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BobG
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Nov20-04, 03:04 PM
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Quote Quote by Machinus
This problem is pretty complicated, and I have fooled around with it in Maple but I don't understand the concepts well enough to work it out yet.

Suppose we have a simple hyperbolic orbit where there is a central mass M and a smaller mass m, and there is more than enough energy in this system for the smaller mass to whiz by and fly back out into space, reasonably deflected. For this problem we have at least two conservation principles to work with, one being total energy

[tex]E=\frac{1}{2}m\vec{v}^2-\frac{GMm}{r}+\frac{ml^2}{2r^2}[/tex]

[tex]\frac{d}{dt}E=0[/tex]

and total angular momentum

[tex]L=mr^2\dot\theta[/tex]

[tex]\frac{d}{dt}L=0[/tex]

I am attempting to solve for maximum [tex]\dot{r}[/tex], with as few variables remaining in the equation as possible.

I have attempted to solve for [tex]\ddot{r}=0[/tex] and simplify that, but it is really complex, due in part to the transformation to polar coordinates:

[tex]\vec{v}^2=\dot{r}^2+r\dot{\theta}^2[/tex]

If anyone knows of a solution or a way to work this out that they think I will be able to do, I would really appreciate the help.
If you're only concerned about the motion of the object, not it's actual energy. You can toss the mass out of the equation and use specific energy per unit of mass. You can also toss out rotational kinetic energy.

[tex]E=\frac{1}{2}v^2-\frac{GM}{r}[/tex]

You can do the same with the angular momentum. You want specific angular momentum per unit of mass. For trajectories, your specific angular momentum is the cross product of the radius and velocity vectors.

[tex]H=\vec{r}\times\vec{v}[/tex]

Something has to be given. Since you're talking about hyperbolic orbits, you want E to be some value greater than 0 (If E=0, you have a parabolic orbit; if E<0, you have a closed elliptical or circular orbit).

Your max velocity is going to depend on the lower limit you set for r (there's some point where r is smaller than the central mass's radius).