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

Transcription

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
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