# When is replacing two bodies with a single one at the centre of mass valid?

Two bodies of mass $$m$$ and $$2m$$(represented as mm) are in a circular orbit around a large body of mass $$M$$. The radius of $$M$$ is $$R$$ and the radius of the circular orbit of $$m$$ and $$2m$$ is $$2R$$. The sense of rotation of $$m$$ and $$2m$$ is the opposite (i.e. they collide after a certain amount of time).
The collision between $$m$$ and $$2m$$ is perfectly inelastic. The question asks to find the theoretical distance of closest approach of the combined mass, to $$M$$. The gravitational effects of $$m$$ and $$2m$$ on each other and on $$M$$ are to be neglected.

Method $$1$$:
Conserve momentum just before and just after the collision. Let the point of collision be $$A$$ and the point of closest approach be $$B$$.
Equate angular momentum of combined mass $$3m$$ at point A to that at point B.
Similarly, equate the total mechanical energies at points A and B.
All these equations give: $$r_{min} = \left(\frac{2R}{17}\right)$$

Method $$2$$:
During the collision, only internal forces act between $$m$$ and $$2m$$, so the movement of their centre of mass ($$CM$$) should not be affected. Hence centre of mass follows the elliptical trajectory that it initially followed. Distance of closest approach would then be equal to the semi-minor axis. $$r_{min} = \left(\frac{2R}{3}\right)$$

Why is Method $$2$$ incorrect? I believe the reason has to do with the loss of kinetic energy in the collision. But as the position and velocity of the $$CM$$ do not change due to the collision, I do not see why the combined mass would not follow the initial trajectory of the $$CM$$.

Method 2 is certainly incorrect, because the elliptical path with $$M$$ at the centre is not a solution of the equations of motion for a particle in an orbit. It is true that the motion of the centre of mass is continuous through the collision, so you can use its position and velocity to find the correct Keplerian orbit which it will follow.
I'm not convinced that Method 1 is correct either, since my application of that the approach just described gives $$8R/5$$.
• Thank you very much. It was silly of me to not notice that M needs to be at a focus and not centre. As for the answer you got, I'm fairly certain that 2R/17 (the given answer) is correct. The calculations are easy to make a mistake in. The final equation comes out to be $17x^2−36x+4 =0$ where $x=\frac{r_{min}}{R}$. Commented Aug 9, 2022 at 14:13
• I persist in thinking the answer is $8R/5$. The position and velocity after the collision are $2R$ and $(2/3)\sqrt{GM/R}$, and at the other apse it's $8R/5$ and $(5/6)\sqrt{GM/R}$. You can easily check this conserves energy $(-5/6)GMm/R$ and angular momentum $4m\sqrt{GMR}$.