Transcript r - Wayne State University

Central-Force Motion
Chapter 8
Prof. Claude A Pruneau
Physics and Astronomy Department
Wayne State University
8.1 Introduction
• Consider motion of two particles affected by a
force connecting the center of the two bodies.
• One of few problems that can be solved
completely.
• Historically important e.g motion of planets,
alpha-particle scattering on nuclei
8.2 Reduced Mass
• Description of a two-particle system
• Discussion restricted to frictionless
(conservative) systems.
8.2 Reduced Mass (cont’d)
• Assume a force is acting between the two particles along a line
joining them.
• Particle positions described in some arbitrary reference frame
as positions r1 and r2, or in terms of the CM frame position, R,
and the relative position vector r = r1 - r2 .
m1
m1
r1
r1
CM
R
r2
Arbitrary Frame
r
CM
R0
r2
m2
m2
CM Frame
Lagrangian for a two-body
system
• Assume the potential energy is only a
function of the distance between the particles,
U=U(r)
• Lagrangian may be written:
2
2
&2  U(r)
L  m1 r&
1  m2 r
1
2
1
2
• Translational motion of the system
uninteresting;
– Use R=0.
Two-body CM Coordinates
r
r
m1r1  m2 r2  0
• We have
m2 r
r1 
r
m1  m2
Solving for r1 and r2 :
r r
r  r1  r2
m1 r
r2  
r
m1  m2
Substitute in the Lagrangian r1 and r2 :
2
2
1
&  U(r)
L  12 m1 r&

1
2 m2 r2
2
L  12 m1
L  12 m1
m2 r& 1
m1 r&
r  2 m2
r  U(r)
m1  m2
m1  m2
m22
m1  m2 
L  12 m1m2
L
2
2
r2
r&  12 m2
m2
m1  m2 
2
m1  m2 
r2
r&  12 m1m2
1 m1m2 r&2
r  U(r)
2 m1  m2
m12
2
r2
r&  U(r)
m1
m1  m2 
2
r2
r&  U(r)
r&2
L   r  U(r)
1
2

m1m2
m1  m2
 reduced mass
2 to 1 reduction
8.3 Conservation Theorems
• Particle of mass  in a central force field described
by the potential function U(r).
• Symmetry implies conservation of angular
momentum.
r r
L  r  p  constant
L
Radius vector and momentum lie
in a plane normal to the angular
momentum vector L.
The problem reduced to 2
dimensions: i.e. along “r” and “q”.
r
p
• Lagrangian


L  12  r&2  r 2&2  U(r)
Lagrangian cyclic in q implies:
Angular momentum, pq, conjugate to q, is a
conserved quantity.
L
d L
p& 
0

dt &
L
2&
p 
 r   constant
&
First integral of motion
• The system’s symmetry permits the
integration of one equation of motion.
• p is the first integral of the motion.
• Denote it
l  r & constant
2
Note
• l can be negative or positive
Interpretation of l as Areal velocity
• The radius vector
sweeps out an area dA
in a time interval dt.
dA  r d
1
2
r (t1 )
2
rd
d
r (t 2 )
The areal velocity is thus
dA 1 2 d 1 2 &
 2r
 2r 
dt
dt
l

 constant
2
r (t)
Kepler’s 2nd law of planetary motion
dA
 constant
dt
• No particular assumption made about the form of U(r)
implies:
• This result is NOT limited to an inverse-square law
force but is valid for all central forces.
• Since the motion of the CM is not interesting, only
one degree of freedom remains to be considered.
• Linear momentum conservation adds nothing new
here…
• Energy conservation provides the only remaining
equation of motion.
T  U  E  constant
E= 12 v 2  U (r)
E= 12  r&2  r 2&2  U(r)

Total Energy

2
1
l
E= 12 r&2 
 U (r)
2
2 r
8.4 Equations of Motion
• Assume U(r) is specified. Solve for dr/dt:
1 l2
2
1
E= 2 r& 
 U (r)
2
2 r


dr
2
l2
r&

E  U (r)  2 2
dt

r
Solving for dt, and integrate to get a solution t = t(r).
Invert it to get r = r(t)
Alternatively, obtain q= q(r), starting with
d dt
&
d 
dr  dr
dt dr
r&


dr
2
l2
r&

E  U (r)  2 2
dt

r
 (r) 

d dt
&
d 
dr  dr
dt dr
r&

l  r 2& constant

 l / r 2 dr

l2 
2   E  U (r) 
2 r 2 

• Inversion of the result (if possible) yields the standard
form (general) solution r = r(t).
• Because l is constant, d/dt is a monotonic function
of time.
• The above integral is in practice possible only for a
limited number of cases…
Remarks
• with F(r) = rn, solutions may be expressed in
terms of elliptic integrals for certain integers
and fractional values of “n”.
• Solution may be expressed in terms of
circular functions for n=1, -2, and –3.
• Case n = 1 is the harmonic oscillator.
• Case n = -2 is the inverse square law.
Solution using Lagrange equations
L d L

