- #1

Kyouran

- 70

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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?

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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