For a central force the position and the force are antiparallel, so rF=0.
r
F
dL
N  rF 0
N 
N is torque
dt
Newton II, angular

dL
0
dt
So, angular momentum, L, is constant
Since the Angular Momentum, L, is constant:
Its magnitude is fixed
Its direction is fixed.
L r p
L
r
• By the definition of the cross product, both the
position and the momentum are perpendicular to
the angular momentum.
• The angular momentum is constant.
• Therefore, the position and momentum are
restricted to a plane. The motion is restricted to a
plane.
p
Show that the conservation of angular momentum
implies that equal areas are swept out in equal times:
r+dr
r
dr
The green shaded area is dA
The parallelogram formed
by r and dr is twice the area
of dA. But, by definition,
the magnitude of rdr is
the area of the
parallelogram.
1
1
dA 
r dr sin 
r  dr 
2
r
2
dr
dr sin 
(height)
Proof of Kepler’s Second Law
dA 
1
r  dr
The area swept out in time dt
r  vdt
But dr is just v dt
2
dA 
1
2
dA

1
dt
2m
dA
L
dt

2m
r p
Divide by dt, and change v to (1/m)p
The angular momentum is constant, so
the rate at which area is swept out is
also constant.
Central Forces in Polar Coordinates
Let’s use the Lagrangian:
Don’t confuse the Lagrangian, “L”,
with the angular momentum, “L”.
L  T V
L
1
2
L

m r eˆ  reˆr
1
2

m r  r
2
2
2

2
V r 
 V r 
Motion is restricted to
a plane, and the
potential is that for a
central force.
The Radial Equation
L
1
2
mr  
2
2
1
2
d  L  L


dt   r   r
mr  V r 
2
L
r
d  L 

  mr
dt   r 
 mr
L
r
 m r  f  r 
2
m r  m r  f  r 
2
The Angular Equation
L
1
mr  
2
2
2
1
2
mr  V r 
L

2
 mr 
2
mr    0

dt
d
2
d  L  L


dt      
L

0
But the term in parenthesis is
just the angular momentum of a
point particle, so conservation
of angular momentum falls out
of the constraints on the
Lagrangian!
From the Lagrangian, we found Newton’s Laws
in Polar Coordinates:
Radial eqn.: m r  m r 
2
r  r 
2
V
r
f (r )
m
Theta eqn.: m r  2 m r  0
r  2 r  0

dt
d
2
r  l
2

r  0
l 
L
m
 r v
f (r )  
V
r
Obtain solutions for SHAPE, don’t need to solve for t
Put it in terms of u and theta:
r 
Let
1
u
u 
Solve for r and its derivatives
in terms of u and theta.
1
r
r 
1
u
2
u 
 1 du d 
u
2
d  dt
 r 
2
du
d
 l
du
d
d 
du 
d u
d u l
2 2 d u
r 

l


l



l

l
u


2
2
2

dt 
d 
d
d r
d
2
2
2
Define the Differential Equation for the Orbit Shape
To reiterate: r 
r 
1
u
2
u 
 1 du d 
u
2
d  dt
1
r  l
2
u
1
u
 r 
2
2
du
d
 l
2

du
d
2
2
d 
du 
d u
d u l
2 2 d u
r 

l


l



l


l
u


2
2
2

dt 
d 
d
d r
d
r  r 
2
f (r )
m
2
l u
2
2
d u
d


1
u
Replace r’s and thetas
2
  lu 
2
2
1
 f  /m
u
d u
d

u  
Now we have a differential eqn. of u and theta
f 1 / u 
2
2
l u m
Now, we plug our force law into the differential
equation we have derived.
The force is:
f (r ) 
k
r
2
d u
d

u  
f 1 / u 
2
k  GMm
2
2

l u m
 ku
2
2
2

l u m
k
ml
2
This equation is similar to a simple harmonic oscillator with
a constant force offset, except the independent variable is
theta not time.
u  A cos       
r 
k
ml
2
1
A cos       
k
ml
2
Manipulate the Shape Equation a Bit More:
1
r 
A cos       
k
ml
2
Choose theta nought so that theta equal to 0 yields the
distance of closest approach   0
o
2
ml / k
r 
1 A
ml
2
cos( 
k
“Look Ma, an ellipse!”
“But Johnny, It doesn’t look like an ellipse???”
“Oh Ma… don’t you know nothin’?”
We want to show that this equation is an ellipse
1
?
2
ml / k
r 
1 A
ml
2
cos( 

