Solutions to Problems in Goldstein,
Classical Mechanics, Second Edition
Homer Reid
December 1, 2001
Chapter 3
Problem 3.1
A particle of mass m is constrained to move under gravity without friction on the
inside of a paraboloid of revolution whose axis is vertical. Find the one-dimensional
problem equivalent to its motion. What is the condition on the particle’s initial
velocity to produce circular motion? Find the period of small oscillations about
this circular motion.
We’ll take the paraboloid to be defined by the equation z = αr2 . The kinetic
and potential energies of the particle are
m 2
(ṙ + r2 θ̇2 + ż 2 )
2
m
= (ṙ2 + r2 θ̇2 + 4α2 r2 ṙ2 )
2
V = mgz = mgαr2 .
T =
Hence the Lagrangian is
L=
m
(1 + 4α2 r2 )ṙ2 + r2 θ̇2 − mgαr2 .
2
This is cyclic in θ, so the angular momentum is conserved:
l = mr2 θ̇ = constant.
1
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
2
For r we have the derivatives
∂L
= 4α2 mrṙ2 + mrθ̇2 − 2mgαr
∂r
∂L
= m(1 + 4α2 r2 )ṙ
∂ ṙ
d ∂L
= 8mα2 rṙ2 + m(1 + 4α2 r2 )r̈.
dt ∂ ṙ
Hence the equation of motion for r is
8mα2 rṙ2 + m(1 + 4α2 r2 )r̈ = 4α2 mrṙ2 + mrθ̇2 − 2mgαr
or
m(1 + 4α2 r2 )r̈ + 4mα2 rṙ2 − mrθ̇2 + 2mgαr = 0.
In terms of the constant angular momentum, we may rewrite this as
m(1 + 4α2 r2 )r̈ + 4mα2 rṙ2 −
l2
+ 2mgαr = 0.
mr3
So this is the differential equation that determines the time evolution of r.
If initially ṙ = 0, then we have
m(1 + 4α2 r2 )r̈ + −
l2
+ 2mgαr = 0.
mr3
Evidently, r̈ will then vanish—and hence ṙ will remain 0, giving circular motion—
if
l2
= 2mgαr
mr3
or
p
θ̇ = 2gα.
So if this condition is satisfied, the particle will execute circular motion (assuming its initial r velocity was zero). It’s interesting to note that the condition on
θ̇ for circular motion is independent of r.
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
3
Problem 3.2
A particle moves in a central force field given by the potential
V = −k
e−ar
,
r
where k and a are positive constants. Using the method of the equivalent onedimensional potential discuss the nature of the motion, stating the ranges of l and
E appropriate to each type of motion. When are circular orbits possible? Find the
period of small radial oscillations about the circular motion.
The Lagrangian is
i
mh 2
e−ar
ṙ + r2 θ̇2 + k
.
2
r
As usual the angular momentum is conserved:
L=
l = mr2 θ̇ = constant.
We have
e−ar
∂L
= mrθ̇2 − k (1 + ar) 2
∂r
r
∂L
= mṙ
∂ ṙ
so the equation of motion for r is
k
e−ar
(1 + ar) 2
m
r
k
e−ar
l2
(1 + ar) 2 .
= 2 3−
m r
m
r
r̈ = rθ̇2 −
The condition for circular motion is that this vanish, which yields
r
e−ar0 /2
k
θ̇ =
(1 + ar0 ) 3/2 .
m
r
(1)
(2)
0
What this means is that that if the particle’s initial θ velocity is equal to the
above function of the starting radius r0 , then the second derivative of r will
remain zero for all time. (Note that, in contrast to the previous problem, in this
case the condition for circular motion does depend on the starting radius.)
