Editing Two-body problem (section)

== Reduction to two independent, one-body problems ==
{{See also|Classical central-force problem#Relation to the classical two-body problem}}
{{See also|Kepler problem}}
{{Duplication|date=June 2019|section=yes|dupe=Classical central-force problem#Relation to the classical two-body problem|note=we keep this article as the primary article for the math part}}

The complete two-body problem can be solved by re-formulating it as two one-body problems: a trivial one and one that involves solving for the motion of one particle in an external [[potential]]. Since many one-body problems can be solved exactly, the corresponding two-body problem can also be solved.

[[File:Two-body Jacobi coordinates.JPG|thumb|300px|[[Jacobi coordinates]] for two-body problem; Jacobi coordinates are <math>\boldsymbol{R}=\frac {m_1}{M} \boldsymbol{x}_1 + \frac {m_2}{M} \boldsymbol{x}_2 </math> and <math>\boldsymbol{r} = \boldsymbol{x}_1 - \boldsymbol{x}_2 </math> with <math>M = m_1+m_2 \ </math>.<ref name=Betounes>{{cite book|title=Differential Equations| author=David Betounes|url=https://archive.org/details/differentialequa0000beto|url-access=registration| isbn=0-387-95140-7 | page=58; Figure 2.15|date=2001|publisher=Springer}}</ref>]]

Let {{math|'''x'''<sub>1</sub>}} and {{math|'''x'''<sub>2</sub>}} be the vector positions of the two bodies, and ''m''<sub>1</sub> and ''m''<sub>2</sub> be their masses. The goal is to determine the trajectories {{math|'''x'''<sub>1</sub>(''t'')}} and {{math|'''x'''<sub>2</sub>(''t'')}} for all times ''t'', given the initial positions {{math|1='''x'''<sub>1</sub>(''t'' = 0)}} and {{math|1='''x'''<sub>2</sub>(''t'' = 0)}} and the initial velocities {{math|1='''v'''<sub>1</sub>(''t'' = 0)}} and {{math|1='''v'''<sub>2</sub>(''t'' = 0)}}.

When applied to the two masses, [[Newton's laws of motion#Newton's second law|Newton's second law]] states that
{{NumBlk||<math display="block">\mathbf{F}_{12}(\mathbf{x}_{1},\mathbf{x}_{2}) = m_{1} \ddot{\mathbf{x}}_{1} </math>|Equation {{EquationRef|1}}}}
{{NumBlk||<math display="block">\mathbf{F}_{21}(\mathbf{x}_{1},\mathbf{x}_{2}) = m_{2} \ddot{\mathbf{x}}_{2} </math>|Equation {{EquationRef|2}}}}

where '''F'''<sub>12</sub> is the force on mass 1 due to its interactions with mass 2, and '''F'''<sub>21</sub> is the force on mass 2 due to its interactions with mass 1. The two dots on top of the '''x''' position vectors denote their second derivative with respect to time, or their acceleration vectors.

Adding and subtracting these two equations decouples them into two one-body problems, which can be solved independently. ''Adding'' equations (1) and ({{EquationNote|2}}) results in an equation describing the [[center of mass]] ([[barycenter]]) motion. By contrast, ''subtracting'' equation (2) from equation (1) results in an equation that describes how the vector {{math|1='''r''' = '''x'''<sub>1</sub> − '''x'''<sub>2</sub>}} between the masses changes with time. The solutions of these independent one-body problems can be combined to obtain the solutions for the trajectories {{math|'''x'''<sub>1</sub>(''t'')}} and {{math|'''x'''<sub>2</sub>(''t'')}}.

=== 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.

===Displacement vector motion (2nd one-body problem)===
Dividing both force equations by the respective masses, subtracting the second equation from the first, and rearranging gives the equation
<math display="block">
\ddot {\mathbf{r}} = \ddot{\mathbf{x}}_{1} - \ddot{\mathbf{x}}_{2} = 
\left( \frac{\mathbf{F}_{12}}{m_{1}} - \frac{\mathbf{F}_{21}}{m_{2}} \right) =
\left(\frac{1}{m_{1}} + \frac{1}{m_{2}} \right)\mathbf{F}_{12}
</math>
where we have again used [[Newton's third law]] {{math|1='''F'''<sub>12</sub> = −'''F'''<sub>21</sub>}} and where {{math|'''r'''}} is the [[Displacement (vector)|displacement vector]] from mass 2 to mass 1, as defined above.

The force between the two objects, which originates in the two objects, should only be a function of their separation {{math|'''r'''}} and not of their absolute positions {{math|'''x'''<sub>1</sub>}} and {{math|'''x'''<sub>2</sub>}}; otherwise, there would not be [[translational symmetry]], and the laws of physics would have to change from place to place. The subtracted equation can therefore be written:
<math display="block">\mu \ddot{\mathbf{r}} = \mathbf{F}_{12}(\mathbf{x}_{1},\mathbf{x}_{2}) = \mathbf{F}(\mathbf{r})</math>
where <math>\mu</math> is the '''[[reduced mass]]'''
<math display="block">\mu = \frac{1}{\frac{1}{m_1} + \frac{1}{m_2}} = \frac{m_1 m_2}{m_1 + m_2}.</math>

Solving the equation for {{math|'''r'''(''t'')}} is the key to the two-body problem. The solution depends on the specific force between the bodies, which is defined by <math>\mathbf{F}(\mathbf{r})</math>. For the case where <math>\mathbf{F}(\mathbf{r})</math> follows an [[inverse-square law]], see the [[Kepler problem]].

Once {{math|'''R'''(''t'')}} and {{math|'''r'''(''t'')}} have been determined, the original trajectories may be obtained
<math display="block">\mathbf{x}_1(t) = \mathbf{R} (t) + \frac{m_2}{m_1 + m_2} \mathbf{r}(t)</math>
<math display="block">\mathbf{x}_2(t) = \mathbf{R} (t) - \frac{m_1}{m_1 + m_2} \mathbf{r}(t)</math>
as may be verified by substituting the definitions of '''R'''  and '''r''' into the right-hand sides of these two equations.