Central Force Motion

Chapter 4

!1!

Central Force Motions

IO Reminder : Polar coordinates

. Equations of motion in some systems aren't separable in

Cartesian coordinates,

but can be # me simple and solvable in a

different choice of coordinates.

ryEo

'

| Polar coordinates ( in ZD )# EI cylindrical ( in 3D )

,5 '=reI= xetxtyej

To Recall (

darted) : n2=x 't

,'

)×

O= Arctaa

%-) endjetjieo' anddaoE= - ei

'± Elisa ⇒fdjfoitoeg

Spherical coordinates : more complicated but same logic .

Zn +

IE, F=sE '

\,e¢10Leg X= nsino cost1 y = n Sind Sino

•,

,

>Y Z

= ncoso cost

→1

¢ \ ! ei'

C ¥ +× Tinto

, .no e± '

+

=|5

'

' nicotine . rjzeat

a not Zio . ¢ }in¢ws•

niisinot Ziosiao + Znoiouoso

!2!

IIO Central Motion and potential

I particle : ¥75 '

In spherical coordinates' '

5'

a he' F= FGIE'

central Ei=Ipy

OEZ To force Hill1 -7 ×

g--

• Central forces are conservative : U= - f 'Fk'Ids '

.For central forces the motion takesplace on a plane :

I - Kei

'

→ Torque F= 5 'xF'

: 5 'since 5 '×ej=o '

=) 4¥ , o with F= 5 '×mF= & constant of motion+

I I'

E plane 1 to % =) motion is effectively ZD !

Moreover : F's (ngxmfnioo = mice E'

=) 726 = cst = § → if n small,

!6! largeConstant of motion

(angular momentum )

Equations of motion : mat . .m|5'

' 902

ziot.no. =/

- od%,

=) Zio + not =L dtp)= o @ we recover io

'

. est

Let Us get rid of o'

usingO' = £

MNZ

!3!

⇒ mti . no' ) = moi - nmmk÷

,

= - ¥⇐ min

'

= . day,

+ mb÷,

= - dodge

with Ueq ( n ) = UG ) tzbm÷ Effective potential

we've mapped the problem onto an effective one dimensional

system :

msi = -

d.Y÷A =) formally solvable !

This equation can be integrated once ( conservation of energy) :

tmn 't Uepp ( s ) = E= cst Id effectivemotion !

=) integrate

#O Motion in central potentials and Kepler 's Problem

So the first step is now to plot Uefp ( n ) :

E± : FG ) : - nk.( gravitation - Cotton ,

Coulomb Ke 9¥)2

UGK -

Kg =\ Ueq , - Kg t ¥mn

Note : E } Ueq for motion to take place

!4!

- Hui Ueed "

Kaska -

Fi€... ":YtIe⇒.

.=±1 Mr mk

!

bad . -- -

!• E70 : Unbounded motion

:Ueq ( nc ) TE ( o : bounded motion

. Uepplnd = E : circular motion : 5=0

Shape of the orbits : n ( tl,

OIH =) nkel trajectory( forget the dependence)

.

*i=¥od¥ . He ⇐ ⇒

t.FM#noGhf+lo1m__.n4mlE- U ) - lihir*

in'

= It ( gb•mln÷ ) =0

'

¥o(mP÷fFo) Formal solution an

any UGI

. fat ⇐ that # × TIED=mE÷¥÷iEnt¥5

so mini . ueeet± dado.

. ÷ftp.T.n-mgn.IFHwhere F=

- olddt

!5!

this equation is a bit nasty : non linear and second order !

To solve it,

we use the following trick :

U -

. lg Change of variable

¥o=¥Its = - theIfo :# '

adios = . EE.

+ ¥ GET

⇒ Etna . ÷ # l'

. n= . tho.tt#Eye2xYayii'

= - Il 's Eat ' 1 =E*Mkl⇒ dotohtu = - pnfqF( %) "

Harmonic oscillator"

with driving force-

EE.Flkl

Kepler problem : F=- khz = - Ku

'

⇒ dlyz,

+ u = my easy !

¢general solution ob homogeneous

⇒ uco ) = my + A cost . oo )equation

µ particular solution

f eccentricity

!6!

Let A=. mpgke ⇒ n(o1=GYmKl

- E cost - I. )

.

E is determined by the system 's energy and angular momentum:

E .

. Emnitueeohl =Emm?÷l¥ol'

. ¥tfn÷= . . .=mztg÷(e

'. i ) ⇒ e=TtE"

' ⇒ " e "

m KiEto : ET ,

E=o : E= 1

E- Em; :{ =o

Orbits :For the Kepler problem

,

the orbits are conic sections

( = hyperbola ,parabola

,ellipse ,

circle )

Ref : Given a point ( focus) and a line ( dinectnix ) :EIIT

PE conic section ⇐ ) d ( P,

focus ) = E d( P,

direct

.in/dinafnix#tP,

d( P, focus )= n

=d( P

,direct.ir/=dtncoso

d F( focus ) n= Ed + En cos 0(⇒ n= Edl- ECOSO

E== : Circle : d→ A,

E→ o,

Ed : R fixed

E=Emin= - M¥2 ,R= QYMK : nco1=R

circle !

!7!

EI : parabola ( E=o ) : d. =n( t use ) = n - ×

⇒ ( dtxl '=r '= city '

⇒

= × 't Zxdt de×= ¥ (y ?dY

anyNY c= fate

#.

=Ea

,ix

: k¥0Iii ×.

( : tic- minor

EIo.li Ellipse / hyperbolaa

axis-

( or { < 11 Ey ,major axis

Eto E > 0

Ed =n - n{ use = s - ex =) e%Hxt=i=x' tyi= Eydhtx't2dx|

⇒ × '( 1- e 4 - zdaixty' =eat

⇒ Keefe '÷r+t÷=i÷⇒ ( x .

ei÷t+y÷=e¥ot+÷t.

#-

Key"

ellipse14

- e'

oeeti: K÷+f÷= I E. × - Feet (closed)

at i÷ , ,ei=¥ :( e- . Feet~ hyperbolaE > l : ¥ -Ye÷= I

, t.EE (open

!8!

#O Kepler 's law

!1! Planets move in elliptical orbits about the Sun where the sun

atone focus ( Moss mpknef

,

: See next chapter )

!2! The area per unit time swept out by a radius vector fromthe son to a planet is constant :

to = est : DA :{ n' do so dff.

= est

nqI2*¥da

!3! The square of a planet 's period is proportional to the cube

of themajor axe ) of the planet 's orbit

.

Pnoob : daft =z£m =) f. It-

.T = 2¥ A over a period

A. tab for an ellipse

=I ,[d÷,g, ,

with ed= lilmk

= IT a a × # ⇒ TEEpm.la?4jIyt2a4(1- EY0

and {D= all - EY , lifnk =\ T '=4T,÷ma3

.a

>

!9!

Summary ( and different derivation of theeq . of )

the trajectory

!1! mini . no' )= - KG ' } mini = . dueeyd

,mrio = l

.

Uepp = - Ent IZmsh

!2! E = { most + Uepp ( r )

⇒ o¥t=±FmeF±FEm2n2

and o¥=odIo=f÷.

'Io

!3! This yields : f.to/mn# = ± @ - Oo )

- TFnEt2m±s - thin constant

!4! set

u=4n⇒ du

= - Ynzdn

The integral become $ : fbdu_ -

= 0 - do

5215 TEETHCan be computed : new - I

= - cost - Oo )Ftc

m-k~

@

Let e= them :

⇒ mf÷ f- = I - E cos Co - Oo )

# s= mkleil- E cos (O - oo )