To find the frequency of small oscillations, let’s suppose the particle is executing a circular orbit with radius r0 (in which case the θ velocity is given by
(2)), and suppose we nudge it slightly so that its radius becomes r = r0 + x,
where x is small. Then (1) becomes
ẍ =
e−ar0
k
e−a[r0 +x]
k
1 + ar0
(1
+
a[r
+
x])
−
0
m
r02
m
[r0 + x]2
(3)
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
4
Since x is small, we may write the second term approximately as
k e−ar0
x
≈
(1
+
ar
+
ax)(1
−
ax)
1
−
2
0
m r02
r0
−ar0
−ar0
k
(1 + ar0 )
e
k e
≈ (1 + ar0 ) 2 +
a
−
a(1
+
ar
)
−
2
x
0
m
r0
m r02
r0
k
e−ar0
k e−ar0
2
2
≈ (1 + ar0 ) 2 −
2a
+
+
a
r
0 x.
m
r0
m r02
r0
The first term here just cancels the first term in (??), so we are left with
2
k e−ar0
2
2a +
+ a r0 x
ẍ =
m r02
r0
The problem is that the RHS here has the wrong sign—this equation is satisfied
by an x that grows (or decays) exponentially, rather than oscillates. Somehow
I messed up the sign of the RHS, but I can’t find where–can anybody help?
Problem 3.3
Two particles move about each other in circular orbits under the influence of gravitational forces, with a period τ . Their motion is suddenly stopped, and they are
then released
and allowed to fall into each other. Prove that they collide after a
√
time τ /4 2.
Since we are dealing with gravitational forces, the potential energy between
the particles is
k
U (r) = −
r
and, after reduction to the equivalent one-body problem, the Lagrangian is
L=
µ 2
k
[ṙ + r2 θ̇2 ] +
2
r
where µ is the reduced mass. The equation of motion for r is
µr̈ = µrθ̇2 −
k
.
r2
(4)
If the particles are to move in circular orbits with radius r0 , (4) must vanish at
r = r0 , which yields a relation between r0 and θ̇:
1/3
k
r0 =
µθ̇2
1/3
kτ 2
(5)
=
4π 2 µ
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
5
where we used the fact that the angular velocity in the circular orbit with period
τ is θ̇ = 2π/τ .
When the particles are stopped, the angular velocity goes to zero, and the
first term in (4) vanishes, leaving only the second term:
r̈ = −
k
.
µr2
(6)
This differential equation governs the evolution of the particles after they are
stopped. We now want to use this equation to find r as a function of t, which
we will then need to invert to find the time required for the particle separation
r to go from r0 to 0.
The first step is to multiply both sides of (6) by the integrating factor 2ṙ:
2ṙr̈ = −
2k
ṙ
µr2
or
d 2
d
ṙ = +
dt
dt
2k
µr
from which we conclude
ṙ2 =
2k
+ C.
µr
(7)
The constant C is determined from the boundary condition on ṙ. This is simply
that ṙ = 0 when r = r0 , since initially the particles are not moving at all. With
the appropriate choice of C in (7), we have
dr
=
ṙ =
dt
2k
µ
=
2k
µ
1/2 r
1/2 r
1
1
−
r
r0
r0 − r
.
rr0
(8)
We could now proceed to solve this differential equation for r(t), but since in
fact we’re interested in solving for the time difference corresponding to given
boundary values of r, it’s easier to invert (8) and solve for t(r):
Z 0 dt
∆t =
dr
dr
r0
Z 0 −1
dr
=
dr
dt
r0
µ 1/2 Z 0 rr 1/2
0
dr
=
2k
r
−
r
0
r0
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
6
We change variables to u = r/r0 , du = dr/r0 :
=
1/2
Z 0
µ 1/2
u
3/2
r0
du
2k
1−u
1
Next we change variables to u = sin2 x, du = 2 sin x cos x dx :
Z 0
µ 1/2
3/2
sin2 x dx
r0
2k
π/2
µ 1/2
3/2 π
=
r0
.
2k
2
=2
Now plugging in (5), we obtain
µ 1/2 kτ 2 1/2 π ∆t =
2k
4π 2 µ
2
τ
= √
4 2
as advertised.
Thanks to Joseph Westlake for correcting a mistake in my earlier solution
to this problem.
Problem 3.6
(a) Show that if a particle describes a circular orbit under the influence of an
attractive central force directed at a point on the circle, then the force varies
as the inverse fifth power of the distance.
(b) Show that for the orbit described the total energy of the particle is zero.
(c) Find the period of the motion.