0.5
-1.5
-1
-0.5
0.5
-0.5
k
-1
1
1.5
Review of Some Basic Ellipse Properties
The ellipse below has the equation: x 2  2 y 2  3
Semiminor Axis
x  2y  3
2
2
2y  3
1
2
y 
3
2
b
0.5
-1.5
-1
-0.5
0.5
1
1.5
-0.5
-1
x  2y  3
2
Semimajor axis:
2
x 3
2
a
x
3
To solve for the eccentricity of an ellipse, use the
defining relationship for the ellipse and solve equations
for two special cases:
r  r   co n stan t
w h en r   r , co n stan t  2 r 
Pythagorean Triangle
When touching the right edge
b  (  a )  co n stan t /4
2
2
 a   a    a   a   co n stan t
(2) co n stan t  2 a
Plug (2) into (1)
(1)
2
b   a   a
2
 
2
b
1  
a
2
1
b
r
2
r
0.5
-1.5
-1
-0.5
0.5
-0.5
-1
a
1
1.5
a
Put our Equation for the Ellipse in the Form of Our Shape Equation
1
r
r

0.5
-1.5
-1
-0.5
a
0.5
-0.5
1
1.5
a
-1
r  r  2a
Defining equation for an ellipse
2
2
2
r   r sin    2  a  r cos 


2
r   r sin     a  r cos   4  ar cos 
2
2
2
2
2
2
r   r  4  a  4  ar cos 
2
2
2
2
r   r  4  a   a  r cos 
2
2

r'
Put our Equation for the Ellipse in the Form of Our Shape Equation
r   r  4  a   a  r cos 
2
2
r  2a  r

From the last slide
From the defining relationship
 2 a  r   r  4  a   a  r cos 
2
2

4 a  r  4 ar  r  4  a  4  ar cos 
2
2
2
2
2
4 a  4 ar  4  a  4  ar cos 
2
2
2
a  r   a   r cos 
2

r  1   cos    a 1  
r 

a 1 
2

1   c os  
2

Define the Latus Rectum, 
r

The latus rectum is the distance to a
focus from a point on the ellipse
perpendicular to the major axis
  a 1  
2a
r    2a
Defining relationship for the ellipse
r  2a  
Solve it for r
r 
2
2
 4 a
2
2
2
4 a  4 a  4  a
2
  a 1  
2
2


Define a Pythagorean relationship
4 a    4 a  
2
2
2
2
 4 a
2
2
Solve the resulting relationship for alpha
Finally, Show that our Central Force Yielded an Elliptical Orbit!
So, our general equation for an ellipse is
r 

a 1 
2


1   cos  

1   cos  
And the solution to our shape equation was
2
ml / k
r 
1 A
ml
2
cos( 
k
They have the same form! We’ve derived Kepler’s
Second Law. The trajectories of planets for a central force
are described by ellipses.
Relations Between Orbit Parameters
r 

a 1 
2

1   cos  

2
ml / k
r 
1 A
ml
2
c os( 
k
  ml / k  l / GM
2
 
Aml
k
2
2
 Al / GM
2

1   cos  
Kepler’s Third Law

 Adt 
A Integration of the area dA/dt over one period gives A (1)
0
A
L
Kepler’s Second Law
2m


0

A dt 
L
 2m
0

dt  
0
l
l
dt 
2
2

 dt 
0
l

Use Kepler II to relate integral and l. (2)
2
Set (1) and (2) equal
A
l
2
  
2A
l
Kepler’s Third Law Derivation Continued
 
2A
2
 ab
l
l
4 a
2
 
2
4

l
 
ml
k

2 a
2

1 
2

2


4 a a 1  
3
l
2
GMm
2
l
2
ml
1
2

l
2

4 a 
2

l
2
GM
2
 
2
4
3
2
2
GM
a
3
Look How Well The Solar System Fits Kepler’s Third Law!
 
