Center of mass - cant find error in derivation

In summary, the poster has a problem with their derivation of the center of mass of a binary system. They have found conflicting results for the distances of the center of mass to the masses and have posted their derivation for help. They also mention introducing a third mass at the center of mass and finding that the center of mass of the entire system does not change. However, the third mass experiences no net acceleration and the poster is questioning the validity of their calculation. They are looking for assistance in identifying any errors in their derivation.
  • #1
Kyouran
70
10
First of all, I am sorry if this is in the wrong subforum; I'm not sure if astrodynamics should be under astronomy. The post is also fairly long. My problem is that in my derivation of the center of mass of a binary system, I get a conflicting result for the distances of the center of mass to the masses. I must have made an error in my derivation, and I guessed the best way to ask for help is to just post the entire derivation, so here goes:

Suppose I have a system of 2 masses (for example, planets). The position vector of the center of mass of such a system with respect to some arbitrary inertial reference frame with origin O is given by:

[tex] \vec{r_{cm}} = \frac{m_1 \vec{r_1} + m_2 \vec{r_2}}{m_1 + m_2} (eq. 1) [/tex]

Now, I can easily prove that this point is located on the line segment connecting the 2 masses:

[tex] \vec{r_{cm}} = \frac{m_1 \vec{r_1}}{m_1 + m_2} + \frac{m_2 \vec{r_2}}{m_1 + m_2} [/tex]
[tex] \vec{r_{cm}} = \frac{m_1}{m_1 + m_2}\vec{r_1} + (1 - \frac{m_1}{m_1 + m_2})\vec{r_2} [/tex]
[tex] \vec{r_{cm}} = \alpha \vec{r_1} + (1 - \alpha)\vec{r_2} [/tex]

Similarly, setting [tex] \beta = 1-\alpha [/tex]
I get:
[tex] \vec{r_{cm}} = (1-\beta) \vec{r_1} + \beta \vec{r_2} [/tex]

From these relationships, I can find the distance between the 2 masses and the center of mass as:

[tex] \vec{r_{cm,1}} = \beta \vec{r_21} \rightarrow \vec{r_{cm,1}} = \frac{m_2}{m_1 + m_2} \vec{r_{21}}[/tex]
[tex] \vec{r_{cm,2}} = \alpha \vec{r_12} \rightarrow \vec{r_{cm,2}} = \frac{m_1}{m_1 + m_2} \vec{r_{12}}[/tex]

Where r_ij is the vector from i to j. Now taking the norm of the vectors and taking into consideration that [tex] ||\vec{r_{12}} || = ||\vec{r_{21}} || [/tex], i get:

[tex] || \vec{r_{cm,1}} || = \frac{m_2}{m_1 + m_2} || \vec{r_{12}} ||[/tex]
[tex] || \vec{r_{cm,2}} || = \frac{m_1}{m_1 + m_2} || \vec{r_{12}} ||[/tex]

Which gives me the relation:

[tex] \frac{|| \vec{r_{cm,1}} ||}{|| \vec{r_{cm,2}} ||} = \frac{m_2}{m_1} (eq. 2) [/tex]

This seems to be logical; if the second mass is heavier then the center of mass will be closer to the second mass. If I now go back to the first equation and differentiate it twice with respect to time I get:

[tex] \frac{d^2\vec{r_{cm}}}{dt^2} = \frac{m_1 \frac{d^2\vec{r_1}}{dt^2} + m_2 \frac{d^2\vec{r_2}}{dt^2}}{m_1 + m_2} [/tex]

The only forces on the masses 1 and 2 is the gravitational force, therefore:

[tex] m_1 \frac{d^2\vec{r_1}}{dt^2} = G \frac{m_1 m_2}{|| \vec{r_{12}} ||^3} \vec{r_{12}} [/tex]
[tex] m_2 \frac{d^2\vec{r_2}}{dt^2} = - G \frac{m_1 m_2}{|| \vec{r_{12}} ||^3} \vec{r_{12}} [/tex]

Therefore, I find that:

[tex] \frac{d^2\vec{r_{cm}}}{dt^2} = \vec{0} (eq. 3) [/tex]

Which means that the center of mass does not accelerate. If I now place a mass m3 at the center of mass of this binary system, the center of mass of the entire system doesn't change, and it can be shown that equations (1), (2) and (3) still hold. Also, it can be shown that this mass will stay at the center of mass, and that it experiences no net acceleration as well. When I thus calculate the forces on this third mass I get:

[tex] m_3 \frac{d^2 \vec{r_{cm}}}{dt^2} = G m_3 ( \frac{m_1}{|| \vec{r_{cm,1}}||^3} \vec{r_{cm,1}} + \frac{m_2}{|| \vec{r_{cm,2}} ||^3} \vec{r_{cm,2}} ) = \vec{0} [/tex]

from which I find:

[tex] \frac{|| \vec{r_{cm,1}}||^2}{|| \vec{r_{cm,2}}||^2} = \frac{m_1}{m_2} (eq. 4) [/tex]

From equation 2 and 4 I then find:

[tex] ||\vec{r_{cm,1}} ||^ 3 = ||\vec{r_{cm,2}}||^3 [/tex]

Now this would imply that the distance of the 2 objects to the center of mass is the same. I know this can't be true, so there must somewhere be an error in my derivation or thinking. I really don't see where this could be, so can anyone point out the mistake I made?
 
