Editing Two-body problem (section)

=== Center of mass motion (1st one-body problem) ===

Let <math>\mathbf{R} </math> be the position of the [[center of mass]] ([[barycenter]]) of the system. Addition of the force equations (1) and (2) yields
<math display="block">m_1 \ddot{\mathbf{x}}_1 + m_2 \ddot{\mathbf{x}}_2 = (m_1 + m_2)\ddot{\mathbf{R}}  = \mathbf{F}_{12} + \mathbf{F}_{21} = 0</math>
where we have used [[Newton's laws of motion|Newton's third law]] {{math|1='''F'''<sub>12</sub> = −'''F'''<sub>21</sub>}} and where
<math display="block">\ddot{\mathbf{R}}  \equiv \frac{m_{1}\ddot{\mathbf{x}}_{1} + m_{2}\ddot{\mathbf{x}}_{2}}{m_{1} + m_{2}}.</math>

The resulting equation:
<math display="block">\ddot{\mathbf{R}}  = 0</math>
shows that the velocity <math>\mathbf{v} = \frac{dR}{dt}</math> of the center of mass is constant, from which follows that the total momentum {{math|''m''<sub>1</sub> '''v'''<sub>1</sub> + ''m''<sub>2</sub> '''v'''<sub>2</sub>}} is also constant ([[conservation of momentum]]). Hence, the position {{math|'''R'''(''t'')}} of the center of mass can be determined at all times from the initial positions and velocities.