particle of mass m moves under an attractive central force with magnitude A/r3
(a)  Find the condition for which it moves with constant radial speed.
(b)  For this special case, find the orbital equation r(Φ), where r and Φ are polar coordinates.

Solution:

• Concepts:
Motion in a central potential
• Reasoning:
The particle moves under an attractive central force.  The motion is in a plane.  Energy E and angular momentum M are conserved.
• Details of the calculation:
(a)  For motion in a central potential we have
L = T - U = ½m[(dr/dt)2 + r2(dΦ/dt)2] - U(r).
E = ½m[(dr/dt)2 + r2(dΦ/dt)2] + U(r) = constant.
M = pΦ = mr2dΦ/dt = constant.
E = ½m(dr/dt)2 + M2/(2mr2) + U(r) = ½m(dr/dt)2 + Ueff(r),
with Ueff(r) = M2/(2mr2) + U(r).
Here U(r) = -A/(2r2).
Lagrange's equations yield
md2r/dt2 - M/(mr3) = f(r).
If d2r/dt2 = 0 then M2/(mr3) = -f(r) = A/r3 ,  M2 = mA.
For this motion E = ½mvr2, with vr constant.
(b)  dr/dt = vr = constant.  dΦ/dt = M/mr2,  dr/dΦ = mr2vr/M.
dr/r2 = (mvr/M)dΦ,  (1/r0 – 1/r) = (mvr/M)(Φ – Φ0).
Choose Φ0 = 0.  Then 1/r = 1/r0 – (mvr/M)Φ,  r = r0/(1 - mvrr0Φ/M).
r = r0/(1 – constant*Φ).
If the constant is positive the particle spirals outward, if it is negative the particle spirals inward.
If the constant is zero, orbit is circular.

#### Problem:

Consider the potential energy function U(r) = kr-1exp(-αr), where k < 0 and α > 0.
(a)  Find the corresponding force F
(b)  Assuming a particle of mass m, subject to this force, moves in a circle of radius a, find its angular momentum M and energy E.  What is the period of the circular motion?

Solution:

• Concepts:
Motion in a central potential, F = f(r)(r/r), f(r) = (-∂U/∂r)
The motion is in a plane.  Energy E and angular momentum M are conserved.
M = mr2dΦ/dt,  E = ½m(dr/dt)2 + M2/(2mr2) + U(r) = ½m(dr/dt)2 + Ueff(r) .
The equation of the orbit is  (M/r2)d/dΦ[(M/mr2)dr/dΦ] - M2/(mr3) = f(r).
• Reasoning:
We are given a central potential.  The motion is in a plane.  We use spherical coordinates and let the motion be in the θ = π/2 plane.  Then the direction of the angular momentum M is perpendicular to this plane and the equations given above can be applied.
• Details of the calculation:
(a)  f(r) = (-dU/dr) = (αk/r + k/r2) exp(-αr) = -(α + 1/r)(|k|/r)exp(-αr),  F = f(r)(r/r).

(b)  Motion in a circle:  dr/dΦ = 0 at r = a.
(M/r2)d/dΦ[(M/mr2)dr/dΦ] - M2/(mr3) = f(r).
The above equation becomes  M2/(mr3) = (α + 1/r)(|k|/r)exp(-αr) at r = a.
Angular momentum:
M2 = ma2(α + 1/a)|k|exp(-αa).
E = ½m(dr/dt)2 + M2/(2mr2) + U(r.
At r = a, we have dr/dt = 0.
Energy:
E = |k|(α/2 - 1/(2a))exp(-αa).
Period:
T = 2π/ω,  ω = dΦ/dt = M/(ma2).

#### Problem:

A particle of mass m moves in a circular orbit of radius r  = a under the influence of the central attractive force
f(r) = -g exp(-br)/r2,
where g and b are positive constants.
(a)  What is the effective potential energy, Ueff(M,r), for radial motion in terms of r and the angular momentum M?  (Your answer may contain an integral.)
(b)  For what values of b will this orbit be stable?
(c)  What is the frequency of small radial oscillation about these stable circular orbits?

