r

F

For a central force the position and the force are anti-parallel, so rF=0.

0N r F dL

Ndt

0dL

dt

So, angular momentum, L, is constant

N is torqueNewton II, angular

Since the Angular Momentum, L, is constant:

Its magnitude is fixedIts direction is fixed.

L r p

• By the definition of the cross product, both the position and the momentum are perpendicular to the angular momentum.

• The angular momentum is constant.• Therefore, the position and momentum are

restricted to a plane. The motion is restricted to a plane.

L

rp

rdr

r+dr

Show that the conservation of angular momentum implies that equal areas are swept out in equal times:

r dr

The green shaded area is dA

The parallelogram formed by r and dr is twice the area of dA. But, by definition, the magnitude of rdr is the area of the parallelogram.

1 1sin

2 2dA r dr r dr

dr sin (height)

Proof of Kepler’s Second Law

1

2dA r dr

1

2dA r vdt

1

2

dAr p

dt m

2

dA L

dt m

The area swept out in time dt

But dr is just v dt

Divide by dt, and change v to (1/m)p

The angular momentum is constant, so the rate at which area is swept out is also constant.

Central Forces in Polar Coordinates

Let’s use the Lagrangian:

L T V Don’t confuse the Lagrangian, “L”,with the angular momentum, “L”.

21

ˆ ˆ2 rL m r e re V r Motion is restricted to

a plane, and the potential is that for a central force. 2 2 21

2L m r r V r

2 2 21 1

2 2L mr mr V r

Lmr

r

d L

mrdt r

2Lmr f r

r

2mr mr f r

d L L

dt r r

The Radial Equation

2 2 21 1

2 2L mr mr V r

2Lmr

0

L

2 0d

mrdt

d L L

dt

The Angular Equation

But the term in parenthesis is just the angular momentum of a point particle, so conservation of angular momentum falls out of the constraints on the Lagrangian!

From the Lagrangian, we found Newton’s Laws in Polar Coordinates:

2 Vmr mr

r

Radial eqn.:

2 0mr mr Theta eqn.:

( )V

f rr

2 0r r

2 0d

rdt

2r l Ll r v

m

2 ( )f rr r

m

Obtain solutions for SHAPE, don’t need to solve for tPut it in terms of u and theta:

1r

uLet

1u

r

22 2

1 1 du d du dur u r l

u u d dt d d

Solve for r and its derivatives in terms of u and theta.

2 2 22 2

2 2 2

d du d u d u l d ur l l l l u

dt d d d r d

To reiterate:

22 2

1 1 du d du dur u r l

u u d dt d d

2 2 22 2

2 2 2

d du d u d u l d ur l l l l u

dt d d d r d

1r

u

2 ( )f rr r

m

2

22 2 21 1/

d ul u lu f m

d u u

22

1r l

u

2

2 2

1/f ud uu

d l u m

Now we have a differential eqn. of u and theta

Replace r’s and thetas

Define the Differential Equation for the Orbit Shape

Now, we plug our force law into the differential equation we have derived.

2( )

kf r

r

k GMmThe force is:

2

2 2

2

2 2 2

1/f u ku

l u

d u ku

d mlm l u m

This equation is similar to a simple harmonic oscillator with a constant force offset, except the independent variable is theta not time.

2cos

ku A

ml 2

1

cosr

kA

ml

2

1

cosr

kA

ml

0o Choose theta nought so that theta equal to 0 yields the distance of closest approach

2

2

/

1 cos(

ml kr

mlA

k

“Look Ma, an ellipse!”“But Johnny, It doesn’t look like an ellipse???”“Oh Ma… don’t you know nothin’?”

Manipulate the Shape Equation a Bit More:

2

2

/

1 cos(

ml kr

mlA

k

-1.5 -1 -0.5 0.5 1 1.5

-1

-0.5

0.5

1

We want to show that this equation is an ellipse

?

-1.5 -1 -0.5 0.5 1 1.5

-1

-0.5

0.5

1

The ellipse below has the equation: 2 22 3x y

a

b

2 2

2

2 3

3

3

x y

x

x

Semimajor axis:

2 2

2

2 3

2 3

3

2

x y

y

y

Semiminor Axis

Review of Some Basic Ellipse Properties

-1.5 -1 -0.5 0.5 1 1.5

-1

-0.5

0.5

1

To solve for the eccentricity of an ellipse, use the defining relationship for the ellipse and solve equations for two special cases:

2 2 2

22 2

2

constant

when , constant 2

( ) constant /4

constant

constant 2

1

r r

r r r

b a

a a a a

a

b a a

b

a

Pythagorean Triangle

When touching the right edge

a

r rb

a

Plug (2) into (1)

(1)

(2)

-1.5 -1 -0.5 0.5 1 1.5

-1

-0.5

0.5

1

r

22 2 2

2 2 2 2 2 2

2 2 2 2

2 2

2

sin 2 cos

sin cos 4 cos

4 4 cos

4 cos

r r a

r r a r

r r a r ar

r r a ar

r r a a r

Defining equation for an ellipse

a a

r

Put our Equation for the Ellipse in the Form of Our Shape Equation

r'

