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Page 1

Solutions to Problems in Goldstein,

Classical Mechanics, Second Edition

Homer Reid

August 22, 2000

Chapter 1

Problem 1.1

A nucleus, originally at rest, decays radioactively by emitting an electron of mo-
mentum 1.73 MeV/c, and at right angles to the direction of the electron a neutrino
with momentum 1.00 MeV/c. ( The MeV (million electron volt) is a unit of energy,
used in modern physics, equal to 1.60 x 10−6 erg. Correspondingly, MeV/c is a
unit of linear momentum equal to 5.34 x 10−17 gm-cm/sec.) In what direction does
the nucleus recoil? What is its momentum in MeV/c? If the mass of the residual
nucleus is 3.90 x 10−22 gm, what is its kinetic energy, in electron volts?

Place the nucleus at the origin, and suppose the electron is emitted in the
positive y direction, and the neutrino in the positive x direction. Then the
resultant of the electron and neutrino momenta has magnitude

|pe+ν | =


(1.73)2 + 12 = 2 MeV/c,

and its direction makes an angle

θ = tan−1
1.73

1
= 60◦

with the x axis. The nucleus must acquire a momentum of equal magnitude
and directed in the opposite direction. The kinetic energy of the nucleus is

T =
p2

2m
=

4 MeV2 c−2

2 · 3.9 · 10−22 gm
·
1.78 · 10−27 gm

1 MeV c−2
= 9.1 ev

This is much smaller than the nucleus rest energy of several hundred GeV, so
the non-relativistic approximation is justified.

1

Page 2

Homer Reid’s Solutions to Goldstein Problems: Chapter 1 2

Problem 1.2

The escape velocity of a particle on the earth is the minimum velocity required
at the surface of the earth in order that the particle can escape from the earth’s
gravitational field. Neglecting the resistance of the atmosphere, the system is con-
servative. From the conservation theorem for potential plus kinetic energy show
that the escape velocity for the earth, ignoring the presence of the moon, is 6.95
mi/sec.

If the particle starts at the earth’s surface with the escape velocity, it will
just manage to break free of the earth’s field and have nothing left. Thus after
it has escaped the earth’s field it will have no kinetic energy left, and also no
potential energy since it’s out of the earth’s field, so its total energy will be zero.
Since the particle’s total energy must be constant, it must also have zero total
energy at the surface of the earth. This means that the kinetic energy it has at
the surface of the earth must exactly cancel the gravitational potential energy
it has there:

1

2
mv2e � G

mMR
RR

= 0

so

v =



(

2GMR
RR

)

=

(

2 � (6:67 � 1011 m3 kg−3 s−2) � (5:98 � 1024 kg)
6:38 � 106 m

)1/2

= 11:2 km/s �
1 m

1:61 km
= 6:95 mi/s:

Page 35

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 one-
dimensional 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

L =
m

2

[

ṙ2 + r2θ̇2
]

+ k
e−ar

r
.

As usual the angular momentum is conserved:

l = mr2θ̇ = constant.

We have

∂L

∂r
= mrθ̇2 − k (1 + ar) e

−ar

r2

∂L

∂ṙ
= mṙ

so the equation of motion for r is

r̈ = rθ̇2 − k
m

(1 + ar)
e−ar

r2

=
l2

m2r3
− k

m
(1 + ar)

e−ar

r2
. (1)

The condition for circular motion is that this vanish, which yields

θ̇ =



k

m
(1 + ar0)

e−ar0/2

r
3/2
0

. (2)

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 exe-
cuting 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 =
k

m

(

1 + ar0
)e−ar0

r20


k

m
(1 + a[r0 + x])

e−a[r0+x]

[r0 + x]2
(3)

Page 36

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 4

Since x is small, we may write the second term approximately as

≈ k
m

e−ar0

r20
(1 + ar0 + ax)(1 − ax)

(

1 − 2 x
r0

)

≈ k
m

(1 + ar0)
e−ar0

r20
+

k

m

e−ar0

r20

(

a − a(1 + ar0) − 2
(1 + ar0)

r0

)

x

≈ k
m

(1 + ar0)
e−ar0

r20
− k

m

e−ar0

r20

(

2a +
2

r0
+ a2r0

)

x.