Solution:

• Concepts:
Motion in a central potential, F = f(r)(r/r), f(r) = -∂U/∂r
Energy E and angular momentum M are conserved.
M = mr2dΦ/dt,  E = ½m(dr/dt)2 + M2/(2mr2) + U(r) = ½m(dr/dt)2 + Ueff(r) .
md2r/dt2 = -∂Ueff(r)/∂r.
• Reasoning:
The force is given.  The particle moves under the action of a central force.
• Details of the calculation:
(a)  U(r) = -∫infrf(r)dr = ∫infr(g exp(-br)/r2)dr.
Ueff(r) = M2/(2mr2) + ∫infr(g exp(-br)/r2)dr.
(b)  For a circular orbit of radius a we need d2r/dt2|a = 0,
therefore ∂Ueff(r)/∂r|a = 0, -f(a) = M2/ma3,  g exp(-ba)/a2 = M2/ma3.
For the orbit to be stable we need a restoring force.  Let r = a + ρ, r >> ρ.
We need md2ρ/dt2 = -αρ.
But md2ρ/dt2 = -∂Ueff/∂ρ = -∂2Ueff/∂ρ2|ρ=0ρ.
We need ∂2Ueff/∂ρ2 > 0 at ρ = 0.
2Ueff/∂ρ2|ρ=0 = (∂2Ueff/∂r2)|r=a = ∂(-M2/(mr3) - f(r))/∂r|a = 3M2/(ma4) - g exp(-ba)[2/a3 + b/a2].
We need 3M2/(ma4) - g exp(-ba)[2/a3 + b/a2]  > 0, or 3g exp(-ba)/a3 - g exp(-ba)[2/a3 + b/a2] > 0.
We therefore need 1/a - b > 0, b < 1/a.
(c)  md2ρ/dt2 = -∂2Ueff/∂ρ2|ρ=0ρ = -[g exp(-ba)/a2][1/a - b]ρ.
ω = (g exp(-ba)/(ma2)][1/a - b])½.
In the limit b = 0 we get the result for the Kepler orbit, ω = (g/(ma3))½, the period Tosc of small radial oscillations is equal to the orbital period Torbit = 2π(g/(ma3)).

Motion with a given orbit

#### Problem:

(a) Find the force law for a central force field which allows a particle to move in a spiral orbit r = kΦ2, where k is a positive constant.
(b) Compute the effective one-dimensional potential energy.
(c) Find the total energy, E, for which this motion is allowed.

Solution:

• Concepts:
Motion in a central potential, F = f(r)(r/r), f(r) = (-dU/dr)
• Reasoning:
For motion under the influence of a central force E and M are conserved.
M = mr2dΦ/dt,  E = ½m(dr/dt)2 + Ueff(r),  Ueff(r) = U(r) + M2/(2mr2).
The equation that relates the orbit to the force is
(M2u2/m)(d2u/dΦ2 + u) = -f(u),  where  u = 1/r.
The orbit is given, we have to find f(r).
• Details of the calculation:
(a)  u = 1/r = 1/(kΦ2) is given.
du/dΦ = -2/(kΦ3),  d2u/dΦ2 = 6/(kΦ4) = 6ku2,
(M2u2/m)(6ku2 + u) = -f(u),
f(r) = -(M2/m)((6k/r4) + (1/r3)).
For a particle of mass m and angular momentum M to have the orbit r = kΦ2 the force must be
f(r) = -(M2/m)((6k/r4) + (1/r3)).
(b)  f(r) = -(M2/m)((6k/r4) + (1/r3)).  What is the effective potential energy?
md2r/dt2 = -f(r) + M2/(mr3) =  -d(U(r) + M2/(2mr2))/dr = -dUeff(r)/dr.
md2r/dt2 = -6M2k/(mr4) - M2/(mr3) + M2/(mr3) = -6M2k/(mr4 ).
dUeff(r)/dr = 6M2k/(mr4),  Ueff(r) = -2M2k/(mr3),  Ueff(∞) = 0.
(c)  What must  be the energy of the particle?
E = ½ m(dr/dt)2 + Ueff(r),
r = kΦ2,
dr/dt = 2kΦdΦ/dt = 2k √(r/k) M/(mr2).
E = ½m(4k2r/k) M2/(m2r4) - 2M2k/(mr3) = 0.
For the particle to have the orbit (r = kΦ2), its total energy must be zero.

