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Two masses in 3d space attract each other with a potential relative to the distance between them. There is also an external force on each particle based on the distance from a origin. I want to find the equations of motion.

I thought about reducing it to the 1 body problem or use Lagrangian formalism to find the equation of motion, however, the external force kind of ruins that approach. Any ideas on how to approach this?

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This sounds like a three body problem, in the limit of one of the bodies being very heavy. As such there won't be a closed form solution except for unusually symmetrical cases. There may be approximate solutions for special cases. Are you after a general solution or would a special case do? – John Rennie Nov 12 '12 at 7:34
A general solution is of course preferred, but a special case would be helpful as well. Thank you for replying! – Splooge Nov 12 '12 at 7:39
You should have a quick Google for three body problem to get some ideas. The special cases tend to be where symmetry reduces the degrees of freedom. – John Rennie Nov 12 '12 at 7:50
Can you be more specific about what the forces are? The setup and solution will depend on the forces a lot. If the external force has a well-defined potential, you can still probably use the Lagangian formalism. For example, there's nothing wrong with describing two blocks attached by a vertical spring falling due to gravity by the Lagrangian $L=\frac{1}{2}(m_{1}{\dot x}_{1}^{2}+m_{2}{\dot x}_{2}^{2})-\frac{1}{2}k(L-(x_{1}-x_{2}))^{2}-g(m_{1}x_{1} +m_{2}x_{2})$ And gravity is certainly an external force there. You just need a well-defined $T-V$ and some technical constraints to do Lagrangians – Jerry Schirmer Nov 12 '12 at 16:56

You simply write the Newton equations: $$m_1\ddot{\vec{r}}_1=\vec{F}_{12}(\vec{r}_1-\vec{r}_2)+\vec{F}_{ext}(\vec{r}_1), \\m_2\ddot{\vec{r}}_2=-\vec{F}_{12}(\vec{r}_1-\vec{r}_2)+\vec{F}_{ext}(\vec{r}_2).$$ Then you can introduce the relative distance $\vec{r}=\vec{r}_1-\vec{r}_2$ and the center of mass position $\vec{R}=(m_1\vec{r}_1+m_2\vec{r}_2)/(m_1+m_2)$. You will obtain a relative motion equation for $\vec{r}$ with a "pumping term" due different accelerations of particle 1 and particle 2 in the external force, and a center of mass equation for $\vec{R}$ that describes the motion of $\vec{R}$ due to external force "pulling" both particles. An exactly soluble case is when the corresponding two-body problem is soluble and the external force is uniform in space: $\vec{F}_{ext}=\vec{F}(t)$.

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