Two-body problem Details, Meaning Two-body problem Article and Explanation Guide

Two-body problem Guide, Meaning , Facts, Information and Description

In mechanics, the two-body problem is a special case of the n-body 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.

Table of contents

1 Statement of problem
2 Sketch of solution
3 Reduction to a single body problem
4 Reduction to two dimensions
5 Change of variables
6 Gravity
7 See also

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 and be the positions of the two bodies, and and be their masses, we have (from Newton's second law):

Note that these two vector equations comprise six scalar differential equations, each of second order. Therefore, we will need twelve () constants of integration.

Sketch of solution

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 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

where

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 one-body 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 two-body 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

We now multiply the first equation by , the second by , and subtract, giving

where

and

is the reduced mass of the system.

The positions of the bodies are m2/(m1+m2) and -m1/(m1+m2) times x, respectively.

Thus we have reduced the problem to a one-body problem.

Reduction to two dimensions

Starting with the one-body differential equation above, we take the cross product with the linear momentum

to get

But the first term on the right is a scalar, and the second term is the angular momentum

which, by conservation of angular momentum is a constant.

always points in the same direction, which means that the plane determined by and 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 r-component of acceleration is

Therefore, we have

Another change of variables is useful: let .

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.

For example, consider two bodies like the Sun orbiting each other:

at a distance of 1 AU: the orbital period is year, the same as the orbital period of the Earth would be if the Sun would have twice its actual mass
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 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 times the usual distance of geosynchronous orbits would be geosynchronous.