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| {{About|the two-body problem in classical mechanics|the career management problem of working couples|two-body problem (career)}}
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| [[Image:orbit5.gif|thumb|right|400px|Two bodies with similar mass orbiting around a common [[Barycentric coordinates (astronomy)|barycenter]] with elliptic orbits.]]
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| In [[classical mechanics]], the '''two-body problem''' is to determine the motion of two point particles that interact only with each other. Common examples include a [[satellite]] orbiting a [[planet]], a [[planet]] orbiting a [[star]], two [[star]]s orbiting each other (a [[binary star]]), and a classical [[electron]] orbiting an [[atomic nucleus]] (although to solve this system correctly a quantum mechanical approach must be used).
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| The two-body problem can be re-formulated 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. By contrast, the [[three-body problem]] (and, more generally, the [[n-body problem|''n''-body problem]] for ''n'' ≥ 3) cannot be solved in terms of first integrals, except in special cases.
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| [[Image:orbit2.gif|thumb|right|200px|Two bodies with a slight difference in [[mass]] orbiting around a common [[Barycentric coordinates (astronomy)|barycenter]]. The sizes, and this particular type of orbit are similar to the [[Pluto]]-[[Charon (moon)|Charon]] system and also to Earth-Moon system in which the center of mass is inside the bigger body instead.]]
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| ==Reduction to two independent, one-body problems==
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| [[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=http://books.google.com/?id=oNvFAzQXBhsC&pg=PA58 |isbn=0-387-95140-7 |page=58; Figure 2.15 |year=2001 |publisher=Springer}}</ref>]]
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| Let '''x'''<sub>1</sub> and '''x'''<sub>2</sub> be the 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 '''x'''<sub>1</sub>(''t'') and '''x'''<sub>2</sub>(''t'') for all times ''t'', given the initial positions '''x'''<sub>1</sub>(''t'' = 0) and '''x'''<sub>2</sub>(''t'' = 0) and the initial velocities '''v'''<sub>1</sub>(''t'' = 0) and '''v'''<sub>2</sub>(''t'' = 0).
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| When applied to the two masses, [[Newton's laws of motion#Newton's second law|Newton's second law]] states that
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| :<math>
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| \mathbf{F}_{12}(\mathbf{x}_{1},\mathbf{x}_{2}) = m_{1} \ddot{\mathbf{x}}_{1} \quad \quad \quad (\mathrm{Equation} \ 1)
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| </math>
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| :<math>
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| \mathbf{F}_{21}(\mathbf{x}_{1},\mathbf{x}_{2}) = m_{2} \ddot{\mathbf{x}}_{2} \quad \quad \quad (\mathrm{Equation} \ 2)
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| </math>
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| 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.
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| 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]] ([[Barycentric coordinates (astronomy)|barycenter]]) motion. By contrast, ''subtracting'' equation (2) from equation (1) results in an equation that describes how the vector '''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 '''x'''<sub>1</sub>(''t'') and '''x'''<sub>2</sub>(''t'').
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| ===Center of mass motion (1st one-body problem)===
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| Addition of the force equations (1) and (2) yields
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| :<math>
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| 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
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| </math>
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| where we have used [[Newton's laws of motion|Newton's third law]] '''F'''<sub>12</sub> = −'''F'''<sub>21</sub> and where
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| :<math>
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| \ddot{\mathbf{R}} \equiv \frac{m_{1}\ddot{\mathbf{x}}_{1} + m_{2}\ddot{\mathbf{x}}_{2}}{m_{1} + m_{2}}
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| </math>
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| :<math>
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| \mathbf{R}
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| </math> is the position of the [[center of mass]] ([[Barycentric coordinates (astronomy)|barycenter]]) of the system.
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| The resulting equation:
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| :<math>
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| \ddot{\mathbf{R}} = 0
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| </math> | |
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| shows that the velocity '''V''' = ''d'''''R'''/''dt'' of the center of mass is constant, from which follows that the total momentum ''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 '''R''' (''t'') of the center of mass can be determined at all times from the initial positions and velocities.
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| ==Two-body motion is planar==
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| The motion of two bodies with respect to each other always lies in a plane (in the [[center of mass frame]]). Defining the [[linear momentum]] '''p''' and the [[angular momentum]] '''L''' by the equations
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| :<math>
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| \mathbf{L} = \mathbf{r} \times \mathbf{p} = \mathbf{r} \times \mu \frac{d\mathbf{r}}{dt}
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| </math>
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| the rate of change of the angular momentum '''L''' equals the net [[torque]] '''N'''
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| :<math>
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| \mathbf{N} = \frac{d\mathbf{L}}{dt} = \dot{\mathbf{r}} \times \mu\dot{\mathbf{r}} + \mathbf{r} \times \mu\ddot{\mathbf{r}} \ ,
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| </math>
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| and using the property of the [[vector cross product]] that '''v''' × '''w''' = '''0''' for any vectors '''''v''''' and '''''w''''' pointing in the same direction,
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| :<math>
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| \mathbf{N} \ = \ \frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F} \ ,
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| </math>
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| with '''F''' = μ ''d'' <sup>2</sup>'''r''' / ''dt'' <sup>2</sup>.
