Numeriska ber äkningar i Naturvetenskap och Teknik

Today’s topic:

Some Celestial Mechanics

F),,( zyx),,( ZYX

Numeriska beräkningar i Naturvetenskap och Teknik

Coordinate systems

Cartesian coordinates

z

x

y

xe

ye

ze

Unit vectors areorthogonal with norm 1


Cylindrical coordinates

x

z

e y

z

eze


Vector- och scalar product in cylindrical coordinates

)1,0,0(ze

)0,cos,sin( e

)0,sin,(cos e

e

ee

cos

sin

sin

cos

x

y

eee

eee

eee

z

z

z

Orthogonal

Right hand system


Spherical coordinates

z

x

ye

e

re


FrT

dtpdF

vmrprL

TFrvmvdtpdrp

dtrd

dtprd

dtLd

0)(

2

2

dtrdmF Force law

Torque

Angular momentum

gives:

Introductory mechanics

vmr

L

Fr

T


The angular momentum is constant in a central force field...


A quantity that does not change with time, i.e. our case does not change along the trajectory of a planet is called a:

CONSTANT OF MOTION

If we can find a quantity whose time derivative is zero that quantity is a constant of motion.

TdtLd

Central force

r

0T

FrT

F

constant0 LdtLd

r x p is orthogonal to r, i.e. r is orthogonal to L which is constant.

0)( prrLr

1 Angular momentum is a constant of motion

2. Motion is in a plane


In order write down the equations of motion we need the acceleration in cylindrical

coordinates.

This problem relies on the calculus you learn in math class!


Velocity in cylindrical coordinates

),sin,cos( zr

),cossin ,sincos(),sin,cos( zzdtd

dtrd

),0,0()0,cos,sin()0,sin, (cos z

Motion in the plane due to central force 0z

eedtrd )cos,sin()sin, (cos

Radial velocity Angular velocity


In the same way… acceleration in cylindrical coordinates

eedtrd )cos,sin()sin, (cos

)()(2

2

edtdee

dtdeee

dtd

dtrd

)( edtdeee

dtde


and also using the same method we can derive


Acceleration in cylindrical coordinates

edtde

dtd och

edtde

dtd )cos,sin()sin,(cos

)0,cos,sin( e

)0,sin,(cos e

Look at this at home!

Acceleration in cylindrical coordinates

)(2

2

edtdeee

dtde

dtrd

eeeeedtrd 22

2

Ins. from above gives that we have TWO components

ee )2()( 2


Equations of motion in the central force system

2

2

dtrdmF ),,(),,( zyxmFFF zyx

this can also be written as:

))2()(( 2zzz ezeemeFeFeF

eedtrd )2()( 22

2

with the acceleration in the plane

00 0


Equations of motion in the plane in cylindrical coordinates

emeF )( 2

em )2(0

02

)2(1)(1 22

dtd

Depends explicitly on the force

Can be integrated without defining F

Now, we note that

0)( 2 dtd

i.e.

Which gives


Sector velocity

y

x

d

ddA 21

dA

22

21

21

dtd

dtdA

zem 2

mL

dtdA

2konstant

21 02

Kepler’s second law

2ml 2ml


Rho direction:Equations of motion in the plane in cylindrical coordinates

emeF )( 2

)( 2 mF

2ml

24

22

2 ml

ml

The angular momentum can be used to switch between rho and phi!

since

)( 24

2

mlmF

mlmF 3

2

We have

Substitution gives:

We have two functions oftime, rho and phi. We want ONE!


The energy is a second constant of motion...


A second constant of motion

A conservative force, i.e. a force with potential

VddF


WHY of interest?

Examples of such forces?


mlmF 3

2

VddF


How to get a first order time derivative out of this?


)2

( 2

2

mlV

ddm

dtd

We think of the chain rule again and multiply by

)2

( 2

2

mlV

dd

dtd

dtdm

)

2( 2

2

mlV

dtdm

These are equal


in the eq. below

)2

( 2

2

mlV

dtdm

mm

dtd

)21( 2

)2

()21( 2

22

mlV

dtdm

dtd

0)

221( 2

22

mlVm

dtd

We now have time derivatives on both sides of this equation!

i.e.

Continue by looking at the left hand side

l.h can be written


Look at this at home!

22

2

22

2

242

2

2

mρmρρm

ml

2ml

Kinetic energy from radial motion

From L constant we have (still)


Potential energy

Kinetic energy from motion in phi

Lets identify the terms!

Solving the equations of motion

One can now either try to integrate with respect to the time, t, or, one can solve with respect to the angle.

Two steps for a ”straightforward” solution.

1. Transform equation to be distance rho as function ofthe angle phi instead of time.

2. Make a 1/rho substitution to create a standard linear diff. eq. with constant coefficients.

mlmF 3

2

2

kF mlmk3

2

2



From second order time derivative to second order derivative in phi:

2ml 2ml dmldt 2

dd

ml

dtd

2

)()( 2222

2

dd

ml

dd

ml

dd

ml

dtd

dtd


Apply it two times

23

2

22 )(

kml

dd

ml

ddl

At this point we have

dd

dd )/1(1

2 but

232

2 )( kumul

ddu

ml

ddlu

2

32

2

222

kumul

dud

mul

/1u

Binet!



Binet’s equation for the kepler case (1/r2 )

22

222

)( kuudud

mul

22

2

)(lkmu

dud

2)cos(lkmAu

Second order diff equation. (solve with characteristic equation!)


Different orbits

2)cos(lkmAu

Reference direction when α is zero

)cos1()cos1(12

2

2

elkm

kmlA

lkm

)cos1(12

ekml


Different orbits

Investigate in the project!

)cos1(12

ekml


Other thoughts:

Which velocity, in which direction, will give circular orbit?

Is there a maximum velocity for a planet to stay in a closed orbit around the Sun?

If the velocity is below the escape velocity, how does different start angles influence the shape of the orbit? Can you create ellipses and circles from the same starting speed?

If a small planet passes close by another planet (e.g. an elliptic orbit that passes close to a jupiter like planet) what will happen. Why? (Voyager slingshots).

If we integrate what should be a closed orbit with bad precision what will happen?


Orbital motionρ(t)

Ekonstant22

1 222 Vmm

)

2(2

2

2

Emlk

m

dE

mlk

m

dt)

2(2

12

0

2

)2

(

12

dE

mlk

mt


Orbital motionρ(t)

0

2

)2

(

12

dE

mlk

mt

This integral can in principle be solved t(ρ) but its inversion ρ(t) is not possible in ”simple functions”. The same is true for the angle as a function of time.


Extra


Mean anomaly (ohmega constant, if e=0)

Actual angle = true anomaly)

Variable substitution...

)cos1( ea

a

Half major axis

Eccentric anomaly

a

tt


After this substitution...

0

3

)cos1( dekmat

Kepler’s third law (can also be found from geometrical considerations)

kmade

kma 2/3

2

0

3

2)cos1(


)sin()cos1(3

0

3

ekmade

kmat

3

2mak

sinet Kepler’s equation

)cos1( ea How find ρ(t)?

Only numerical solution

Gives ρ for this t!

Generally at time t


Two body problem

For two interacting bodies the mass above is substituted by the so-called reduced mass

21

21

mmmm

Three body problem...

Many tried to solve it (Poincare and others) but no solutionexists in simple analytical form. Power series expansions exist.The problem has a very interesting background story. As an example, find and read on your own the story behind the Mittag-Leffler prize.


drdzddV 21

Notera att volymelementet i cylinderkoordinater är:

x

z

d

d

dz

y


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Numeriska ber äkningar i Naturvetenskap och Teknik