Problems

Problem 1:

Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital radius of 4.22´105 km.  From these data, determine the mass of Jupiter.

Solution:

Concepts, principles, relations that apply to the problem:
Motion in a central potential,  Kepler's third law
Why do they apply?
Two astronomical objects pull each other towards their common center of mass.  Each object accelerates towards the center of mass.  The acceleration of object 1 is a1 = Gm2/R2, and the acceleration of object 2 is a2 = Gm1/R2.  If each object has velocity perpendicular to the direction of its acceleration and v12/R1 = a1, v22/R2 = a2, with R1 and R2 being the distances of object 1 and object 2 from the CM, then both objects orbit their common CM in circular orbits.  If object 1 is much more massive than object 2, then the CM lies very close to the center of object 1.  Then R2 is approximately equal to R and we can write Gm1/R = v22.  If object 2 is in a circular orbit about a much more massive object 1, its speed is given by this formula.  We may write v2 = 2pR/T2, where T2 is the period of object 2.  We then have Gm1/R = (2pR/T2)2, or T22 = (4p2/Gm1)R3.
How do they apply?
Let object 1 be Jupiter, and object 2 be Io.  Then
T22 = (4p2/Gm1)R3.
m1 = (4p2/GT22)R3.
= 4p2(4.22´108m)3/((6.67´10-11Nm2/kg2)(1.77´24´60´60s)2) = 1.9´1027kg.
Details of the calculation:
None

Problem 2:

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, principles, relations that apply to the problem:
Motion in a central potential, F = f(r)(r/r), f(r) = (-U/r)
Why do they apply?
f(r) = (-U/r) = -kr,  U(r) = (1/2)kr2.  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 = (1/2)mv2 + (1/2)kR2.
How do they apply?
For circular motion at constant speed we have |F| = mv2/R = kR.
v = Rdf/dt,  mR(df/dt)2 = kR.  (df/dt)2 = k/m.  v = R(k/m)1/2.
The energy is E = T + U = (1/2)mv2 + (1/2)kR2 = (1/2)mR2(k/m) + (1/2)kR2 = kR2.
Details of the calculation:
None

Problem 3:

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, principles, relations that apply to the problem:
Motion in a central potential,  Lagrange's equations
Why do they apply?
We are asked to set up Lagrange's equations of motion in plane polar coordinates for motion in a central potential subject to drag.
How do they apply?
L = T - U = (1/2)m[(dr/dt)2 + r2(df/dt)2] - U(r).
The generalized coordinates are r and f.
If a problem involves forces that cannot be derived from a potential, the Lagrange's equations become
,
where the Qi are the generalized forces not derivable from a potential.  The Qi are defined through
Qi = F×r/qi.
Qr = Fr,  Qf = F×r/¶f = rFf.
.
F = -mk(dr/dt),  Fr = -mk(dr/dt),  Ff = -mkr(df/dt).
Qr = -mk(dr/dt),  Qf = -mkr2(df/dt).
Details of the calculation:
Lagrange's equations yield the equations of motion.
.
The angular momentum decays exponentially.

.

.

Problem 4:

Consider the potential energy function U(r) = kr-1exp(-ar), where k < 0 and a > 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, principles, relations that apply to the problem:
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 = mr2df/dt,
.
The equation of the orbit is
.
Why do they apply?
We are given a central potential.  The motion is in a plane.  We use spherical coordinates and let the motion be in the q = p/2 plane.  Then the direction of the angular momentum M is perpendicular to this plane and the equations given above can be applied.
How do they apply?
(a)  f(r) = (-dU/dr) = (ak/r + k/r2) exp(-ar) = -(a + 1/r)(|k|/r)exp(-ar),  F = f(r)(r/r).
Details of the calculation:
Motion in a circle:  dr/df = 0 at r = a.
.
becomes  M2/(mr3) = (a + 1/r)(|k|/r)exp(-ar) at r = a.
Angular momentum:
M2 = ma2(a + 1/a)|k|exp(-aa).
 
is evaluated at r = a, dr/dt = 0.

Energy:
E = |k|(a/2 - 1/(2a))exp(-aa).

Period:
T = 2p/w,  w = df/dt = M/(ma2).

