{{Short description|Equation giving the form of a central force}} {{Classical mechanics}}
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 <math>r</math> as a function of angle <math>\theta</math>. For the Binet equation, the orbital shape is instead more concisely described by the reciprocal <math>u = 1/r</math> as a function of <math>\theta</math>. Define the specific angular momentum as <math>h=L/m</math> where <math>L</math> is the angular momentum and <math>m</math> is the mass. The Binet equation, derived in the next section, gives the force in terms of the function <math> u(\theta) </math>: <math display="block">F(u^{-1}) = -m h^2 u^2 \left(\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^2}+u\right).</math>
==Derivation== Newton's second law for a purely central force is <math display="block">F(r) = m \left(\ddot{r}-r\dot{\theta }^2\right).</math>
The conservation of angular momentum requires that <math display="block">r^{2}\dot{\theta } = h = \text{constant}.</math>
Derivatives of <math>r</math> with respect to time may be rewritten as derivatives of <math>u=1/r</math> with respect to angle: <math display="block">\begin{align} &\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{align} </math>
Combining all of the above, we arrive at <math display="block">F = m\left(\ddot{r}-r\dot{\theta }^2\right) = -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)</math>
The general solution is <ref>{{Cite book |last=Goldstein|first=Herbert |title=Classical mechanics |date=1980 |publisher=Addison-Wesley Pub. Co |isbn=0-201-02918-9 |location=Reading, Mass.|oclc=5675073}}</ref> <math display="block">\theta = \int_{r_0}^r \frac{\mathrm dr}{r^2\sqrt{\frac{2m}{L^2} (E-V) - \frac{1}{r^2}}} + \theta_0</math> where <math>(r_0, \theta_0)</math> is the initial coordinate of the particle, <math>V</math> the potential energy and <math>E</math> the total energy (<math>E=T+V</math>).
==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 <math display="block">-k u^2 = -m h^2 u^2 \left(\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^{2}}+u\right)</math> <math display="block">\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^{2}}+u = \frac{k}{mh^2} \equiv \text{constant}>0.</math>
If the angle <math>\theta</math> is measured from the periapsis, then the general solution for the orbit expressed in (reciprocal) polar coordinates is <math display="block">l u = 1 + \varepsilon \cos\theta.</math>
The above polar equation describes conic sections, with <math>l</math> the semi-latus rectum (equal to <math>h^2/\mu = h^2m/k</math>) and <math>\varepsilon</math> the orbital eccentricity.
==== Relativistic ==== The relativistic equation derived for a De Sitter–Schwarzschild metric is <ref>{{Cite journal |last=Alvarez-Perez |first=Jose Luis |title=The equation of Binet in classical and relativistic orbital mechanics |journal=European Journal of Physics, vol. 7, n. 3 |publisher=IOP Publishing |year=2026 |doi=10.1088/1361-6404/ae5af7 |url=https://iopscience.iop.org/article/10.1088/1361-6404/ae5af7 |access-date=2026-05-13 |arxiv=2512.07485 }}</ref> <math display="block"> \frac{\mathrm{d}^{2}u}{\mathrm{d}\phi^{2}}+u= \begin{cases} \dfrac{r_s c^2}{2h^2} +\dfrac{3}{2}r_s u^2 -\dfrac{\Lambda c^2}{3h^2u^3} & \text{(particle)} \\[1.2em] \dfrac{3}{2}r_s u^2 & \text{(photon)} \end{cases}</math> where <math>c</math> is the speed of light, <math>r_s</math> is the Schwarzschild radius and <math>\Lambda</math> is the cosmological constant. And for Reissner–Nordström metric we will obtain <math display="block"> \frac{\mathrm{d}^{2}u}{\mathrm{d}\phi^{2}}+u= \begin{cases} \dfrac{r_s c^2}{2h^2} +\dfrac{3}{2}r_s u^2 -\dfrac{GQ^{2}}{4\pi\varepsilon_0 c^{4}} \left(\dfrac{c^2}{h^2}u+2u^3\right) & \text{(particle)} \\[1.2em] \dfrac{3}{2}r_s u^2 -\dfrac{GQ^{2}}{2\pi\varepsilon_0 c^{4}}u^3 & \text{(photon)} \end{cases} </math> where <math>Q</math> is the electric charge and <math>\varepsilon_0</math> 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 <math display="block">l \, \frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^2} = - \varepsilon \cos \theta.</math>
The force law is therefore <math display="block">F = -mh^{2}u^{2} \left(\frac{- \varepsilon \cos \theta}{l}+\frac{1 + \varepsilon \cos \theta}{l}\right)=-\frac{m h^2 u^2}{l}=-\frac{m h^2}{l r^2},</math> which is the anticipated inverse square law. Matching the orbital <math>h^2/l = \mu</math> to physical values like <math>GM</math> or <math>k_e q_1 q_2/m</math> reproduces Newton's law of universal gravitation or Coulomb's law, respectively.
