kepler problem 开 普 勒 问 题
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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. - PowerPoint PPT PresentationTRANSCRIPT
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 )
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