Transcript Weeks_4

MA4248 Weeks 4-5.
Topics Motion in a Central Force Field, Kepler’s
Laws of Planetary Motion, Couloumb Scattering
Mechanics developed to model the universe
- follow the seasons, predict eclipses and comets,
compute the position of the moon and planets
Babylonians (2000-300BC) – arithmetical models
Claudius Ptolemy (85-165AD) – geometric models
based on epicycles that prevailed for 1400 years !
http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Ptolemy.html
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PTOLEMAIC THEORY
Earth
Planet
Earth is fixed,
each planet moves in a circular epicycle whose
center moves in a circle with center near the Earth.
2
REVOLUTION
Nicolaus Copernicus (1473-1543) – produced a
heliocentric (versus geocentric) theory of cosmology
Galileo Galilei (1564-1642) – questioned authority
- refuted Aristotle’s claim that heavy bodies fall faster
- championed Copernican theory over Ptolemaic
- sentenced to house arrest by the dreaded Inquisition
Tycho Brahe (1546-1601) – observational astronomer
- Danish King helped him build 37-foot quadrant
- compiled, over 20 years, most accurate records
- Emperor Rudolph II sponsored his move to Prague
and collaboration with Kepler
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GEOMETRY OF ELLIPSES
2a-r
r

2ea
a = semi-major axis (half of horizontal diameter)
e = eccentricity
foci  angle
2
2
2 2
(2a  r )  r  4e a  4ear cos 
2
p / r  1  e cos , p  a (1  e )
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ALGEBRA OF ELLIPSES
( x , y)
2a-r
r

2ea
( x, y)  (r cos , r sin )
2
2
pe 
p
2 
2
(1  e ) x 

y


2
2
1 e 
1 e

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KEPLER’S LAWS
Johann Kepler (1571-1630) – mathematician who
believed in “the simplicity and harmonious unity of
the universe” (quote page 323 David Burton)
I. Each planet moves around the sun in an ellipse,
with the sun at one focus.
II. The radius vector from the sun to the planet
sweeps out equal areas in equal intervals of time.
III. The squares of the periods of any two planets
are proportional to the cubes of the semimajor
axes of their respective orbits: T ~ a 3/2.
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ANGULAR MOMENTUM AND TORQUE
The angular momentum , torque (about any fixed
point) of a body with movement,
force
is


 

d
r
L  r  (m ) Torque  r  F
dt
(vector cross-products is orthogonal to both vectors
and its magnitude equals the area of the parallelogram)

 

2
d
d
r
d
r
d
r
 L   (m )  r  (m 2 )
dt
dt
dt
dt


2


d
r
 r  (m 2 )  r  F  Torque 7
dt
CENTRAL FORCE
A force acting on a body is central if its direction is
along the line connecting the body to a fixed point.

r

p r
body
p
fixed point

F
central force
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CENTRAL FORCE
For a central force

  
Torque  r  F  r  (F r )  0
therefore d 
L  Torque  0
dt

Therefore the angular momentum L is constant.
 
Therefore, since r  L  0 (why?), the body
moves in a plane. When does it move in a line?
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POLAR COORDINATES
Construct an orthonormal coordinate system
r̂

r  xx̂  yŷ
ŷ
θ̂
x̂
Define polar coordinates x  r cos , y  r sin 
and unit-vector valued functions (for r > 0)
r̂ ( r , )  (cos , sin ), ˆ ( r , )  (  sin , cos )
ˆ
dr̂ 
d


 (  sin , cos )   ˆ ,
  r̂
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dt
dt
POLAR COORDINATES
Therefore, the velocity of a particle is

dr  d (rr̂)  dr r̂  r dr̂  r r̂  r θ θ̂
dt dt
dt
dt
and its acceleration is
2

 

d r  r  r θ 2 r̂  r θ  2r θ θ̂
2
dt
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CENTRAL FORCE
Therefore, if a body moves in a central force, then

2
d
r
F  F r̂  m 2 
dt
2

m r  r θ r̂  m r θ  2r θ θ̂
2

m r  r θ  F

hence
and







2
d
1



m r θ  2r θ 
mr θ  0.
r dt
Remark


2
F   | F |, L  | L |  mr θ
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KEPLER’S SECOND LAW
Remark 1: Since L is constant and
dA  1 (r)(rθ )  L
dt
2
2m
equals the rate at which the radius vector sweeps out
Area, Kepler’s Second Law holds for any central force
Remark 2: If F is conservative F   dV(r ) dr
2

and θ  L mr , mr   dVeff (r ) dr
where
Veff (r)  L 2m r  V(r)
2
2
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INTEGRATING THE EQUATION
Remark 3:We can obtain a differential equation for r
dVeff ( r ) dr
dVeff ( r )
dr
mr  

dt
dr
dt
dt


2
1


1
E  2 mr  r  V(r )  2 mr  Veff (r )
dt 
mdr
2( E  Veff ( r ))
In general, the integral on the right
will not be an elementary function
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GRAVITATIONAL & COULOMB FORCES
Remark 4: If
2
F   k r   dV dr, V(r )   k r
then the effective potential
Veff (r)  L 2m r
2
Veff
has this graph -->
for k > 0 (what
if k < 0 ?)
2
k
r
E3
r
E1
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INTEGRATING THE EQUATION
Remark 5:We can also substitute the identity
dr
dr
L

r  θ 
d
d mr 2
1
to obtain
 2 2mE 1 2mk 
dθ   r


dr
2
2
2 
L
r
Lr 

let u = 1/r
2


mk
2
L
E
u  2 1  1 
cos(
θ

θ
)
0

2
L 
mk

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KEPLER’S FIRST LAW
Remark 6: This has the form of a conic section
p r  1  e cos (θ  θ 0 )
2
with semi-latus-rectum p  L mk
and eccentricity
e  1  2L E mk
2
E  E0   mk 2L , e  0
o
Ellipse
E

E

0
,
0

e

1
r 0
E  0, e  1 Parabola
0  E, 1  e  1 Hyperbola
2
2
2
Circle
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KEPLER’S THIRD LAW
Remark 8: The semi-major axis is determined by the energy
a   k 2E
Remark 9: The rate of area swept out
dA  L  1 ka 1  e 2 
dt 2m 2 m
Remark 10: The area of the ellipse (b semi-major axis)
A  πab 
o
r
Remark 11: The period
πa
2
1 e
2
τ  2π m k a
32
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KEPLER’S EQUATION
Remark 12: The true anomaly is the angle
α  θ  θ0
from pericenter and the eccentric anomaly is
r - a  -ea cos ψ
Remark 13: Integrating the equations of motion
for r yields Kepler’s (transcendental) equation
2π
ψ
e
sin
ψ

(t

t
)
0
τ
o
r
Remark
14: For Earth’s orbit e  0.016732
o
θ 0  102.85 fromVernal Equinox
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SUPERINTEGRABILITY
Remark 15: A mechanical system is integrable if
you can express the state as a function of time (even
an non-elementary function). This requires constants
of motion (such as angular momentum) and is a very
special condition. In very special cases additional
constants of motion exist that ensure closed orbits.
These superintegrable systems include motion in a
central –k/r (k > 0) force where the
Laplace-Runge-Lenz
vector defined by
o



r

K  m r  L  mk r̂
is a constant of motion (Problem 12, page 25)
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