Solution 9

Vector Method

Let the location of the masses be described by position vectors $\vec{r}_1, \vec{r}_2,$ and $\vec{r}_3$ with the origin at the Center of Mass (COM). The mutual separation between any two particles is $d$.

1. Forces on $m_1$

The net gravitational force on $m_1$ is the vector sum of forces from $m_2$ and $m_3$.

$$ \vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} $$

Using the vector form of gravitational force ($|\vec{r}_i – \vec{r}_j| = d$):

$$ \vec{F}_1 = \frac{Gm_1m_2(\vec{r}_2 – \vec{r}_1)}{|\vec{r}_2 – \vec{r}_1|^3} + \frac{Gm_1m_3(\vec{r}_3 – \vec{r}_1)}{|\vec{r}_3 – \vec{r}_1|^3} $$

Since the denominator is $d^3$ in both terms:

$$ \vec{F}_1 = \frac{Gm_1}{d^3} [ m_2(\vec{r}_2 – \vec{r}_1) + m_3(\vec{r}_3 – \vec{r}_1) ] $$ $$ \vec{F}_1 = \frac{Gm_1}{d^3} [ m_2\vec{r}_2 – m_2\vec{r}_1 + m_3\vec{r}_3 – m_3\vec{r}_1 ] \quad \text{…(i)} $$

2. Using the Center of Mass Property

Since the origin is at the COM, by definition:

$$ m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3 = 0 $$ $$ \implies m_2\vec{r}_2 + m_3\vec{r}_3 = -m_1\vec{r}_1 $$

Substitute this back into equation (i):

$$ \vec{F}_1 = \frac{Gm_1}{d^3} [ (-m_1\vec{r}_1) – m_2\vec{r}_1 – m_3\vec{r}_1 ] $$ $$ \vec{F}_1 = -\frac{Gm_1}{d^3} (m_1 + m_2 + m_3) \vec{r}_1 $$

3. Centripetal Force

For the system to rotate with angular velocity $\omega$ without changing shape, this gravitational force must provide the necessary centripetal force pointing towards the origin:

$$ \vec{F}_{centripetal} = -m_1 \omega^2 \vec{r}_1 $$

Equating the forces:

$$ -m_1 \omega^2 \vec{r}_1 = -\frac{Gm_1(m_1 + m_2 + m_3)}{d^3} \vec{r}_1 $$