#### Problem:

Under the influence of a central force f(r), a particle follows the trajectory described by r = a/(Φ+1)2, where a is a constant.
(a)  Find the force of f(r).
(b)  At Φ = 0, the particle receives an impulse which reduce to zero its radial component of the velocity (vr) and which doubles its transverse component on the velocity (vf).  What is the path of the particle following this impulse?

Solution:

• Concepts:
Motion in a central potential, F = f(r)(r/r), f(r) = (-dU/dr)
• Reasoning:
For motion under the influence of a central force E and M are conserved.
M = mr2dΦ/dt,  E = ½m(dr/dt)2 + Ueff(r),  Ueff(r) = U(r) + M2/(2mr2).
The equation that relates the orbit to the force is
(M2u2/m)(d2u/dΦ2 + u) = -f(u),  where  u = 1/r.
The orbit is given, we have to find f(r).
• Details of the calculation:
(a)  u = (Φ+1)2/a is given,  d2u/dΦ2 = (2/a).
f(u) = -(M2u2/m) ((2/a) + u),
f(r) = -(M2/ma) [(2/r2) + (a/r3)].
(b)  At Φ = 0 and r = a we have vΦ = a(dΦ/dt),  M = ma2dΦ/dt.
The impulse doubles vΦ and therefore doubles the angular momentum.
After the impulse the equation of the orbit is
((2M)2u2/m)(d2u/dΦ2 + u) = (M2u2/m)((2/a) + u)  or
d2u/dΦ2 = 1/(2a) + u/4 - u  =  1/(2a) - ¾u.
[The angular momentum of the particle has changed because of the impulse, but the force is still the same.]
We want to solve this second order differential equation for u.
Let u = Asin(B(Φ + δ)) + C,   d2u/dΦ2 =  - AB2sin(B(Φ + δ)) = - B2u + B2C.
Then B2 = ¾,  B2C = 1/(2a),  C = 2/(3a),  1/r = A sin(√¾(Φ + δ)) + 2/(3a).
Initial conditions:
At r = a, Φ = 0,  1/a = Asin(√¾δ) + 2/(3a),  Asin(√¾δ) = 1/(3a).
At  r = a, dr/dt = 0.
r = 1/[A sin(√¾(Φ + δ)) + (2/(3a))].
dr/dt = (M/mr2)dr/dΦ .
dr/dt |r=a =  0, dr /dΦ|Φ=0 = 0.
dr/dΦ = - r2√¾Acos(√¾(Φ + δ)).
At Φ = 0 we have
√¾ Acos(√¾δ) = 0.  √¾δ = π/2.
Therefore from above:
A = 1/(3a),  1/r = [1/(3a)]sin(√¾Φ + π/2) + 2/(3a).
r = 3a/[sin(√¾Φ + π/2) + 2/(3a)] describes the path of the particle after the impulse.

#### Problem:

A particle of mass m moves with angular momentum M in a central potential V(r).  It moves in a elliptical orbit with semi-major axis a and semi-minor axis b.  The center of the ellipse is at the origin.  Find the potential V(r), the magnitude of the central force f(r) and the energy of the particle

Solution:

• Concepts:
Motion in a central potential, F = f(r)(r/r), f(r) = -dU/dr
• Reasoning:
For motion under the influence of a central force E and M are conserved.
M = mr2dΦ/dt,  E = ½m(dr/dt)2 + Ueff(r),  Ueff(r) = U(r) + M2/(2mr2).
The equation that relates the orbit to the force is
(M2u2/m)(d2u/dΦ2 + u) = -f(u),  where  u = 1/r.
The orbit is given, we have to find f(r).
• Details of the calculation:
equation of an ellipse: x2/a2 + y2/b2 = 1.

