Download - Today’s topic: Some Celestial Mechanics F Numeriska beräkningar i Naturvetenskap och Teknik
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Today’s topic:
Some Celestial Mechanics
F
),,( zyx),,( ZYX
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Coordinate systems
Cartesian coordinates
z
x
y
xe
ye
ze
Unit vectors areorthogonal with norm 1
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Cylindrical coordinates
x
z
e y
z
eze
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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
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Spherical coordinates
z
x
ye
e
re
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FrT
dt
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
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The angular momentum is constant in a central force field...
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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.
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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
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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!
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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
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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
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and also using the same method we can derive
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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!
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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
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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
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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
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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
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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!
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The energy is a second constant of motion...
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A second constant of motion
A conservative force, i.e. a force with potential
Vd
dF
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WHY of interest?
Examples of such forces?
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A second constant of motion
m
lmF
3
2
Vd
dF
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How to get a first order time derivative out of this?
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A second constant of motion
)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
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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
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Look at this at home!
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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)
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Potential energy
Kinetic energy from motion in phi
Lets identify the terms!
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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
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Solving the equations of motion
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
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Apply it two times
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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!
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Solving the equations of motion
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!)
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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
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Different orbits
Investigate in the project!
)cos1(
12
ekm
l
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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?
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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
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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.
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Extra
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Mean anomaly (ohmega constant, if e=0)
Actual angle = true anomaly)
Variable substitution...
)cos1( ea
a
Half major axis
Eccentric anomaly
a
tt
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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(
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)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
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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.
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drdzddV 2
1
Notera att volymelementet i cylinderkoordinater är:
x
z
d
d
dz
y
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