Editing Two-body problem (section)

== Energy of the two-body system ==

If the force {{math|'''F'''('''r''')}} is [[Conservative force|conservative]] then the system has a [[potential energy]] {{math|''U''('''r''')}}, so the total [[Mechanical energy|energy]] can be written as
<math display="block">E_\text{tot} = \frac{1}{2} m_1 \dot{\mathbf{x}}_1^2 + \frac{1}{2} m_2 \dot{\mathbf{x}}_2^2 + U(\mathbf{r}) = \frac{1}{2} (m_1 + m_2) \dot{\mathbf{R}}^2  + {1 \over 2} \mu \dot{\mathbf{r}}^2 + U(\mathbf{r})</math>

In the center of mass frame the [[Kinetic energy#Frame of reference|kinetic energy]] is the lowest and the total energy becomes
<math display="block">E = \frac{1}{2} \mu \dot{\mathbf{r}}^2 + U(\mathbf{r})</math>
The coordinates {{math|'''x'''<sub>1</sub>}} and {{math|'''x'''<sub>2</sub>}} can be expressed as
<math display="block"> \mathbf{x}_1 = \frac{\mu}{m_1} \mathbf{r}</math>
<math display="block"> \mathbf{x}_2 = - \frac{\mu}{m_2} \mathbf{r}</math> 
and in a similar way the energy ''E'' is related to the energies {{math|''E''<sub>1</sub>}} and {{math|''E''<sub>2</sub>}} that separately contain the kinetic energy of each body: 
<math display="block">\begin{align}
E_1 & = \frac{\mu}{m_1} E = \frac{1}{2} m_1 \dot{\mathbf{x}}_1^2 + \frac{\mu}{m_1} U(\mathbf{r}) \\[4pt]
E_2 & = \frac{\mu}{m_2} E = \frac{1}{2} m_2 \dot{\mathbf{x}}_2^2 + \frac{\mu}{m_2} U(\mathbf{r}) \\[4pt]
E_\text{tot} & = E_1 + E_2
\end{align}</math>