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{{Short description|Motion problem in classical mechanics}}
In ], the '''two-body problem''' is a special case of the ] that admits a ] solution. The most commonly encountered version of the problem, involving an ] ], is encountered in ] and the ] of the ] ]. This problem was first solved by ].
{{About|the two-body problem in classical mechanics|the relativistic version|Two-body problem in general relativity|the career management problem of working couples|Two-body problem (career)}}
{{multiple image
|direction=horizontal
|align=right
|width1=200
|width2=100
|image1=orbit5.gif
|image2=orbit2.gif
|footer='''Left:''' Two bodies of similar ] orbiting a common ] external to both bodies, with ]s. This model is typical of ].<br>'''Right:''' Two bodies with a "slight" difference in mass orbiting a common barycenter. Their sizes and this type of orbit are similar to the ] (in which the barycenter is external to both bodies), as well as the ]–] system (in which the barycenter is internal to the larger body).
}}
{{Astrodynamics}}


In ], the '''two-body problem''' is to calculate and predict the motion of two massive bodies that are orbiting each other in space. The problem assumes that the two bodies are ]s that interact only with one another; the only force affecting each object arises from the other one, and all other objects are ignored.
This article deals with the general case where it is not assumed that one body has a much smaller mass than the other one.


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


A simpler "one body" model, the "]", 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).
We restrict ourselves to the ] case, with forces that depend only on the positions of the bodies and obey the ].


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 ], allowing it to be solved completely, and giving a solution simple enough to be used effectively.
Letting <math>\mathbf{x}_{1}</math> and <math>\mathbf{x}_{2}</math> be the positions of the two bodies, and <math>m_{1}</math> and <math>m_{2}</math> be their masses, we have (from ]):
:<math>\mathbf{F}_{12}(\mathbf{x}_{1},\mathbf{x}_{2}) = m_{1} \ddot\mathbf{x}_{1}</math>
:<math>\mathbf{F}_{21}(\mathbf{x}_{1},\mathbf{x}_{2}) = m_{2} \ddot\mathbf{x}_{2}</math>
Note that these two ] equations comprise six ] ]s, each of second order. Therefore, we will need twelve (<math>6 \times 2</math>) constants of ].


By contrast, the ] (and, more generally, the ] for ''n''&nbsp;≥&nbsp;3) cannot be solved in terms of first integrals, except in special cases.
==Sketch of solution==


== Results for prominent cases ==
We start by taking advantage of Newton's third law to reduce the two-body problem to two equivalent one-body problems, one for the ] of the system, and one for the relative motion of the two bodies.


=== Gravitation and other inverse-square examples ===
We can identify ] combinations of the dependent variables to decouple the equations. Adding the differential equations, we get
:<math>m_{1}\ddot\mathbf{x}_{1} + m_{2}\ddot\mathbf{x}_{2} = (m_{1} + m_{2})\ddot\mathbf{x}_{cm} = 0</math>
where
:<math>\mathbf{x}_{cm} \equiv \frac{m_{1}\mathbf{x}_{1} + m_{2}\mathbf{x}_{2}}{m_{1} + m_{2}}</math>
is the position of the ] (]) of the system. Integration shows:
*the total momentum is constant (])
*the center of mass remains at rest, or moves in a straight line at a constant ] (see also ]); this provides six of the constants of integration.


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).
Next, we notice that because of ], the equivalent one-body problem is really a two dimensional problem. This provides two more constants. At this point it is convenient to switch to ].


Under the force of ], 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 ]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.
This is as far as we can go for the general problem. We focus on the inverse square law force, as the most important case of the two-body problem.


For the derivation of the solutions to the problem, see ] or ].
==Reduction to a single body problem==


In principle, the same solutions apply to macroscopic problems involving objects interacting not only through gravity, but through any other attractive ] obeying an ], with ] 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.
Using the strong form of Newton's third law, as well as the fact that the magnitude of the force depends only on the distance between the bodies, we have that
:<math>\mathbf{F}_{12}(\mathbf{x}_{1},\mathbf{x}_{2}) = - \mathbf{F}_{21}(\mathbf{x}_{1},\mathbf{x}_{2}) = -F(|\mathbf{x}_{1} - \mathbf{x}_{2}|) \frac{\mathbf{x}_{1} - \mathbf{x}_{2}}{|\mathbf{x}_{1} - \mathbf{x}_{2}|}</math>


