Motion of reduced mass under inverse square force:

Let, gravitational force is taken as an example of inverse square force.

Now the force of attraction due to the gravitational force between the two mass points \( m_1 \) and \( m_2 \) is given by

\( \vec{F}(r)=-G\frac{m_1m_2}{r^2}\hat{r}\tag{1} \)

where G is the gravitational constant.

If \( \mu=\frac{m_1m_2}{m_1+m_2} \) be the reduced mass of the system of these two masses, then equation (1) will be

\( \mu\cdot\frac{d^2\vec{r}}{dt^2}=-G\frac{m_1m_2}{r^2}\hat{r} \)

or, \( \frac{m_1m_2}{m_1+m_2}\ddot{\vec{r}}=-G\frac{m_1m_2}{r^2}\hat{r} \)

or, \( \ddot{\vec{r}}=-G\frac{M}{r^2}\hat{r}\tag{2} \)

where, \( M=m_1+m_2 \) is the total mass of the system.

Now the equation (2) represents the equation of motion of a particle of unit mass at a vector distance \( \vec{r} \) ( here it is the distance between the two mass points) from a fixed mass M exerting a force of attraction on it.