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| 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 '''r''' × '''F''' = '''0''' and the [[conservation of angular momentum|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'''.
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| == Laws of Conservation of Energy for each of two bodies for arbitrary potentials ==
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| In system of the center of mass for arbitrary potentials
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| :<math>~U_{12} = U(\mathbf{r}_1 - \mathbf{r}_2) </math>
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| :<math>~U_{21} = U(\mathbf{r}_2 - \mathbf{r}_1) </math>
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| the value of [http://vv-voronkov.spb.ru/EN/two-lawe.html energies] of bodies not change:
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| :<math>~E_1 = m_1 \frac{v_1^2}{2} + \frac{m_2} {m_1+m_2} U_{12} = Const_1(t) </math>
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| :<math>~E_2 = m_2 \frac{v_2^2}{2} + \frac{m_1} {m_1+m_2} U_{21} = Const_2(t) </math>
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| ==Central forces==
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| {{main|Classical central-force problem}}
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| For many physical problems, the force '''F'''('''r''') is a [[central force]], i.e., it is of the form
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| :<math>\mathbf{F}(\mathbf{r}) = F(r)\hat{\mathbf{r}}</math>
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| where r = |'''r'''| and '''r̂''' = '''r'''/r is the corresponding [[unit vector]]. We now have: | |
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| :<math>
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| \mu \ddot{\mathbf{r}} = {F}(r) \hat{\mathbf{r}} \ ,
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| </math>
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| where ''F(r)'' is negative in the case of an attractive force.
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| ==Work==
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| The total work done in a given time interval by the forces exerted by two bodies on each other is the same as the work done by one force applied to the total relative displacement.
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| ==See also==
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| * [[Kepler orbit]]
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| * [[Energy drift]]
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| * [[Equation of the center]]
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| * [[Euler's three-body problem]]
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| * [[Gravitational two-body problem]]
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| * [[Kepler problem]]
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| * [[n-body problem|''n''-body problem]]
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| * [[Virial theorem]]
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| * [[Two-body problem (career)]]
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| ==References==
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| {{reflist|1}}
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| ==Bibliography==
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| * {{cite book | author = [[Lev Landau|Landau LD]], [[Evgeny Lifshitz|Lifshitz EM]] | year = 1976 | title = Mechanics | edition = 3rd. | publisher = Pergamon Press | location = New York | isbn = 0-08-029141-4}}
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| * {{cite book | author = [[Herbert Goldstein|Goldstein H]] | year = 1980 | title = [[Classical Mechanics (textbook)|Classical Mechanics]] | edition = 2nd. | publisher = Addison-Wesley | location = New York | isbn = 0-201-02918-9}}
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| ==External links==
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| * [http://scienceworld.wolfram.com/physics/Two-BodyProblem.html Two-body problem] at [[ScienceWorld|Eric Weisstein's World of Physics]]
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| {{DEFAULTSORT:Two-Body Problem}}
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| [[Category:Concepts in physics]]
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| [[Category:Orbits]]
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| [[Category:Classical mechanics]]
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The Easter Eggstravaganza is less than two weeks away. Remember it staying held towards the 11th of April. You will require to attend the now even bigger event. It will be hosted at fresh Eggstravaganza site, Freedom College in Oakley.
Being culturally sensitive are going to make the trip more comfortable for travelers and it is a good idea to also carry footwear that is comfy. May also the all inclusive holidays, may also offer additional advises in this regard.
Green clothing can be boring on its own. So how can you improve your style and continue to be Irish? Automobiles design and deep symbology, stainless steel Celtic jewelry can become your accessory!
Next, check also should the free website template matches your chosen niche. Don't go with free web page templates that look all too serious whenever you are selling baby products. Each morning same light, don't opt for cute templates if you are preparing to target a elder audience in the commercial demographic.
A national day off is to be able to and leads that St. Patrick's Day couldn't come any easily. Annually anticipated, March 17 is marked in every calendar as being a day to rest and make merry.
Outdoor Venue - If have chosen an outdoor venue to be a garden clearly farmhouse, specific there are a couple of pedestal fans and greenery around so that the temperature remains fashionable.
After nailing the end of the first piece of plywood together with stud I worked my way in the board while holding it up. The other end for the plywood was hanging journey house and bouncing a new in the breeze. Once i lined down the nail-gun resistant to the complex wood grain I realized I'd a mistake. I didn't really know the place that the next stud was. Extremely not good enough to hit it having a nail because of the other side of this large piece of plywood.
As doable ! see, you will discover several hiking safety tips adhere to when venturing through the forest. You want to be selected remain as safe as is feasible. Always prepare for the worst, even when it is not going to occur.
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