Kepler - Classical Mechanics - Lecture Slides, Slides of Classical Mechanics

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Kepler

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Inverse

Square

Force

Force

can

be

derived

from

a

potential.^ –^

^ <

0 for

attractive

force

Choose

constant

of

integration

so

V (

)^ =

r V

m^2

r^1

int F (^2) r 2

R^

m^1

int F 1

r = r

- r 1 2

V^ r

r Qr^

^

^2

2 1

2 1^ m m

mm 

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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.

r r r V T L  

^

^

(^

(^22) 2 1 2

r J r r V T E

^

^

2 2 1 2 2 1 2

)^ constant

(^

(^22) 2 1 2

^

^

r r

T^

^

r V

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Apsidal

Position

Elliptical

orbits

have

stable

apses.

-^

Kepler’s

first

law

-^

Minimum

and

maximum

values

of^ r.

-^

Other

orbits

only

have

a

minimum. The

energy

is^

related

to

e:

-^

Set

r^ =

r ,^2

no

velocity

cos (^1) ( 1 1

e es r^

 r s

r^1

es^ e r^2 r^

^21

es^ e r^

^11

12 2 ) 2 2 (^1) (

EJ 

e^

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Effective

Potential

Treat

problem

as

a^ one

dimension

only.

-^

Just

radial

r^ term.

Minimum

in potential

implies

bounded

orbits.

-^

For

^ >

0,^

no^

minimum

-^

For

E^

^ 0,

unbounded

r J r Veff

^

^

2 2 2

eff r^

V

T

r J r r E^

^

2 2 1 2 2 1 2

Veff 0

r

Veff 0

r

unbounded

possiblybounded

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