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$$\mu =\frac{{m}_{1}{m}_{2}}{{m}_{1}+{m}_{2}}$$
is known as “reduced mass”. It has the same unit as that of “mass”. It represents the “effect” of two bodies, if we want to treat “two body” system as “one body” system.
In the nutshell, we can treat motion of “two body” system along a straight line as “one body” system, which has a mass equal to “μ” and acceleration equal to relative acceleration, “ ${a}_{r}$ ”.
The two bodies are approaching towards each other. Hence, magnitude of velocity of approach is given by :
$${v}_{r}={v}_{1}+{v}_{2}$$
This is the expression of the magnitude of velocity of approach. We can write corresponding vector equation for velocity of approach in relation to reference direction. In this module, however, we will avoid vector notation or vector interpretation to keep the discussion simplified.
The kinetic energy of the “two body” system is given as the sum of kinetic energy of individual bodies,
$$K=\frac{1}{2}{m}_{1}{v}_{1}^{2}+\frac{1}{2}{m}_{2}{v}_{2}^{2}$$
We can write this expression of kinetic energy in terms of relative velocity i.e. velocity of approach. For this, we need to express individual speeds in terms of relative speed as :
$${v}_{r}={v}_{1}+{v}_{2}$$
But,
$${m}_{1}{v}_{1}={m}_{2}{v}_{2}$$
$$\Rightarrow {v}_{2}=\frac{{m}_{1}{v}_{1}}{{m}_{2}}$$
Substituting in the expression of relative velocity,
$$\Rightarrow {v}_{r}={v}_{1}+\frac{{m}_{1}{v}_{1}}{{m}_{2}}$$
$$\Rightarrow {v}_{1}=\frac{{m}_{2}{v}_{r}}{{m}_{1}+{m}_{2}}$$
Similarly,
$$\Rightarrow {v}_{2}=\frac{{m}_{1}{v}_{r}}{{m}_{1}+{m}_{2}}$$
Now, putting these expressions of individual speed in the equation of kinetic energy :
$$\Rightarrow K=\frac{{m}_{1}{m}_{2}^{2}{v}_{r}^{2}}{{\left({m}_{1}+{m}_{2}\right)}^{2}}+\frac{{m}_{2}{m}_{1}^{2}{v}_{r}^{2}}{{\left({m}_{1}+{m}_{2}\right)}^{2}}$$
$$\Rightarrow K=\frac{{m}_{1}{m}_{2}{v}_{r}^{2}}{{\left({m}_{1}+{m}_{2}\right)}^{2}}\left({m}_{1}+{m}_{2}\right)$$
$$\Rightarrow K=\frac{{m}_{1}{m}_{2}{v}_{r}^{2}}{{m}_{1}+{m}_{2}}$$
$$\Rightarrow K=\frac{1}{2}\mu {v}_{r}^{2}$$
This result also indicates that we can treat “two body” system as “one body” system from the point of view of kinetic energy, as if the body has reduced mass of “μ” and speed equal to the magnitude of relative velocity, “ ${v}_{r}$ ”.
Problem 1: Two masses “ ${m}_{1}$ ” and “ ${m}_{2}$ ” are initially at rest at a great distance. At a certain instant, they start moving towards each other, when released from their positions. Considering absence of any other gravitational field, calculate velocity of approach when they are at a distance “r” apart.
Solution : The bodies are at large distance in the beginning. There is no external gravitational field. Hence, we can consider initial gravitational energy of the system as zero (separated y infinite distance). Also, the bodies are at rest in the beginning. The initial kinetic energy is also zero. In turn, initial mechanical energy of the system is zero.
Let “ ${v}_{r}$ ” be the velocity of approach, when the bodies are at a distance “r” apart. Applying conservation of mechanical energy, we have :
$${K}_{i}+{U}_{i}={K}_{f}+{U}_{f}$$
$$\Rightarrow 0+0=\frac{1}{2}\mu {v}_{r}^{2}-\frac{G{m}_{1}{m}_{2}}{r}$$
$$\Rightarrow {v}_{r}^{2}=\frac{2G{m}_{1}{m}_{2}}{\mu r}=\frac{2G{m}_{1}{m}_{2}\left({m}_{1}+{m}_{2}\right)}{{m}_{1}{m}_{2}r}$$
$$\Rightarrow {v}_{r}=\sqrt{\left\{\frac{2G\left({m}_{1}+{m}_{2}\right)}{r}\right\}}$$
From the discussion above, we conclude the followings about the motion of “two body” system along a straight line :
1: Each body follows a straight line trajectory.
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: Magnitude of gravitational force is same for two bodies, but they vary as the distance between them changes.
5: We can treat “two body” system equivalent to “one body” system by using concepts of (i) reduced mass “μ” (ii) relative velocity, “ ${v}_{r}$ ” and (iii) relative acceleration “ ${a}_{r}$ ”.
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