Definitions

# Two-body problem

[too-bod-ee]

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 stars orbiting each other (a binary star), and a classical electron orbiting an atomic nucleus.

The two-body problem can be re-formulated as two independent one-body problems, which involve 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 for n>3) cannot be solved, except in special cases.

## Reduction to two independent, one-body problems

Let x1 and x2 be the 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





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.

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 = x1 − x2 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)

Addition of the force equations (1) and (2) yields


m_{1}ddotmathbf{x}_{1} + m_{2}ddotmathbf{x}_{2} = (m_{1} + m_{2})ddotmathbf{x}_{cm} = mathbf{F}_{12} + mathbf{F}_{21} = 0

where we have used Newton's third law F12 = −F21 and where


mathbf{x}_{cm} equiv frac{m_{1}mathbf{x}_{1} + m_{2}mathbf{x}_{2}}{m_{1} + m_{2}}

is the position of the center of mass (barycenter) of the system. The resulting equation


ddotmathbf{x}_{cm} = 0

shows that the velocity vcm 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 xcm(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


ddot mathbf{r} = ddotmathbf{x}_{1} - ddotmathbf{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 should only be a function of r and not of their absolute positions x1 and x2; otherwise, physics would not have translational symmetry, i.e., the laws of physics would change from place to place. Therefore, the subtracted equation can be written


mu ddotmathbf{r} = mathbf{F}_{12}(mathbf{x}_{1},mathbf{x}_{2}) = mathbf{F}(mathbf{r})

where $mu$ is the reduced mass


mu = frac{1}{frac{1}{m_{1}} + frac{1}{m_{2}}} = frac{m_{1}m_{2}}{m_{1} + m_{2}}

This equation for r(t) is the essential problem to be solved in a two-body problem; general solution methods are described below.

Once xcm(t) and r(t) have been determined, the original trajectories may be obtained


mathbf{x}_{1}(t) = mathbf{x}_{cm}(t) + frac{m_{2}}{m_{1} + m_{2}} mathbf{r}(t)


mathbf{x}_{2}(t) = mathbf{x}_{cm}(t) - frac{m_{1}}{m_{1} + m_{2}} mathbf{r}(t)

as may be verified by substituting the definitions of xcm 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). Defining the linear momentum p and the angular momentum L by the equations


mathbf{L} = mathbf{r} times mathbf{p} = mathbf{r} times mu frac{dmathbf{r}}{dt}

the rate of change of the angular momentum L equals the net torque N


frac{dmathbf{L}}{dt} = dotmathbf{r} times mudotmathbf{r} + mathbf{r} times muddotmathbf{r}

# mathbf{r} times mathbf{F}

mathbf{N}.

Since most physical forces obey Newton's strong third law of motion—which says 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.

## General solution for central forces

For many physical problems, the force F(r) is a central force, i.e., it is of the form

$mathbf\left\{F\right\}\left(mathbf\left\{r\right\}\right) = F\left(r\right)hat\left\{mathbf\left\{r\right\}\right\}$
where r = |r| and = r/r is the corresponding unit vector.

In this case it is often useful to switch to polar coordinates, since the motion is planar. The radial component of the displacement vector equation can be written


F(r) = mu(ddot{r} - r dot{theta}^{2}) = mufrac{d^{2}r}{dt^{2}} - mu r omega^{2} = mufrac{d^{2}r}{dt^{2}} - frac{L^{2}}{mu r^{3}}

where ω is the angular velocity and L = μ r2 ω is the magnitude of the angular momentum L, which is constant by the conservation of angular momentum. Therefore the independent variable can be changed from t to θ


frac{d}{dt} = frac{L}{mu r^{2}} frac{d}{dtheta}

giving the new equation of motion


frac{L}{r^{2}} frac{d}{dtheta} left(frac{L}{mu r^{2}} frac{dr}{dtheta} right)- frac{L^{2}}{mu r^{3}} = F(r).