The dynamical system of a two-body system under the influence of torque turns out to be a ].<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 ===
We now multiply the first equation by <math>\frac{1}{m_{1}}</math>, the second by <math>\frac{1}{m_{2}}</math>, and subtract, giving
:<math>\ddot\mathbf{x}_{1} - \ddot\mathbf{x}_{2} = -(\frac{1}{m_{1}} + \frac{1}{m_{2}})F(|\mathbf{x}_{1} - \mathbf{x}_{2}|) \frac{\mathbf{x}_{1} - \mathbf{x}_{2}}{|\mathbf{x}_{1} - \mathbf{x}_{2}|}</math>
or
:<math>\mu \ddot\mathbf{x} = -F(|\mathbf{x}|)\frac{\mathbf{x}}{|\mathbf{x}|}</math>
where
:<math>\mathbf{x} \equiv \mathbf{x}_{1} - \mathbf{x}_{2}</math>
and
:<math>\mu \equiv \frac{1}{\frac{1}{m_{1}} + \frac{1}{m_{2}}} = \frac{m_{1} m_{2}}{m_{1} + m_{2}}</math>
is the ] of the system.


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.
The positions of the bodies are <math>\frac{m_2}{m_1+m_2}\mathbf{x}</math> and <math>-\frac{m_1}{m_1+m_2}\mathbf{x}</math>, respectively.


]s in an atom are sometimes described as "orbiting" its ], following an ] of ] (this is the source of the term "]"). However, electrons don't actually orbit nuclei in any meaningful sense, and ] 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.
Thus we have reduced the problem to a one-body problem.


==Reduction to two dimensions== == 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 ]. Since many one-body problems can be solved exactly, the corresponding two-body problem can also be solved.
Starting with the one-body differential equation above, we take the cross product with the ]
:<math>\mathbf{p} = \mu \dot\mathbf{x}</math>
to get
:<math>\mu \ddot\mathbf{x} \times \mathbf{p} = -\frac{F(|\mathbf{x}|)}{|\mathbf{x}|}(\mathbf{x} \times \mathbf{p})</math>
But the first term on the right is a scalar, and the second term is the ]
:<math>\mathbf{L} = \mathbf{x} \times \mathbf{p}</math>
which, by ] is a constant.
:<math>\ddot\mathbf{x} \times \mathbf{p}</math>
always points in the same direction, which means that the plane determined by <math>\ddot\mathbf{x}</math> and <math>\dot\mathbf{x}</math> is always the same, and the motion is therefore in a plane. This provides two more constants of integration.


] 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>]]
==Change of variables==


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)}}.
Having reduced the problem to two dimensions, at this point it is convenient to switch to polar coordinates. In polar coordinates, the vector differential equation reduces to a scalar equation, due to the fact that the force, and therefore the acceleration, is always toward the origin.


When applied to the two masses, ] states that
It can be shown that <var>r</var>-component of acceleration is
{{NumBlk||<math display="block">\mathbf{F}_{12}(\mathbf{x}_{1},\mathbf{x}_{2}) = m_{1} \ddot{\mathbf{x}}_{1} </math>|Equation {{EquationRef|1}}}}
:<math>\ddot{r} - r \dot{\theta}^{2}</math>
{{NumBlk||<math display="block">\mathbf{F}_{21}(\mathbf{x}_{1},\mathbf{x}_{2}) = m_{2} \ddot{\mathbf{x}}_{2} </math>|Equation {{EquationRef|2}}}}
Therefore, we have
:<math>\mu(\ddot{r} - r \dot{\theta}^{2}) = F(r)</math>
Another change of variables is useful: let <math>u \equiv \frac{1}{r}</math>.


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.
==Newtonian Gravity==


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 ] (]) 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'')}}.
Applying the gravitational formula we get that the position of the first body with respect to the second is governed by the same differential equation as the position of a very small body orbiting a body with a mass equal to the sum of the two masses, because m1.m2/&mu;=m1+m2.


