PHY6426/Fall 2007: CLASSICAL MECHANICS
HOMEWORK ASSIGNMENT #3
due by 9:35 a.m. Wed 09/17
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly.
1. Goldstein, Problem 3.10. In addition to the parameters specified in the formulation of the problem, assume also
that the aphelion distance of the planet’s orbit is known. (35 points)
Solution
otations: M = mass of the planet, L = the angular momentum, gravitational potential U (r) = −k/r, k > 0.
Eccentricity of an elliptic orbit
e =
r
α = −
1+
EL2
2EL2
≈
1
+
≡1−α→
M k2
M k2
EL2
.
M k2
(1)
At aphelion (as well as at perehelion), the radial velocity (ṙ) is equal to zero, therefore right before the collision,
the energy is related to the angular momentum via
E=
k
L2
− ,
2
2M ra
ra
where ra is the distance to aphelion.
Momentum conservation
M va + mvc = (M + m)v 0 ,
where va = L/M ra is the tangential velocity of the planet at aphelion, vc is the velocity of the comet, and v 0 is
the velocity of the planet plus comet after the collision. As M m,
v0 =
L
m
+
vc .
M ra
M
Energy after the collision must be equal to zero for the planet to be on a parabolic trajectory. Neglecting m
compared to M,
E0 =
k
M v 02
−
=0
2
ra
or
M v 02
L2
=
−E
2
2M ra2
L m
L2
L2
+
v
=
−E
c
2M ra2
ra M
2M ra2
L m
αM k 2
vc = −E =
,
ra M
L2
where in the last line I used Eq.(1). The velocity of the comet is then
vc =
αM 2 k 2 ra
mL3
2
φ’
φ
θ
φmax
FIG. 1:
and
Tc =
α2 M 4 k 4 ra2
1
mvc2 =
.
2
2mL6
This form can be simplified if one uses the formula for the aphelion distance to exclude L
L2
p
=
1−e
M kα
= ra M kα
= ra3 M 3 k 3 α3
ra =
L2
L6
(2)
and
Tc =
1 α2 M 4 k 4 ra2
M 1 k
=
.
2 mra3 M 3 k 3 α3
2m α ra
Check the units: k/ra is energy, the rest of the result is dimensionless. It seems that the results for Tc diverges
as α−1 for α → 0. However, the aphelion goes to infinity in the same limit, so the divergence is cancelled. Using
Eq. (2), one can re-express the result in terms of the angular momentum
Tc =
M 2 k2
.
2mL2
2. Goldstein, Problem 3.15. The zenith at a given point on the Earth’s surface is the direction pointing away from
the direction of the force of gravity at that location. (30 points)
Solution
Speed v 0 is found from the energy conservation
mv 2
k
mv 02
=
− →
2
2
R
p
v0 =
v 2 − 2k/mR
where R is the radius of the Earth and m is the mass of the meteor. An inspection of the figure shows that
φ0 = φmax + θ − π, where φmax is the angle to the asymptote of the hyperbola at infinity and θ is the latitude
3
of the point of impact. From the equation of orbit,
p
r =
cos φmax
e cos (φmax ) + 1
= −1/e
=∞→
φmax = cos−1 (−1/e) ,
where e =
q
1+
2EL2
mk2
and
0
−1
φ = θ − π + cos
From energy conservation
E=

1
− q
1+
2EL2
mk2

.
k
mv 2
−
2
R
whereas
L = mvR sin φ.
Finally,

φ0 = θ − π + cos−1 − q
1+
1
m(mv 2 −2k/R)v 2 R2 sin2 φ
k2

.
3. A particle moves in the 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 conditions for the trajectories to be bounded/unbounded. [13 points]
b) When is the circular motion possible? [10 points]
c) Find the period of small radial oscillations about the circular trajectory. [12 points]
Solution
It is convenient to introduce dimensionless effective potential energy
u(x) ≡ aUef f /k = −
e−x
β
+ 2,
x
2x
where x ≡ r/a ≥ 0 and β ≡ M 2 /2mka ≥ 0, and also a dimensionless total energy
ε ≡ Ea/k.
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.
(3)
4
Graphically, this equation is solved in Fig. 2. 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.(3) 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+ 5
1
≈ 0.84...
1 + 5 2 + 5 exp −
y0 =
4
2
If β > y0 , equation (3) 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. 3. 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. 4, I am plotting u (x) near the
minimum for β = 0.1. Fig.5 shows the same function but near the maximum. Figure 5, shows u (x) for β = 0.4
near the maximum. 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=
mṙ2
1
k
+ mr2 φ̇2 − e−r/a
2
2
r
and the equation of motion is given by
mr̈ = Feff (r) ,
where
M2
d
−
mr3
dr
Fef f (r) =
k −r/a
e
,
r
where we have used the angular momentum conservation M = mr 2 φ̇. 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 ≡
x = r/a
p
ma3 /k
and
f (x) =
d
γ
−
3
x
dx
e−x
x
,
with
γ=
M2
.
kma
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,
0
0.2
0.4
0.6
0.8
1
1
2
x
3
4
5
5
FIG. 2: Graphical solution of the equation x (x + 1) exp (−x) = β
x (τ ) = xc + y.
Substituting this into the equation of motion and expanding f (xc + y) = f (xc ) + f 0 (xc ) y = f 0 (xc ) y, we get
d2 y
= f 0 (xc ) y ≡ −ω 2 y,
dτ 2
where ω 2 = −f 0 (xc ) . The period of oscillations is then
s
1
T = 2πτ0
0
−f (xc )
0
20
40
60
80
0.2
0.4
0.6
x
0.8
1
1.2
6
FIG. 3: u (x) for β = 1.0.
[1] In Nuclear Physics, the Yukawa potential is used to describe 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 the 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.
1
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
x
0
–0.5
–1
FIG. 4: u (x) for β = 0.4 near the minimum.
0.5
–1.5
7
0.002
14
12
10
8
6
x
0
–0.002
–0.004
FIG. 5: u (x) for β = 0.4 near the maximum.
4
8