0
r dt r&
Lagrange equation for “r”


U
2
&
&
&
 r  r  
 F(r)
r
1
Use variable change u 
r
du
1 dr
1 dr dt
1 r&
 2
 2
 2
d
r d
r dt d
r &
Remember
l   r 2&
l
&
 2
r
Compute…
l
&
 2
r
du
1 r&

  2   r&
d
l
r &
d 2u
d    dt d   
 r&&

 r& 
 r&  
2


d  l  d dt  l 
l &
d
d 2u
 r&&
2 2

  2 r r&&
2
l  l 
d
l
 r 2 
• Solving
l2
2
d
u
2
r&&  2 u

d 2
r&2 
l2

3
u
2


U
2
&
 F(r)
• Substitute back into  r&& r  
r
d 2  1 1
r 2
  2 F r
2  
d  r  r
l

• Which is useful if one wishes to find the force
law that produces a particular orbit r=r().
Example 8.1 – Log-spiral
Find the force law for a central-force field that allows a particle to
a
move in a logarithmic spiral orbit given by, r  ke , where k and
a are constants.
d  1
d  ea  a ea
Solution


d  r  d  k 
k
d 2  1  a 2 ea a 2


2  
k
r
d  r 
Calculate
Now use
d 2  1 1
r 2
  2 F r
d 2  r  r
l

1

To find:
1
 r 2   d 2  1  1   r 2   a 2 1 
F r   2   2       2  
 
r
 l   d  r  r   l   r


l2
  3 a2 1
r
Force is Attractive and Inverse cube!
Example 8.2 – r(t), (t)
Determine the functions r(t) and (t) for the problem in Ex 8.1.
Solution:
Start with:
l
l
&
  2  2 2a
r
k e
Rearrange, integrate:
e
2a
l
d 
dt
2
k
Answer:
e
2a
l
d   2 dt
k

1  2a lt
 (t) 
ln  2  C 
2a   k

e2a
lt

 C'
2
2a
k
Similarly for r(t), remember
And write
Answer (2):
l
l

2
r
 k 2 e2a
r2
2a lt
2a
e 
C
2
2
k
k
r(t) 
2a lt

 k 2C
Where l and C are determined by the initial conditions
Example 8.3 – Total Energy
What is the total energy of the orbit of the previous two examples?
Solution:
Need U…
l 2



l2
1
U (r)  
a2 1 2
2
r
l
l
&
  2  2 2a
r
k e
&


dr
U r    Fdr 
a 1  3

r
2
r  kea
d d dr
l

 2
dt dr dt r
dr l
l
al
a
r&
 2 a ke 
2
d r
r
r

lim U r  0
r
2
1  al 
l
E    

2
2  r 
2r
Given the reference
2

2r 2

lim U r  0
r
 0
l2 a 2  1
8.5 Orbits in a central field
• Radial velocity of a particle in central field


dr
2
l2
r&

E  U (r)  2 2
dt

r
•
Vanishes at the roots of the radical


l2
E  U (r)  2 2  0

r
2
l2
E  U (r) 
0
2
2 r
E  U (r) 
•
•
•
•
l
2
0
2 r
Vanishing of dr/dt implies turning points
Two roots in general: rmin and rmax.
Motion confined to an annular region between
rmin and rmax.
Certain combinations of E and l may lead to a
single root: one then has a circular motion,
and dr/dt = 0 at all times.
2
• Periodic motion in U(r) implies the orbit is
closed; I.e. loops on itself after a certain
number of excursions about the center of
force.
• The change in  while going from rmin to rmax
is a function of the potential and need not be
180o.
• It can be calculated!
• Because the motion is symmetric in time:
 

rmax
rmin
 
 l / r 2 dr

l2 
2   E  U (r) 
2
2

r


• Path closed only if Dq is a rational fraction of
2p.
• Dq= 2p(a/b) where a and b are integers.
• In this case, after b periods the particle will
have completed a revolutions and returned to
its original position.
n1
U
(r)

r
• For
a closed noncircular path
exists only for n = -2 or +1.
8.6 Centrifugal Energy and Effective Potential
• In dr/dt, d/dt, …, we have
l2
E  U (r) 
2
2 r
• Where each term has the dimension of energy.
• Remember that
• Write
l
2
l   r 2&
1 2 &2
 r 
2
2
2r
l2
1 2 &2
 r 
• Interpret
2
2
2r
l2
energy” U c 
2r 2
as a “potential
U c
l2
2
&
F





