PHY4222/Spring 08: CLASSICAL MECHANICS II
HOMEWORK ASSIGNMENT #4: SOLUTIONS
due by 12:50 p.m. Wed 02/06
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly. Every (algebraic) final result
must be supplemented by a check of units. Without such a check, no more
than 75% of the credit will be given even for an otherwise correct solution.
1. Thornton & Marion, Problem 8-10. [30 points]
Solution
For the bounded motion in the 1/r 2 force,
1
T = − U.
2
min
This result is easy to prove for a circular orbit, when E = Ueff
. Indeed,
M2
|k|
+
r
2mr2
|k|
M2
= 2 −
=0
r
mr3
M2
=
m |k|
Ueff = −
dUeff
dr
rmin
mk 2
2M 2
mk 2
U = U (rmin ) = − 2
M
mk 2
mk 2
min
= Ueff
=−
E = T +U =T −
2
2M 2
M 2
2
mk
mk
1
1
T =
− 2 =− U
=−
2
2M
2
M
2
min
Ueff
= Ueff (rmin ) = −
When the Sun’s mass changes, the kinetic energy of Earth remains the same, whereas is potential energy is
halved. Then
1
1
1
E1 = T1 + U1 = T0 + U0 = − U0 + U0 = 0,
2
2
2
where 0 and 1 refers to stages before and after the mass change, correspondingly. The new orbit is a parabola.
2. Thornton & Marion, Problem 3-43. [30 points]
Solution
a) Effective potential energy
Ueff =
M2
1
k (r − a)2 +
.
2
2mr2
b) The radius of a circular orbit is found from
dUeff
= 0
dr
1
M2
k (r − a)2 =
2
mr3
2
Since it is a motion with constant angular velocity ω0 = φ̇, it is convenient here to re-write M as M = mr 2 φ̇ =
mr2 ω0 so that
k (r0 − a) = mr0 ω02
r r
k r0 − a
ω0 =
m
r0
Ueff =
M2
1
k (r − a)2 +
2
2mr2
c) The frequency of small oscillations is obtained by expanding Ueff in Taylor series about r = r0
0
Ueff (r) = Ueff (r0 ) + (r − r0 ) Ueff
(r0 ) +
1
00
(r − r0 )2 Ueff
(r0 ) + . . .
2
The first term is a constant which can be absorbed into a reference energy, the second term vanishes because of
the condition for a circular orbit. Differentiating:
00
Ueff
(r0 ) = k + 3
m2 r04 ω02
M2
=k+3
= k + 3mω02
4
mr0
mr04
Total energy
E=
mṙ2
mṙ2
1
2
+ Ueff (r) =
+
k + 3mω02 (r − r0 ) ,
2
2
2
from where the frequency of small oscillations is
r
r r
r
k r0 − a
k
k
k 4r0 − 3a
2
ω=
+ 3ω0 =
+3
=
.
m
m
m r0
m
r0
3. A particle moves in a central force field given by the Yukawa potential[1]
k
U (r) = − exp(−r/a),
r
where k and a are positive.
a) Determine the nature of trajectories (bounded vs unbounded), depending on the values of total energy and
angular momentum. [20 points]
b) When is the circular motion possible? [10 points]
c) Find the period of small radial oscillations about the circular trajectory. [10 points]
Hints
a) it is convenient to work with the dimensionless radial distance x = r/a and dimensionless total ε = Ea/k
and effective potential u = Ueff a/k energies.
b) if you do everything right, the equation for the extrema of the effective potential energy must be of the
following form
x (x + 1) e−x = β,
where β ≡ Mz2 /2mak. Although there is no analytic solution to this equation, you can establish the condition
for this equation to have either zero or two roots.
Solution
It is convenient to introduce dimensionless effective potential energy
u(x) ≡ aUef f /α = −
β
e−x
+ 2,
x
2x
3
where x ≡ r/a ≥ 0 and β ≡ M 2 /2mαa ≥ 0, and also a dimensionless total energy
ε ≡ Ea/α.
The turning points of a trajectory are given by the roots of the equation
u(x) = ε.
Let us find a derivative of function u (x) :
u0 (x) = e−x /x2 + e−x /x − β/x3 .
The extrema of the effective potential energy are determined from u0 (x) = 0 or
x (x + 1) e−x = β.
(1)
Although we cannot solve this equation in elementary functions, we can determine a condition for this equation
to have roots. Indeed, equating the derivative of the left-hand side of Eq.(1) to zero, we see that the left-hand
side has a maximum at
√
1+ 5
.
x0 =
2
The maximum value of the LHS is
√ !
√ √ 1
1+ 5
y0 =
1 + 5 2 + 5 exp −
≈ 0.84...
4
2
If β > y0 , equation (1) has no roots.
