Why in binary star system the 2 bodies revolve around their center of mass? Can you prove it mathematically?
Is it just an observation?
I know the force between them is $\frac{Gm_1m_2}{r^2}$.
Centripetal force on any one of them is $\frac{mv^2}{r_1}$, where $r_1$ is radius of curvature.
Why is that radius of curvature about center of mass only? Why it can't be any other point?
 A: If we start from the fact that the orbits of the two bodies have equal time periods, this means they have equal angular velocity $\omega$
From $a=\omega^2r$, where $a$ is the acceleration, this means that $\frac{a}{r}$ is the same for each body.
so if $r_1$ is the distance from the orbital centre for body $m_1$ and similar for the $2$ subscripts
$$\frac{Gm_2}{r_1(r_1+r_2)^2} = \frac{Gm_1}{r_2(r_1+r_2)^2}$$
The left term is the $\frac{a}{r}$ for body $m_1$, and the right term is for body $m_2$,
then $$m_2r_2 = m_1r_1$$
and this equation is true only if the orbital centre is also the centre of mass.
A: Given two bodies, you have their positions $\mathbf{r}_1=(x1,\,y1,\,z1),\,\mathbf{r}_2=(x_2,\,y_2,\,z_2)$ and velocities $\dot{\mathbf{r}}_1,\,\dot{\mathbf{r}}_2$. The force of gravity only depends on $\mathbf{r}_1-\mathbf{r}_2$, and there's a force that body $1$ experiences $\mathbf{F}_1$ and a force that body $2$ experiences $\mathbf{F}_2$. These forces should be equal and opposite due to Newton's third law and also due to the nature of gravitational force.
Since the force depends on the difference, we notice that everything will behave the same even if we add some offset $\mathbf{r}_c$ to the vectors $\mathbf{r}_{1},\mathbf{r}_{2}$.
If we let $(m_1 + m_2)\mathbf{r}_c = m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2$, this is a nice choice because $(m_1 + m_2)\ddot{\mathbf{r}}_c = m_1 \ddot{\mathbf{r}}_1 + m_2 \ddot{\mathbf{r}}_2 = \mathbf{F}_1 + \mathbf{F}_2 = 0$. $\mathbf{r_{c}}$ will be constant because the forces $F_{1}$ and $F_{2}$ were equal and opposite.
For convenience, now I'll use the position vectors
$\mathbf{s_{1}} = \mathbf{r_{1} - r_{c}}$.
$\mathbf{s_{2}} = \mathbf{r_{2} - r_{c}}$.
$\mathbf{s_{c}} = \mathbf{r_{c} - r_{c}} = 0 $.
I'm now doing everything in the center-of-mass frame, where the center of mass takes coordinate 0.
$m_1 \mathbf{s_{1}} + m_2 \mathbf{s_{2}} = m_1 r_1 - m_1 r_c + m_2 r_2 - m_2 r_c = 0$.
This is the best part, because now s1 and s2 (our positions for particles 1 and 2) are related, and I can solve for one to get rid of the other.
Let's define a reduced mass $\mu = \frac{m_{1}m_{2}}{m_{1}+m_{2}}$
$F = \frac{Gm_{1}m_{2}}{\left(\mathbf{r_{1}}-\mathbf{r_{2}}\right)^{2}}
= \frac{Gm_{1}m_{2}}{\left(\mathbf{s_{1}}-\mathbf{s_{2}}\right)^{2}}
= \frac{Gm_{1}m_{2}}{\left(\mathbf{s_{1}}+\frac{m_{1}}{m_{2}}\mathbf{s_{1}}\right)^{2}}
=G \frac{m_{1}}{m_{1}+m_{2}}\frac{m_{2}}{m_{1}+m_{2}}\frac{m_{2}^{2}}{s_{1}^{2}}
= G \mu^{2}\frac{m_{2}}{m_{1}}\frac{1}{s_{1}^{2}}
$
Note that $\Delta{}s = s_{1}-s_{2} = s_{1} + \frac{m_{1}}{m_{2}}s_{1} = \frac{m_{1}+m_{2}}{m_{2}}s_{1} = \frac{m_{1}s_{1}}{\mu}$ so that
$F = \frac{G m_{1}m_{2}}{\Delta{}s^{2}}$
Whether you write the Force as depending on $s_{1}, s_{2}, or \Delta{}s$, you now have an equation relating the acceleration of a position to the position itself, rather than the difference of two positions $r_{1}-r_{2}$.
The solution to this vector differential equation is an ellipse, hence binary star systems will always be an ellipse around the center of mass, because such transformations and choices of coordinates are always possible and binary star systems as assumed will always obey these basic equations.
