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The trajectory of two body system depends on the initial velocities of the bodies and their relative mass. If the mass of the bodies under consideration are comparable, then bodies move around their “center of mass” along two separate circular trajectories. This common point about which two bodies revolve is also known as “barycenter”.
In order to meet the requirement imposed by laws of motion and conservation laws, the motion of two bodies executing circular motion is constrained in certain ways.
Since external force is zero, the acceleration of center of mass is zero. This is the first constraint. For easy visualization of this constraint, we consider that center of mass of the system is at rest in a particular reference frame.
Now, since bodies are moving along two circular paths about "center of mass", their motions should be synchronized in a manner so that the length of line, joining their centers, is a constant . This is required; otherwise center of mass will not remain stationary in the chosen reference. Therefore, the linear distance between bodies is a constant and is given by :
$$r={r}_{1}+{r}_{2}$$
Now this condition can be met even if two bodies move in different planes. However, there is no external torque on the system. It means that the angular momentum of the system is conserved. This has an important deduction : the plane of two circular trajectories should be same.
Mathematically, we can conclude this, using the concept of angular momentum. We know that torque is equal to time rate of change of angular momentum,
$$\frac{dL}{dt}=r\times F$$
But, external torque is zero. Hence,
$$\Rightarrow r\times F=0$$
It means that “ r ” and “ F ” are always parallel. It is only possible if two planes of circles are same. We, therefore, conclude that motions of two bodies are coplanar. For coplanar circular motion, center of mass is given by definition as :
$${r}_{cm}=\frac{-{m}_{1}{r}_{1}+{m}_{2}{r}_{2}}{{m}_{1}+{m}_{2}}=0$$
$$\Rightarrow {m}_{1}{r}_{1}={m}_{2}{r}_{2}$$
Taking first differentiation with respect to time, we have :
$$\Rightarrow {m}_{1}{v}_{1}={m}_{2}{v}_{2}$$
Now dividing second equation by first,
$$\Rightarrow \frac{{m}_{1}{v}_{1}}{{m}_{1}{r}_{1}}=\frac{{m}_{2}{v}_{2}}{{m}_{2}{r}_{2}}$$
$$\Rightarrow \frac{{v}_{1}}{{r}_{1}}=\frac{{v}_{2}}{{r}_{2}}$$
$$\Rightarrow {\omega}_{1}={\omega}_{2}=\omega \phantom{\rule{1em}{0ex}}\left(say\right)$$
It means that two bodies move in such a manner that their angular velocities are equal.
The gravitational force on each of the bodies is constant and is given by :
$$F=\frac{G{m}_{1}{m}_{2}}{{\left({r}_{1}+{r}_{2}\right)}^{2}}=\frac{G{m}_{1}{m}_{2}}{{r}^{2}}$$
Since gravitational force provides for the requirement of centripetal force in each case, it is also same in two cases. Centripetal force is given by :
$${F}_{C}={m}_{1}{r}_{1}{\omega}^{2}={m}_{2}{r}_{2}{\omega}^{2}=\frac{G{m}_{1}{m}_{2}}{{r}^{2}}$$
Each body moves along a circular path. The gravitational force on either of them provides the centripetal force required for circular motion. Hence, centripetal force is :
$${m}_{1}{r}_{1}{\omega}^{2}=\frac{G{m}_{1}{m}_{2}}{{\left({r}_{1}+{r}_{2}\right)}^{2}}$$
$$\Rightarrow {\omega}^{2}=\frac{G{m}_{2}}{{r}_{1}{\left({r}_{1}+{r}_{2}\right)}^{2}}$$
Let the combined mass be “M”. Then,
$$M={m}_{1}+{m}_{2}$$
Using relation ${m}_{1}{r}_{1}={m}_{2}{r}_{2}$ , we have :
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