Transcript Kepler

Kepler
Inverse Square Force

Force can be derived from a
potential.
k < 0 for attractive force
•
F2int
m2

m1m2
m1  m2

Choose constant of
integration so V() = 0.
r = r1 – r2
r2
Qr 
m1
R
F1int
r1
V
k
r2
k
r

V
r
Kepler Lagrangian


The Lagrangian can be
expressed in polar
coordinates.
L is independent of time.
• The total energy is a
constant of the motion.
• Orbit is symmetrical about
an apse.
T  12  (r 2  r 2 2 )
k
V
L  T  V  12  (r 2  r 2 2 ) 
r
k
2
J
k
E  T  V  12 r  12 2 
r
r
2
r
Kepler Orbits

The right side of the orbit
equation is constant.
•
•
•
•

Equation is integrable
Integration constants: e, 0
e related to initial energy
Phase angle corresponds to
orientation.
The substitution can be
reversed to get polar or
Cartesian coordinates.
d 2u
Qr
k

u




d 2
J 2u 2
J2
u
1
u
r
k
J
2
[1  e cos(   0 )]
J2
s
ke
1 1
 [1  e cos(   0 )]
r es
r  e( s  r cos(   0 ))
Conic Sections

r
s
• 0 init orientation (set to 0)
• s is the directrix.

focus
r  e( s  r cos  )
The orbit equation describes
a conic section.

The constant e is the
eccentricity.
•
•
•
•
sets the shape
e < 1 ellipse
e =1 parabola
e >1 hyperbola
Apsidal Position


Elliptical orbits have stable
apses.
1 1
 (1  e cos  )
r es
• Kepler’s first law
• Minimum and maximum
values of r
• Other orbits only have a
minimum
r1 
The energy is related to e:
r1
• Set r = r2, no velocity
e  (1 
2 EJ 2
k
2
)
1
2
es
1 e
r2 
r
s

es
1 e
r2
Angular Momentum

The change in area between
orbit and focus is dA/dt
dr
r
• Related to angular velocity


The change is constant due
to constant angular
momentum.
This is Kepler’s 2nd law
A  12 rr  12 r 2
J  r 2
J
A 
2
Period and Ellipse

r
r1
s

The area for the whole
ellipse relates to the period.
• semimajor axis: a=(r1+r2)/2.
r2

This is Kepler’s 3rd law.
• Relation holds for all orbits
• Constant depends on , k
A  a
A  a
2
3
2
1  e  a
2

J2
k
 2a
3
A

T    2a 2 
A
k

2
3
2
2 EJ 2
k 2
J


2
k
Effective Potential
Treat problem as a one
dimension only.

J2 k
E  r 
  Tr  Veff
2
r
r
1
2
• Just radial r term.
Minimum in potential implies
bounded orbits.

• For k > 0, no minimum
• For E > 0, unbounded
Veff
0
Veff
r
unbounded
r
0
possibly
bounded
Veff
2
1
2
J2
k


2r 2 r