If two objects of different masses are held at a distance $d$ and then I let them go, they will meet at the center of mass of the particle system due to mutual gravitational attraction
My question (which may be rather silly) is that why do they meet a the center of mass, why not anywhere else? Is there some sort of mathematical proof behind this?
If the center of mass is not moving, then because there are no external forces it must remain in place. But if the bodies meet somewhere else, then the place where they meet would then be the center of mass, which is a contradiction since the center of mass must remain where it was at first!
Edit: A (hopefully) clearer explanation.
Suppose that at $t=0$ the center of mass (CM) is at some point $A$. It is important that the two bodies be initially at rest, because if it is initally at rest then it must be always at rest, which implies that the CM will always be at $A$. The two planets, particles, whatever start going towards each other and at some time $t=T$ they crash into each other at some point $B$. Now at $t=T$ all the mass is concentrated at point $B$: the two bodies have formed one single, bigger body. This means that the CM must be at $B$, since that's where all the mass is.
We knew that at $t=0$ the CM was at $A$, and that at $t=T$ it was at $B$. Since we said that the CM can't move, the only option is that $A=B = \text{CM}$.
Let two particles be $A$ and $B$ at x-coordinates $x_A$ and $x_B$ respectivly having masses $m_A$ and $m_B$ respectivily at time $0$ sec. And centre of mass $x_{C.M}=R$ at $t=0$
Let us assume the initial velocities of $A$ and $B$ be 0 in lab's frame.
Centre of mass will be at $(m_A+m_B)R=m_Ax_A+m_Bx_B \tag{1}$.
differentiating both sides twice:
$(m_A+m_B)a_{C.M}=m_Aa_A+m_Ba_B \tag{2}$
Let $F_A$ and $F_B$ be the forces on $A$ and $B$ respectivily. Gravitational force on $A$ and $B$ are same in magnitude and opposite in direction.
$F_B=-F_A$
So acceleration of $A$ is $a_A=F_A/m_A$ and acceleration of $B$ is $a_B=-F_A/m_B$. Putting these values in eqn (2) we got:
$a_{C.M}=0$
Now since the initial velocities of $A$ and $B$ were $0$ so eqn(1)'s derivative directly implies that initial velocity of $C.M$ is $0$.
Since $a_{C.M}$ is 0 so centre of mass remains at $R$ whatever the positions of $A$ and $B$ becomes.
Let gravitational force acts for a time $t$. After time $t$ the positions of $A$ and $B$ becomes $x_A(t)$ and $x_B(t)$
Integrating both sides of eqn 2 within limits $0$ to $t$ we got:
$m_A(v_A(t)-v_A(0))=-m_B(v_B(t)-v_B(0)) \tag{3}$
But $v_A(0)$ and $v_B(0)$ are velocities at $t=0$ so $v_A(0)=v_B(0)=0$.
So eqn 3 becomes:
$m_Av_A(t)=-m_Bv_B(t)$
Again integrating within limits $0$ to $t$:
$m_Ax_A(t)+m_Bx_B(t)=m_Ax_A+m_Bx_B \tag{4}$
When $A$ and $B$ collides $x_A(t)=x_B(t)=R^{'}$
Now we wish to find $R^{'}$
From eqn 4 we got :
$m_AR^{'}+m_BR^{'}=m_Ax_A+m_Bx_B$
$\implies R^{'}=\dfrac{m_Ax_A+m_Bx_B}{m_A+m_B}$
From eqn 1
$R^{'}=R$
Q.E.D
Let us go to the system where the center of mass is always at rest and at (0,0,0). The definitions of the center of mass is
where R are the coordinates of the center of mass and r_i the coordinates of the masses m_i and R is (0,0,0) in the center of mass at rest system.
So we have
m_1***r_1=m2***r_2 where r are the distances from the (0,0,0)
This means that by definition, and construction in this example, no matter what makes r_1 and r_2 move towards or away from each other, the center of mass is still at (0,0,0).
In the lab system R may be wandering about due to the forces acting on the masses, but the center of mass is a point describing the collective motion of particles.
