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{{Short description|Motion problem in classical mechanics}}

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{{Astrodynamics}}
In [[Physics:Classical mechanics|classical mechanics]], the '''two-body problem''' is to predict the motion of two massive objects which are abstractly viewed as [[Physics:Point particle|point particle]]s. The problem assumes that the two objects interact only with one another; the only force affecting each object arises from the other one, and all other objects are ignored.

The most prominent example of the classical two-body problem is the gravitational case (see also [[Physics:Kepler problem|Kepler problem]]), arising in astronomy for predicting the orbits (or escapes from orbit) of objects such as [[Astronomy:Satellite|satellite]]s, [[Astronomy:Planet|planet]]s, and stars. A two-point-particle model of such a system nearly always describes its behavior well enough to provide useful insights and predictions.

A simpler "one body" model, the "[[Physics:Classical central-force problem|central-force problem]]", treats one object as the immobile source of a force acting on the other. One then seeks to predict the motion of the single remaining mobile object. Such an approximation can give useful results when one object is much more massive than the other (as with a light planet orbiting a heavy star, where the star can be treated as essentially stationary).

However, the one-body approximation is usually unnecessary except as a stepping stone. For many forces, including gravitational ones, the general version of the two-body problem can be reduced to a pair of one-body problems, allowing it to be solved completely, and giving a solution simple enough to be used effectively.

By contrast, the [[Physics:Three-body problem|three-body problem]] (and, more generally, the [[Physics:N-body problem|''n''-body problem]] for ''n''&nbsp;≥&nbsp;3) cannot be solved in terms of first integrals, except in special cases.

== Results for prominent cases ==

=== Gravitation and other inverse-square examples ===

The two-body problem is interesting in astronomy because pairs of astronomical objects are often moving rapidly in arbitrary directions (so their motions become interesting), widely separated from one another (so they will not collide) and even more widely separated from other objects (so outside influences will be small enough to be ignored safely).

Under the force of [[Physics:Gravity|gravity]], each member of a pair of such objects will orbit their mutual center of mass in an elliptical pattern, unless they are moving fast enough to escape one another entirely, in which case their paths will diverge along other planar [[Conic section|conic section]]s. If one object is very much heavier than the other, it will move far less than the other with reference to the shared center of mass. The mutual center of mass may even be inside the larger object.

For the derivation of the solutions to the problem, see [[Physics:Classical central-force problem|Classical central-force problem]] or [[Physics:Kepler problem|Kepler problem]].

In principle, the same solutions apply to macroscopic problems involving objects interacting not only through gravity, but through any other attractive [[Scalar potential|scalar force field]] obeying an [[Inverse-square law|inverse-square law]], with [[Physics:Coulomb's law|electrostatic attraction]] being the obvious physical example. In practice, such problems rarely arise. Except perhaps in experimental apparatus or other specialized equipment, we rarely encounter electrostatically interacting objects which are moving fast enough, and in such a direction, as to avoid colliding, and/or which are isolated enough from their surroundings.

The dynamical system of a two-body system under the influence of torque turns out to be a [[Sturm–Liouville theory|Sturm-Liouville equation]].<ref>{{cite journal |last1=Luo |first1=Siwei |title=The Sturm-Liouville problem of two-body system |journal=Journal of Physics Communications | date=22 June 2020 |volume=4 |issue=6 |page=061001 |doi=10.1088/2399-6528/ab9c30|bibcode=2020JPhCo...4f1001L |doi-access=free }}</ref>

=== Inapplicability to atoms and subatomic particles ===

Although the two-body model treats the objects as point particles, classical mechanics only apply to systems of macroscopic scale. Most behavior of subatomic particles ''cannot'' be predicted under the classical assumptions underlying this article or using the mathematics here.

[[Physics:Electron|Electron]]s in an atom are sometimes described as "orbiting" its [[Physics:Atomic nucleus|nucleus]], following an [[Physics:Bohr model|early conjecture]] of [[Biography:Niels Bohr|Niels Bohr]] (this is the source of the term "[[Physics:Atomic orbital|orbital]]"). However, electrons don't actually orbit nuclei in any meaningful sense, and [[Physics:Quantum mechanics|quantum mechanics]] are necessary for any useful understanding of the electron's real behavior. Solving the classical two-body problem for an electron orbiting an atomic nucleus is misleading and does not produce many useful insights.

== Reduction to two independent, one-body problems ==


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 [[Physics:Potential|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|[[Physics:Jacobi coordinates|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, [[Physics: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 [[Physics:Center of mass|center of mass]] ([[Barycenter|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 [[Physics:Center of mass|center of mass]] ([[Barycenter|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 [[Physics: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 ([[Physics:Conservation of momentum|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 [[Physics:Translational symmetry|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 '''[[Physics:Reduced mass|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|inverse-square law]], see the [[Physics:Kepler problem|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.

== 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 [[Physics:Angular momentum|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 [[Physics:Reduced mass|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 [[Physics:Torque|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 [[Physics: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 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|perpendicular]] to the constant vector {{math|'''L'''}}.

== Energy of the two-body system ==

If the force {{math|'''F'''('''r''')}} is [[Physics:Conservative force|conservative]] then the system has a [[Physics:Potential energy|potential energy]] {{math|''U''('''r''')}}, so the total [[Physics: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 [[Physics: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>

== Central forces ==
{{main|Physics:Classical central-force problem}}

For many physical problems, the force {{math|1='''F'''('''r''')}} is a [[Physics:Central force|central force]], i.e., it is of the form
<math display="block">\mathbf{F}(\mathbf{r}) = F(r)\hat{\mathbf{r}}</math>
where {{math|1=''r'' = {{abs|'''r'''}}}} and {{math|1='''r̂''' = '''r'''/''r''}} is the corresponding [[Unit vector|unit vector]]. We now have:
<math display="block">\mu \ddot{\mathbf{r}} = {F}(r) \hat{\mathbf{r}} \ ,</math>
where ''F''(''r'') is negative in the case of an attractive force.

==See also==
* [[Energy drift]]
* [[Astronomy:Equation of the center|Equation of the center]]
* [[Astronomy:Euler's three-body problem|Euler's three-body problem]]
* [[Astronomy:Kepler orbit|Kepler orbit]]
* [[Physics:Kepler problem|Kepler problem]]
* [[Physics:N-body problem|''n''-body problem]]
* [[Physics:Three-body problem|Three-body problem]]
* [[Physics:Virial theorem|Virial theorem]]

==References==
{{Reflist|30em}}

==Bibliography==

* {{cite book | author = Landau LD |  author2 = Lifshitz EM |  date = 1976 | title = Mechanics | edition = 3rd. | publisher = Pergamon Press | location = New York | isbn = 0-08-029141-4 | url-access = registration | url = https://archive.org/details/mechanics00land }}
* {{cite book | author = Goldstein H |  date = 1980 | title = Classical Mechanics | edition = 2nd. | publisher = Addison-Wesley | location = New York | isbn = 0-201-02918-9}}

== External links ==

* [http://scienceworld.wolfram.com/physics/Two-BodyProblem.html Two-body problem] at Eric Weisstein's World of Physics


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