Twobody problem
From Academic Kids

In mechanics, the twobody problem is a special case of the nbody problem that admits a closed form solution. The most commonly encountered version of the problem, involving an inverse square law force, is encountered in celestial mechanics and the Bohr model of the hydrogen atom. This problem was first solved by Isaac Newton.
This article deals with the general case where it is not assumed that one body has a much smaller mass than the other one.
Contents 
Statement of problem
We restrict ourselves to the classical case, with forces that depend only on the positions of the bodies and obey the strong form of Newton's third law.
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 Newton's second law):
 <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 vector equations comprise six scalar differential equations, each of second order. Therefore, we will need twelve (<math>6 \times 2<math>) constants of integration.
Sketch of solution
We start by taking advantage of Newton's third law to reduce the twobody problem to two equivalent onebody problems, one for the center of mass of the system, and one for the relative motion of the two bodies.
We can identify linear 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 center of mass (barycenter) of the system. Integration shows:
 the total momentum is constant (conservation of momentum)
 the center of mass remains at rest, or moves in a straight line at a constant velocity (see also Motion of the center of mass); this provides six of the constants of integration.
Next, we notice that because of conservation of angular momentum, the equivalent onebody problem is really a two dimensional problem. This provides two more constants. At this point it is convenient to switch to polar coordinates.
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 twobody problem.
Reduction to a single body problem
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>
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 reduced mass of the system.
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.
Thus we have reduced the problem to a onebody problem.
Reduction to two dimensions
Starting with the onebody differential equation above, we take the cross product with the linear momentum
 <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 angular momentum
 <math>\mathbf{L} = \mathbf{x} \times \mathbf{p}<math>
which, by conservation of angular momentum 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.
Change of variables
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.
It can be shown that rcomponent of acceleration is
 <math>\ddot{r}  r \dot{\theta}^{2}<math>
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>.
Newtonian Gravity
Applying the gravitational formula we get that the position of the first body with respect to the second is governed by the same differential eqation 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/μ=m1+m2.
Assume:
 the vector r is the position of one body relative to the other (above called x)
 r, v, the semimajor axis a, and the specific relative angular momentum h are defined accordingly (hence r is the distance)
 <math>\mu={G}(m_1+m_2)\,<math> the standard gravitational parameter (the sum of those for each mass)
where:
 <math>m_1<math> and <math>m_2<math> are the masses of the two bodies.
Then:
 the orbit equation 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 similar conic sections; the same ratios apply for the velocities, and, without the minus, for the angular momentum with respect to the barycenter and for the kinetic energies
 for circular orbits <math>rv^2 = r^3 \omega^2 = 4 \pi^2 r^3/T^2 = \mu<math>
 for elliptic orbits: <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 parabolic trajectories <math>r v^2<math> is constant and equal to <math>2 \mu<math>
 h is the total angular momentum divided by the reduced mass
 the specific orbital energy 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 semimajor axis times the absolute value of the specific orbital energy
For example, consider two bodies like the Sun orbiting each other:
 the reduced mass is one half of the mass of one Sun (one quarter of the total mass)
 at a distance of 1 AU: the orbital period 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 EarthSun 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 EarthSun 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[3]{2} \approx 1.26<math> AU: the orbital period is 1 year, the same as the orbital period of the Earth around the Sun
Similarly, a second Earth at a distance from the Earth equal to <math>\sqrt[3]{2}<math> times the usual distance of geosynchronous orbits would be geosynchronous.
General Relativistic Gravity
In the general theory of relativity 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 precessing 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 Einstein, and the rigorous two body problem was later solved by Howard Percy Robertson.
Examples
 a binary star, e.g. Alpha Centauri (approx. the same mass)
 a double planet, e.g. Pluto with its moon Charon (mass ratio 0.147)
 a binary asteroid, e.g. 90 Antiope (approx. the same mass)