2
4
2
GM
a
3
Earth presumably fits this rule… Using Earth years for time,
and Astronomical Units (1 AU = 1 Earth-Sun distance) for
distance renders the constants equal to 1.
Universality of Gravitation: Dark Matter
Keplerian Motion vs. Constant Density Sphere Motion
Keplerian Motion: All the mass of the galaxy
would be assumed to be focused very close to
2
the origin.
M nucleus m
v
G
r
v
m
2
r
G M nucleus
r
Velocity curve, Assuming a Constant Density Sphere All the Way Out
4
Only the mass interior to
4
3
3
M  r 

r
m
2
the star in question acts.
mv
3
G 3
2
4
Density, assuming a
3 
r
r
  M /  R 
uniform density sphere.
v
3

4
v  r G 
3
r
Are Central Forces Conservative?
F 
1  

1  

hF

gF
u





w
v 
gh   v
w
fh   w

 fFu  

u

 hFw   v 

1  

gF



v
fg   u
v
u  r, v   , w  
f  1, g  r , h  r sin 
 


F 
sin  F  
 F   r 


r sin    


1
 


 Fr    r sin  F    

r sin    
r

1

 
 r sin  F  

r sin    r

1

 Fr    


 fFu   w 

Central Forces Are Conservative!
 


F 
 0    0  r

r sin    


1
 



 r sin  f  r     0    

r sin    
r

1

 
0 

r sin    r

1

F
 r    0

Energy Equation of an Orbit in a
Central Field
v r r 
2
1
2
2

2
m r r 
2
2
2
2
r 
  V  r   E  constant
2


1
2  du 
2
1
m l 

u

V
u
 E


2
  d  



1
u
  lu
r  l
2
du
d
2
r  l u
2
2
d u
d
2
Orbital Energies in an inverse
Square Law
V r   
k
 ku
r
 du 2

E  ku
2

 u  
2
d

(1
/
2
)
m
l




2
  du 

2
m l 
  u   ku  E
2
  d  

1
2
2
2E
2 ku
 du 
2



u


2
2
d

ml
ml


1
d 
2E
ml
2

du
2 ku
ml
2
u
2
Integrating Orbital Energy Equation
d 
1
c  bu  au
a   1, b 
2k
ml
  o 

2
du
2
2E
, c
ml
1
c  bu  au


1
1
   o  cos  

1


2
du 
2
1
a


 2u
2
ml

2

4k
8E


2 4
2
m l
ml 
2k
cos
1

b  2 au 


2
b  4 ac 

Integrating Orbital Energy Equation Continued


1
1
   o  cos  

1




 2u
2
ml

2

4k
8E


2 4
2
m l
ml 
2k
k  um l
cos     o   
2
k  2 Eml
2
2
k 1  2 E m l / k cos     o    k  um l
2
2
um l  k  k 1  2 E m l / k co s     o 
2
2
2
2
r 
ml / k
1
1  2 E m l / k cos     o 
2
2
2
Total Energy of an Orbit
2
r 
 
ml / k
1  2 E m l / k cos     o 
1
2
1
2E
2
2
To fit earlier form for an ellipse
ml / k
k
 
ml
2

 a 1 
k
  1
2
2E
2

Earlier Relations

k



a
E 
2E

k
k
2a
Total energy of the orbit.
Ellipses, Parabolas, Hyperbolas
2
r 
 
ml / k
1
1  2 E m l / k cos     o 
2
1  2Eml / k
2
2
2
E < 0 Ellipse or Circle
E=0 Parabola
E>0 Hyperbolic
e<1
e=1
e>1
Maximum Velocity for An
elliptical Orbit
mv
2
G
Mm
2
v 
2
 E
r
2G M
r
 E 0
Limits of Radial Motion
E
m 2 l 
 r  2   V r  E
2 
r 
2
1.5
2
r  U r   E
2
U r 
ml
2
2r
2
2
2r
2
1
0.5
5
-0.5
-1
m
ml
-1.5
 V r
U is the effective potential
U (r )
10
V (r )
15
20
Limits Continued
U r   E  0
ml
2
2r
2

k
E 0
r
m l  2 kr  2 E r  0
2
2
 2 E r  2 kr  m l  0
2
r1,0 
r1,0 
2k 
2
4k  8Eml
2
4 E
k
k  2Eml
2
2 E
2
2