=== Center of mass motion (1st one-body problem) ===
Assume:
*the vector '''''r''''' is the position of one body relative to the other (above called '''x''')
*''r'', ''v'', the ] ''a'', and the ] ''h'' are defined accordingly (hence ''r'' is the distance)
*<math>\mu={G}(m_1+m_2)\,</math> the ] (the sum of those for each mass)
where:
*<math>m_1</math> and <math>m_2</math> are the masses of the two bodies.


Let <math>\mathbf{R} </math> be the position of the ] (]) of the system. Addition of the force equations (1) and (2) yields
Then:
<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>
*the ] applies; recalling that the positions of the bodies are m2/(m1+m2) and -m1/(m1+m2) times '''r''', respectively, we see that the two bodies' orbits are ] ]s; the same ratios apply for the velocities, and, without the minus, for the ] with respect to the barycenter and for the kinetic energies
where we have used ] {{math|1='''F'''<sub>12</sub> = −'''F'''<sub>21</sub>}} and where
*for ]s <math>rv^2 = r^3 \omega^2 = 4 \pi^2 r^3/T^2 = \mu</math>
<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>
*for ]s: <math>4 \pi^2 a^3/T^2 = \mu</math> (with ''a'' expressed in AU and ''T'' in years, and with ''M'' the total mass relative to that of the Sun, we get <math>a^3/T^2 = M</math>)
*for ] <math>r v^2</math> is constant and equal to <math>2 \mu</math>
*''h'' is the total ] divided by the ]
*the ] formulas apply, with specific potential and kinetic energy and their sum taken as the totals for the system, divided by the reduced mass; the kinetic energy of the smaller body is larger; the potential energy of the whole system is equal to the potential energy of one body with respect to the other, i.e. minus the energy needed to escape the other if the other is kept in a fixed position; this should not be confused with the smaller amount of energy one body needs to escape, if the other body moves away also, in the opposite direction: in that case the total energy the two need to escape each other is the same as the aforementioned amount; the conservation of energy for each mass means that an increase of kinetic energy is accompanied by a decrease of potential energy, which is for each mass the inner product of the force and the change in position relative to the barycenter, not relative to the other mass
*for elliptic and hyperbolic orbits <math>\mu</math> is twice the semi-major axis times the absolute value of the specific orbital energy


The resulting equation:
For example, consider two bodies like the Sun orbiting each other:
<math display="block">\ddot{\mathbf{R}} = 0</math>
*the reduced mass is one half of the mass of one Sun (one quarter of the total mass)
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 (]). Hence, the position {{math|'''R'''(''t'')}} of the center of mass can be determined at all times from the initial positions and velocities.
*at a distance of 1 AU: the ] is <math>{1\over 2} \sqrt{2}</math> year, the same as the orbital period of the Earth would be if the Sun would have twice its actual mass; the total energy per kg reduced mass (90 MJ/kg) is twice that of the Earth-Sun system (45 MJ/kg); the total energy per kg total mass (22.5 MJ/kg) is one half of the total energy per kg Earth mass in the Earth-Sun system (45 MJ/kg)
*at a distance of 2 AU (each following an orbit like that of the Earth around the Sun): the orbital period is 2 years, the same as the orbital period of the Earth would be if the Sun would have one quarter of its actual mass
*at a distance of <math>\sqrt{2} \approx 1.26</math> AU: the orbital period is 1 year, the same as the orbital period of the Earth around the Sun


===Displacement vector motion (2nd one-body problem)===
Similarly, a second Earth at a distance from the Earth equal to <math>\sqrt{2}</math> times the usual distance of ]s would be geosynchronous.
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 ] {{math|1='''F'''<sub>12</sub> = −'''F'''<sub>21</sub>}} and where {{math|'''r'''}} is the ] 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 ], and the laws of physics would have to change from place to place. The subtracted equation can therefore be written:
==General Relativistic Gravity==
<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 ''']'''
<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 ], see the ].
In the ] gravity behaves somewhat differently, but, to a first approximation for weak fields, the effect is to slightly strengthen the gravity force at small separations. Kepler's First Law is modified so that the orbit is a ] ellipse, its major and minor axes rotating slowly in the same sense as the oribital motion. The law of conservation of angular momentum still applies (Kepler's Second Law). Kepler's Third Law would in principle be altered slightly, but in practice, the only way to measure the sum of the masses is by applying that Law as it stands, so there is effectively no change. These results were first obtained approximately by ], and the rigorous two body problem was later solved by ].


