Solution

The gravitational three-body Lagrangian is

L = \frac{1}{2} \sum_{i = 1}^{3} m_{i} ∣ \dot{r}_{i} ∣^{2} + G (\frac{m _{1} m _{2}}{∣ r _{1} - r _{2} ∣} + \frac{m _{1} m _{3}}{∣ r _{1} - r _{3} ∣} + \frac{m _{2} m _{3}}{∣ r _{2} - r _{3} ∣}) .

Put

M_{12} = m_{1} + m_{2}

and

M = M_{12} + m_{3}

. Solving the definitions of the Jacobi coordinates for three particles gives

r_{1} r_{2} r_{3} = c + \frac{m _{3}}{M} b + \frac{m _{2}}{M _{12}} a, = c + \frac{m _{3}}{M} b - \frac{m _{1}}{M _{12}} a, = c - \frac{M _{12}}{M} b .

Substitution into the kinetic energy makes all cross terms cancel:

T = \frac{1}{2} α ∣ \dot{a} ∣^{2} + \frac{1}{2} β ∣ \dot{b} ∣^{2} + \frac{1}{2} γ ∣ \dot{c} ∣^{2},

where

α = \frac{m _{1} m _{2}}{m _{1} + m _{2}}, β = \frac{( m _{1} + m _{2} ) m _{3}}{m _{1} + m _{2} + m _{3}}, γ = m_{1} + m_{2} + m_{3} .

The pair separations are

r_{1} - r_{2} = a, r_{1} - r_{3} = b + \frac{m _{2}}{M _{12}} a, r_{2} - r_{3} = b - \frac{m _{1}}{M _{12}} a .

V (a, b) = - \frac{G m _{1} m _{2}}{∣ a ∣} - \frac{G m _{1} m _{3}}{b + \frac{m _{2}}{M _{12}} a} - \frac{G m _{2} m _{3}}{b - \frac{m _{1}}{M _{12}} a},

which is independent of

c

. Thus

c

\frac{d}{d t} (γ \dot{c}) = 0, \ddot{c} = 0 .

The center of mass moves uniformly because the isolated system has no external force.

Finally, substituting the inverse coordinate transformation into

L = \sum_{i} m_{i} r_{i} \times \dot{r}_{i}

again cancels every cross term and yields

L = α a \times \dot{a} + β b \times \dot{b} + γ c \times \dot{c} .

Solved by gpt-5.6-sol high.

 Ancestors (10)