This means that there are no extrema of the effective potential energy curve, which is a monotonically decreasing
function of the distance. On the other hand, if β < y0 , there are necessarily two roots, corresponding to a
minimum and maximum of u (x) . A plot of u (x) for β = 1.0 is shown in Fig. 2. In this case the motion is
unbounded and the turning point correspond to the intersection of the u (x) curve by line u = ε > 0. If β < y 0 ,
it is hard to see both the minimum and maximum on the same plot. In Fig. 3, I am plotting u (x) near the
minimum for β = 0.1. In Fig.4 is the same function but near the maximum. Figure 5, shows a cartoon for
u (x) , in which the maximum is blown up. Several situations are possible. For energies above the maximum
value of the potential curve, the motion is unbounded. For ε > 0 but below the maximum, the motion can be
bounded or unbounded depending on whether the particle starts off to the left of the maximum or to the right,
correspondingly. For ε < 0 but above the minimum, the motion is bounded. Finally, when the energy is equal
to the minimum, the trajectory is a circle.
Now, let’s study the stability of a circular orbit. The Lagrangian of our system is
L=
M2
α
mṙ2
+
+ e−r/a
2
2mr2
r
and the equation of motion is given by
mr̈ = F (r) ,
where
F (r) =
∂
∂r
M2
α −r/a
.
+
e
2mr2
r
It is convenient to re-write the e.o.m. in the dimensionless form
d2 x
= f (x) ,
dτ 2
where
τ ≡ t/τ0 , τ0 ≡
p
ma3 /α
4
1
0.8
0.6
0.4
0.2
0
0
2
4
6
8
x
x*(x+1)*exp(-x)
beta=1.0
beta=0.4
FIG. 1: Graphical solution of the equation x (x + 1) exp (−x) = β
and
f (x) =
d
dx
β
e−x
+
2
2x
x
.
On a circular orbit, r = const. Suppose that the radius of the orbit is rc , then f (xc ) = 0,where xc = r/rc .
When a small perturbation is applied in the radial direction,
x (τ ) = xc + δ (τ ) .
Substituting this into the equation of motion, we get
d2 δ
= f 0 (xc ) δ ≡ −ω 2 δ,
dτ 2
where ω 2 = −f 0 (xc ) . The period of oscillations is then
s
T = 2πτ0
1
−f 0 (x
c)
4. Bonus: A shell is fired from the ground along the Earth’s radius.
a) Find an implicit time-dependence of the trajectory, i.e., the dependence of time on position r, for all possible
values of the shells’s total energy E, i.e., E > 0, E < 0, and E = 0. [15 points]
5
50
40
30
20
10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
x
beta=1.0
FIG. 2: u (x) for β = 1.0.
b) Obtain the asymptotic forms of the explicit time-dependence, r (t) , for E → 0 and E → +∞. [10 points]
c) In what limit is the free-fall motion recovered? [5 points]
Solution
From energy conservation
α
mṙ2
− , α>0
2
r r
2 α
ṙ =
+E .
m r
E =
Introducing two scales
τ≡
s
mα2
2 |E|3
and q ≡ |E| /α,
the equation is integrated to
t=τ
where ± corresponds to E
Z
qr
qR
√
dx x
√
,
1±x
(2)
0. Let’s first discuss the case of E > 0. Now we need to solve the integral
6
0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
x
-0.2
-0.4
FIG. 3: u (x) for β = 0.4 near the minimum.
I=
R
√
√ √
dx x/ 1 ± x. Changing the varible to y = 1 + x gives
Z p
I =2
y 2 − 1dy
which, via another change of variables y = cosh z reduces to
Z
Z
1
2
I = 2 sinh zdz = (cosh 2z − 1) dz = sinh 2z − z = sinh z cosh z − z.
2
Now we need to express this result via the original variable x.
ez + e−z
t + t−1
=
→ t2 − 2ty + 1 = 0 →
2
2
p
t = ez = y ± y 2 − 1
y = cosh z =
We interested in positive z, hence t > 1 and
p
t = y + y2 − 1
p
√
√
z = ln y + y 2 − 1 = ln x + x + 1
The product sinh z cosh z is re-arranged as follows
q
p
√ p
sinh z cosh z = y sinh cosh−1 y = y cosh2 cosh−1 y − 1 = y y 2 − 1 = x (x + 1).
7
0.005
x
2
4
6
8
10
12
0
-0.005
-0.01
-0.015
FIG. 4: u (x) for β = 0.4 near the maximum.
FIG. 5: A cartoon of u (x) with both minimum and maximum.
Thus,
I=
p
√
√
x (x + 1) − ln x + x + 1 .
Finally, evaluating the prime function at the limits, we obtain
#
"
√
√
p
p
qr + qr + 1
t=τ
qr (qr + 1) − qR (qR + 1) − ln p
√
qR + qR + 1
√
Now, let’s consider the limits. For E → 0, variables qr and qR are small. The
terms are expanded to order
x5/2 :
p
√
√
1
1
1
x(1 + x) = x 1 + x + . . . = x + x3/2 − x5/2 + . . .