Once {{math|'''R'''(''t'')}} and {{math|'''r'''(''t'')}} have been determined, the original trajectories may be obtained
==Examples==
<math display="block">\mathbf{x}_1(t) = \mathbf{R} (t) + \frac{m_2}{m_1 + m_2} \mathbf{r}(t)</math>
*a ], e.g. ] (approx. the same mass)
<math display="block">\mathbf{x}_2(t) = \mathbf{R} (t) - \frac{m_1}{m_1 + m_2} \mathbf{r}(t)</math>
*a ], e.g. ] with its moon ] (mass ratio 0.147)
as may be verified by substituting the definitions of '''R''' and '''r''' into the right-hand sides of these two equations.
*a ], e.g. ] (approx. the same mass)

== Two-body motion is planar ==

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

Proof: Defining the ] {{math|'''p'''}} and the ] {{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 ] 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 ] {{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 ] 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 ]) that the force between two particles acts along the line between their positions, it follows that {{math|1='''r''' × '''F''' = '''0'''}} and the ] (conserved). Therefore, the displacement vector {{math|'''r'''}} and its velocity {{math|'''v'''}} are always in the plane ] to the constant vector {{math|'''L'''}}.

== Energy of the two-body system ==

If the force {{math|'''F'''('''r''')}} is ] then the system has a ] {{math|''U''('''r''')}}, so the total ] 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 ] 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}) \\
E_2 & = \frac{\mu}{m_2} E = \frac{1}{2} m_2 \dot{\mathbf{x}}_2^2 + \frac{\mu}{m_2} U(\mathbf{r}) \\
E_\text{tot} & = E_1 + E_2
\end{align}</math>

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

For many physical problems, the force {{math|1='''F'''('''r''')}} is a ], 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 ]. We now have:
<math display="block">\mu \ddot{\mathbf{r}} = {F}(r) \hat{\mathbf{r}} \ ,</math>
where {{math|''F''(''r'')}} is negative in the case of an attractive force.


==See also== ==See also==
* ]
* ]
* ]
* ]
* ]
* ]
* ]
* ]

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

==Bibliography==

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

== External links ==

* at ]


{{Authority control}}
*]
*]


{{DEFAULTSORT:Two-Body Problem}}
]
] ]
]
]

Latest revision as of 03:40, 2 December 2024

Motion problem in classical mechanics This article is about the two-body problem in classical mechanics. For the relativistic version, see Two-body problem in general relativity. For the career management problem of working couples, see Two-body problem (career). Left: Two bodies of similar mass orbiting a common barycenter external to both bodies, with elliptic orbits. This model is typical of binary stars.
Right: Two bodies with a "slight" difference in mass orbiting a common barycenter. Their sizes and this type of orbit are similar to the Pluto–Charon system (in which the barycenter is external to both bodies), as well as the EarthMoon system (in which the barycenter is internal to the larger body).
Part of a series on
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In classical mechanics, the two-body problem is to calculate and predict the motion of two massive bodies that are orbiting each other in space. The problem assumes that the two bodies are point particles that 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 Kepler problem), arising in astronomy for predicting the orbits (or escapes from orbit) of objects such as satellites, planets, 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 "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 three-body problem (and, more generally, the n-body problem for n ≥ 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 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 sections. 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 Classical central-force problem or Kepler problem.

In principle, the same solutions apply to macroscopic problems involving objects interacting not only through gravity, but through any other attractive scalar force field obeying an inverse-square law, with 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 equation.

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.

Electrons in an atom are sometimes described as "orbiting" its nucleus, following an early conjecture of Niels Bohr (this is the source of the term "orbital"). However, electrons don't actually orbit nuclei in any meaningful sense, and 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

See also: Classical central-force problem § Relation to the classical two-body problem See also: Kepler problem
This section duplicates the scope of other articles, specifically Classical central-force problem#Relation to the classical two-body problem. Please discuss this issue and help introduce a summary style to the section by replacing the section with a link and a summary or by splitting the content into a new article. (June 2019)

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.