2

2

2 2 2 2 2

2 2 2

2

2

2

2

2

4 cos

4 cos

4 4 4 4 cos

4 4 4 4 cos

cos

1 cos 1

2

1

1 os

2

c

r a a r

r a a r

a r ar r a ar

a ar a ar

a r a r

r a

a

r

r

r

r a

a r

Put our Equation for the Ellipse in the Form of Our Shape Equation

From the last slide

From the defining relationship

-1.5 -1 -0.5 0.5 1 1.5

-1

-0.5

0.5

1The latus rectum is the distance to a focus from a point on the ellipse perpendicular to the major axis

21a

2 2 2 2

2 2 2

2 2

2

2

2

2

2

4

1

4

4

2

4

4

4

4

r a

a

a

a a

r

r

a a

a

a

a

Defining relationship for the ellipse

Solve it for r

Define a Pythagorean relationship

Solve the resulting relationship for alpha

r

2a

Define the Latus Rectum,

21

1 cos 1 cos

ar

2

2

/

1 cos(

ml kr

mlA

k

So, our general equation for an ellipse is

And the solution to our shape equation was

They have the same form! We’ve derived Kepler’s Second Law. The trajectories of planets for a central force are described by ellipses.

Finally, Show that our Central Force Yielded an Elliptical Orbit!

2 2

22

/ /

/

ml k l GM

AmlAl GM

k

21

1 cos 1 cos

ar

2

2

os(

/

1 cr

m

ml

lk

k

A

Relations Between Orbit Parameters

2

LA

m

0

Adt A

0 0 0 02 2 2 2

L l l lAdt dt dt dt

m

2

2

l AA

l

Integration of the area dA/dt over one period gives A (1)

Kepler’s Second Law

Use Kepler II to relate integral and l. (2)

Set (1) and (2) equal

Kepler’s Third Law

2 22 2 12

A ab a

l l l

2 4 2 2 3 2 2 32

2 2 2

4 1 4 1 4a a a a

l l l

2 2 2ml ml l

k GMm GM 2

2 34a

GM

Kepler’s Third Law Derivation Continued

22 34

aGM

Earth presumably fits this rule… Using Earth years for time, and Astronomical Units (1 AU = 1 Earth-Sun distance) for distance renders the constants equal to 1.

Look How Well The Solar System Fits Kepler’s Third Law!

Universality of Gravitation: Dark Matter

3

3

4

34

/3

M r

M R

Only the mass interior to the star in question acts.

Density, assuming a uniform density sphere.

Keplerian Motion: All the mass of the galaxy would be assumed to be focused very close to

the origin.2

2nucleus

nucleus

M m vG m

r r

GMv

r

Velocity curve, Assuming a Constant Density Sphere All the Way Out

32

23

4

3

4r mmv

Gr r

v r G

Keplerian Motion vs. Constant Density Sphere Motion

r

v

1 1 1w v u w v uF hF gF u fF hF v gF fF w

gh v w fh w u fg u v

, ,

1, , sin

u r v w

f g r h r

1sin

sin

1sin

sin

1sin

sin

r

r

F

F

F F F rr

r Fr r

r Fr r

Are Central Forces Conservative?

10 0

sin

1sin 0

sin

10

sin0r

rr

r f rr

F

r

Fr r

Central Forces Are Conservative!

Energy Equation of an Orbit in a Central Field

2 2 2 2

2 2

22

2

2 1

1constant

2

1

2

duml u V u E

d

v r r

m r r V r E

2

22 2

2

1r

u

lu

dur l

d

d ur l u

d

Orbital Energies in an inverse Square Law

kV r ku

r

22 21

2

duml u ku E

d

2

22

2

22

2 2

2 2

(1/ 2)

2

1

2 2

2

du E ku

d duE ku

uml ml

ud ml

du E kuu

d ml ml

Integrating Orbital Energy Equation

2

2 2

2

21

2

2 4

2

2

1

1

2 21, ,

1

221

cos1

1 2cos

4

4 8

o

o

d duc bu au

k Ea b c

ml ml

duc bu

b au

a b acau

ku

ml

k Em l ml

2

21

2

2 4 2

2

2 2

2 2 2

2 2

2

2

2

/

221

cos1 4 8

cos2

1 2 / cos

1 2 / co

/ cos

s

1 1 2

o

o

o

o

o

ku

ml

k Em l ml

k uml

k Eml

k Eml k k uml

uml k k Eml k

ml kr

Eml k

Integrating Orbital Energy Equation Continued

2

2

2

2

2 2

2

/

/

1 1 2 / cos

2

21

1

21

2

o

ml kE

k

a

E

kE

ml kr

Eml k

ka

E

l

a

m

k

k

Total Energy of an Orbit

To fit earlier form for an ellipse

Earlier Relations

Total energy of the orbit.

