Two-Body Dynamics

Let us consider the motion of a dynamical system that consists of two freely moving and mutually interacting point objects. Suppose that our first object is of mass $m_1$, and is located at displacement ${\bf r}_1$. Likewise, our second object is of mass $m_2$, and is located at displacement ${\bf r}_2$. Let the first object exert a force ${\bf f}_{21}$ on the second. By Newton's third law, the second object exerts an equal and opposite force, ${\bf f}_{12} = -{\bf f}_{21}$, on the first. (See Section 1.2.4.) Suppose that there are no other forces in the problem. The equations of motion of our two objects are thus

$$m_1\,\frac{d^{\,2}{\bf r}_1}{dt^{\,2}} = -{\bf f},$$ (1.432)

$$m_2\,\frac{d^{\,2}{\bf r}_2}{dt^{\,2}} ={\bf f},$$ (1.433)

where ${\bf f} = {\bf f}_{21}$.

The center of mass of our system is located at

$${\bf R} = \frac{m_1\,{\bf r}_1+ m_2\,{\bf r}_2}{m_1 + m_2}.$$ (1.434)

(See Section 1.4.2.) Hence, we can write

$${\bf r}_1 = {\bf R} -\frac{m_2}{m_1+m_2}\,{\bf r},$$ (1.435)

$${\bf r}_2 = {\bf R} + \frac{m_1}{m_1+m_2}\,{\bf r},$$ (1.436)

where ${\bf r}= {\bf r}_2-{\bf r}_1$. Substituting the previous two equations into Equations (1.432) and (1.433), and making use of the fact that the center of mass of an isolated system does not accelerate (see Section 1.4.2), we find that both equations yield

$$\mu\,\frac{d^{\,2}{\bf r}}{d t^{\,2}} = {\bf f},$$ (1.437)

where

$$\mu = \frac{m_1\,m_2}{m_1+ m_2}$$ (1.438)

is called the reduced mass. Hence, we have effectively converted our original two-body problem into an equivalent one-body problem. In the equivalent problem, the force ${\bf f}$ is the same as that acting on both objects in the original problem (modulo a minus sign). However, the mass, $\mu$, is different, and is less than either of $m_1$ or $m_2$ (which is why it is called the “reduced” mass).