Jacobi coordinates for two-body problem; Jacobi coordinates are R = m 1 M x 1 + m 2 M x 2 {\displaystyle {\boldsymbol {R}}={\frac {m_{1}}{M}}{\boldsymbol {x}}_{1}+{\frac {m_{2}}{M}}{\boldsymbol {x}}_{2}} and r = x 1 x 2 {\displaystyle {\boldsymbol {r}}={\boldsymbol {x}}_{1}-{\boldsymbol {x}}_{2}} with M = m 1 + m 2   {\displaystyle M=m_{1}+m_{2}\ } .

Let x1 and x2 be the vector positions of the two bodies, and m1 and m2 be their masses. The goal is to determine the trajectories x1(t) and x2(t) for all times t, given the initial positions x1(t = 0) and x2(t = 0) and the initial velocities v1(t = 0) and v2(t = 0).

When applied to the two masses, Newton's second law states that

F 12 ( x 1 , x 2 ) = m 1 x ¨ 1 {\displaystyle \mathbf {F} _{12}(\mathbf {x} _{1},\mathbf {x} _{2})=m_{1}{\ddot {\mathbf {x} }}_{1}} Equation 1
F 21 ( x 1 , x 2 ) = m 2 x ¨ 2 {\displaystyle \mathbf {F} _{21}(\mathbf {x} _{1},\mathbf {x} _{2})=m_{2}{\ddot {\mathbf {x} }}_{2}} Equation 2

where F12 is the force on mass 1 due to its interactions with mass 2, and F21 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 (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 r = x1x2 between the masses changes with time. The solutions of these independent one-body problems can be combined to obtain the solutions for the trajectories x1(t) and x2(t).

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

Let R {\displaystyle \mathbf {R} } be the position of the center of mass (barycenter) of the system. Addition of the force equations (1) and (2) yields m 1 x ¨ 1 + m 2 x ¨ 2 = ( m 1 + m 2 ) R ¨ = F 12 + F 21 = 0 {\displaystyle 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} where we have used Newton's third law F12 = −F21 and where R ¨ m 1 x ¨ 1 + m 2 x ¨ 2 m 1 + m 2 . {\displaystyle {\ddot {\mathbf {R} }}\equiv {\frac {m_{1}{\ddot {\mathbf {x} }}_{1}+m_{2}{\ddot {\mathbf {x} }}_{2}}{m_{1}+m_{2}}}.}

The resulting equation: R ¨ = 0 {\displaystyle {\ddot {\mathbf {R} }}=0} shows that the velocity v = d R d t {\displaystyle \mathbf {v} ={\frac {dR}{dt}}} of the center of mass is constant, from which follows that the total momentum m1 v1 + m2 v2 is also constant (conservation of momentum). Hence, the position 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 r ¨ = x ¨ 1 x ¨ 2 = ( F 12 m 1 F 21 m 2 ) = ( 1 m 1 + 1 m 2 ) F 12 {\displaystyle {\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}} where we have again used Newton's third law F12 = −F21 and where r is the 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 r and not of their absolute positions x1 and x2; 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: μ r ¨ = F 12 ( x 1 , x 2 ) = F ( r ) {\displaystyle \mu {\ddot {\mathbf {r} }}=\mathbf {F} _{12}(\mathbf {x} _{1},\mathbf {x} _{2})=\mathbf {F} (\mathbf {r} )} where μ {\displaystyle \mu } is the reduced mass μ = 1 1 m 1 + 1 m 2 = m 1 m 2 m 1 + m 2 . {\displaystyle \mu ={\frac {1}{{\frac {1}{m_{1}}}+{\frac {1}{m_{2}}}}}={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}.}

Solving the equation for r(t) is the key to the two-body problem. The solution depends on the specific force between the bodies, which is defined by F ( r ) {\displaystyle \mathbf {F} (\mathbf {r} )} . For the case where F ( r ) {\displaystyle \mathbf {F} (\mathbf {r} )} follows an inverse-square law, see the Kepler problem.