The first term here just cancels the first term in (??), so we are left with

ẍ =
k

m

e−ar0

r20

(

2a +
2

r0
+ a2r0

)

x

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 grav-
itational 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

U(r) = −k
r

and, after reduction to the equivalent one-body problem, the Lagrangian is

L = µ
2

[ṙ2 + r2θ̇2] +
k

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:

r0 =

(

k

µθ̇2

)1/3

=

(

kτ2

4π2µ

)1/3

(5)

Page 69

Homer Reid’s Solutions to Goldstein Problems: Chapter 9 10

of some set of four variables of which two are old variables and two are new.
After some hacking, I arrived at the set {x;Q1; py; Q2}. In terms of this set, the
remaining quantities are

y =

(

1

2
x− 1

�2
py

)

cotQ1 +
1


Q2 (9)

px =

(

�2

4
x− 1

2
py

)

cotQ1 −


2
Q2 (10)

P1 =

(

�2x2

8
− 1

2
xpy +

1

2�2
p2y

)

csc2Q1 (11)

P2 =


2
x+

1


py (12)

We now seek a generating function of the form F (x;Q1; py; Q2): This is of mixed
type, but can be related to a generating function of pure F1 character according
to

F1(x;Q1; y;Q2) = F (x;Q1; py; Q2) − ypy:
Then the principle of least action leads to the condition

pxẋ+ pyẏ = P1Q̇1 + P2Q̇2 +
@F

@x
ẋ+

@F

@py
ṗy +

@F

@Q1
Q̇1 +

@F

@Q2
Q̇2 + yṗy + pyẏ

from which we obtain

px =
@F

@x
(13)

y = − @F
@py

(14)

P1 = −
@F

@Q1
(15)

P2 = −
@F

@Q2
: (16)

Doing the easiest first, comparing (12) and (16) we see that F must have
the form

F (x;Q1; py; Q2) = −


2
xQ2 −

1


pyQ2 + g(x;Q1; py): (17)

Plugging this in to (14) and comparing with (14) we find

g(x;Q1; py) =

(

−1
2
xpy +

1

2�2
p2y

)

cotQ1 + (x;Q1): (18)

Plugging (17) and (18) into (13) and comparing with (10), we see that

@

@x
=
�2

4
x cotQ1

Page 70

Homer Reid’s Solutions to Goldstein Problems: Chapter 9 11

or

(x;Q1) =
�2x2

8
cotQ1: (19)

Finally, combining (19), (18), (17), and (15) and comparing with (11) we see
that we may simply take �(Q1) ≡ 0: The final form of the generating function
is then

F (x;Q1; py; Q2) = −
(



2
x+

1


py

)

Q2 +

(

�2x2

8
− 1

2
xpy +

1

2�2
p2y

)

cotQ1

and its existence proves the canonicality of the transformation.
Turning now to the solution of the problem, we take the B field in the z

direction, i.e. B = B0k̂; and put

A =
B0
2

(

− y î + x ĵ
)

:

Then the Hamiltonian is

H(x; y; px; py) =
1

2m

(

p −
q

c
A

)2

=
1

2m

[

(

px +
qB0
2c

y

)2

+

(

py −
qB0
2c

x

)2
]

=
1

2m

[

(

px +
�2

2
y

)2

+

(

py −
�2

2
x

)2
]

where we put �2 = qB=c. In terms of the new variables, this is

H(Q1; Q2; P1; P2) =
1

2m

[

(




2P1 cosQ1

)2

+
(




2P1 sinQ1

)2
]

=
�2

m
P1

= !cP1

where !c = qB=mc is the cyclotron frequency. From the Hamiltonian equations
of motion applied to this Hamiltonian we see thatQ2; P1; and P2 are all constant,
while the equation of motion for Q1 is

Q̇1 =
@H

@P1
= !c −→ Q1 = !ct+ �

for some phase �. Putting r =


2P1=�; x0 = P2=�, y0 = Q2=� we then have

x = r(sin!ct+ �) + x0; px =
m!c

2
[r cos(!ct+ �) − y0]

y = r(cos!ct+ �) + y0; py =
m!c

2
[r sin(!ct+ �) + x0]

in agreement with the standard solution to the problem.

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