r

• The associated force is: c
r
r 3
• Traditionally called a centrifugal force.
– Although it is, STRICTLY SPEAKING, NOT A
FORCE
– but rather a pseudo-force.
– We continue to use the term nonetheless…
• The term l 2 2  r 2 can then be interpreted as
the centrifugal potential energy, and included
with U(r) to define an effective potential
energy.
2
l
V (r)  U (r) 
2r 2
• V(r) is a fictitious potential that combines the
real or actual potential U(r) with the energy
term associated with the angular motion
about the center of force.
• For an inverse-square law central-force
motion, one gets:
2
k
l
V (r) 

2
r
2r
100
80
60
l2
1 2&

r 
2
2
2 r
40
20
0
-20
-40
-60
V (r)
k

r
V ()  0
Energy
V (r)
Turning point(s)
(apsidal distances)
unbound
r3
r4
E1
1 2
r&
2
r
r1 r2
bound
E2
E3
• Values of E less than Vmin    k 2l do not result in
physically real motion; given velocity is imaginary.
2
2
• Techniques illustrated here are used in modern
atomic, molecular and nuclear physics (but in the
context of QM).
8.7 Planetary Motion – Kepler’s Problem
• Consider the specific case of an inverse-square force
law.
 l / r 2 dr
 (r)  
 constant

k
l2 
2  E  
r 2 r 2 

 
• Integral soluble with variable substitution u=1/r.
• Define the origin of  so r is a minimum.
cos 
l2 1
1
k r
2El 2
1
k 2
cos 
l2 1
1
k r
2El 2
1
k 2
• Define constants
l2
a
k
• Then one can re-write:
2El 2
  1
k
l2 1
2El 2
 1 1
cos
2
k r
k
• To get the equation of a conic section
with one focus at the origin
a
r
 1  cos
a
r
 1  cos
• The quantity, , is called eccentricity, and
• 2a is termed the latus rectum of the orbit.
• Conic sections are formed by the
intersection of a plane and a cone.
• More specifically … by the loci of points
(formed by a plane) where the ratio of the
distance from a fixed point (the focus) to a
fixed line (called the directrix) is a constant.
Hyperbola, >1
Parabola, 1
Ellipse, 0<<1
Directrix
For parabola
Circle, =0
a
r
 1  cos
• q=0 corresponds to a pericenter, i.e. rmin, whereas
rmax corresponds to the apocenter.
• The general term for turning points is apsides.
• Planetary Motion:
• Major axis
a
k
a
• Minor axis
b
1 
2
a
1 
2


2E
l
2 E
 
rmin  a 1   
 
rmax  a 1   
a
1 
a
1 
a
b
P
a
P
a
• Period of elliptic motion:
2
dt 
dA
l
T
A
2
 dt  l  dA
0
0
2
T
A
l
A   ab
2
2
k
• The area of an ellipse is: T 
 ab 

l
l 2E
• The period is then….
• Noting b 
aa
• One also finds:
T  k
2
4

 3
2
T 
a
k

2
E
l
2 E
3/ 2
Kepler’s Third Law
• Given the gravitational force: F(r)  
Gm1m2
r2
k
 2
r
2 3
2 3
4

a

4

a
• The square of the period: T 2 

Gm2
G m1  m2


• Where the last approx is realized for m1 << m2.
• Kepler’s statement is correct only if the mass m1 of a
planet can be neglected with respect to the mass m2
of the sun.
• Correction needed for Jupiter given that it is
1/1000 of the mass of the Sun.
Kepler’s Laws
1. Planets move in elliptical orbits about the
sun with the sun at one focus.
2. The area per unit time swept out by a radius
vector from the sun to a planet is constant.
3. The square of a planet’s period is
proportional to the cube of the major axis of
the planet’s orbit.
Example 8.4
Halley’s comet, which passed around the sun early in 1986, moves
in a highly elliptical orbit with an eccentricity of 0.967 and a period
of 76 years. Calculate its minimum and maximum distances from
1
the sun.
2 2


Solution: a  Gmsun


2
 4 
2



365day 24hr 3600s 
11 Nm
30
  6.67  10
1.99  10 kg  76 yr
2 
yr day hr 

kg



2
4




a  2.68  1012 m

We thus find:



m 1 0.967  5.27  10
rmin  2.68  1012 m 1 0.967  8.8  1010 m
rmax  2.68  1012
10
m
2







1
2
8.8 Orbital Dynamics