Once R(t) and r(t) have been determined, the original trajectories may be obtained x 1 ( t ) = R ( t ) + m 2 m 1 + m 2 r ( t ) {\displaystyle \mathbf {x} _{1}(t)=\mathbf {R} (t)+{\frac {m_{2}}{m_{1}+m_{2}}}\mathbf {r} (t)} x 2 ( t ) = R ( t ) m 1 m 1 + m 2 r ( t ) {\displaystyle \mathbf {x} _{2}(t)=\mathbf {R} (t)-{\frac {m_{1}}{m_{1}+m_{2}}}\mathbf {r} (t)} 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 p and the angular momentum L of the system, with respect to the center of mass, by the equations L = r × p = r × μ d r d t , {\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} =\mathbf {r} \times \mu {\frac {d\mathbf {r} }{dt}},}

where μ is the reduced mass and r is the relative position r2r1 (with these written taking the center of mass as the origin, and thus both parallel to r) the rate of change of the angular momentum L equals the net torque N N = d L d t = r ˙ × μ r ˙ + r × μ r ¨   , {\displaystyle \mathbf {N} ={\frac {d\mathbf {L} }{dt}}={\dot {\mathbf {r} }}\times \mu {\dot {\mathbf {r} }}+\mathbf {r} \times \mu {\ddot {\mathbf {r} }}\ ,} and using the property of the vector cross product that v × w = 0 for any vectors v and w pointing in the same direction,

N   =   d L d t = r × F   , {\displaystyle \mathbf {N} \ =\ {\frac {d\mathbf {L} }{dt}}=\mathbf {r} \times \mathbf {F} \ ,} with F = μdr/dt.

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

Energy of the two-body system

If the force F(r) is conservative then the system has a potential energy U(r), so the total energy can be written as E tot = 1 2 m 1 x ˙ 1 2 + 1 2 m 2 x ˙ 2 2 + U ( r ) = 1 2 ( m 1 + m 2 ) R ˙ 2 + 1 2 μ r ˙ 2 + U ( r ) {\displaystyle 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} )}

In the center of mass frame the kinetic energy is the lowest and the total energy becomes E = 1 2 μ r ˙ 2 + U ( r ) {\displaystyle E={\frac {1}{2}}\mu {\dot {\mathbf {r} }}^{2}+U(\mathbf {r} )} The coordinates x1 and x2 can be expressed as x 1 = μ m 1 r {\displaystyle \mathbf {x} _{1}={\frac {\mu }{m_{1}}}\mathbf {r} } x 2 = μ m 2 r {\displaystyle \mathbf {x} _{2}=-{\frac {\mu }{m_{2}}}\mathbf {r} } and in a similar way the energy E is related to the energies E1 and E2 that separately contain the kinetic energy of each body: E 1 = μ m 1 E = 1 2 m 1 x ˙ 1 2 + μ m 1 U ( r ) E 2 = μ m 2 E = 1 2 m 2 x ˙ 2 2 + μ m 2 U ( r ) E tot = E 1 + E 2 {\displaystyle {\begin{aligned}E_{1}&={\frac {\mu }{m_{1}}}E={\frac {1}{2}}m_{1}{\dot {\mathbf {x} }}_{1}^{2}+{\frac {\mu }{m_{1}}}U(\mathbf {r} )\\E_{2}&={\frac {\mu }{m_{2}}}E={\frac {1}{2}}m_{2}{\dot {\mathbf {x} }}_{2}^{2}+{\frac {\mu }{m_{2}}}U(\mathbf {r} )\\E_{\text{tot}}&=E_{1}+E_{2}\end{aligned}}}

Central forces

Main article: Classical central-force problem

For many physical problems, the force F(r) is a central force, i.e., it is of the form F ( r ) = F ( r ) r ^ {\displaystyle \mathbf {F} (\mathbf {r} )=F(r){\hat {\mathbf {r} }}} where r = |r| and = r/r is the corresponding unit vector. We now have: μ r ¨ = F ( r ) r ^   , {\displaystyle \mu {\ddot {\mathbf {r} }}={F}(r){\hat {\mathbf {r} }}\ ,} where F(r) is negative in the case of an attractive force.

See also

References

  1. Luo, Siwei (22 June 2020). "The Sturm-Liouville problem of two-body system". Journal of Physics Communications. 4 (6): 061001. Bibcode:2020JPhCo...4f1001L. doi:10.1088/2399-6528/ab9c30.
  2. David Betounes (2001). Differential Equations. Springer. p. 58; Figure 2.15. ISBN 0-387-95140-7.

Bibliography

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