Problem GPS 3.13 If the center of force is located at the origin of an

Goldstein Problem 3-6
Problem GPS 3.13
If the center of force is located at the origin of an x-y coordinate system, then the orbit is a circle
that passes through the origin as, for example, drawn below. This circle has radius a, and its
origin is at (a, 0). The equation that describes it is (x!a)2 + y2 = a2 . In terms of polar coordinates
measured from the center of force, we have x = r cosθ and y = r sinθ . After substituting for x
and y in the equation of the circle, we obtain the representation of the orbit in polar coordinates
as r = 2a cosθ . Note that a full orbit is traversed as θ varies from !π/2 to π/2 instead of the
usual 0 to 2π.
Problem Goldstein 3-6
(a) Start with the differential equation of the orbit in terms of the variable
u = 1/r,
l2 2
u
m
µ
¶
1
d2 u
2 + u = −f ( u ) ,
dθ
(1)
where l is the angular momentum, m is the mass of the particle, and
u = (2a cos θ)−1 .
(2)
tan θ
du
=
,
dθ
2a cos θ
(3)
·
¸
2 tan2 θ
1
d2 u
1
+
=
.
2a cos θ
cos θ
dθ2
(4)
The required derivatives are
and
After we substitute Eqs.(2) and (4) into (1), we obtain
l2
8a2 l2 5
l2
1
1
1
(1 + tan2 θ) = − u2
=
−
u .
f( ) = − u2
u
m a cos θ
m a cos3 θ
m
(5)
Thus, in terms of r, the force is
k
,
r5
(6)
8a2 l2
.
m
(7)
f(r) = −
where the force constant k is
k=
(b) The potential corresponding to Eq.(6) is
V (r) = −
k
.
4r4
(8)
Thus, the total energy of this orbit is
E=
·2
k
m ·2
(r + r 2 θ ) − 4 .
2
4r
(9)
With
r = 2a cos θ ,
1
(10)
we find
·
·
r = −2aθ sin θ .
(11)
After substituting Eqs.(10) and (11) into Eq.(9), we find
·2
E = 2ma2 θ −
k
,
4r4
(12)
but from the conservation of angular momentum we also know that
·
l = mr2 θ .
(13)
Thus, Eq.(12) simplifies to
E=
1
4r4
µ
8a2 l2
−k
m
¶
=0,
(14)
which is seen to equal zero after substituting for k using Eq.(7).
(c) The period of the motion may be found from the conservation of angular
momentum, Eq.(13), in the form
·
l
dA
r2 θ
=
=
,
2
2m
dt
(15)
where A is the area of the orbit, in this case, πa2 . Integrating Eq.(15) over a
full period τ gives
τl
= πa2 ,
2m
(16)
τ = 2πma2 /l .
(17)
and solving for τ gives
(d) Start with the definitions of x and y in polar coordinates, to obtain
·
·
·
·
·
·
x = r cos θ − rθ sin θ ,
(18)
and
y = r sin θ + rθ cos θ .
·
(19)
·
Then, substitute for r and θ, using Eqs.(11) and (13) to find
·
x=−
l
tan θ ,
ma
2
(20)
and
·
y=
l
(1 − tan2 θ) .
2ma
(21)
We can find the speed either from
·2
·2
v = (x + y )1/2 ,
(22)
or, equivalently from
·2
·2
·
v = (r + r2 θ )1/2 = 2aθ.
·
(23)
·
·
After substituting for x and y or for θ, we obtain
v=
l
sec2 θ .
2ma
(24)
At the center of force θ equals either π/2 or −π/2, for which sec θ and tan θ are
· ·
infinite. Thus, it is clear that x, y, and v are also infinite at this point.
3