The effective force for Schwarzschild coordinates is<ref>{{Cite web |title=The first-order orbital equation |url=http://chaos.swarthmore.edu/courses/PDG07/AJP/AJP000352.pdf |archive-url=https://web.archive.org/web/20120426001843/http://chaos.swarthmore.edu/courses/PDG07/AJP/AJP000352.pdf |archive-date=2012-04-26 |access-date=2026-03-23 |website=Am. J. Phys.|date=2007|last=D’Eliseo|first=Maurizio M.|publisher=American Association of Physics Teachers|via=chaos.swarthmore.edu|url-status=usurped}}</ref> <math display="block">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).</math> 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<ref>{{Cite arXiv |eprint = astro-ph/0306611|last1 = Behera|first1 = Harihar | title = A flat space-time relativistic explanation for the perihelion advance of Mercury|last2 = Naik|first2 = P. C|year = 2003 }}</ref>).
In the parameterized post-Newtonian formalism we will obtain <math display="block">F = -\frac{GMm}{r^2} \left(1+(2+2\gamma-\beta)\left(\frac{h}{rc}\right)^2\right).</math> where <math>\gamma = \beta = 1</math> for the general relativity and <math>\gamma = \beta = 0</math> in the classical case.
===Cotes spirals=== An inverse cube force law has the form <math display="block">F(r) = -\frac{k}{r^3}.</math>
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 <math display="block">\frac{\mathrm{d}^2 u}{\mathrm{d}\theta^2}+u=\frac{k u}{m h^2} = C u.</math>
The differential equation has three kinds of solutions, in analogy to the different conic sections of the Kepler problem. When <math>C < 1</math>, the solution is the epispiral, including the pathological case of a straight line when <math>C = 0</math>. When <math>C = 1</math>, the solution is the hyperbolic spiral. When <math>C > 1</math> 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 <math>D</math> is <math display="block">D \, u(\theta)= \sec \theta.</math>
Differentiating <math>u</math> twice and making use of the Pythagorean identity gives <math display="block">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 = 2 D^3 u^3-D \, u.</math>
The force law is thus <math display="block">F = -mh^2u^2 \left( 2 D^2 u^3- u + u\right) = -2mh^2D^2u^5 = -\frac{2mh^2D^2}{r^5}.</math>
Note that solving the general inverse problem, i.e. constructing the orbits of an attractive <math>1/r^5</math> force law, is a considerably more difficult problem because it is equivalent to solving <math display="block">\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta ^{2}}+u=Cu^3</math>
which is a second order nonlinear differential equation.
==See also== {{Portal|Astronomy|Physics}} *{{slink|Bohr–Sommerfeld quantization#Relativistic orbit}} *Classical central-force problem *General relativity *Two-body problem in general relativity *Bertrand's theorem
==References== {{Reflist}}
{{DEFAULTSORT:Binet Equation}} Category:Classical mechanics