Kepler Problem
开 普 勒 问 题
行星运动的描述——运动学
地球人的观点
Sun path; analemma; star trails
历史的回顾: 地心说；日心说；开普勒
开普勒问题的文化方面
开普勒问题的物理
地球人的观点
Sun path; analemma; star trails
Sun path by Justin Quinnell
2010 Analemma 地球仪8字曲线 by Tamas Ladanyi, TWAN
Star trails by Harold Davis
星迹
历史的回顾: 地心说；日心说；开普勒
The frontispiece to
Galileo’s Dialogue
Concerning the Two
World Systems (1632).
According to the labels,
Copernicus is to the right,
with Aristotle and Ptolemy
at the left; Copernicus was
drawn with Galileo’s face,
however
Claudius Ptolemy
(127—152 working in
Alexandria, Egypt )
Nicolaus Copernicus
(1473—1543)
epicycle 本轮
equant
eccentric
earth
deferent
均轮
Ptolemy system(70 circles)
均轮
偏心
本轮
Ptolemy system (70 circles)
Copernicus system (46 circles)
Galileo Galilei (1564—1642)
Tycho Brahe (1546—1601)
Johannes Kepler (1571—1630)
Issac Newton (1642 — 1727)
Kepler’s
nested set:
Saturn—Jupiter—Mars—Earth—Venus—Mercury
cube
dodecahedron
octahedron
tetrahedron
icosahedron
“Pythagorean” or “Platonic” solids
•Mysterium Cosmographorum (Cosmic Mystery) (1596)
•Harmony of the World (1619).
Kepler's laws
(1) The orbit of each planet about the Sun is an ellipse
with the Sun at one focus.
(the law of orbit);
(2) The line joining any planet and the Sun sweeps
out equal areas in equal times.
(the law of areas);
(3) The square of the period of revolution of a planet
about the Sun is proportional to the cube of the
planet's mean distance of the Sun.
(the law of period or Harmonic law)
开普勒问题的文化方面
The task of deducing Kepler’s laws from Newton’s laws
is called the Kepler Problem.
Its solution is one of the crowning achievements on
Western thought. It is part of our cultural heritage
just as
Beethoven’s symphonies
or Shakespeare’s plays
or the ceiling of the Sistine Chapel
are part of our heritage.
— The Mechanical Universe p498, CIT
4.4 Kepler problem and *scattering
m1,r1
r  r1  r2
m2 ,r2
m1m2
f ~G 3
r
m 1 r1   f r  r
m 2 r2  f r  r
O
Mathematician vs Physicist
r  r1  r2
m1 r1  m2 r2  0
 1
1 
r1  r2  
 f r  r

m m 
2 
 1
Mathematician
standard procedure of
differential equation
Physicist
m, r; M, rC
separate tendency & relative
motion
central force, conservation of L
r, p co-planar, L  m r 2 k
L
areal “velocity”
 constant
2m
 can never change sign.
r1  r1 t 
r 2  r 2 t 
1st integrals, conservation of E
effective potential
classification of orbits (not r1, r2 )
Thanks
dxlu@nju.edu.cn
1
Define
m1 r1  m2 r2  0
 1
1 
r  
 f r  r

m m 
2 
 1
m1  m2  mC
m1 r1  m2 r 2  mC rC
1
1
1


m1 m2 m
rC  0
m r   f (r) r
reduced mass
Mass?
Position vector?
force?
Which particle’s eq?
2
M  r  f r  r  0
Lr pC
Conservation of L
• r and p are co-planar
•L


2

 r  m r er  r e  m r  k
• areal “velocity”
2
M  r  f r  r  0
Lr pC
Conservation of L
• r and p are co-planar
•L


2

 r  m r er  r e  m r  k
1
rd  r
2
2
M  r  f r  r  0
Lr pC
Conservation of L
• r and p are co-planar
•L


2

 r  m r er  r e  m r  k
• areal “velocity”
dA 1 r d r
L


 constant
dt 2 dt
2m
•  can never change sign.
3
1st integrals



m r  r  2   f r r
2 



m r  2r   0 m r   L

dr dr
 1 2
r  dr 
 dt  dr  r  d r 
dt dt
2 
2
2
2




L
L
L
2
 r  dr  r  2   dr   2 3 dr  d 2 2 
m r
 mr 
 2m r 
m1m2 


 f (r ) r  dr  F  dr  dU d  G r   d 



v2


1
k
2
22
m r  r    E
2
r
k

r
4
Effective potential
1
2 2
mr 
2
2
1
1
L
2
2



E  m r 

U
r
 m r  U eff r 
2
2
2m r
2
2
U eff
k
r  

2m r 2 r
L
k  ? r0



2
r  ? r
2m r
2
L
hyperbola
Total energy
Kinetic energy?
parabola
ellipse
circle
11
5
r (t )
r  r ( )
2
1
L
2
E  m r 
 U r 
2
2
2m r
dr d
L dr
r 

2
d dt m r d
12
 2 
 
L
  E 
 U r  
2
2m r
 m 
 
2
d 


L
12
2




2
L


2
 U r  
m r  E 
2
2m r
 m 
 
L

2 2
mr 
 m
12
2

L
k  
 
E 
2
r  
2m r

dr
L

2 1
mr  2
 m
12

L2 2mk  
2mE  2 

r  
r

dr r
dr

L

m k L
2mk m k 
 2 
2mE  2   2 
r
L
L 

 r
2
2
2
2
1

d 
12
2
r
2 2


m k  L mk 
2mE  2   
 
L
 r L  

L

 x 
~
 d arccos  
2
2
 a 

a x
dx
2
12
1
d 
r
  arccos
L mk

r
L
1
 0
 m 2k 2
 2
 2  2 mE 
 L

p
p
1
1
r
cos   0  
 r
2
2
e
L
L
1  2 mE 2 2
p
m k
mk
p
r
1  e cos(   0 )


k
E   1 e
2p
2
p
r
1  e cos(   0 )
2
For ellipse orbit
p
p
rmin 
r
 rmax
1 e
1 e
rmin  rmax
p
a

2
2
1 e
b  ap
L
p
mk


k
E   1 e
2p
eccentricity
e=0
0<e<1
energy
k
E
2p
k
E
2a
orbit
circle
ellipse
e=1
E=0
parabola
e>1
E>0
hyperbola
2
L2
p
mk
dA
L

A
dt 
t
 dt
2m
2m  3 2
2m
 a b 
T
a
p
L
L
mm
1
2m 3 2
L
32

a
 2 a
L
mGmm
(m  m) Gmm
2
T  2 a
32
1
1
32
 2 a
( Kepler )
G m  m 
Gm
*Hyperbola orbit
1 2
E  m v0
2
L  m v0b
Scattering angle
  2    2   
p
r
1  e cos
cos   1 e

k
tan   cot  
2 Eb
2
 cos 

 cot  
sin  

Eq.(4.4.17)
1
1
e 1 2
e
1
2

2E
2E
e 1
2
k

L
2mE 2 2
m k
1
m
2
2 2
v0 b
m
Assignment: 4.7, 4.12, *4.14