(d) Find ẋ, ẏ, and v as a function of angle around the circle and show that all
three quantities are infinite as the particle goes through the center of force.
Let’s suppose the center of force is at the origin, and that the particle’s orbit
is a circle of radius R centered at (x = R, y = 0) (so that the leftmost point
of the particle’s origin is the center of force). The equation describing such an
orbit is
√
r(θ) = 2R(1 + cos 2θ)1/2
7
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
so
u(θ) =
1
1
.
=√
r(θ)
2R(1 + cos 2θ)1/2
(9)
Differentiating,
du
sin 2θ
= √
dθ
2R(1 + cos 2θ)3/2
2 cos 2θ
1
du
sin2 2θ
= √
+
3
dθ
(1 + cos 2θ)5/2
2R (1 + cos 2θ)3/2
1
1
2 cos 2θ + 2 cos2 2θ + 3 sin2 2θ .
= √
5/2
2 2R (1 + cos 2θ)
(10)
Adding (9) and (10),
1
d2 u
+u= √
(1 + cos 2θ)2 + 2 cos 2θ + 2 cos2 2θ + 3 sin2 2θ
2
5/2
dθ
2R(1 + cos 2θ)
1
=√
[4 + 4 cos 2θ]
2R(1 + cos 2θ)5/2
4
=√
2R(1 + cos 2θ)3/2
= 8R2 u3 .
(11)
The differential equation for the orbit is
m d
1
d2 u
+
u
=
−
V
2
2
dθ
l du
u
(12)
Plugging in (11), we have
m d
8R u = − 2
V
l du
2 3
1
u
so
1
2l2R2 4
V
=−
u
u
m
−→
V (r) = −
2l2 R2
mr4
(13)
so
f (r) = −
8l2R2
mr5
(14)
which is the advertised r dependence of the force.
(b) The kinetic energy of the particle is
T =
m 2
[ṙ + r2 θ̇2 ].
2
(15)
8
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
We have
r=
√
2R(1 + cos 2θ)1/2
r2 = 2R2 (1 + cos 2θ)
√
sin 2θ
θ̇
ṙ = 2R
(1 + cos 2θ)1/2
ṙ2 = 2R2
sin2 2θ 2
θ̇
1 + cos 2θ
Plugging into (15),
2 2
T = mR θ̇
2 2
= mR θ̇
sin2 2θ
+ 1 + cos 2θ
1 + cos 2θ
sin2 2θ + 1 + 2 cos 2θ + cos2 2θ
1 + cos θ
= 2mR2 θ̇2
In terms of l = mr2 θ̇, this is just
2R2 l2
mr4
But this is just the negative of the potential energy, (13); hence the total particle
energy T + V is zero.
=
(c) Suppose the particle starts out at the furthest point from the center of force
on its orbit, i.e the point x = 2R, y = 0, and that it moves counter-clockwise
from this point to the origin. The time required to undergo this motion is half
the period of the orbit, and the particle’s angle changes from θ = 0 to θ = π/2.
Hence we can calculate the period as
Z π/2
dt
dθ
τ =2
dθ
0
Z π/2
dθ
=2
θ̇
0
Using θ̇ = l/mr2 , we have
m
=2
l
Z
π/2
r2 (θ) dθ
0
Z
4R2 m π/2
(1 + 2 cos 2θ + cos2 2θ) dθ
l
0
4R2 m 3π
=
·
l
4
3πR2 m
=
.
l
=
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
9
Problem 3.8
(a) For circular and parabolic orbits in an attractive 1/r potential having the same
angular momentum, show that the perihelion distance of the parabola is one
half the radius of the circle.
(b) Prove that in the same central force
√ as in part (a) the speed of a particle at
any point in a parabolic orbit is 2 times the speed in a circular orbit passing
through the same point.
(a) The equations describing the orbits are
 2
l



mk r=
1
l2



mk 1 + cos θ
(circle)
(parabola.)
Evidently, the perihelion of the parabola occurs when θ = 0, in which case
r = l2 /2mk, or one-half the radius of the circle.
