# Binet equation

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The Binet equation, derived by Jacques Philippe Marie Binet, provides the form of a central force given the shape of the orbital motion in plane polar coordinates. The equation can also be used to derive the shape of the orbit for a given force law, but this usually involves the solution to a second order nonlinear ordinary differential equation. A unique solution is impossible in the case of circular motion about the center of force.

## Equation

The shape of an orbit is often conveniently described in terms of relative distance $r$ as a function of angle $\theta$ . For the Binet equation, the orbital shape is instead more concisely described by the reciprocal $u=1/r$ as a function of $\theta$ . Define the specific angular momentum as $h=L/m$ where $L$ is the angular momentum and $m$ is the mass. The Binet equation, derived in the next section, gives the force in terms of the function $u(\theta )$ :

$F({u}^{-1})=-mh^{2}u^{2}\left({\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u\right).$ ## Derivation

Newton's Second Law for a purely central force is

$F(r)=m({\ddot {r}}-r{\dot {\theta }}^{2}).$ The conservation of angular momentum requires that

$r^{2}{\dot {\theta }}=h={\text{constant}}.$ Derivatives of $r$ with respect to time may be rewritten as derivatives of $u=1/r$ with respect to angle:

{\begin{aligned}&{\frac {\mathrm {d} u}{\mathrm {d} \theta }}={\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {1}{r}}\right){\frac {\mathrm {d} t}{\mathrm {d} \theta }}=-{\frac {\dot {r}}{r^{2}{\dot {\theta }}}}=-{\frac {\dot {r}}{h}}\\&{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}=-{\frac {1}{h}}{\frac {\mathrm {d} {\dot {r}}}{\mathrm {d} t}}{\frac {\mathrm {d} t}{\mathrm {d} \theta }}=-{\frac {\ddot {r}}{h{\dot {\theta }}}}=-{\frac {\ddot {r}}{h^{2}u^{2}}}\\\end{aligned}} Combining all of the above, we arrive at

$F=m({\ddot {r}}-r{\dot {\theta }}^{2})=-m\left(h^{2}u^{2}{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+h^{2}u^{3}\right)=-mh^{2}u^{2}\left({\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u\right)$ ## 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

$-ku^{2}=-mh^{2}u^{2}\left({\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u\right)$ ${\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {k}{mh^{2}}}\equiv {\text{constant}}>0.$ If the angle $\theta$ is measured from the periapsis, then the general solution for the orbit expressed in (reciprocal) polar coordinates is

$lu=1+\varepsilon \cos \theta .$ The above polar equation describes conic sections, with $l$ the semi-latus rectum (equal to $h^{2}/\mu =h^{2}m/k$ ) and $\varepsilon$ the orbital eccentricity.

#### Relativistic

The relativistic equation derived for Schwarzschild coordinates is

${\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {r_{s}c^{2}}{2h^{2}}}+{\frac {3r_{s}}{2}}u^{2}$ where $c$ is the speed of light and $r_{s}$ is the Schwarzschild radius. And for Reissner–Nordström metric we will obtain

${\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {r_{s}c^{2}}{2h^{2}}}+{\frac {3r_{s}}{2}}u^{2}-{\frac {GQ^{2}}{4\pi \varepsilon _{0}c^{4}}}\left({\frac {c^{2}}{h^{2}}}u+2u^{3}\right)$ where $Q$ is the electric charge and $\varepsilon _{0}$ is the vacuum permittivity.

### Inverse Kepler problem

Consider the inverse Kepler problem. What kind of force law produces a noncircular elliptical orbit (or more generally a noncircular conic section) around a focus of the ellipse?

Differentiating twice the above polar equation for an ellipse gives

$l\,{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}=-\varepsilon \cos \theta .$ The force law is therefore

$F=-mh^{2}u^{2}\left({\frac {-\varepsilon \cos \theta }{l}}+{\frac {1+\varepsilon \cos \theta }{l}}\right)=-{\frac {mh^{2}u^{2}}{l}}=-{\frac {mh^{2}}{lr^{2}}},$ which is the anticipated inverse square law. Matching the orbital $h^{2}/l=\mu$ to physical values like $GM$ or $k_{e}q_{1}q_{2}/m$ reproduces Newton's law of universal gravitation or Coulomb's law, respectively.

The effective force for Schwarzschild coordinates is

$F=-GMmu^{2}\left(1+3\left({\frac {hu}{c}}\right)^{2}\right)=-{\frac {GMm}{r^{2}}}\left(1+3\left({\frac {h}{rc}}\right)^{2}\right)$ .

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).

In the parameterized post-Newtonian formalism we will obtain

$F=-{\frac {GMm}{r^{2}}}\left(1+(2+2\gamma -\beta )\left({\frac {h}{rc}}\right)^{2}\right)$ .

where $\gamma =\beta =1$ for the general relativity and $\gamma =\beta =0$ in the classical case.

### Cotes spirals

An inverse cube force law has the form

$F(r)=-{\frac {k}{r^{3}}}.$ 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

${\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u={\frac {ku}{mh^{2}}}=Cu.$ The differential equation has three kinds of solutions, in analogy to the different conic sections of the Kepler problem. When $C<1$ , the solution is the epispiral, including the pathological case of a straight line when $C=0$ . When $C=1$ , the solution is the hyperbolic spiral. When $C>1$ 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 $D$ is

$D\,u(\theta )=\sec \theta .$ Differentiating $u$ twice and making use of the Pythagorean identity gives

$D\,{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}=\sec \theta \tan ^{2}\theta +\sec ^{3}\theta =\sec \theta (\sec ^{2}\theta -1)+\sec ^{3}\theta =2D^{3}u^{3}-D\,u.$ The force law is thus

$F=-mh^{2}u^{2}\left(2D^{2}u^{3}-u+u\right)=-2mh^{2}D^{2}u^{5}=-{\frac {2mh^{2}D^{2}}{r^{5}}}.$ Note that solving the general inverse problem, i.e. constructing the orbits of an attractive $1/r^{5}$ force law, is a considerably more difficult problem because it is equivalent to solving

${\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2}}}+u=Cu^{3}$ which is a second order nonlinear differential equation.