Inclassical mechanics, theKepler problem is a special case of thetwo-body problem, in which the two bodies interact by acentral force that varies in strength as theinverse square of the distance between them. The force may be either attractive or repulsive. The problem is to find the position or speed of the two bodies over time given theirmasses,positions, andvelocities. Using classical mechanics, the solution can be expressed as aKepler orbit using sixorbital elements.

The Kepler problem is named afterJohannes Kepler, who proposedKepler's laws of planetary motion (which are part ofclassical mechanics and solved the problem for the orbits of the planets) and investigated the types of forces that would result in orbits obeying those laws (calledKepler's inverse problem).^[1]

For a discussion of the Kepler problem specific to radial orbits, seeRadial trajectory.General relativity provides more accurate solutions to the two-body problem, especially in stronggravitational fields.

The inverse square law behind the Kepler problem is the most important central force law.^[1]: 92The Kepler problem is important incelestial mechanics, sinceNewtonian gravity obeys aninverse square law. Examples include a satellite moving about a planet, a planet about its sun, or two binary stars about each other. The Kepler problem is also important in the motion of two charged particles, sinceCoulomb’s law ofelectrostatics also obeys aninverse square law.

The Kepler problem and thesimple harmonic oscillator problem are the two most fundamental problems inclassical mechanics. They are theonly two problems that have closed orbits for every possible set of initial conditions, i.e., return to their starting point with the same velocity (Bertrand's theorem).^[1]: 92

The Kepler problem also conserves theLaplace–Runge–Lenz vector, which has since been generalized to include other interactions. The solution of the Kepler problem allowed scientists to show that planetary motion could be explained entirely by classical mechanics andNewton’s law of gravity; the scientific explanation of planetary motion played an important role in ushering in theEnlightenment.

The Kepler problem begins with the empirical results ofJohannes Kepler arduously derived by analysis of the astronomical observations ofTycho Brahe. After some 70 attempts to match the data to circular orbits, Kepler hit upon the idea of theelliptic orbit. He eventually summarized his results in the form ofthree laws of planetary motion.^[2]

What is now called the Kepler problem was first discussed byIsaac Newton as a major part of his Principia. His "Theorema I" begins with the first two of his three axioms orlaws of motion and results in Kepler's second law of planetary motion. Next Newton proves his "Theorema II" which shows that if Kepler's second law results, then the force involved must be along the line between the two bodies. In other words, Newton proves what today might be called the "inverse Kepler problem": the orbit characteristics require the force to depend on the inverse square of the distance.^[3]: 107

Thecentral forceF between two objects varies in strength as theinverse square of the distancer between them:

${\displaystyle \mathbf{F}=\frac{k}{r^2}\hat{\mathbf{r}}}$

wherek is a constant and${\displaystyle \hat{\mathbf{r}}}$ represents theunit vector along the line between them.^[4] The force may be either attractive (k < 0) or repulsive (k > 0). The correspondingscalar potential is:

${\displaystyle V(r)=\frac{k}{r}}$

The equation of motion for the radius${\displaystyle r}$ of a particle of mass${\displaystyle m}$ moving in acentral potential${\displaystyle V(r)}$ is given byLagrange's equations

${\displaystyle m\frac{d^{2}r}{d{t}^2}-mr{\omega }^2=m\frac{d^{2}r}{d{t}^2}-\frac{L^2}{m{r}^3}=-\frac{dV}{dr}}$

${\displaystyle \omega \equiv \frac{d\theta }{dt}}$ and theangular momentum${\displaystyle L=m{r}^2\omega }$ is conserved. For illustration, the first term on the left-hand side is zero for circular orbits, and the applied inwards force${\displaystyle \frac{dV}{dr}}$ equals thecentripetal force requirement${\displaystyle mr{\omega }^2}$, as expected.

IfL is not zero the definition ofangular momentum allows a change of independent variable from${\displaystyle t}$ to${\displaystyle \theta }$

${\displaystyle \frac{d}{dt}=\frac{L}{m{r}^2}\frac{d}{d\theta }}$

giving the new equation of motion that is independent of time

${\displaystyle \frac{L}{r^2}\frac{d}{d\theta }\left(\frac{L}{m{r}^2}\frac{dr}{d\theta }\right)-\frac{L^2}{m{r}^3}=-\frac{dV}{dr}}$

The expansion of the first term is

${\displaystyle \frac{L}{r^2}\frac{d}{d\theta }\left(\frac{L}{m{r}^2}\frac{dr}{d\theta }\right)=-\frac{2L^2}{m{r}^5}{\left(\frac{dr}{d\theta }\right)}^2+\frac{L^2}{m{r}^4}\frac{d^{2}r}{d{\theta }^2}}$

This equation becomes quasilinear on making the change of variables${\displaystyle u\equiv \frac{1}{r}}$ and multiplying both sides by${\displaystyle \frac{m{r}^2}{L^2}}$

${\displaystyle \frac{du}{d\theta }=\frac{-1}{r^2}\frac{dr}{d\theta }}$

${\displaystyle \frac{d^{2}u}{d{\theta }^2}=\frac{2}{r^3}{\left(\frac{dr}{d\theta }\right)}^2-\frac{1}{r^2}\frac{d^{2}r}{d{\theta }^2}}$

After substitution and rearrangement:

${\displaystyle \frac{d^{2}u}{d{\theta }^2}+u=-\frac{m}{L^2}\frac{d}{du}V\left(\frac{1}{u}\right)}$

For an inverse-square force law such as thegravitational orelectrostatic potential, thescalar potential can be written

${\displaystyle V(\mathbf{r})=\frac{k}{r}=ku}$

The orbit${\displaystyle u(\theta )}$ can be derived from the general equation

${\displaystyle \frac{d^{2}u}{d{\theta }^2}+u=-\frac{m}{L^2}\frac{d}{du}V\left(\frac{1}{u}\right)=-\frac{km}{L^2}}$

whose solution is the constant${\displaystyle -\frac{km}{L^2}}$ plus a simple sinusoid

${\displaystyle u\equiv \frac{1}{r}=-\frac{km}{L^2}\left[1+e\cos (\theta -{\theta }_0)\right]}$

where${\displaystyle e}$ (theeccentricity) and${\displaystyle {\theta }_0}$ (thephase offset) are constants of integration.

This is the general formula for aconic section that has one focus at the origin;${\displaystyle e=0}$ corresponds to acircle, corresponds to an ellipse,${\displaystyle e=1}$ corresponds to aparabola, and${\displaystyle e>1}$ corresponds to ahyperbola. The eccentricity${\displaystyle e}$ is related to the totalenergy${\displaystyle E}$ (cf. theLaplace–Runge–Lenz vector)

${\displaystyle e=\sqrt{1+\frac{2E{L}^2}{k^{2}m}}}$

Comparing these formulae shows that corresponds to an ellipse (all solutions which areclosed orbits are ellipses),${\displaystyle E=0}$ corresponds to aparabola, and${\displaystyle E>0}$ corresponds to ahyperbola. In particular,${\displaystyle E=-\frac{k^{2}m}{2L^2}}$ for perfectlycircular orbits (the central force exactly equals thecentripetal force requirement, which determines the required angular velocity for a given circular radius).

For a repulsive force (k > 0) onlye > 1 applies.

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