<< Chapter < Page | Chapter >> Page > |
$$\Rightarrow M=\frac{{m}_{2}{r}_{2}}{{r}_{1}}+{m}_{2}={m}_{2}\left(\frac{{r}_{1}+{r}_{2}}{{r}_{1}}\right)$$
$$\Rightarrow {m}_{2}=\frac{M{r}_{1}}{{r}_{1}+{r}_{2}}$$
Substituting in the equation, involving angular velocity,
$$\Rightarrow {\omega}^{2}=\frac{GM{r}_{1}}{{r}_{1}{\left({r}_{1}+{r}_{2}\right)}^{3}}=\frac{GM}{{r}^{3}}$$
$$\Rightarrow \omega =\sqrt{\left(\frac{GM}{{r}^{3}}\right)}$$
This expression has identical form as for the case when a body revolves around another body at rest along a circular path (compare with “Earth – satellite” system). Here, combined mass “M” substitutes for the mass of heavier mass at the center and sum of the linear distance replaces the radius of rotation.
The linear velocity is equal to the product of the radius of circle and angular velocity. Hence,
$${v}_{1}=\omega {r}_{1}$$
$${v}_{2}=\omega {r}_{2}$$
We can easily find the expression for time period of revolution as :
$$T=\frac{2\pi}{\omega}=\frac{2\pi {r}^{\frac{3}{2}}}{\sqrt{\left(GM\right)}}$$
This expression also has the same form as for the case when a body revolves around another body at rest along a circular path (compare with “Earth – satellite” system). Further squaring on either side, we have :
$$\Rightarrow {T}^{2}\propto {r}^{3}$$
Here, we set out to find moment of inertia of the system about the common axis passing through center of mass and perpendicular to the plane of rotation. For this, we consider each of the bodies as point mass. Note that two bodies are rotating about a common axis with same angular velocity. Clearly, MI of the system is :
$$I={m}_{1}{r}_{1}^{2}+{m}_{2}{r}_{2}^{2}$$
We can express individual distance in terms of their sum using following two equations,
$$r={r}_{1}+{r}_{2}$$
$${m}_{1}{r}_{1}={m}_{2}{r}_{2}$$
Substituting for “ ${r}_{1}$ ” in the equation or "r", we have :
$$\Rightarrow r=\frac{{m}_{2}{r}_{2}}{{m}_{1}}+{r}_{2}={r}_{2}\left(\frac{{m}_{1}+{m}_{2}}{{m}_{1}}\right)$$
$$\Rightarrow {r}_{2}=\frac{r{m}_{1}}{{m}_{1}+{m}_{2}}$$
Similarly, we can express, “ ${r}_{1}$ ” as :
$$\Rightarrow {r}_{1}=\frac{r{m}_{2}}{{m}_{1}+{m}_{2}}$$
Substituting for “ ${r}_{1}$ ” and “ ${r}_{2}$ ” in the expression of moment of inertia,
$$\Rightarrow I=\frac{{m}_{1}{m}_{2}^{2}{r}^{2}}{{\left({m}_{1}+{m}_{2}\right)}^{2}}+\frac{{m}_{2}{m}_{1}^{2}{r}^{2}}{{\left({m}_{1}+{m}_{2}\right)}^{2}}$$
$$\Rightarrow I=\frac{{m}_{1}{m}_{2}{r}^{2}}{{\left({m}_{1}+{m}_{2}\right)}^{2}}X\left({m}_{1}+{m}_{2}\right)$$
$$\Rightarrow I=\frac{{m}_{1}{m}_{2}{r}^{2}}{{m}_{1}+{m}_{2}}$$
$$\Rightarrow I=\mu {r}^{2}$$
This expression is similar to the expression of momemnt of inertia of a particle about an axis at a perpendicualr distance, "r". It is, therefore, clear that “Two body” system orbiting around center of mass can be treated as “one body” system by using concepts of net distance “r” and reduced mass “μ”.