Problem 5:

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, principles, relations that apply to the problem:
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 and the angular momentum are conserved.
.
.
Why do they apply?
A potential energy function U(r) is given.  The particle moves under the action of a central force.
How do they apply?
(a)  For a circular orbit with radius a we need r = a, dr/dt = 0, d2r/dt2 = 0.
.
.
The period for the circular motion is t = 2p/w, w = df/dt = M/ma2 = 2a(k/m)1/2.
t = (p/a)(m/k)1/2.
Details of the calculation:
(b)  For a stable orbit we need a restoring force.  Let r = a + r, r >> r.
We need md2r/dt2 = -ar = -Ueff(r)/¶r = -(Ueff/¶r)|r=0 - (2Ueff/¶r2)|r=0r,
or 2Ueff/¶r2 > 0 at r = 0.

.
.

The orbit with radius a is stable, a = 24ka2.  The period of small radial oscillation about a is
.

Problem 6:

(a) Find the force law for a central force field which allows a particle to move in a spiral orbit r = kf2, 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, principles, relations that apply to the problem:
Motion in a central potential, F = f(r)(r/r), f(r) = (-dU/dr)
Why do they apply?
For motion under the influence of a central force E and M are conserved.
M = mr2,  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/df2 + u) = -f(u),  where  u = 1/r.
The orbit is given, we have to find f(r).
How do they apply?
(a)  u = 1/r = 1/(kf2) is given. 
du/df = -2/(kf3),  d2u/df2 = 6/(kf4) = 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 = kf2 the force must be
f(r) = -(M2/m)((6k/r4) + (1/r3)).
Details of the calculation:
(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),
,
E = ½m(4k2r/k) M2/(m2r4) - 2M2k/(mr3) = 0.
For the particle to have the orbit (r = kf2), its total energy must be zero.

Problem 7:

Under the influence of a central force f(r), a particle follows the trajectory described by r = a/(f+1)2, where a is a constant.
(a)  Find the force of f(r).
(b)  At f = 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, principles, relations that apply to the problem:
Motion in a central potential, F = f(r)(r/r), f(r) = (-dU/dr)
Why do they apply?
For motion under the influence of a central force E and M are conserved.
M = mr2,  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/df2 + u) = -f(u),  where  u = 1/r.
The orbit is given, we have to find f(r).
How do they apply?
(a)  u = (f+1)2/a is given,  d2u/df2 = (2/a).
 f(u) = -(M2u2/m) ((2/a) + u),
f(r) = -(M2/ma) [(2/r2) + (a/r3)].
Details of the calculation:
(b)  At f = 0 and r = a we have vf = a(df/dt),  M = ma2df/dt.
The impulse doubles vf and therefore doubles the angular momentum.
After the impulse the equation of the orbit is
((2M)2u2/m)(d2u/df2 + u) = (M2u2/m)((2/a) + u)  or
d2u/df2 = 1/(2a) + u/4 - u  =  1/(2a) - (3/4)u.
[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(f + d)) + C,   d2u/df2 =  - AB2sin(B(f + d)) = - B2u + B2C.
Then B2 = ¾,  B2C = 1/(2a),  C = 2/(3a),  1/r = A sin(Ö(3/4)(f + d)) + 2/(3a).
Initial conditions:
 At r = a, f = 0,  1/a = Asin(Ö(3/4)d) + 2/(3a),  Asin(Ö(3/4)d) = 1/(3a).
At  r =a, dr/dt = 0. 
r = 1/[A sin(Ö(3/4)(f + d)) + (2/(3a))].
 dr/dt = (M/mr2)dr/df .
dr/dt |r=a =  0, dr /df|f=0 = 0.
dr/df = - r2Ö(3/4)Acos(Ö(3/4)(f + d)).
At f = 0 we have
Ö(3/4) Acos(Ö(3/4)d) = 0.  Ö(3/4)d = p/2.
Therefore from above:
A = 1/(3a),  1/r = [1/(3a)]sin(Ö(3/4)f + p/2) + 2/(3a).
r = 3a/[sin(Ö(3/4)f + p/2) + 2/(3a)] describes the path of the particle after the impulse.

Problem 8:

Calculate the differential and total scattering cross section for scattering a particle off a fixed, "hard" sphere of radius R.