(b) For the parabola, we have
l2
sin θ
ṙ =
θ̇
mk (1 + cos θ)2
sin θ
= rθ̇
1 + cos θ
(16)
so
v 2 = ṙ2 + r2 θ̇2
sin2 θ
2 2
+1
= r θ̇
(1 + cos θ)2
2
2
2 2 sin θ + 1 + 2 cos θ + cos θ
= r θ̇
(1 + cos θ)2
1
2 2
= 2r θ̇
1 + cos θ
2mkr3 θ̇2
l2
2k
=
mr
=
(17)
in terms of the angular momentum l = mr2 θ̇2 . On the other hand, for the circle
ṙ = 0, so
k
l2
(18)
v 2 = r2 θ̇2 = 2 2 =
m r
mr
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
10
where we used that fact that, since this is a circular orbit, the condition k/r =
l2 /mr2 is satisfied. Evidently (17) is twice (18) for the same
√ particle at the
same point, so the unsquared speed in the parabolic orbit is 2 times that in
the circular orbit at the same point.
Problem 3.12
At perigee of an elliptic gravitational orbit a particle experiences an impulse S (cf.
Exercise 9, Chapter 2) in the radial direction, sending the particle into another
elliptic orbit. Determine the new semimajor axis, eccentricity, and orientation of
major axis in terms of the old.
The orbit equation for elliptical motion is
r(θ) =
a(1 − 2 )
.
1 + cos(θ − θ0 )
(19)
For simplicity we’ll take θ0 = 0 for the initial motion of the particle. Then
perigee happens when θ = 0, which is to say the major axis of the orbit is on
the x axis.
Then at the point at which the impulse is delivered, the particle’s momentum
is entirely in the y direction: pi = pi ĵ. After receiving the impulse S in the radial
(x) direction, the particle’s y momentum is unchanged, but its x momentum is
now px = S. Hence the final momentum of the particle is pf = S î+pi ĵ. Since the
particle is in the same location before and after the impulse, its potential energy
is unchanged, but its kinetic energy is increased due to the added momentum:
Ef = Ei +
S2
.
2m
(20)
Hence the semimajor axis length shrinks accordingly:
af = −
k
k
ai
=−
=
.
2Ef
2Ei + S 2 /m
1 + S 2 /(2mEi )
(21)
Next, since the impulse is in the same direction as the particle’s distance from
the origin, we have ∆L = r × ∆p = 0, i.e. the impulse does not change the
particle’s angular momentum:
Lf = Li ≡ L.
(22)
With (20) and (22), we can compute the change in the particle’s eccentricity:
r
2Ef L2
f = 1 +
mk 2
r
L2 S 2
2Ei L2
+ 2 2.
(23)
= 1+
2
mk
m k
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
11
What remains is to compute the constant θ0 in (19) for the particle’s orbit after
the collision. To do this we need merely observe that, since the location of the
particle is unchanged immediately after the impulse is delivered, expression (19)
must evaluate to the same radius at θ = 0 with both the “before” and “after”
values of a and :
af (1 − 2f )
ai (1 − 2i )
=
1 + i
1 + f cos θ0
or
(
)
1 af (1 − 2f )
cos θ0 =
−1 .
f
ai (1 − i )
Problem 3.13
A uniform distribution of dust in the solar system adds to the gravitational attraction of the sun on a planet an additional force
F = −mCr
where m is the mass of the planet, C is a constant proportional to the gravitational
constant and the density of the dust, and r is the radius vector from the sun to the
planet (both considered as points). This additional force is very small compared to
the direct sun-planet gravitational force.
(a) Calculate the period for a circular orbit of radius r0 of the planet in this combined field.
(b) Calculate the period of radial oscillations for slight disturbances from this circular orbit.
(c) Show that nearly circular orbits can be approximated by a precessing ellipse
and find the precession frequency. Is the precession the same or opposite
direction to the orbital angular velocity?
(a) The equation of motion for r is
l2
+ f (r)
mr3
k
l2
− 2 − mCr.
