<p>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 <a href="https://en.wikipedia.org/wiki/Bertrand%27s_theorem" rel="nofollow">Bertrand's theorem</a>).  There will be no 'transient' part of the solution.</p>

CC BY-SA 3.0
2015-05-14T19:49:03.980Z
2015-05-14T19:49:03.980Z
DilithiumMatrix
<p><strong>There is no transient part</strong></p>

<p>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.</p>

<p>However there is another way of solving the newtonian equations, in the case of a two body problem, and that is <em>Kepler's laws of planetary motion</em>. 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.</p>

CC BY-SA 3.0
2015-11-03T17:19:35.527Z
2015-11-03T17:19:35.527Z
James K
<p>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?</p>

CC BY-SA 3.0
2015-05-14T19:49:39.000Z
Note that initial conditions need both position and momentum vectors --- i.e. there are 6 degrees of freedom.
DilithiumMatrix
<orbit>
Two body transient solution
Patrick Cummings
# Two body transient solution

(Asked by: username_0 on 2015-05-12)

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?

**Tags:** orbit
> **Comments:**
> **username_1**: Note that initial conditions need both position and momentum vectors --- i.e. there are 6 degrees of freedom.  

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## 2 answers

### Answer by username_1 on 2015-05-14. Score: 1

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](https://en.wikipedia.org/wiki/Bertrand%27s_theorem)). There will be no 'transient' part of the solution.


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### Answer by username_2 on 2015-11-03. Score: 0

**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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