Solution:

Concepts, principles, relations that apply to the problem:
The scattering cross section, scattering from a spherically symmetric potential
Why do they apply?
The potential has spherical symmetry, V(r) = ¥, r < R;  V(r) = 0, r >R.
How do they apply?
For scattering off a central potential we have s(q) = |(b/sinq)(db/dq)|,  q = p - 2f0.

Referring to the figure above we have
b/R = sina,
q = p - 2a.
a = (p - q)/2.
b = Rsin(p/2 - q/2) = Rcos(q/2).
db/dq = (R/2)sin(q/2).
s(q) = (b/sinq)((R/2)sin(q/2) = R2cos(q/2)sin(q/2)/(2sinq) = R2/4.
stotal = òs(q)dW, = òòs(q)sin(q)dqdf = pR2 = total elastic scattering cross section.
Details of the calculation:
None

Problem 9:

A fixed force center scatters a particle of mass m and initial velocity u0 according to the force law f(r) = k/r3.  Determine the differential scattering cross section.

Solution:

Concepts, principles, relations that apply to the problem:
A particle moving in a central potential
U = U(r),  F = f(r)(r/r), f(r) = (-U/r).  U = k/(2r2).  E > 0 for scattering.
Why do they apply?
We are asked to find the differential scattering cross section for a particle being scattered by a central force.
How do they apply?
fr = -U/r = k/r3.
s(q) = |(b/sinq)(db/dq)|  q = p -2f0 if k > 0.
We need to find q as a function of b.
Details of the calculation:
 U = k/(2r2),  U(u) = ku2/2,  u = 1/r.

for motion in a central potential.
umax = (E/(b2E + k/2))1/2.

Let k > 0, repulsive potential:
.
,  since
.
q = p - pb/(b2 + k/(2E))1/2,  (p - q)2 = (pb)2/(b2 + k/(2E)),
b2 = [k/(2E)](p - q)2/[q(2p - q)].
db2/dq  = 2bdb/dq  = -(k/E)p2(p - q)/[q2(2p - q)2].
s(q) = (k/(2E))p2(p - q)/[q2(2p - q)2sinq].

Let k < 0, attractive potential:
If b2E < |k/2| we find no solution for umax, there is no turning point, the particle spirals into the origin.  Its angular velocity increases beyond any limit, such that mr2df/dt = M stays constant.
If b2E > |k/2| we have f0 = (bp/2)/(b2 + (k/(2E)))1/2 > p/2.
f0 may be much larger than 2p, the particle may turn around the force center several times.

Problem 10:

A target of Aluminum (A = 27) with an aerial density of 1 mg/cm2 is positioned perpendicular to a 0.5 mA beam of deuterons (Z = 1).  If the cross section for producing protons at an angle of 30o with respect to the beam direction is 15mb/sr, (1mb = 10-27 cm2), how many protons per second will be incident on a 1 cm2 detector facing the target at 30o and 10 cm from the target?

Solution:

Concepts, principles, relations that apply to the problem:
The cross section
Why do they apply?
We are given the cross section for producing a proton in a single deuteron - Aluminum nucleus collision.
We are asked to find the total number of protons hitting a detector per second.
How do they apply?
If we have not one, but many scattering centers we distinguish the following limiting cases:
(a)  small beam, big target:

(# of particles scattered into the solid angle dW)/s
= [(# of beam particles)/s] ´ [(# of target particles)/area ] ´ s(W)dW.

(b)  big beam, small target:

(# of particles scattered into the solid angle dW)/s
= [(# of beam particles)/(area ´ s)] ´ (# of target particles) ´ s(W)dW.
Details of the calculation:
In this  problem we have case (a).
(# of beam particles)/s = (0.5´10-6C/s)/(1.6´10-19C) = 3.125´10-12/s.
(# of aluminum atoms)/cm2 = (1mg/cm2)/(27´1.6´10-27kg) = 2.29´1019/cm2.
What is dW?
r2dW = (10 cm2)dW = 1 cm2,  dW = 0.01 sr.
# of protons/s = (3.125´10-12/s) ´ (2.29´1019/cm2) ´ (15´10-27cm2) ´ 0.01 = 1 ´ 104/s.