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MechanicsPhysics 151

Lecture 6Kepler Problem

(Chapter 3)

What We Did Last Time

Discussed energy conservationDefined energy function h Conserved ifConditions for h = E

Started discussing Central Force ProblemsReduced 2-body problem into central force problem

Problem is reduced to one equationUsed angular momentum conservationEnergy conservation gives

Now we must solve this

2

3 ( )lmr f rmr

= +

0L t∂ ∂ =

22

2

1 ( ) const2 2m lE r V r

mr= + + =

Goals for Today

Analyze qualitative behavior of central-force problemSolutions: bounded or unbounded

Determined by the “shape” of the potential

Solve the Kepler problemGet the shape of the orbit

As if we don’t know yet…Derive Kepler’s 3rd Law

Period of rotation is proportional to the 3/2 power of the major axis

2

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

2 ( )2

lr E V rm mr⎛ ⎞

= − −⎜ ⎟⎝ ⎠

2

2( ) ( )2

lV r V rmr

′ ≡ + Quasi potential including the centrifugal force

( )E V r′>2

( ) 02

mr E V r′= − >2

( )2

mrE V r′= +

Inverse-Square Force

Consider an attractive 1/r2 force

Gravity or electrostatic force

1/r2 force dominates at large rCentrifugal force dominates at small rA dip forms in the middle

r( )V r′

2( ) kf rr

= − ( ) kV rr

= −

2

2( )2

k lV rr mr

′ = − +

kr

−

2

22lmr

Unbounded Motion

Take V’ similar to 1/r2 caseOnly general features are relevant

E = E1 r > rminParticle can go infinitely far

r

( )V r′

1E

2E

3E

212

mr

Arrive from r = ∞

Turning point

Go toward r = ∞

E V ′= 0r =A 1/r2 force would make a hyperbola

1 min( )E V r′=

3

Bounded Motion

E = E2 rmin < r < rmaxParticle is confined between two circles

r

( )V r′

1E

2E

3E

212

mrGoes back and forth between

two radii

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

A 1/r2 force would make an ellipse

Circular Motion

E = E3 r = r0 (fixed)Only one radius is allowed

Classification into unbounded, bounded and circular motion depends on the general shape of V’

Not on the details (1/r2 or otherwise)

r

( )V r′

1E

2E

3E

Stays on a circle

0r

0( )E V r′=

0r =

0constr r= =

Another Example

Attractive r-4 forceV’ has a bumpParticle with energy E may be either bounded or unbounded, depending on the initial r

3

aVr

= − 4

3afr

= −2

3 22a lVr mr

′ = − +

r

V ′

2

22lmr

V

E

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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

rstable

r

unstable

E

E

2

02

mr E V ′= − = 0dVmrdr′

= − =

constr =

00 and r r r= =

0r

Stable circular orbit requires2

2 0d 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 orbitSwitch from dt to dθ

Switch from r to

( )r r θ=

2l mr θ=2

d l ddt mr dθ

=

2

3 0l dVmrmr dr

− + =2

2 2 3 0l d l dr l dVr d mr d mr drθ θ

⎛ ⎞ − + =⎜ ⎟⎝ ⎠

2

1 1du d drd d r r dθ θ θ

⎛ ⎞= = −⎜ ⎟⎝ ⎠

1u r≡

2d dudr du

= −

const

Orbit Equation

Solving this equation gives the shape of the orbitNot that it’s easy (How could it be?)Will do this for inverse-square force later

One more useful knowledge can be extracted without solving the equation

2 1

2 2

( ) 0udVd u mud l duθ

+ + =Switching variables

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Symmetry of Orbit

Equation is even, or symmetric, in θReplacing θ with –θ does not change the equationSolution u(θ) must be symmetric if the initial condition isChoosing θ = 0 at t = 0, θ –θ makes

Orbit is symmetric at angles where du/dθ = 0

2 1

2 2

( ) 0udVd u mud l duθ

+ + =

(0) (0)u u→ (0) (0)du dud dθ θ

→ −OK OK if (0) 0dudθ

=

Symmetry of Orbit

Orbit is symmetric about every turning point = apsidals

That’s why I didn’t care too much about the sign ofSolve the orbit between a pair of apsidal points Entire orbit is known

Now it’s time to solve the equation0du

dθ=

Orbit is invariant under reflection about apsidal vectors

r

Solving Orbit Equation

Integrating the diff eqn will give energy conservationOne can use energy conservation to save effort

