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Kepler's laws and trajectories



 
 
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  #1  
Old March 25th 04, 09:34 PM
tetrahedron
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Default Kepler's laws and trajectories

I have been taught that the motion of a two-body system can take the
form of an ellipse, a parabola or a hyperbola, but how can I derive
these trajectories explicitly in polar and eventually cartesian
coordinates?

Henceforth is r the distance between the bodies, M and m the
respective masses, etc.

mu*r'' = mu*r*(theta')^2 + G*M*m/r^2 ; mu := M*m/(M+m)

mu*r'' - mu*r*(theta')^2 - G*M*m/r^2 = 0 ; L := mu*r^2*r'

mu*r'' - L^2/(mu*r^3) - G*M*m/r^2 = 0 ; *r'

mu*r''*r' - L^2*r'/(mu*r^3) - G*M*m*r'/r^2 = 0 ; integrate

mu*(r')^2/2 + L^2/(2*mu*r^2) + G*M*m/r = E

Even if I solve the last DE, then im stuck with an inverse function.
The problem is analogous if I try to solve for the angle theta. Is
there a way to obtain r(t) and theta(t), that is a parametrization of
the trajectory in polar coordinates? The rest would follow
immediately. Thanks for any help =)
  #2  
Old March 26th 04, 10:42 AM
Bjoern Feuerbacher
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Default Kepler's laws and trajectories

tetrahedron wrote:

I have been taught that the motion of a two-body system can take the
form of an ellipse, a parabola or a hyperbola,


Only true if the interaction force follows an inverse square law (or if
the force increases linearly with distanc, IIRC) - but according to the
thread title, you seem to know that.


but how can I derive
these trajectories explicitly in polar and eventually cartesian
coordinates?

Henceforth is r the distance between the bodies, M and m the
respective masses, etc.

mu*r'' = mu*r*(theta')^2 + G*M*m/r^2 ; mu := M*m/(M+m)


Apparently you made a sign error - the force of gravitation tends to
decrease the radius, hence it's sign has to be negative here.


mu*r'' - mu*r*(theta')^2 - G*M*m/r^2 = 0 ; L := mu*r^2*r'

mu*r'' - L^2/(mu*r^3) - G*M*m/r^2 = 0 ; *r'

mu*r''*r' - L^2*r'/(mu*r^3) - G*M*m*r'/r^2 = 0 ; integrate

mu*(r')^2/2 + L^2/(2*mu*r^2) + G*M*m/r = E


Yes, there is definitely a sign error above, which propagated until
here. The gravitational potential energy has to be negative!


Even if I solve the last DE, then im stuck with an inverse function.
The problem is analogous if I try to solve for the angle theta. Is
there a way to obtain r(t) and theta(t), that is a parametrization of
the trajectory in polar coordinates? The rest would follow
immediately. Thanks for any help =)


IIRC, the trick was to write L as L = m omega r^2 and omega as dphi/dt.
Then, you can change the dr/dt in your differential equation above into
a dr/dphi, and the resulting differential equation can be solved
straightforwardly. You then get an equation r = r(phi), which can be
identified with the known equations of an ellipse, a parabola or a
hyperbola, depending on the parameters.

HTH!


Bye,
Bjoern
  #3  
Old March 27th 04, 06:31 AM
John Schutkeker
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Default Kepler's laws and trajectories

Bjoern Feuerbacher wrote in
:

tetrahedron wrote:

The motion of a two-body system can take the
form of an ellipse, a parabola or a hyperbola.
How can I derive
these trajectories explicitly in polar and eventually cartesian
coordinates?
If I solve the DE, I'm stuck with an inverse function.
The problem is analogous if I try to solve for the angle theta. Is
there a way to obtain r(t) and theta(t)?


IIRC, the trick was to write L as L = m omega r^2 and omega as dphi/dt.
Then, you can change the dr/dt in your differential equation above into
a dr/dphi, and the resulting differential equation can be solved
straightforwardly. You then get an equation r = r(phi), which can be
identified with the known equations of an ellipse, a parabola or a
hyperbola, depending on the parameters.


This is the classic analysis that derives the ellipse, and Plummer's
textbook on mechanics presents it quite elegantly, although you sound savvy
enough to work it through yourself, given what Bjorn just said. But don't
waste time trying to derive Cartesian expressions. The problem is
intrinsically polar, and the endless supply of square roots you get from
going to Cartesian geometry complexitizes the problem so much that any
meaning becomes quite well concealed in the algebraic mess.

As it happens, the time dependence of the trajectory has *never* been
solved, even though the problem has lingered for 400 years! In fact, an
NSF Presidential Young Investigator at MIT, a superstar professor named
Jack Wisdom, is currently working aggressively on the problem, and he is
generally accepted as having the best mind for celestial mechanics since
Poincare.

If you've got $25 to spend, and are interested in a very readable textbook
on the state of the art, buy JMA Danby's book on Celestial Mechanics from
the Willmann-Bell website, willbell.com. If you decide to save the $ and
use library loan instead, make sure you get the 1992 edition. Danby very
readably presents the state of the art in terms of senior level vector
mechanics, so it's a *lot* more accessible to the amateur than Wisdom's
work, which is based on perturbation theory, Lagrangians and Hamiltonians,
stuff that I'm still trying to teach myself.

To make a long story short, the equation you derived, t(theta) *cannot* be
inverted into an expression for theta(t), because in canonical form, the
task reduces to inverting Sin(E)+E=C*t, which is commonly known to be
impossible to invert. I wasted two years of my life getting to this point,
and the entire community of researchers seems to be in denial of this fact.
Pick up "Solving Kepler's Equation over Three Centuries," for an amazing
compendium of work that refuses to accept that the non-invertability of
that well-known equation. I've never seen so many hammerheads publish
pointless analyses in the same place.

The problem is that the people see the roadblock bass-ackwardly. They
think that accepting non-invertability is admitting defeat, whereas it is
really just clearing the decks so that you can start hunting for a truly
creative idea. I can't say any more about this until I publish it, but I
like to think that I might have a magic idea to break the logjam.

If you want to get into deeper math than vector mechanics, invest $65 in
Wisdom and Sussman's book, "Structure and Interpretation of Classical
Mechanics," from the MIT Press. You may also want to supplement that with
Florian Scheck's "Mechanics: From Newton's Laws to Deterministic Chaos"
MIT's textbook for 8.09, their undergraduate class in Lagrangian mechanics.

The biggest problem with Wisdom's book is that it introduces a slightly
revised notation for writing Lagrangians and Hamiltonians, even though
there's nothing truly gained by doing that, except to make the equations
easier to write, and slightly harder to read. I'd stick with the standard
notation to get started, before trying to read Wisdom. Scheck is $15
cheaper, although I expect that it may have nothing useful on Celestial
Mechanics.

Personally, I think that Newton's Laws are enough to solve this problem,
and that introducing Lagrangians causes more confusion than clarity,
although a certain amount of inventiveness is required to implement Newton,
as well.

The same is *not* true for perturbation theory, which looks like an
absolutely indispensable tool.

My friend, you have chosen to work on the single most important problem in
physics today, and I wish you good luck and happy hunting. Feel free to
write me directly for further discussion. Just take "-nospam" off my
address in the message header.

JS
 




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