This equation becomes quasilinear on making the change of variables u = 1/r


frac{d^{2}u}{dtheta^{2}} + u = -frac{mu}{L^{2}u^{2}} F(1/u)

## Bertrand's theorem

Bertrand showed that only two types of forces result in closed orbits, F(r) = αr (a linear force) and F(r) = α/r2 (an inverse square law). A closed orbit is one that is re-entrant: it returns to its starting position after a finite time with exactly the same velocity; hence, it executes exactly the same motion over and over again. One criterion for this is that period for oscillating radially must equal a rational number times the period for rotating around the orbit. This commensurability of period may be true in special cases for other force laws, but it is generally true for the two special laws cited above.

## Solving central-force problems in terms of known functions

This equation for u(θ) can be solved numerically for nearly any central force F(1/u). However, only a handful of forces result in formulae for u in terms of known functions of θ. Such problems are said to be integrable, since they relate to solving an integral


theta = int^{r} frac{dr}{r^{2}} frac{1}{sqrt{C - u^{2} - frac{2}{h^{2}} int^{r} F(r) dr}}

where C is a constant of integration. This integral derives from integrating the equation for u once


left(frac{du}{dtheta} right)^{2} = C - u^{2} - frac{2}{h^{2}} int^{r} F(r) dr

If the force is a power law, i.e., if F(r) = G rn, then u can be expressed in terms of circular functions and/or elliptic functions if and only if n equals 1, -2, -3 (circular functions) and -7, -5, -4, 0, 3, 5, -3/2, -5/2, -1/3, -5/3 and -7/3 (elliptic functions). Similarly, there are only eight possible linear combinations of power laws that give solutions in terms of circular and elliptic functions


F(r) = Ar^{-3} + Br


F(r) = Ar^{-3} + Br^{-2}


F(r) = Ar^{-3} + Br + Cr^{3} + Dr^{5}


F(r) = Ar^{-3} + Br + Cr^{-5} + Dr^{-7}


F(r) = Ar^{-3} + Br^{-2} + Cr + D


F(r) = Ar^{-3} + Br^{-2} + Cr^{-4} + Dr^{-5}


F(r) = Ar^{-3} + Br^{-2} + Cr^{-3/2} + Dr^{-5/2}


F(r) = Ar^{-3} + Br^{-1/3} + Cr^{-5/3} + Dr^{-7/3}

The first two are special cases of the next four, and always result in circular functions.

## Newton's theorem of revolving orbits

The term r−3 occurs in all the force laws above, indicating that the addition of the inverse-cube force does not influence the solubility of the problem in terms of known functions. More generally, Newton showed that, with adjustments in the initial conditions, the addition of such a force does not affect the radial motion of the particle, but multiplies its angular motion by a constant factor. An extension of Newton's theorem was discovered in 2000 by Mahomed and Vawda.

## Bonnet's theorem

Bonnet's theorem states that if the same orbit would be produced by n different types of forces under different initial conditions of velocity, then the same orbit can be produced by a ''linear combination of the same forces, if the initial velocity condition is chosen judiciously.

## Inverse-square force laws: the Kepler problem

If $F$ is an inverse-square law central force such as gravity or electrostatics in classical physics


F = frac{alpha}{r^{2}} = alpha u^{2}

for some constant $alpha$ (negative for an attractive force, positive for a repulsive one), the trajectory equation becomes linear


frac{d^{2}u}{dtheta^{2}} + u = -frac{alpha mu}{L^{2}}.

The solution of this equation is


u(theta) equiv frac{1}{r(theta)} = -frac{alpha mu}{L^{2}} + A cos(theta - theta_{0})

where $A$ and $theta_\left\{0\right\}$ are constants, $Age0$ and for $alpha > 0$ (repulsive force) the additional requirement $Agefrac\left\{alpha mu\right\}\left\{L^\left\{2\right\}\right\}$. This solution, applicable for values of $theta$ for which u > 0, shows that the orbit is a conic section, i.e., an ellipse, a hyperbola or parabola, depending on whether $A$ is less than, greater than, or equal to $-frac\left\{alpha mu\right\}\left\{L^\left\{2\right\}\right\}$, and a straight line for $alpha = 0$.

This inverse-square law case of the two-body problem is called the Kepler problem.

## Work

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.