Switch variables

2 1

2 2

( ) 0udVd u mud l duθ

+ + =

2 2

2 ( )2 2

mr lE V rmr

= + +2

2

2 ( )2

lr E V rm mr⎛ ⎞

= − −⎜ ⎟⎝ ⎠

Integrate this…1

22 2

2 ( )2 umVdu mE ud l lθ

= − − −

l durm dθ

= −

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Inverse Square Force

Look it up in a math text book and find

Just substitute α, β and γOr…

2 2

2 2arccos4

dx xx x

β γγα β γ β αγ

⎛ ⎞+⎜ ⎟= −⎜ ⎟−+ + −⎝ ⎠

∫

2

kfr

= −kVr

= − 22 2

2 2du mE mku ud l lθ

= − + −

2 222 2mE mku

l l

du du

θ= −+ −

∫ ∫

Working It Out Yourself

( )2 22 2

2 4 2

2 2

2 42

2 2

2 4

2 22 2 2

22

2

1

1

sinsin

mE mku mE m k mkl l l l l

mE m kmkl ll

mE m kl l

du dudu u

du

u

d

θ

ω ω ωω

= − = −+ − + − −

= −+ ⎛ ⎞−⎜ ⎟−

⎜ ⎟+⎝ ⎠

= − = −

∫ ∫ ∫

∫

∫

Define as cosω

2 2

2 42 sinmE m k

l ldu dω ω= +

2

2 2

2 42

cos cos( )mkl

mE m kl l

uω θ θ

−′= − =

+Solve this for u = 1/r

Solution

This matches the general equation of a conic

e is eccentricity

2

2 2

1 21 1 cos( )mk Elur l mk

θ θ⎛ ⎞

′= = + + −⎜ ⎟⎜ ⎟⎝ ⎠

( )1 1 cos( )C er

θ θ ′= + −

circleE = e = 0

ellipseE < 0e < 1parabolaE = 0e = 1hyperbolaE > 0e > 1

One focus is at the origin

2

22mk

l−

Matches the qualitative

classification of the orbits

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Energy and Eccentricity

E = 0 separates unbounded and bounded orbits

Borderline = Parabola

Circular orbit requires

r( )V r′

kr

−

2

22lmr

Parabola

Hyperbola

Ellipse

Circle0

2

2 30 0

0r

dV k ldr r mr′

= − =

2

0 20 0

( )2

k lV r Er mr

′ = − + =

2

22mkE

l= −

e > 1 hyperbolaθ’ is the turning point (perihelion)cos(θ – θ’) > –1/e limits θ

e = 1 parabola

Unbound Orbits ( )1 1 cos( )C er

θ θ ′= + −

θ ′

θ θ ′−r

θ ′r

Bound Orbits

Ends of the major axis areLength of the major axis

( )1 1 cos( )C er

θ θ ′= + −

θ ′r

( )1/ 1r C e= ±

1 1 12 (1 ) (1 ) 2

kaC e C e E

⎛ ⎞= + = −⎜ ⎟+ −⎝ ⎠

2

2

21 Elemk

= +

2

mkCl

=

Major axis is given by the total energy E

2a

2b

Minor axis is2

212

lb a emE

= − = −

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Rotation Period

We know that the areal velocity is constant

Express τ in terms of a

2kaE

= −2

2lbmE

= −2 2

38l kA abmE

π π= = −

212 2

dA lrdt m

θ= =

Area of the orbit

Period ofrotation

2

32mkdAA

dt Eτ π= = −

3/ 22 mak

τ π= Period of rotation is proportional to the 3/2 power of the major axis

Kepler’s Third Law of Planetary Motion

Kepler’s Third Law

Kepler’s third law is not exactThe reason: reduced massk is given by the gravity

Period of rotation becomes

Coefficient is same for all planets only if M >> m

1 1 1M mµ

= + planet

Sun

3/ 22 mak

τ π=

2 2

Mm kf Gr r

= − = −

3/ 2 3/ 212 2( )

a ak G M mµτ π π= =

+

k GMm=

Time Dependence

So far we dealt with the shape of the orbit: r = r(θ)We don’t have the full solutions r = r(t) and θ = θ(t)

Why aren’t we doing it?It’s awfully complicated

Not that bad to get t = t(θ) See Goldstein Section 3.8Inverting to θ = θ(t) impossiblePhysicists spent centuries calculating approximate solutions

Already got physically interesting features of the solutionLeave it to the computers

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Summary

Studied qualitative behavior of the orbitsBounded or unbounded Shape of

Derived orbit equation from the eqn of radial motionr (or u = 1/r) as a function of θ

Analyzed the Kepler ProblemSolved the orbitConic depending on EFor elliptic orbit, major axis depends only on EKepler’s third law of planetary motion

2

2 2

1 21 1 cos( )mk Elr l mk

θ θ⎛ ⎞

′= + + −⎜ ⎟⎜ ⎟⎝ ⎠

2kaE

= −

2

2( ) ( )2

lV r V rmr

′ ≡ +