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Center of mass - cant find error in derivation

  1. May 24, 2015 #1
    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?
    Last edited: May 24, 2015
  2. jcsd
  3. May 24, 2015 #2
    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.
    Last edited: May 24, 2015
  4. May 24, 2015 #3
    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
    Last edited: May 24, 2015
  5. May 24, 2015 #4
    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]
  6. May 24, 2015 #5
    sorry.. i understand the mistake before... but just as i tried to post u posted along!
  7. May 24, 2015 #6
    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.
  8. May 24, 2015 #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.
  9. May 24, 2015 #8
    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.
  10. May 24, 2015 #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.
  11. May 24, 2015 #10
    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.
  12. May 24, 2015 #11
    you are right.. why it moves afterwards? it has a net force acting on it.. this is what i said.
  13. May 24, 2015 #12
    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.
  14. May 24, 2015 #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, thats the idea of gravity: everything falls towards the center of mass.
    Last edited: May 24, 2015
  15. May 24, 2015 #14

    D H

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    Staff Emeritus
    Science Advisor

    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.
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