Today’s topic: Some Celestial Mechanics F Numeriska beräkningar 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

e

e

cos

sin

sin

cos

x

y

eee

eee

eee

z

z

z

Orthogonal

Right hand system


Spherical coordinates

z

x

ye

e

re


FrT

dt

pdF

vmrprL

TFrvmvdt

pdrp

dt

rd

dt

prd

dt

Ld

0

)(

2

2

dt

rdmF 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.

Tdt

Ld

Central force

r

0T

FrT

F

constant0 Ldt

Ld

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( zzdt

d

dt

rd

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

Motion in the plane due to central force 0z

eedt

rd )cos,sin()sin, (cos

Radial velocity Angular velocity


In the same way… acceleration in cylindrical coordinates

eedt

rd )cos,sin()sin, (cos

)()(2

2

edt

dee

dt

deee

dt

d

dt

rd

)( edt

deee

dt

de


and also using the same method we can derive


Acceleration in cylindrical coordinates

edt

de

dt

doch

edt

de

dt

d )cos,sin()sin,(cos

)0,cos,sin( e

)0,sin,(cos e

Look at this at home!

Acceleration in cylindrical coordinates

)(2

2

edt

deee

dt

de

dt

rd

eeeeedt

rd 22

2

Ins. from above gives that we have TWO components

ee )2()( 2


Equations of motion in the central force system

2

2

dt

rdmF ),,(),,( zyxmFFF zyx

this can also be written as:

))2()(( 2zzz ezeemeFeFeF

eedt

rd)2()( 2

2

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

dt

d

Depends explicitly on the force

Can be integrated without defining F

Now, we note that

0)( 2 dt

d

i.e.

Which gives


Sector velocity

y

x

d

ddA 2

1dA

22

2

1

2

1

dt

d

dt

dA

zem 2

m

L

dt

dA

2konstant

2

1 02

Kepler’s second law

2ml 2m

l


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

emeF )( 2

)( 2 mF

2m

l24

22

2 m

l

m

l

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

since

)(24

2

m

lmF

m

lmF

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

Vd

dF


WHY of interest?

Examples of such forces?


m

lmF

3

2

Vd

dF


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


)2

(2

2

m

lV

d

dm

dt

d

We think of the chain rule again and multiply by

)2

(2

2

m

lV

d

d

dt

d

dt

dm

)

2(

2

2

m

lV

dt

dm

These are equal


in the eq. below

)2

(2

2

m

lV

dt

dm

mm

dt

d)

2

1( 2

)2

()2

1(

2

22

m

lV

dt

dm

dt

d

0)

22

1(

2

22

m

lVm

dt

d

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

mρ

mρ

ρm

m

l

2m

l

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.

m

lmF

3

2

2

kF

m

lm

k3

2

2



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

2m

l 2ml dmldt 2 d

d

m

l

dt

d2

)()(2222

2

d

d

m

l

d

d

m

l

d

d

m

l

dt

d

dt

d


Apply it two times

23

2

22)(

k

m

l

d

d

m

l

d

dl

At this point we have

d

d

d

d )/1(12

but

232

2 )( kum

ul

d

du

m

l

d

dlu

2

32

2

222

kum

ul

d

ud

m

ul

/1u

Binet!



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

22

222

)( kuud

ud

m

ul

22

2

)(l

kmu

d

ud

2)cos(l

kmAu

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


Different orbits

2)cos(l

kmAu

Reference direction when α is zero

)cos1()cos1(1

2

2

2

e

l

km

km

lA

l

km

)cos1(

12

ekm

l


Different orbits

Investigate in the project!

)cos1(

12

ekm

l


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 Vm

m

)2

(2

2

2

Em

lk

m

d

Emlk

m

dt

)2

(2

12

0

2

)2

(

1

2d

Emlk

mt


Orbital motionρ(t)

0

2

)2

(

1

2d

Emlk

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( dek

mat

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

k

made

k

ma 2/32

0

3

2)cos1(


)sin()cos1(3

0

3

ek

made

k

mat

3

2

ma

k

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

mm

mm

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 2

1

Notera att volymelementet i cylinderkoordinater är:

x

z

d

d

dz

y


Today’s topic: Some Celestial Mechanics F Numeriska beräkningar i Naturvetenskap och Teknik.

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