# 4.13 Two body system - linear motion  (Page 2/3)

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Since no external force is applied, the subsequent motion due to internal gravitational force does not change the position of center of mass in accordance with second law of motion. The bodies simply move towards each other such that center of mass remains at rest.

The two bodies move along a straight line joining their centers. The line of motion also also passes through center of mass. This has one important implication. The plane containing one body and “center of mass” and the plane containing other body and “center of mass” are same. It means that motions of two bodies are “coplanar” with line joining the centers of bodies and center of mass. The non-planar motions as shown in the figure below are not possible as motion is not along the line joining the centers of two bodies.

Let suscripts "1" and "2" denote two bodies. Also, let “ ${r}_{1}$ ”, “ ${v}_{1}$ ”, “ ${a}_{1}$ ” and “ ${r}_{2}$ ”, “ ${v}_{2}$ ”, “ ${a}_{2}$ ” be the magnitudes of linear distance from the center of mass, speeds and magnitudes of accelerations respectively of two bodies under consideration. Also let “center of mass” of the system is the origin of reference frame. Then, by definition of center of mass :

${r}_{cm}=\frac{-{m}_{1}{r}_{1}+{m}_{2}{r}_{2}}{{m}_{1}+{m}_{2}}$

But “center of mass” lies at the origin of the reference frame,

$⇒{r}_{cm}=\frac{-{m}_{1}{r}_{1}+{m}_{2}{r}_{2}}{{m}_{1}+{m}_{2}}=0$

$⇒{m}_{1}{r}_{1}={m}_{2}{r}_{2}$

Taking first differentiation of position with respect to time, we have :

$⇒{v}_{cm}=\frac{-{m}_{1}{v}_{1}+{m}_{2}{v}_{2}}{{m}_{1}+{m}_{2}}=0$

$⇒{m}_{1}{v}_{1}={m}_{2}{v}_{2}$

Taking first differentiation of velocity with respect to time, we have :

$⇒{a}_{cm}=\frac{-{m}_{1}{a}_{1}+{m}_{2}{a}_{2}}{{m}_{1}+{m}_{2}}=0$

$⇒{m}_{1}{a}_{1}={m}_{2}{a}_{2}$

Considering only magnitude and combining with Newton’s law of gravitation,

${F}_{12}={F}_{21}=\frac{G{m}_{1}{m}_{2}}{{\left({r}_{1}+{r}_{2}\right)}^{2}}$

Since distance of bodies from center of mass changes with time, the gravitational force on two bodies is equal in magnitude at a given instant, but varies with time.

## Newton’s second law of motion

We can treat “two body” system equivalent to “one body” system by stating law of motion in appropriate terms. For example, it would be interesting to know how force can be related to the relative acceleration with which two bodies are approaching towards each other. Again, we would avoid vector notation and only consider the magnitudes of accelerations involved. The relative acceleration is sum of the magnitudes of individual accelerations of the bodies approaching towards each other :

${a}_{r}={a}_{1}+{a}_{2}$

According to Newton's third law, gravitational force on two bodies are pair of action and reaction and hence are equal in magnitude. The magnitude of force on each of the bodies is related to acceleration as :

$F={m}_{1}{a}_{1}={m}_{2}{a}_{2}$

$⇒{m}_{1}{a}_{1}={m}_{2}{a}_{2}$

We can note here that this relation, as a matter of fact, is same as obtained using concept of center of mass. Now, we can write magnitude of relative acceleration of two bodies, “ ${a}_{r}$ ”, in terms of individual accelerations is :

$⇒{a}_{r}={a}_{1}+\frac{{m}_{1}{a}_{1}}{{m}_{2}}$

$⇒{a}_{r}={a}_{1}\left(\frac{{m}_{1}+{m}_{2}}{{m}_{2}}\right)$

$⇒{a}_{1}=\frac{{m}_{2}{a}_{r}}{{m}_{1}+{m}_{2}}$

Substituting in the Newton’s law of motion,

$⇒F={m}_{1}{a}_{1}=\frac{{m}_{1}{m}_{2}{a}_{r}}{{m}_{1}+{m}_{2}}$

$⇒F=\mu {a}_{r}$

Where,

$⇒\mu =\frac{{m}_{1}{m}_{2}}{{m}_{1}+{m}_{2}}$

## Reduced mass

The quantity given by the expression :

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