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${\displaystyle {\textbf {F))={\frac {d}{dt))(m{\textbf {v)))}$ Second law of motion
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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 ${\displaystyle r}$ as a function of angle ${\displaystyle \theta }$. For the Binet equation, the orbital shape is instead more concisely described by the reciprocal ${\displaystyle u=1/r}$ as a function of ${\displaystyle \theta }$. Define the specific angular momentum as ${\displaystyle h=L/m}$ where ${\displaystyle L}$ is the angular momentum and ${\displaystyle m}$ is the mass. The Binet equation, derived in the next section, gives the force in terms of the function ${\displaystyle u(\theta )}$:

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

Derivation

Newton's Second Law for a purely central force is

$${\displaystyle F(r)=m\left({\ddot {r))-r{\dot {\theta ))^{2}\right).}$$

The conservation of angular momentum requires that

$${\displaystyle r^{2}{\dot {\theta ))=h={\text{constant)).}$$

Derivatives of ${\displaystyle r}$ with respect to time may be rewritten as derivatives of ${\displaystyle u=1/r}$ with respect to angle:

$${\displaystyle {\begin{aligned}&{\frac {\mathrm {d} u}{\mathrm {d} \theta ))={\frac {\mathrm {d} }{\mathrm {d} t))\left({\frac {1}{r))\right){\frac {\mathrm {d} t}{\mathrm {d} \theta ))=-{\frac {\dot {r)){r^{2}{\dot {\theta ))))=-{\frac {\dot {r)){h))\\&{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2))}=-{\frac {1}{h)){\frac {\mathrm {d} {\dot {r))}{\mathrm {d} t)){\frac {\mathrm {d} t}{\mathrm {d} \theta ))=-{\frac {\ddot {r)){h{\dot {\theta ))))=-{\frac {\ddot {r)){h^{2}u^{2))}\end{aligned))}$$

Combining all of the above, we arrive at

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

The general solution is ^[1]

$${\displaystyle \theta =\int _{r_{0))^{r}{\frac {\mathrm {d} r}{r^{2}{\sqrt ((\frac {2m}{L^{2))}(E-V)-{\frac {1}{r^{2))))))}+\theta _{0))$$

${\displaystyle (r_{0},\theta _{0})}$

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

$${\displaystyle -ku^{2}=-mh^{2}u^{2}\left({\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2))}+u\right)}$$

$${\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2))}+u={\frac {k}{mh^{2))}\equiv {\text{constant))>0.}$$

If the angle ${\displaystyle \theta }$ is measured from the periapsis, then the general solution for the orbit expressed in (reciprocal) polar coordinates is

$${\displaystyle lu=1+\varepsilon \cos \theta .}$$

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

Relativistic

The relativistic equation derived for Schwarzschild coordinates is^[2]

$${\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2))}+u={\frac {r_{s}c^{2)){2h^{2))}+{\frac {3r_{s)){2))u^{2))$$

${\displaystyle c}$

${\displaystyle r_{s))$

$${\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2))}+u={\frac {r_{s}c^{2)){2h^{2))}+{\frac {3r_{s)){2))u^{2}-{\frac {GQ^{2)){4\pi \varepsilon _{0}c^{4))}\left({\frac {c^{2)){h^{2))}u+2u^{3}\right)}$$

${\displaystyle Q}$

${\displaystyle \varepsilon _{0))$

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

$${\displaystyle l\,{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2))}=-\varepsilon \cos \theta .}$$

The force law is therefore

$${\displaystyle F=-mh^{2}u^{2}\left({\frac {-\varepsilon \cos \theta }{l))+{\frac {1+\varepsilon \cos \theta }{l))\right)=-{\frac {mh^{2}u^{2)){l))=-{\frac {mh^{2)){lr^{2))},}$$

${\displaystyle h^{2}/l=\mu }$

${\displaystyle GM}$

${\displaystyle k_{e}q_{1}q_{2}/m}$

The effective force for Schwarzschild coordinates is^[3]

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

In the parameterized post-Newtonian formalism we will obtain

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

${\displaystyle \gamma =\beta =1}$

${\displaystyle \gamma =\beta =0}$

Cotes spirals

An inverse cube force law has the form

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

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

$${\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2))}+u={\frac {ku}{mh^{2))}=Cu.}$$

The differential equation has three kinds of solutions, in analogy to the different conic sections of the Kepler problem. When ${\displaystyle C<1}$, the solution is the epispiral, including the pathological case of a straight line when ${\displaystyle C=0}$. When ${\displaystyle C=1}$, the solution is the hyperbolic spiral. When ${\displaystyle C>1}$ 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 ${\displaystyle D}$ is

$${\displaystyle D\,u(\theta )=\sec \theta .}$$

Differentiating ${\displaystyle u}$ twice and making use of the Pythagorean identity gives

$${\displaystyle D\,{\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2))}=\sec \theta \tan ^{2}\theta +\sec ^{3}\theta =\sec \theta (\sec ^{2}\theta -1)+\sec ^{3}\theta =2D^{3}u^{3}-D\,u.}$$

The force law is thus

$${\displaystyle F=-mh^{2}u^{2}\left(2D^{2}u^{3}-u+u\right)=-2mh^{2}D^{2}u^{5}=-{\frac {2mh^{2}D^{2)){r^{5))}.}$$

Note that solving the general inverse problem, i.e. constructing the orbits of an attractive ${\displaystyle 1/r^{5))$ force law, is a considerably more difficult problem because it is equivalent to solving

$${\displaystyle {\frac {\mathrm {d} ^{2}u}{\mathrm {d} \theta ^{2))}+u=Cu^{3))$$

which is a second order nonlinear differential equation.