Mechanics Kepler Problem, Lecture Notes - Physics, Study notes of Mechanics

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Mechanics

Physics 151

Lecture 6

Kepler Problem

(Chapter 3)

What We Did Last Time „

Discussed energy conservation

„

Defined energy function

h

Å

Conserved if

„

Conditions for

h = E

„

Started discussing Central Force Problems

„

Reduced 2-body problem into central force problem

„

Problem is reduced to one equation

„

Used angular momentum conservation

„

Energy conservation gives

„

Now we must solve this

2

3

l

mr

f^

r

mr

L

t

2

2

2

const

m

l

E

r

V r

mr

Qualitative Behavior „

Integrating the radial motionisn’t always easy

„

More often impossible…

„

You can still tell general behavior by looking at

„

Energy

E

is conserved, and

E

V’

must be positive

„

Plot

V’

r

) and see how it intersects with

E

2

2

l

r

E

V r

m

mr

2

2

l

V

r

V r

mr

′^

Quasi potential including

the centrifugal force

E

V

r ′

2

mr

E

V

r ′

2

mr

E

V

r ′

Inverse-Square Force „

Consider an attractive 1/

r

2

force

„

Gravity or electrostatic force

„

r

2

force dominates at large

r

„

Centrifugal force dominates atsmall

r

„

A dip forms in the middle

r

V

r ′

2

k

f

r

r

k

V r

r

2

2

k

l

V

r

r

mr

′^

k r 2 −

2

l mr

Bounded Motion „

E = E

2

Æ

r

min

r

r

max

„

Particle is confined between twocircles

r

V

r ′

1 E

2 E

3 E

2

1 2

mr



Goes back andforth between

two radii

Orbit may or may not beclosed. (This one isn’t)

A 1/

r

2

force would

make an ellipse

Circular Motion „

E = E

3

Æ

r

r

0

(fixed)

„

Only one radius is allowed

„

Classification into unbounded, bounded and circular motiondepends on the general shape of

V’

„

Not on the details (1/

r

2 or otherwise)

r

V

r ′

1 E

2 E

3 E

Stays on a circle

0 r

0 (

E

V

r ′

r

0

const

r

r

Stable Circular Orbit „

Circular orbit occurs at the bottom of a dip of

V’

„

Top of a bump works in theory,but it is unstable

„

Initial condition must be exactly

r

stable

r

unstable

E E

2

mr

E

V

dV

mr

dr

const

r

0

0 and

r

r

r

0 r

Stable circular orbit requires

2

2

d Vdr

′^

Orbit Equation „

We have been trying to solve

r

r

t

) and

θ

θ

t

„

We are now interested in the shape of the orbit

„

Switch from

dt

to

d

„

Switch from

r

to

r

r

θ

2

l

mr

2

d

l^

d

dt

mr

d

2

3

l

dV

mr

mr

dr

2

2

2

3

l^

d

l^

dr

l

dV

r

d

mr

d

mr

dr

θ

θ

2

du

d

dr

d

d

r

r

d

θ

θ

θ

u

r

2

d

d

u

dr

du

const

Symmetry of Orbit „

Equation is even, or symmetric, in

θ

„

Replacing

with –

does not change the equation

„

Solution

u

) must be symmetric if the initial condition is

„

Choosing

= 0 at

t

Æ

makes

„

Orbit is symmetric at angles where

du

/ d

2

1

2

2

u

dV

d u

m

u

d

l

du

θ

u

u

du

du

d

d

θ

θ

OK

OK if

du d

θ

Symmetry of Orbit „

Orbit is symmetric about everyturning point = apsidals

„

That’s why I didn’t care too much aboutthe sign of

„

Solve the orbit between a pair of apsidalpoints

Æ

Entire orbit is known

„

Now it’s time to solve the equation

du d

θ

Orbit is invariant under reflection

about apsidal vectors

 r

Inverse Square Force

„

Look it up in a math text book and find

„

Just substitute

and

„

Or…

2

2

arccos

dx

x

x

x

β

γ

γ

α

β

γ

β

αγ

∫

2 k

f

r

k

V

r

2

2

2

du

mE

mku

u

d

l

l

θ

2

2

2

2

2

mE

mku

l^

l du

d

u

θ

∫

∫

Working It Out Yourself

(

)

2

2

2

2

2

4

2

2

2

2

4

2

2

2

2

4

2

2

2

2

2

2

2

2

sin sin

mE

mku

mE

m k

mk

l^

l^

l^

l^

l

mE

m k

mk

l^

l^

l mE

m k

l^

l

du

du

d

u

u

du

u

d

θ

ω

ω

ω

ω

∫

∫

∫

∫

∫

Define as cos

ω

2

2

2

4

2

sin

mE

m k

l^

l

du

d ω

ω

2

2

2

2

4

2

cos

cos(

mkl mE

m k

l^

l u

ω

θ

θ

Solve this for

u

= 1/

r

Energy and Eccentricity „

E

= 0 separates unbounded and

bounded orbits

„

Borderline = Parabola

„

Circular orbit requires

r

V

r ′

k r 2 −

2

l mr

Parabola Hyperbola

Ellipse

Circle

0

2

2

3

0

0

r

dV

k

l

dr

r

mr

′^

2

0

2

0

0

k

l

V

r

E

r

mr

′^

(^22)

mk^2

E

l

„

e

Æ

hyperbola

„

is the turning point (perihelion)

„

cos(

e

limits

„

e

Æ

parabola

Unbound Orbits

(

)

cos(

C

e

r

θ

θ ′

θ ′

θ

θ ′ −

r

θ

r