Gravity 引力

Model of solar system 太阳系模型
Ptolemy’s geocentric model

Kepler's Platonic solid model of the Solar system

Kepler‘s laws of planetary motion 开普勒定律
1. The Law of Orbits: All planets move in elliptical orbits, with the sun
at one focus.
p
r
1  e cos 
p  a(1  e 2 )
p
b
1  e2
2. The Law of Areas: A line that connects a planet to the sun sweeps
out equal areas in equal times.
Area velocity
is constant
d
dA
L  mrv  mr
 2m
C
dt
dt
2
dA ab

dt
T
3. The Law of Periods: The square of the period of any planet is proportional to
the cube of the semimajor axis of its orbit.
From Newton’s law, we have
Newton‘s laws of universal gravitation 万有引力定律
If the orbit is a circle, we have
Cavendish experiment
Gravitational force of a spherical shell 球壳的引力
Geometric relation
Outside spherical shell
Inside spherical shell
Gravity force inside a ball with uniform density
Force along the tunnel
Fx  
Gm' m x
Gm' m
r


x
3
3
R
r
R
d 2x
Gm' m
m 2 
x
3
dt
R

Gm'
R3
Same as simple
harmonic oscillator
T  2
Gm'
 2R
3
R
d 2x
2


x0
2
dt
Gm'
R
Gravitational potential
Gravity potential on earth’s surface
For neutron star
Compare to fusion energy
Compare to energy value of gasoline
E
 4 107 J / kg
m
Orbiting speed 第一宇宙速度
escape speed of earth 第二宇宙速度
escape speed of sun 第三宇宙速度
1 2
Mm 1
2GM s
GM s
v 

 12.3
mv3  G
 mv 2
Rs
Rs
2
R
2
When escape speed is the grater than speed of light, we get blackhole
引力场可以被局部消除
equivalence principle 等效原理
Metric 度规
3D Euclidean
Minkowski
General relativity
引力几何化
hdv  
Gm' m
dz
2
( R  z)
m  hv / c 2
GR result
Two body problem
define
Equation of motion
Effective potential
U eff (r ) 
L
k

2r 2 r
Solve for orbit

1
 L
L
d     
d 
2
2
2
 L  2k  L   r 
 k   L k   r 
2 E    
2 E       
 
r
L
r
 
 
 L  r L 

1
1
a x
2
2
dx   
1
1  ( x / a)
2
d ( x / a)  arccos( x / a)
cos(   0 ) 
 L 
e  1  2 E  
 k 
2
L2
p
k
r
 L 
1  2 E  
 k 
p
1  e cos(   0 )
Thus we obtain Kepler’s first law from gravity law.
p  a(1  e 2 )
 L2  1
   1
 k  r
2

p / r 1
e
dA ab

dt
T
dA L

dt 2 
T  2
ab
L
 2
b 2  ap
ab

1
 2a 3 / 2
 2a 3 / 2
k
G(m'm)
kp
Thus we obtain Kepler’s third law.
Alternative method
(r12  r22 ) E  k (r1  r2 )
1
k 1
k
E  mv12   mv22 
2
r1 2
r2
E
v1r1  v2 r2
E
k
1
k
 mv 2 
2a 2
a
L  mbv
v2 
k
ma
L2
b 
a
mk
2
k
2a
v
v2
v1
Time evolution: Kepler equation
We want to solve for
ab
A(t ) 
t
T
Need to find
A
   (t )
 for given A
b1 2
1

 a E  ac sin E 
a2
2

a cos E  c  r cos   c 
E  e sin E 
p cos 
1  e cos 
2
t
T
cos E 
e  cos 
1  e cos 
1 e

E


1
tan 2
e(1  tan 2 )  1  tan 2
2
2 
2
2  1 e
E


1 e

1  tan 2
1  tan 2  e(1  tan 2 ) 1 
tan 2
2
2
2
1 e
2
1  tan 2
tan
E
1 e


tan
2
1 e
2
(1)
Alternative method
1  2
1 
p2
ab
A(t )   r d '  
d

'

t
2
0
0
2
2 (1  e cos  ' )
T
dx
2
a b
x

arctan
tan
 a  b cos x a 2  b 2
ab
2
Change variables
t  tan( x / 2)
Take derivative with respect to a
dx
2a
a b
x

arctan
tan

 (a  b cos x) 2 (a 2  b 2 )3 / 2
ab
2
2b
x
tan
(a  b)( a  b) 2
2
 a b
x

1 
tan 
2
 ab
2
Take a=1, b=e, and plug in the integral limit


0
d '
2
1 e


arctan
tan

2
2 3/ 2
(1  e cos  ' )
(1  e )
1 e
2
2e
1 e

tan
(1  e 2 ) 3 / 2 1  e
2
 1 e
x
1  
tan 
2
 1 e
2


0
d '
2
1 e


arctan
tan

(1  e cos  ' ) 2 (1  e 2 ) 3 / 2
1 e
2
Define
E  2 arctan
2e
1 e

tan
(1  e 2 ) 3 / 2 1  e
2
 1 e
x
1  
tan 
2
 1 e
1 e

tan
1 e
2

1 
p2
p2
1
2e tan( E / 2) 


d

'

E

2 0 (1  e cos  ' ) 2
2 (1  e 2 )3 / 2 
1  tan 2 ( E / 2) 

ab
ab
( E  e sin E ) 
t
2
T
E  e sin E 
2
t
T
2
Scattering 散射

1
e2 1
k
k
 2
L 2 E v0 b

 L 
e  1  2 E  
 k 
2
Gravitational field 引力场
For spherical shell, we have
Inside the shell
outside the shell

g f 4r 2  0


g f 4r 2  4GMer

gf 0

GM 
g f   2 er
r
Proof of Gauss’s Law
We only have to prove it for a mass point.
Gauss’s divergence theorem: If F is a continuously differentiable
vector field defined on a neighborhood of V, then we have

 Gm  
g  3 ( xi  yj  zk )
r
Gravity field is
  x  1 3x 2
 3 3  5
x  r  r
r
r  x2  y2  z 2
 3 3( x 2  y 2  z 2 ) 

  0
  g  Gm 3 
5
r
r

An arbitrary surface can be replaced by a spherical surface
For spherical surface

2 Gm
g

dS


4

r
 Gm
2

r
Gravity field strength and potential generated by m is
z  r cos
we can expand
by to small parameter
r/a
Taylor expansion formula
(1  x) a  1  ax 
a (a  1) 2
x 
2
2
Gm 
r
r 
V (r )  
1  2  cos     
a 
a
 a  
1/ 2
2
2 2



Gm   r 
1r  3 r 

r 

1    cos      2  cos       
a  a
2  a  8   a 
 a  



2

Gm 
1r 
 r

2

1    cos     (3 cos   1)  
a 
2a

 a

Gm
V1   2 z
a
 Gm 
g 2 k
a
This is a uniform field strength
Gradient in spherical coordinate
Tidal force of moon and sun