kepler problem 开 普 勒 问 题

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)

Ptolemy system(70 circles)

均轮

本轮epicycle

deferent

equanteccentricearth

均轮

本轮

偏心

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).

(1) The orbit of each planet about the Sun is an ellipse with the Sun at one focus. (the law of orbit);

Kepler's laws

(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

11,rm

22 ,rm

21 rrr

O

rr 1 1 rfm

rr 2 2 rfm

0 2211 rr mm

rrr 11

212 1 rf

mm

21 rrr

321~

r

mmGf

Mathematician vs Physicist

Mathematician Physicist, r; M, rC

separate tendency & relative motion

standard procedure of differential equation

t1 1 rr

central force, conservation of L

kL 2 rr, p co-planar,areal “velocity” constant

2

L

can never change sign.

1st integrals, conservation of Eeffective potentialclassification of orbits

t2 2 rr (not r1, r2 )

dxlu@nju.edu.cn

Thanks

0 2211 rr mm

rr 11

21 rf

mm

1

0C rrr )( rf

C21 mmm Define

reduced massCC2 21 1 rrr mmm

111

21

mm

Mass?Position vector?force? Which particle’s eq?

2 0 rrM rf

CprL • r and p are co-planar

• areal “velocity”

keer 2 rrr r • L

Conservation of L

2

CprL

rr d2

1

• r and p are co-planar

keer 2 rrr r • L

Conservation of L

0 rrM rf

2

CprL

• areal “velocity”

constant2d

d

2

1

d

d

L

t

rr

t

A

• can never change sign.

• r and p are co-planar

keer 2 rrr r • L

Conservation of L

0 rrM rf

3 rrfrr 2

tt

r

t

rrr d

d

d

d

dd

rr d2

1st integrals

Er

krr 2 22

2

1 v2

rr d

2

2

1d r

rr

Lr d

2

2

22

2

32

2

2dd

r

Lr

r

L

rFrrrf dd)(

r

k

r

mmG dd 21Ud

0 2 rr 2 r L

rUr

LrE

2

22

2

2

1

rUr

2

1eff

2

4

r

k

r

LrU

2

2

eff2

Effective potential

r

r

r

k

r

L 0

?

?

2 2

2

2 2

2

1 r

hyperbola

parabola

ellipse

circle

Total energy

Kinetic energy?

11

5 )(tr )(rr

d

d

d

d

d

d2

r

r

L

t

rr

rUr

LrE

2

22

2

2

1

21

2

2

2

2

rU

r

LE

r

rUr

LEr

Ld

2

2

d21

2

22

r

r

k

r

LEr

Ld

2

2

21

2

22

21

2

2

2

2

rU

r

LE

r

r

r

k

r

LEr

Ld

2

2

1

21

2

2

22

r

rk

r

LE

L 1d

2

2

21

2

2

a

x

xa

xarccosd

d~

22

r

L

k

r

L

L

kE

L 1d

2

212

2

22

2

22

L

k2

22

L

k

02

1

2

22

2

arccos

EL

k

Lk

rL

erp

k

LE

rp

1

21

1cos

22

20

)(cos1 0

e

pr

k

Lp

2

p

keE

21 2

11 maxmin re

pr

e

pr

2maxmin

12 e

prra

For ellipse orbit

apb

k

Lp

2

)(cos1 0

e

pr

eccentricity energy orbit

e = 0 circle

0 < e < 1 ellipse

e = 1 E = 0 parabola

e > 1 E > 0 hyperbola

p

kE

2

a

kE

2

p

keE

21 2

mmGaT

1 2 23 )Kepler(

1 2 23

mGa

tL

tt

AA

2d

d

d

baL

T 2

paL

232

mmG

La

L

223

2

mmGmm

mma

1

)( 2 23

k

Lp

2

*Hyperbola orbit

202

1vE

bL 0v

e1cos

cos1 e

pr

22

Scattering angle

22

02

2b

E

k

v

cot

2tan

Eb

k

2

sin

coscot

22

2

2

1

k

LE

Eq.(4.4.17)

2E

2

11

1

ee

1

12

e

Assignment: 4.7, 4.12, *4.14