=
mr3
r
mr̈ =
(24)
For a circular orbit at radius r0 this must vanish:
0=
l2
k
− 2 − mCr0
3
mr0
r0
(25)
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
−→ l =
−→ θ̇ =
12
q
mkr0 + m2 Cr04
q
1
l
mkr0 + m2 Cr04
=
mr02
mr02
s
r
k
mCr03
1
+
=
mr03
k
s
mCr03
k
1
+
≈
mr03
2k
Then the period is
r mCr03
m
2π
3/2
1−
≈ 2πr0
k
2k
θ̇
2
Cτ0
= τ0 1 −
8π 2
τ=
3/2 p
where τ0 = 2πr0
m/k is the period of circular motion in the absence of the
perturbing potential.
(b) We return to (24) and put r = r0 + x with x r0 :
k
l2
−
− mC(r0 + x)
m(r0 + x)3
(r0 + x)2
l2
k
x
x
≈
− 2 1−2
− mCr0 − mCx
1−3
mr03
r0
r0
r0
mẍ =
Using (25), this reduces to
3l2
2k
mẍ = − 4 + 3 − mC x
mr0
r0
or
ẍ + ω 2 x = 0
with
2k
3l2
ω=
−
−C
4
2
m r0
mr03
1/2
2
k
2l
−
=
m2 r04
mr03
1/2
where in going to the last line we used (25) again.
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
13
Problem 3.14
Show that the motion of a particle in the potential field
V (r) = −
h
k
+ 2
r
r
is the same as that of the motion under the Kepler potential alone when expressed
in terms of a coordinate system rotating or precessing around the center of force.
For negative total energy show that if the additional potential term is very small
compared to the Kepler potential, then the angular speed of precession of the elliptical orbit is
2πmh
Ω̇ = 2 .
l τ
The perihelion of Mercury is observed to precess (after corrections for known planetary perturbations) at the rate of about 4000 of arc per century. Show that this
precession could be accounted for classically if the dimensionless quantity
η=
k
ka
(which is a measure of the perturbing inverse square potential relative to the gravitational potential) were as small as 7 × 10−8 . (The eccentricity of Mercury’s orbit
is 0.206, and its period is 0.24 year).
The effective one-dimensional equation of motion is
L2
k
2h
− 2+ 3
3
mr
r
r
k
L2 + 2mh
+ 2
=
mr3
r
L2 + 2mh + (mh/L)2 − (mh/L)2
k
=
+ 2
mr3
r
k
[L + (mh/L)]2 − (mh/L)2
+ 2
=
3
mr
r
mr̈ =
If mh L2 , then we can neglect the term (mh/L)2 in comparison with L2 , and
write
[L + (mh/L)]2
k
mr̈ =
+ 2
(26)
mr3
r
which is just the normal equation of motion for the Kepler problem, but with
the angular momentum L augmented by the additive term ∆L = mh/L.
Such an augmentation of the angular momentum may be accounted for by
14
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
augmenting the angular velocity:
2
L = mr θ̇
mh
mh
2
L 1 + 2 = mr θ̇ 1 + 2
L
L
−→
= mr2 θ̇ + mr2 Ω̇
where
mh
2πmh
=
L2 τ
L2 θ̇
is a precession frequency. If we were to go back and work the problem in the
reference frame in which everything is precessing with angular velocity Ω̇, but
there is no term h/r2 in the potential, then the equations of motion would come
out the same as in the stationary case, but with a term ∆L = mr2 Ω̇ added to
the effective angular momentum that shows up in the equation of motion for r,
just as we found in (26).
To put in the numbers, we observe that
m
2π
(h)
Ω̇ =
τ
L2
mka
2π
h
=
τ
L2
ka
1
2π
h
=
τ
1 − e2
ka
Ω̇ =
so
τ Ω̇
h
= (1 − e2 )
ka
2π
= (1 − e2 )τ fprec
where in going to the third-to-last line we used Goldstein’s equation (3-62), and
in the last line I put fprec = Ω̇/2π. Putting in the numbers, we find
h
= (1 − .2062 ) · 0.24 yr · 4000
ka
= 7.1 · 10−8 .
1◦
360000
1 revolution
360◦
1 century−1
100 yr−1
yr−1
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
15
Problem 3.22
In hyperbolic motion in a 1/r potential the analogue of the eccentric anomaly is F
defined by
r = a(e cosh F − 1),
where a(1 − e) is the distance of closest approach. Find the analogue to Kepler’s
equation giving t from the time of closest approach as a function of F .
