Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
1  
If you try you can solve it mathematically yourself very easily or even you can logically visulise it. –  Dvij Feb 16 at 5:16

3 Answers 3

up vote 1 down vote accepted

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
share|improve this answer
    
This answer was amazing. Just asking, the COM will remain in the same place and also two particles will meet at their COM only when there are no external forces? –  Eliza Feb 16 at 14:48
    
@Eliza The $C.M$ will remain stationary it was an assumption(because initial velocities of A and B were 0) e.g C.M of a swinging cork ball does have an initial speed. The thing is even if $C.M$ moves with a constant speed A and B will collide at the $C.M$ Second if there is an external force say a heavy force acting on A then the $C.M$ will also accelerate and finally A and B will collide at the shifted C.M. . –  user31782 Feb 16 at 15:14
    
@anumpam: so what ever the situation, two objects will meet a their $C.M$? In the case where initial velocity is zero and no external force, $C.M$ remains at the same place and thus $A$ and $B$ will meet there. When ext. forces are involved, there will be a shift in $C.M$ but the object will still meet there.... I am just checking if I got the right understanding :) –  Eliza Feb 16 at 15:22
    
@Eliza I am not an expert but i guess this is right. Infact when i was writing the answer i got messed up and opened my school physics book. You got me right Whatever happen if A and B collide then the coordinate of the event where collision occurs is the C.M's coordinate by definition. –  user31782 Feb 16 at 15:44

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 $A=B = \text{CM}$.

share|improve this answer
    
I understand that the position of center of mass will not change when the two objects come towards each other but my question is that WHY do they meet at the COM? –  Eliza Feb 16 at 13:49
    
@Eliza: See edit, let me know if it helps. –  Javier Badia Feb 16 at 14:27

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

center of mass

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.