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]

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

where ${\displaystyle c}$ is the speed of light and ${\displaystyle r_{s}}$ is the Schwarzschild radius. And for Reissner–Nordström metric we will obtain

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

where ${\displaystyle Q}$ is the electric charge and ${\displaystyle \varepsilon _{0}}$ 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

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

The force law is therefore

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

which is the anticipated inverse square law. Matching the orbital ${\displaystyle h^{2}/l=\mu }$ to physical values like ${\displaystyle GM}$ or ${\displaystyle k_{e}q_{1}q_{2}/m}$ reproduces Newton's law of universal gravitation or Coulomb's law, respectively.

The effective force for Schwarzschild coordinates is^[3]

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

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]).

In the parameterized post-Newtonian formalism we will obtain

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

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

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.