We start with Goldstein’s equation (3.65):
r Z r
m
dr
q
t=
2 r0
k
l2
r − 2mr 2 + E
r Z r
m
r dr
q
=
.
2 r0 Er2 + kr − l2
2m
(27)
With the suggested substitution, the thing under the radical in the denominator of the integrand is
Er2 + kr −
l2
l2
= Ea2 (e2 cosh2 F − 2e cosh F + 1) + ka(e cosh F − 1) −
2m
2m
l2
2
2 2
2
= Ea e cosh F + ae(k − 2Ea) cosh F + Ea − ka −
2m
It follows from the orbit equation that, if a(e − 1) is the distance of closest
approach, then a = k/2E. Thus
k 2 e2
k 2 e2
l2
cosh2 F −
−
4E 4E
2m k2
2El2
2
2
=
e cosh F − 1 +
4E
mk 2
k 2 e2
k 2 e2 =
cosh2 F − 1 =
sinh2 F = a2 e2 E sinh2 F.
4E
4E
=
Plugging into (27) and observing that dr = ae sinh F dF , we have
r
r
Z
m F
ma2
[e(sinh F − sinh F0 ) − (F − F0 )]
t=
a(e cosh F − 1) dF =
2E F0
2E
and I suppose this equation could be a jumping-off point for numerical or other
investigations of the time of travel in hyperbolic orbit problems.
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
16
Problem 3.26
Examine the scattering produced by a repulsive central force f = kr−3 . Show that
the differential cross section is given by
σ(Θ)dΘ =
k
(1 − x)dx
2E x2 (2 − x)2 sin πx
where x is the ratio Θ/π and E is the energy.
The potential energy is U = k/2r2 = ku2 /2, and the differential equation
for the orbit reads
m dU
mk
d2 u
+u=− 2
=− 2 u
2
dθ
l du
l
or
d2 u
mk
+ 1+ 2 u=0
2
dθ
l
with solution
u = A cos γθ + B sin γθ
where
(28)
r
mk
.
(29)
l2
We’ll set up our coordinates in the way traditional for scattering experiments:
initially the particle is at angle θ = π and a great distance from the force center,
and ultimately the particle proceeds off to r = ∞ at some new angle θs . The
first of these observations gives us a relation between A and B in the orbit
equation (28):
γ=
u(θ = π) = 0
1+
−→
−→
A cos γπ + B sin γπ = 0
A = −B tan γπ.
(30)
The condition that the particle head off to r = ∞ at angle θ = θs yields the
condition
A cos γθs + B sin γθs = 0.
Using (30), this becomes
− cos γθs tan γπ + sin γθs = 0
or
− cos γθs sin γπ + sin γθs cos γπ = 0
−→
sin γ(θs − π) = 0
−→
γ(θs − π) = π
Homer Reid’s Solutions to Goldstein Problems: Chapter 3
17
or, in terms of Goldstein’s variable x = θ/π,
γ=
1
.
x−1
(31)
Plugging in (29) and squaring both sides, we have
1+
1
mk
=
.
l2
(x − 1)2
Now l = mv0 s = (2mE)1/2 s with s the impact parameter and E the particle
energy. Thus the previous equation is
1+
k
1
=
2Es2
(x − 1)2
or
s2 = −
k (x − 1)2
.
2E x(x − 2)
Taking the differential of both sides,
k 2(x − 1)
(x − 1)2
(x − 1)2
dx
2s ds = −
− 2
−
2E x(x − 2) x (x − 2) x(x − 2)2
k 2x(x − 1)(x − 2) − (x − 1)2 (x − 2) − x(x − 1)2
=−
2E
x2 (x − 2)2
2(1 − x)
k
.
=−
2E x2 (x − 2)2
The differential cross section is given by
σ(θ)dΩ =
| s ds |
.
sin θ
Plugging in (32), we have
σ(θ)dΩ =
as advertised.
(1 − x)
k
dx
2E x2 (x − 2)2 sin θ
(32)