The bodies move about the same axis with the same sense of rotation. The angular momentum of the system, therefore, is algebraic sum of individual angular momentums.
$$L={L}_{1}+{L}_{2}={m}_{1}{r}_{1}^{2}\omega +{m}_{2}{r}_{2}^{2}\omega $$
Substituting for “ ${r}_{1}$ ” and “ ${r}_{2}$ ” with expressions as obtained earlier,
$$\Rightarrow L=\frac{{m}_{1}{m}_{2}^{2}{r}^{2}\omega}{{\left({m}_{1}+{m}_{2}\right)}^{2}}+\frac{{m}_{2}{m}_{1}^{2}{r}^{2}\omega}{{\left({m}_{1}+{m}_{2}\right)}^{2}}$$
$$\Rightarrow L=\frac{{m}_{1}{m}_{2}{r}^{2}\omega}{{\left({m}_{1}+{m}_{2}\right)}^{2}}X\left({m}_{1}+{m}_{2}\right)$$
$$\Rightarrow L=\frac{{m}_{1}{m}_{2}{r}^{2}\omega}{\left({m}_{1}+{m}_{2}\right)}$$
$$\Rightarrow L=\mu {r}^{2}\omega $$
This expression is similar to the expression of angular momemntum of a particle about an axis at a perpendicualr distance, "r". Once again, we see that “Two body” system orbiting around center of mass can be treated as “one body” system by using concepts of net distance “r” and reduced mass “μ”.
The kinetic energy of the system is equal to the algebraic sum of the kinetic energy of the individual body. We write expression of kinetic energy in terms of angular velocity – not in terms of linear velocity. It is so because angular velocity is same for two bodies and can, therefore, be used to simplify the expression for kinetic energy. Now, kinetic energy of the system is :
$$K=\frac{1}{2}{m}_{1}{r}_{1}^{2}{\omega}^{2}+\frac{1}{2}{m}_{2}{r}_{2}^{2}{\omega}^{2}$$
Substituting for “ ${r}_{1}$ ” and “ ${r}_{2}$ ” with expressions as obtained earlier,
$$\Rightarrow K=\frac{{m}_{1}{m}_{2}^{2}{r}^{2}{\omega}^{2}}{2{\left({m}_{1}+{m}_{2}\right)}^{2}}+\frac{{m}_{2}{m}_{1}^{2}{r}^{2}{\omega}^{2}}{2{\left({m}_{1}+{m}_{2}\right)}^{2}}$$
$$\Rightarrow K=\frac{{m}_{1}{m}_{2}{r}^{2}{\omega}^{2}}{2{\left({m}_{1}+{m}_{2}\right)}^{2}}X\left({m}_{1}+{m}_{2}\right)$$
$$\Rightarrow K=\frac{{m}_{1}{m}_{2}{r}^{2}{\omega}^{2}}{2\left({m}_{1}+{m}_{2}\right)}$$
$$\Rightarrow K=\frac{1}{2}\mu {r}^{2}{\omega}^{2}$$
This expression of kinetic energy is also similar to the expression of kinetic energy of a particle rotating about an axis at a perpendicualr distance, "r". Thus, this result also substantiates equivalence of “Two body” system as “one body” system, using concepts of net distance “r” and reduced mass “μ”.
Problem 1 : In a binary star system, two stars of “m” and “M” move along two circular trajectories. If the distance between stars is “r”, then find the total mechanical energy of the system. Consider no other gravitational influence on the system.
Solution : Mechanical energy of the system comprises of potential and kinetic energy. Hence,
$$E=\frac{1}{2}\mu {r}^{2}{w}^{2}-\frac{GMm}{r}$$
We know that angular velocity for “two body” system in circular motion is given by :
$$\Rightarrow \omega =\sqrt{\left\{\frac{G\left(M+m\right)}{{r}^{3}}\right\}}$$
Also, reduced mass is given by :
$$\mu =\frac{Mm}{M+m}$$
Putting in the expression of mechanical energy,
$$\Rightarrow E=\frac{mM{r}^{2}G\left(m+M\right)}{2\left(m+M\right){r}^{3}}-\frac{GMm}{r}$$
$$\Rightarrow E=\frac{GMm}{2r}-\frac{GMm}{r}$$
$$\Rightarrow E=-\frac{GMm}{2r}$$
Thus, we conclude the following :
1: Each body follows a circular path about center of mass.
2: The line joining centers of two bodies pass through center of mass.
3: The planes of two motions are in the same plane. In other words, two motions are coplanar.
4: The angular velocities of the two bodies are equal.
5: The linear distance between two bodies remains constant.
6: Magnitude of gravitational force is constant and same for two bodies.
7: Magnitude of centripetal force required for circular motion is constant and same for two bodies.
8: Since linear velocity is product of angular velocity and distance from the center of revolution, it may be different if the radii of revolutions are different.
9: We can treat two body system with an equivalent one body system by using concepts of (i) combined mass, “M”, (ii) net distance “r” and (iii) reduced mass “μ”.
Notification Switch
Would you like to follow the 'Physics for k-12' conversation and receive update notifications?