# What can be said about the trajectory of a particle subject to more than one central force?

What can be said about the classical trajectory of a massive particle subject to more than just one fixed central forces F=k/r², eg. there are two fixed points placed in space to which the particle is attracted.

Lets say it starts from a point "far" away (distance $$r_0$$)from the attractive center with some initial velocity $$v_0$$. When it gets closer, potential energy becomes more negative and kinetic energy more positive.

Will the particle finally escape with a straight line into far space with $$v_0$$ but other direction for infinity (this would be possible energetically) or can it be "caught" be the two objects so that it never leaves the system in a straight line (also possible energetically)?

For a two body system the answer is simple and depends clearly on what we call "e": it gives a closed elliptic (cought) trajectory for $$e<1$$ or a hyperbolic (escaping) one for $$e>0$$ .

Is there a generalization for more than one attractors in terms of some parameter like energy or angular momentum? Even when I can regard the two objects s a single object far away, this doesn't mean that the trajectory behaves like a hyperbola in the limit of big distances, because there are disturbances near the bodies which makes it hard to calculate a final escape direction.

This image shows symbolically what I mean by "escaping":

For fixed points, you can continue to rely on Energy, which is conserved. The total specific energy (ignoring the particle mass itself) would be

$$e = KE + PE_1 + PE_2 = \frac{1}{2}v^{2} - \frac{GM_{1}}{r_{1}} - \frac{GM_{2}}{r_{2}}$$ where $$M_{1,2}$$ are the fixed point masses and $$r_{1,2}$$ are distances from the particle to the fixed point masses.

Clearly to reach infinity, this must satisfy $$e \ge 0$$ since at infinity the potential terms are $$0$$.

Euler's three-body problem provides an analytic solution to 2 fixed masses with 1 massless particle in the form of elliptic functions.

This works because there is nothing time dependent about the potentials, so Energy is conserved.

As a side note, the full 3-body problem where Mass 1 and Mass 2 are NOT fixed is much more complex. You can think of it multiple ways: time-translation symmetry for the potentials from the particle's point of view is broken so energy conservation is broken. The potentials would become time-dependent ($$r_{1,2}(t)$$) since masses 1,2 can move so energy conservation is broken. And as a whole system of 3 particles, of course they would be able to exchange energy so that the particle's energy can increase or decrease through exchange.

Lastly, there are statements you can make about the restricted 3-body problem where your particle has negligible mass. An intuitive example is Lagrange points, where certain points are stable. A less intuitive example is using the Jacobi integral to compute zero-velocity surfaces which a particle cannot cross since $$v^2 < 0$$ would be required.

• e≥0 is clear when infinity is the starting point. But can it be, that a point comes from infinity and never goes out to infinity again? Since time reversal is always a solution for a problem in classical mechanics, this would mean that such a point was bounded for infinite times and suddenly jumps out to finally reach infinity. Anyway, e>0 for a two body problem yields hyperbolic behavior and can be easily solved. Does e>0 for a many body problem the same? Dec 1, 2022 at 22:26
• Clearly to reach infinity, this must satisfy e≥0 - yes, this is trivial - but if e>0 WILL it ever reach infinity? If yes, by WHAT argument? If no, it depends on what? By only taking energy conservation into account, it would be no contradiction, if it doesn't reach infinity again. Dec 1, 2022 at 22:37
• Are there "stable" trajectories when M1 and M2 are fixed? Can this be solved analytically like Keplers Problem can? Dec 1, 2022 at 22:40
• I edited to briefly refer to euler's three-body problem en.wikipedia.org/wiki/Euler%27s_three-body_problem Dec 1, 2022 at 23:13