The shape of an orbit is often conveniently described in terms of relative distance as a function of angle . For the Binet equation, the orbital shape is instead more concisely described by the reciprocal as a function of . Define the specific angular momentum as where is the angular momentum and is the mass. The Binet equation, derived in the next section, gives the force in terms of the function :
Derivatives of with respect to time may be rewritten as derivatives of with respect to angle:
Combining all of the above, we arrive at
The general solution is [1] where is the initial coordinate of the particle.
Examples
Kepler problem
Classical
The traditional Kepler problem of calculating the orbit of an inverse square law may be read off from the Binet equation as the solution to the differential equation
If the angle is measured from the periapsis, then the general solution for the orbit expressed in (reciprocal) polar coordinates is
Differentiating twice the above polar equation for an ellipse gives
The force law is therefore
which is the anticipated inverse square law. Matching the orbital to physical values like or reproduces Newton's law of universal gravitation or Coulomb's law, respectively.
The effective force for Schwarzschild coordinates is[3]
where the second term is an inverse-quartic force corresponding to quadrupole effects such as the angular shift of periapsis (It can be also obtained via retarded potentials[4]).
The shapes of the orbits of an inverse cube law are known as Cotes spirals. The Binet equation shows that the orbits must be solutions to the equation
The differential equation has three kinds of solutions, in analogy to the different conic sections of the Kepler problem. When , the solution is the epispiral, including the pathological case of a straight line when . When , the solution is the hyperbolic spiral. When the solution is Poinsot's spiral.
Off-axis circular motion
Although the Binet equation fails to give a unique force law for circular motion about the center of force, the equation can provide a force law when the circle's center and the center of force do not coincide. Consider for example a circular orbit that passes directly through the center of force. A (reciprocal) polar equation for such a circular orbit of diameter is
Note that solving the general inverse problem, i.e. constructing the orbits of an attractive force law, is a considerably more difficult problem because it is equivalent to solving
which is a second order nonlinear differential equation.