-4 -2 2 4

-4

-2

2

4

E < 0 Ellipse or Circle e<1E=0 Parabola e=1E>0 Hyperbolic e>1

2

2 2

2

2

/

1 1 2 / cos

1 2 /

o

ml kr

E

E

m k

ml k

l

Ellipses, Parabolas, Hyperbolas

2

2

22

0

mv MmG E

rGM

v Er

Maximum Velocity for An elliptical Orbit

5 10 15 20

-1.5

-1

-0.5

0.5

1

1.5

Limits of Radial Motion

2

2

2

2

2

2

2

2

2

mr E

m

lV r

r

U r

mlU r

r E

V rr

E2

22

ml

r

( )V r( )U r

U is the effective potential

2

2

2 2

2 2

2 2

1,0

2 2

1,0

0

02

2 2 0

2 2 0

2 4 8

4

2

2

U r E

ml kE

r r

ml kr Er

Er kr ml

k k Emlr

E

k k Emlr

E

Limits Continued

2 2

1,0

2

min 2

min

2

2

2

2o

k k Emlr

E

kE

mlk

rE

Minimum Orbital Energy

Radical ought to remain real for elliptical orbits.

Value of E for which the radical is 0

Extremes of motion merge to one value.

Scattering and Bound States are All There Are!

E > 0 E < 0

2

2 2 11

2

duml u V u E

d

Energy Equation for a Central Force Again

kV ku

r

QqV Qqu

r

For attractive inverse square For repulsive inverse square

k Qq

2

2 2 11

2

duml u V u E

d

The Scattering Calculation Proceeds Exactly Like the Bound

State Calculation

22 21

2

duml u ku E

d

22 2

1

2 2d du

E kuu

ml ml

But, we’re going to let k go to –Qq after we integrate

2

2

2 2

2

1

2 2

1

1,

2Ac

1o

4

,

so

d duc bu au

k Ea b c

ml ml

duc bu a

b a

au

u

a b c

Integrate Both the Bound State And the Scattering Problem

2

2 2

22

2 2

2

2

2

2

22 22 2

2

2

2

2

2

Acos f u1

2

41

2

4

1 1

22 1144

4 4

4 4 44 4 4

Acos f u

1 ( )

4

2

2 4

1 4 2

b auf u

b ac

b aub aub acb ac

b ac

f ud

du f u

af u

b ac

f u

b ac

a aub b

b ac

a u abu acac a u abu

f ud a b ac

b acdu a auf u

bu c

2

2 2

2 2

2

1 1 2Acos

A4

4

cos

bu c

d b au a

du b ac au

b audu

aau

b

bu c b a

u c

c

Verify Integral in 6.10.4

2

2

2

2

2

2 2 4 2

2

1

2Acos

4

2 4 cos

2 4 cos

2cos

1 2Acos

4o

o

o

o

o

duc bu au

b u

b c

u b b c

u b b c

k k Eu

ml m

b au

a c

l

b a

ml

Solve Both Bound State and Scattering Problem for u

2

2 2 2

22

2 2

2

2

2

2

2

2cos

1 21 1 c

21 1

os

21 1 cos

cos

o

o

o

o

Qq Qq Eu

ml ml ml

Qq E ml

r ml ml Qq

mlkr

Eml

mlQq

rEm

k

l

Qq

Solution to the Scattering Problem

Solution to the Bound State Problem

Solve Both Problems for r

Scattering in an Inverse Square Field

2

2

2

2

d u Qqu

d

d u ku

d ml ml

2 2

1

co ss

1

corr

kA

QqA

mml l

2

2 2 2

/

1 1 2 / cos o

ml qQr

Eml Q q

Solving the energy

equation for r

attractive

RepulsiveK goes to -Qq

b

so

o

rmin

A Drawing of the Scattering Problem

A small range of impact parameter

scattering center

Trajectory

2

2 2 2

/

1 1 2 / cos o

ml qQr

Eml Q q

Impact parameter b

b

so

o

rmin

A small range of impact parameter

scattering center

Trajectory

Impact parameter b

2

2 2 2

/

1 1 2 / cos o

ml qQr

Eml Q q

0

cos 0

cos 2 cos cos

2 0 asymptotes

o

o o o o

o

r

r

r

or

Asymptotes

min

2o

s o

r

2

2 2 2

/

1 1 2 / cos o

ml qQr

Eml Q q

(1/ 2)2

2

2 2 2

2 2 2

(1/ 2)2

2

2

2

1 1 2 / cos 2 0

1cos

1 2 /

2tan

2

1

2

1tan tan

2

2cot

2

o o

o

o

s o

o s

o ss

Eml Q q

Eml Q q

Eml

Q q

Eml

Q q

(1/ 2)2

2 2

2Eml

Q q

1

2

2 2

21

Eml

Q q

o

/2-o

Angles