r2cos2Φ/a2 + r2sin2Φ/b2 = 1.
r2 = a2b2/(b2cos2Φ + a2sin2Φ),  u2 =  (b2cos2Φ + a2sin2Φ)/(a2b2) = (b2 + (a2 - b2)sin2Φ)/(a2b2).
u2 = (1/a2 + A2sin2Φ), with A2 = (a2 - b2)/(a2b2).
u = (1/a2 + A2sin2Φ)½.
(M2u2/m)(d2u/dΦ2 + u) = -f(u) yields f(u).
du/dΦ = A2sinΦcosΦ/u,  d2u/dΦ2 = A2(cos2Φ - sin2Φ)/u - (A2sinΦcosΦ/u2)(du/dΦ).
u2(d2u/dΦ2 + u) = (1/u)[u2A2(cos2Φ - sin2Φ) - (A4sin2Φcos2Φ) + u4]
= (1/u)[(1/a2 + A2sin2Φ)A2(cos2Φ - sin2Φ) - (A4sin2Φcos2Φ) + (1/a2 + A2sin2Φ)2]
= (1/u)[ A2/a2 + 1/a4].
f(u) = (M2/m)[A2/a2 +1/a4]/u,  f(r) = -(M2/m)[A2/a2 + 1/a4]r = -(M2/m)[r/(a2b2)].
We have a 3-dimensional harmonic oscillator.
f(r) = -kr, U(r) = ½kr2, k = (M2/m)/(a2b2).
(For a circular orbit a = b = r, M = mωr2.  Therefore k = mω2.)
At r = a dr/dt = 0, E = Ueff(r),
E = ½ka2 + M2/(2ma2) = (M2/(2ma2))(1/b2 + 1/a2).

Motion in a central potential with other forces present

#### Problem:

A particle of mass m moves in a plane under the influence of a central force of potential V(r) and also of a linear viscous drag -mk(dr/dt).  Set up Lagrange's equations of motion in plane polar coordinates and show that the angular momentum decays exponentially.

Solution:

• Concepts:
Motion in a central potential,  Lagrange's equations
• Reasoning:
We are asked to set up Lagrange's equations of motion in plane polar coordinates for motion in a central potential subject to drag.
• Details of the calculation:
L = T - U = ½m[(dr/dt)2 + r2(dΦ/dt)2] - U(r).
The generalized coordinates are r and Φ.
If a problem involves forces that cannot be derived from a potential, the Lagrange's equations become
d/dt(∂L/∂(dqi/dt)) -  ∂L/∂qi = Qi,
where the Qi are the generalized forces not derivable from a potential.  The Qi are defined through
Qi = r/∂qi.
Here Qr = Fr,  QΦ = r/∂Φ = rFΦ.
[dr = dr (r/r) + rdΦ (Φ/Φ),  ∂r/∂Φ = r (Φ/Φ)]
F = -mk(dr/dt),  Fr = -mk(dr/dt),  FΦ = -mkr(dΦ/dt).
Qr = -mk(dr/dt),  QΦ = -mkr2(dΦ/dt).

Lagrange's equations yield the equations of motion.
d/dt(∂L/∂(dΦ/dt)) = -mkr2(dΦ/dt),
d/dt(mr2(dΦ/dt)) = dM/dt = -kM.
M = M0exp(-kt).
The angular momentum decays exponentially.
d/dt(∂L/∂(dr/dt)) -  ∂L/∂r = Qr,
md2r/dt2 - mr(dΦ/dt)2 + ∂U/∂r = -km dr/dt.
md2r/dt2 - M2/(mr3) + ∂U/∂r = -km dr/dt.
md2r/dt2 - M02exp(-2kt)./(mr3) + ∂U/∂r = -km dr/dt.