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  • #2
Kyouran said:
If I now place a mass m3 at the center of mass of this binary system, the center of mass of the entire system doesn't change,

why not?

well,, if i now take the center of mass frame in which origin is at the center of mass, then position of center of mass is:

(m1r1 + m2r2 + m3*0)/(m1+m2+m3) which is a different position from that of earlier center of mass

EDIT: the conclusion is wrong since the above equation actually gives answer =0 which implies center of mass remains there itself even after the introduction of 3rd mass.
 
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  • #3
Kyouran said:
Also, it can be shown that this mass will stay at the center of mass, and that it experiences no net acceleration as well
How can m3 experience no force? if m3 is between m1 ans m2, if experiences a mutually opposite gravitational force which are not equal,

EDIT: but i think i hold this argument
 
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  • #4
ash64449 said:
why not?

Intuitively, it shouldn't change. But, mathematically. If you substitute:

[tex] \vec{r_3} = \frac{m_1 \vec{r_1} + m_2 \vec{r_2}}{m_1 + m_2} [/tex]

into the equation for the center of mass of a three body system:

[tex] \vec{r_{cm}} = \frac{m_1 \vec{r_1} + m_2 \vec{r_2} + m_3 \vec{r_3}}{m_1 + m_2 + m_3} [/tex]

and you do some algebra magic (I just checked with maple), you will find:

[tex] \vec{r_{cm}} = \frac{m_1 \vec{r_1} + m_2 \vec{r_2} + m_3 \vec{r_3}}{m_1 + m_2 + m_3} = \frac{m_1 \vec{r_1} + m_2 \vec{r_2}}{m_1 + m_2} [/tex]
 
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  • #5
sorry.. i understand the mistake before... but just as i tried to post u posted along!
 
  • #6
ash64449 said:
How can m3 experience no force? if m3 is between m1 ans m2, if experiences a mutually opposite gravitational force which are not equal,

EDIT: but i think i hold this argument

Well, if I differentiate r3 with respect to time twice, it comes out zero according to my derivation, so somehow the net force on it must be zero.
 
  • #7
Even if you introduce m3 in the center of mass position, center of mass of the new system remains constant.

But you can't say that m3 will not accelerate just because it is kept there. m3 here experience only mutually opposite gravitational force which are not equal.. hence it will accelerate.

however everything will move in such a manner that center of mass remains there again in the new system since sum total of all forces experienced by all masses is zero.
 
  • #8
Kyouran said:
Well, if I differentiate r3 with respect to time twice, it comes out zero according to my derivation, so somehow the net force on it must be zero.

did you differentiate r3 w.r.t time twice in your force on m3=0 derivation or Rcm w.r.t time twice? when i read your derivation ,i see latter.
 
  • #9
Never... force on m3 cannot be zero... i think that is why an error has come in your derivation. it makes sense why that error has come.
 
  • #10
ash64449 said:
did you differentiate r3 w.r.t time twice in your force on m3=0 derivation or Rcm w.r.t time twice? when i read your derivation ,i see latter.

Maybe the issue is that I assumed that r3 stays in the center of mass, and maybe it is only valid at t = 0.

EDIT: I find it weird though that a mass placed at the center of mass would start to move away from it, but it could be if m1 and m2 move accordingly.
 
  • #11
Kyouran said:
Maybe the issue is that I assumed that r3 stays in the center of mass, and maybe it is only valid at t = 0.

you are right.. why it moves afterwards? it has a net force acting on it.. this is what i said.
 
  • #12
Kyouran said:
I find it weird though that a mass placed at the center of mass would start to move away from it, but it could be if m1 and m2 move accordingly.

why weird? yes,m1 and m2 will move accordingly in such a manner that center of mass remains where it is?

motion of m1 and m2 different in first case and second case. why?

because in the second case m3 also produces a force on them which produces the different motion. but center of mass remain where it is in both the cases.
 
  • #13
Well, because i thought everything is supposed to orbit the center of mass of the system, from which it would seem that an object which is placed at the center of mass would orbit it with a radius of 0, meaning it would stay there. After all, that's the idea of gravity: everything falls towards the center of mass.
 
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  • #14
Kyouran said:
After all, that's the idea of gravity: everything falls towards the center of mass.
That is *not* the idea of gravity. The idea of gravity is that everything falls towards everything else, and except for the Newtonian two body point mass problem, that in general is not toward the center of mass.
 

Related to Center of mass - cant find error in derivation

What is the center of mass and why is it important?

The center of mass is a point in a system where the mass of the object or system can be considered to be concentrated. It is important in physics because it helps us understand the motion and behavior of objects and systems.

How do you calculate the center of mass?

The center of mass can be calculated by taking the sum of the individual masses multiplied by their respective distances from a reference point, divided by the total mass of the system. This can be expressed mathematically as:

xcm = (m1x1 + m2x2 + ... + mnxn) / (m1 + m2 + ... + mn)

Where xcm is the center of mass, mn is the mass of the nth object, and xn is the distance from the reference point to the nth object.

What is the error in the derivation of the center of mass formula?

The error in the derivation of the center of mass formula is assuming that the system is in a uniform gravitational field. This is not always the case, and in situations where the gravitational field is not uniform, the center of mass formula may not accurately represent the true center of mass.

Can the center of mass be outside of an object?

Yes, the center of mass can be outside of an object. This can happen when the object has an irregular shape or when the mass is not distributed evenly throughout the object. In these cases, the center of mass may be located at a point outside of the physical boundaries of the object.

How is the center of mass related to an object's stability?

The center of mass is directly related to an object's stability. An object is considered stable when its center of mass is located directly over its base of support. If the center of mass is not over the base of support, the object may topple over. This is why objects with a low center of mass, such as a wide-based pyramid, are more stable than objects with a high center of mass, such as a tall cylinder.

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