Editing Two-body problem (section)

== Two-body motion is planar ==

The motion of two bodies with respect to each other always lies in a plane (in the [[center of mass frame]]).

Proof: Defining the [[linear momentum]] {{math|'''p'''}} and the [[angular momentum]] {{math|'''L'''}} of the system, with respect to the center of mass, by the equations
<math display="block">\mathbf{L} = \mathbf{r} \times \mathbf{p} = \mathbf{r} \times \mu \frac{d\mathbf{r}}{dt},</math>

where {{mvar|μ}} is the [[reduced mass]] and {{math|'''r'''}} is the relative position {{math|'''r'''<sub>2</sub> − '''r'''<sub>1</sub>}} (with these written taking the center of mass as the origin, and thus both parallel to {{math|'''r'''}}) the rate of change of the angular momentum {{math|'''L'''}} equals the net [[torque]] {{math|'''N'''}}
<math display="block">\mathbf{N} = \frac{d\mathbf{L}}{dt} = \dot{\mathbf{r}} \times \mu\dot{\mathbf{r}} + \mathbf{r} \times \mu\ddot{\mathbf{r}} \ ,</math>
and using the property of the [[vector cross product]] that {{math|1='''v''' × '''w''' = '''0'''}} for any vectors {{math|'''v'''}} and {{math|'''w'''}} pointing in the same direction,

<math display="block"> \mathbf{N} \ = \ \frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F} \ ,</math>
with {{math|1='''F''' = ''μ''&thinsp;''d''{{i sup|2}}'''r'''/''dt''{{i sup|2}}}}.

Introducing the assumption (true of most physical forces, as they obey [[Newton's laws of motion|Newton's strong third law of motion]]) that the force between two particles acts along the line between their positions, it follows that {{math|1='''r''' × '''F''' =  '''0'''}} and the [[conservation of angular momentum|angular momentum vector {{math|'''L'''}} is constant]] (conserved). Therefore, the displacement vector {{math|'''r'''}} and its velocity {{math|'''v'''}} are always in the plane [[perpendicular]] to the constant vector {{math|'''L'''}}.