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In this website, it states that if we have a two particles system and measure from centre of mass, then the following equation holds:

$$m_1 r_1 = m_2 r_2$$

where $m_1, m_2$ are masses of the two objects and $r_1, r_2$ are distances from centre of mass to the two objects.

Question: How to obtain the above equation?

Centre of mass is defined to be the weighted sum of all moments. So it is not surprising that centre of mass can be expressed as follows:

$$x_{cm}= \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$

where $x_1, x_2$ are distances from a reference point to the two masses.

However, I have no idea on how to obtain $m_1 r_1 = m_2 r_2$. It seems to me that ratio of masses equals to ratio of distances.

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  • $\begingroup$ In the center of mass frame, $x_{cm}=0$. The letter $r$ in this context implies distance, always a positive number. $\endgroup$ – garyp Dec 13 '16 at 17:21
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I think the existing answers are making this a lot more complicated than it needs to. You are correct that the equation for the position of the center of mass is,

$$x_{cm}= \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$

If you then take the center of mass as the origin, you set $x_{cm} = 0$, and thus $$m_1 |x_1| = m_2 |x_2|.$$ In the context of your question, it seems $r_1$ and $r_2$ refer to the distance specifically (i.e. an always positive number).

If you were dealing with vectors and displacement, then you would want: $$m_1 x_1 = - m_2 x_2$$ which shows that one mass needs to be placed on the opposite end of the first.

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  • $\begingroup$ Okay. If I take center of mass as origin, wouldn't I obtain $m_1 x_1 = -m_2 x_2$ according to your first formula? $\endgroup$ – Idonknow Dec 14 '16 at 0:35
  • $\begingroup$ @Idonknow yes, that's true. I tried to clarify my answer a bit. $\endgroup$ – DilithiumMatrix Dec 14 '16 at 1:01
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I am not so sure about that expression; a definition of the CM system is that $$m_1 r_1^{cm}+m_2 r_2^{cm}=0.$$So $$m_1 r_1^{cm}=-m_2 r_2^{cm}$$ holds. This can also be shown in the laboratory system with the $x_{cm}$ definition (which directly follows form the first eq. I wrote).

Looking at the picture on the page I would say they use $r_i$ as positive distance form the angle point and not as coordinates. If you use $r_i$ as coordinates it has to be $$m_1 r_1=-m_2 r_2 \Leftrightarrow m_1=-m_2 r_2/r_1$$ and one $r_i$ will be negative so $m_i$ stays positive. If you use distances the minus sign would have to drop out to ensure positive masses.

To be honest that page is at best incomplete; they introduce $r_i$ as "where $r_1$ and $r_2$ locate the masses" without drawing them into their figure and without specifying what they mean with "locate".

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For this it's helpful to have vectors, which are (roughly speaking) lists of numbers satisfying the addition law $[a, b, c] + [d, e, f] = [a + d, b + e, c + f]$ and a scalar multiplication law $k[a, b, c] = [ka, kb, kc].$

These can be thought of as points in 3D space if you choose a specific point as "origin" $\vec 0 = [0, 0, 0]$ and 3 specific orthogonal directions as the axes $\hat x = [1, 0, 0],$ $\hat y = [0, 1, 0]$, $\hat z = [0, 0, 1].$ And as usual those two properties of addition and scalar multiplication are enough to define subtraction, $\vec a - \vec b = \vec a + (-1)\vec b.$