2
2
8
Logs are also expanded
ln
√
2
√
√
√
1
1
1
1 2
1 2
1 2
1 √
x + x + 1 = ln 1 + x + x − x . . . =
x+ x− x −
x+ x− x
2
8
2
8
2
2
8
8
3
4
1 √
1
1
1 2
1 √
1 2
+
x+ x− x
x+ x− x
−
+...
3
2
8
4
2
8
√
3
1
= x − x3/2 + x5/2 + . . .
6
40
Combining two expansions,
p
√
x (x + 1) − ln
x+
√
2
1
x + 1 = x3/2 − x5/2 + . . .
3
5
Thus, in the limit E → 0, the relation between t and r reduces to
t/τ =
1
2
1
2
3/2
5/2
3/2
5/2
(qr) − (qr) − (qR) + (qR) + . . .
3
5
3
5
(3)
Introduce the leading order solution of the inverse relation r0 (t) , satisfying
t/τ =
2
2
3/2
3/2
(qr0 ) − (qR)
3
3
or
r0 =
3t
2τ q 3/2
+R
3/2
2/3


3t
+R
=  q 2
3/2
mα
2 2|E|
3 (|E| /α)
2/3
3/2 
"
= p
3t
2m/α
+R
3/2
#2/3
(4)
This is the result we obtained in the class for E = 0. In order to find a correction to r 0 due to small but finite
E, substitute
r = r0 + δr, |δr| r0
into Eq.(3) and expand to leading order in δr :
2
1
2
1
(q [r0 + δr])3/2 − (q [r0 + δr])5/2 − (qR)3/2 + (qR)5/2
3
5
3
5
√
2
2
1
1
5/2
3/2
5/2
≈ (qr0 )2/3 + q 3/2 r0 δr − (qr0 ) − (qR) + (qR) .
3
5
3
5
t/τ =
Notice that we neglected δr in the second term, as this would correspond to next order in E. Using the definition
of r0 , we cancel the corresponding terms in the equation
√
1
1
5/2
5/2
0 = q 3/2 r0 δr − (qr0 ) + (qR)
5
5
or
5/2
δr =
5/2
q r0 − R5/2
|E| r0 − R5/2
=
.
√
√
5
r0
5α
r0
Thus,
5/2
r (t) = r0 +
|E| r0 − R5/2
,
√
5α
r0
where r0 is given in Eq.(4). For long times, when r R, the result is simplified further
r0 ≈
and
r (t) =
3t
p
2m/α
3t
p
!2/3
2m/α
|E|
+
5α
!2/3
3t
p
2m/α
!4/3
+...
9
The opposite limit of E → ∞ corresponds to qr, qR 1. Expanding I for x 1, we obtain
I =x−
1
ln x + . . .
2
or
t/τ = q (r − R) −
1
r
ln + . . .
2 R
The leading-order solution corresponds to a free motion
r0 = R + vt,
where v =
p
2E/m. Finding a correction, we obtain
r = R + vt +
α
vt
+...
ln 1 +
2 |E|
R
The case E < 0 is considered
√ similarly, except for one has to make a trigonometric rather than a hypertrigonometric substitution: y = 1 − x and y = sin z. Then
h
i
p
√
I = − sin−1 1 − x + x (1 − x)
and
t/τ = sin−1
p
1 − qR +
p
qR (1 − qR) − sin−1
p
p
1 − qr − qr (1 − qr).
Although the result looks quite different from what we obained for positivie energies, it is easy to check that
their expansions for small |E| are the same except for the signs
t/r =
1
2
1
2
3/2
5/2
3/2
5/2
(qr) + (qr) − (qR) − (qR) + . . .
3
5
3
5
The rest of the analysis can be performed as before.
Certainly, one could have obtained the results for E → 0, regardless of its sign by expanding the integral for t
[Eq.(2)] in x. Because both limits are small for q → 0,
Z qr
i 1h
i
√
1
2h
3/2
3/2
5/2
5/2
x 1 ∓ x dx =
(qr) − (qR)
(qr) − (qR)
I≈
∓
+...
2
3
5
qR
Likewise, the result for E → +∞ can be obtained by expanding the integrand in series for x 1
Z qr √ x
1
1
r
√
1−
dx = q (r − R) − ln + . . .
I≈
x
2x
2 R
qR
[1] In Nuclear Physics, Yukawa potential is used to describe the forces between nucleons (protons and neurons) mediated via
the exchange of π-mesons. Incidentally, it is also the potential of a point charge immersed into a conducting medium,
e.g., electrolyte or metal. In this context, it is known as Debye-Hückel or screened Coulomb potential potential. At small
distances (r a), the exponential term can be neglected, and the potential reduces to the bare Coulomb potential of a
point charge. At large distances (r a), the potential falls off much faster than the Coulomb potential. This is the effect of
counter-charges that the test charge pulls from the medium. These charge form a cloud surrounding the test charge, which
shields or screens the test charge, so that the net charge of the composite system is equal to zero.