I'm trying to analyze the motion of the particles which exert the gravitational force each other. Let $M_1$, $M_2$ be the masses of the particles, and the equation of motion of particle $M_1$

$$ F=G\frac{M_1M_2}{r^2}=M_1\ddot{r} \\ \ddot{r}r^2-GM_2=0 $$

I find it doubtful that the simple motion has difficult equation of motion.

Should I just use some form of successive approximation?

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    $\begingroup$ en.wikipedia.org/wiki/Two-body_problem $\endgroup$ – Michael Brown May 31 '13 at 4:46
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    $\begingroup$ You're right that the resulting motion is not complicated. See Michael Brown's comment. Amazingly though, there is no general, closed-form solution for 3 or more bodies (en.wikipedia.org/wiki/N-body_problem) $\endgroup$ – Brandon Enright May 31 '13 at 4:49
  • $\begingroup$ Hi @user25245. Are you talking about the special situation of a 1D radial free fall, or are you talking about the more generic situation where the two point masses are orbiting each other? $\endgroup$ – Qmechanic May 31 '13 at 12:43

I suggest to see the scenario physically first, and proceed from there.

Here, the gravitational force does not change the distance from the center, but it provides the centripetal force for the circular motion (or any conic section for that matter, based on initial values).

Assuming it is a circular motion for now, we can just get right into the mathematical description of the motion.

For such systems, masses move in a circle with their COM at the center.

enter image description here

Because there is no other force in system, the COM does not accelerate. It is free to move at a constant velocity though, but this is (usually) irrelevant when analyzing the system.

Because they move in a circle about the COM, just get the distances from their COM, and equate the gravitational force to the centripetal force. This way you have fully described the motion.

Example for $M_2:$

$$ \dfrac{GM_1}{\underbrace{R}_{\text{distance between bodies}}} = \dfrac{{v_2^2}}{\underbrace{R_{2_{cm}}}_{\text{distance of 2 from COM}}} $$

Similar for $M_1$.

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    $\begingroup$ The COM of a system of particles will not accelerate if there is no external force, so the COM has a constant velocity (in astronomy it is also often called the barycenter). But for the sake of a reference frame, the velocity of the COM is often chosen to be zero. $\endgroup$ – fibonatic May 31 '13 at 8:47

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