Now we indeed define $$\vec x_\text{cm} = \frac{m_1 \vec x_1 + m_2 \vec x_2}{m_1 + m_2},$$ and the two vectors $$\vec r_1 = \vec x_1 - \vec x_\text{cm},\\ \vec r_2 = \vec x_2 - \vec x_\text{cm}.$$There are probably several ways to prove the property that you're interested in. Probably the most easy one is a little abstract; computing$$(m_1 + m_2) \vec r_1 = (m_1 + m_2) \vec x_1 - m_1 \vec x_1 - m_2 \vec x_2 = m_2 (\vec x_1 - \vec x_2).$$The choice of $\vec r_1$ on the left hand side was arbitrary so doing the same with $\vec r_2$ we get $m_1(\vec x_2 - \vec x_1),$ which differ by a scalar factor of $-m_2/m_1$ hence $$(m_1 + m_2) \vec r_1 = \frac{-m_2}{m_1} \big( (m_1 + m_2) \vec r_2 \big),$$and cancelling out the $m_1 + m_2$ gives simply $$m_1 \vec r_1 = -m_2 \vec r_2.$$ This is the correct vector expression, and the minus sign simply says that if you go one way to get from the center-of-mass to the one mass, then you must go the exact opposite way to get from the center-of-mass to the other. (In other words, the center of mass lies somewhere in the middle of the line segment connecting the masses. If this minus sign were not there it would mean that it would not be in the middle of the line segment but somewhere on the line.)

Anyway, they're not necessarily being 100% sloppy by skipping this important minus sign: there is then an operation called the length or magnitude or norm of a vector, which is sort of the 3D version of the Pythagorean theorem and an extension of the "absolute value" function on the scalars; if $\vec x = [a, b, c]$ then we define $|\vec x| = \sqrt{a^2 + b^2 + c^2}.$ It is very common to simply refer to this as $x$, dropping the "vector hat" (or boldface in many books) that makes it look like a vector. The specific relation that you quoted is using this operation to get rid of the minus sign, and assuming that $m_{1,2} > 0,$ to find that $$m_1 r_1 = m_1 |\vec r_1| = |m_1 \vec r_1| = \big|(- m_2) \vec r_2\big| = m_2 |\vec r_2| = m_2 r_2.$$

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This is not mathematical answer. This is inituitive. (Edited. Now better.)

Well first of all $m_1$ and $m_2$ are the center of masses of the two objects. So, $m_1$ and $m_2$ are two point masses where the masses of object$_1$ and object$_2$ are concentrated respectively.

The center of mass of the two objects lies somewhere in the line joining the point masses $m_1$ and $m_2$.

Imagine there is an in-extensible mass-less string joining $m_1$ and $m_2$ with $m_1$ on left end and $m_2$ on the right end. Now think of $m_1$ as mass per unit length of the thick part (see figure below) and $m_2$ as mass per unit length of the thin part (as they are point masses so we can assume $m_1$ and $m_2$ as mass per unit length). If $m_1=m_2$ we know that the center of mass coincides with the center of the string and hence $r_1=r_2$ where $r_1$ and $r_2$ are distances of $m_1$ and $m_2$ from the center of mass (!and not from the center of the string). If $m_1>m_2$ the center of mass shifts towards the left i.e., towards $m_1$ and hence $r_1<r_2$.

Now suppose that $m_1>m_2$ and the center of mass is J (in the figure below). If we start adding tiny $m_1$ masses from the left end and $m_2$ masses from the right end along the string without disturbing the center of mass we end up making a rod-like structure like that shown in the figure. The center of mass is at J, the joint of the thick and thin part of the rod. Since $m_1$ and $m_2$ are mass per unit length as assumed earlier so $m_1r_1$(mass per unit length times distance) is the mass of the thick part and $m_2r_2$ is the mass of the thin part and they must be equal.

Hence, $m_1r_1=m_2r_2$ and the center of mass is located at the joint. enter image description here

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A good way to think about it is to express the centripetal force as $F=mrw^2$ (or Mr Omega squared). What you do is explain why omega is the same and so it cancels out since the grav force is the same on both.

If the grav force didn't pass through the centre of mass the other planet/star would either accelerate or decelerate to compensate. As a result both orbit the CoM in a circular orbit and the (line of action) of the forces are in line with each other. The only way this can happen is if they orbit exactly Pi rads out of phase with the same omega.

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  • $\begingroup$ Welcome on Physics SE :) We have a help section for typesetting formulas :) $\endgroup$ – Sanya Dec 13 '16 at 17:52
  • $\begingroup$ There is really no need to apologise - this site is not easy to get used to and we have most of us (including me) been disoriented and happy about advice at first :) $\endgroup$ – Sanya Dec 13 '16 at 21:02

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