Let us consider a large body of mass \( m_1 \) attracts a very small body of mass \( m_2 \). Now the motion of the reduced mass has the consequences given below,

(1)

The reduced mass of the system of masses \( m_1 \) and \( m_2 \) is

\( \mu=\frac{m_1m_2}{m_1+m_2} \)

Since \( m_2 \) is very small compared to the mass \( m_1 \), so \( m_2 \) can be ignored as compared to \( m_1 \).

Therefore, the reduced mass of the system will be

\( \mu=\frac{m_1m_2}{m_1} \)

or, \( \mu=m_2 \)

So the reduced mass is practically equal to the mass of the small body.

(2)

Let us consider \( \vec{r_1} \) and \( \vec{r_2} \) be the position vectors of the masses \( m_1 \) and \( m_2 \) respectively, then we can write

\( m_1\vec{r_1}=m_2\vec{r_2} \)

or, \( \vec{r_1}=\frac{m_2}{m_1}\vec{r_2} \)

Since the mass \( m_2 \) is very small as compared to the mass \( m_1 \) so \( \frac{m_2}{m_1}\approx{0} \).

therefore, \( \vec{r_1}\approx{0} \)

So the centre of mass of the system of these two mass particles coincides with the centre of the larger body.