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In the two body problem, if each body is given an initial condition as a momentum vector, what is the transient part of the solution as it settles into the steady state orbit?

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  • $\begingroup$ My stackoverflow.com/questions/7613946/… is probably not helpful. $\endgroup$ – user21 May 13 '15 at 3:11
  • $\begingroup$ Note that initial conditions need both position and momentum vectors --- i.e. there are 6 degrees of freedom. $\endgroup$ – DilithiumMatrix May 14 '15 at 19:49
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Your question is very unclear. As long as the orbit is bound (i.e. the total energy is negative), then there is a unique closed orbit solution (see Bertrand's theorem). There will be no 'transient' part of the solution.

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There is no transient part

In solving a linear ode y'' + ay' + by = f(x), one finds that the solution is the superimposition of a general solution of y''+ay'+by=0, and a particular solution of the ode. In the situation in which the ode represents damped harmonic motion, the general solution part is said to be transient. This is a particular solution to this ODE, and not a general technique for solving all differential equations, and does not apply to the ode that describes newtonian gravity.

However there is another way of solving the newtonian equations, in the case of a two body problem, and that is Kepler's laws of planetary motion. These give a complete solution, and there is not transient part. Given a body's initial state, you can immediately find it's orbit using Kepler's laws. The body will be in an elliptical orbit, and return to its starting point, having made one complete orbit.

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