Minimum Orbital Energy
r1,0 
k
2
2 E
E m in  
ro  
k  2Eml
k
2 E m in
Radical ought to remain real
for elliptical orbits.
2
2ml
k
2
2
Value of E for which the radical is 0
Extremes of motion merge to one value.
Scattering and Bound States are
All There Are!
E>0
E<0
Energy Equation for a Central
Force Again
2


du


2
2
1
m l 

u

V
u
 E


2
  d  

1

For attractive inverse square
V 
k

For repulsive inverse square
  ku
V 
r
Qq
r
k  Q q
 Q qu
The Scattering Calculation
Proceeds Exactly Like the Bound
State Calculation
2


du

2 
2
1
m l 

u

V
u
 E


2
  d  

1


2


du

2 
2
m l 
  u   ku  E
2
  d  

1
1
d 
2E
ml
2

du
2 ku
ml
2
u
2
But, we’re going to let k go
to –Qq after we integrate
Integrate Both the Bound State
And the Scattering Problem
d 
1
c  bu  au
a   1, b 
2k
ml
  o 

2
du
2
2E
, c
ml
1
c  bu  a u
2
du 
2

b  2au 
A cos  

2
a
b  4ac 

1
Verify Integral in 6.10.4
d
du
A cos  f  u    
1  f u 
2
b  2 au
f u   
b  4 ac
2
2a
 f u  
b  4 ac
2
1
1  f (u )
f u 
1

2
1


b  2 au 
1 

2
b  4 ac 

b  4 ac
2
1
d
du
2
A cos  f  u    

au  bu  c
2
du 

b  4 ac
f u 
1  f u 
 
 
 

b  4 ac 2  a
2
a
au  b u  c
2

b  2 au 
A cos  

2
a
b  4ac 

1

2
2a
2
2
 4 a u  4 abu  4 ac
2
2

d 
b  2 au
 A cos  
2
du 
b  4 ac

1
2
2
b  4 ac   b  4 a u  4 abu 
2
b  4 ac
b  4 ac
2
2
 b  2 au 

2
2 a
b  4 ac
2
au  bu  c
2
au  bu  c
2
Solve Both Bound State and
Scattering Problem for u
  o 

1
c  bu  au

   o  A cos  

du 
2


2
b  4c 
b  2u
2u  b 
b  4 c cos     o 
2u  b 
b  4 c cos     o 
u 
k
ml
2

b  2 au 
A cos  

2
a
b  4ac 

1
2
2

k
2
2 4
m l

2E
ml
2
cos     o 
Solve Both Problems for r
u 
Qq
ml
2
2
2E
 Q q 

cos     o 

2 
2
ml
 ml 


Qq 

1 
2

r
ml

2

2
2E  ml 

1
 c o s    o 
2 

ml  Qq 

1
ml
2
Qq
r 
1 
1
2 Eml
2
Qq 
2
ml
cos     o 
2
k
r 
1
1
2 Eml
k
2
Solution to the Scattering Problem
cos     o 
Solution to the Bound State Problem
Scattering in an Inverse Square
Field
attractive
2
d u
d
r 

2
k
u 
ml
2
d u

d
1
A co s       
ml
Repulsive
K goes to -Qq
2
1
 r 
k
ml

u  
Qq
A co s       
2
Qq
ml
2
2
Solving the energy
equation for r
r 
m l / qQ
1 
1  2 E m l / Q q cos     o 
2
2
2
A Drawing of the Scattering Problem
2
r 
m l / qQ
1 
1  2 E m l / Q q cos     o 
2
2
2
s
Trajectory
o
rmin
A small
range of
impact
parameter
o
b
Impact parameter b
scattering center
Asymptotes
2
r 
m l / qQ
1 
1  2 E m l / Q q cos     o 
2
2
2
 0 r
Trajectory
cos  0   o   r  
cos  2 o   o   cos   o   cos   o   r  
  2 o or 0  asym ptotes
s
o
o
rmin
A small
range of
impact
parameter
b
Impact parameter b
scattering center
   o  rm in
 s    2 o
Angles
2
r 
m l / qQ
1 
1  2 E m l / Q q cos     o 
2
2
2
 2 Eml 2 
1

2 2 
 Q q 
1  2 E m l / Q q cos  2 o   o   0
1 
2
2
1
cos   o  
tan   o 
2
1  2 Eml / Q q
2
 2 Eml 2 

2 2 
Q
q 

2
2
(1 / 2 )
o
1
 s    2 o
o 
1
2

/2-o
s 

tan   o   tan 

 2 Eml 

2 2 
Q
q 

2
2
1
  s   2 Eml 

    s    cot     2 2 
2

 2   Q q 
(1 / 2 )
(1 / 2 )