Classical central-force problem
In classical mechanics, the central-force problem is to determine the motion of a particle in a single central potential field. A central force is a force (possibly negative) that points from the particle directly towards a fixed point in space, the center, and whose magnitude only depends on the distance of the object to the center. In many important cases, the problem can be solved analytically, i.e., in terms of well-studied functions such as trigonometric functions. The solution of this problem is important to classical mechanics, since many naturally occurring forces are central. Examples  include gravity and electromagnetism as described by Newton's law of universal gravitation and Coulomb's law, respectively. The problem is also important because some more complicated problems in classical physics (such as the two-body problem with forces along the line connecting the two bodies) can be reduced to a central-force problem. Finally, the solution to the central-force problem often makes a good initial approximation of the true motion, as in calculating the motion of the planets in the Solar System. Basics. The essence of the central-force problem is to solve for the position r of  a particle moving under the influence of a central force F, either as a function of time "t" or as a function of the angle φ relative to the center of force and an arbitrary axis. Definition of a central force. A conservative central force F has two defining properties. First, it must drive particles either directly towards or directly away from a fixed point in space, the center of force, which is often labeled O. In other words, a central force must act along the line joining O with the present position of the particle. Second, a conservative central  force depends only on the distance "r" between O and the moving particle; it does not depend explicitly on time or other descriptors of position. This two-fold definition may be expressed mathematically as follows. The center of force O can be chosen as the origin of a coordinate system. The vector r joining O to the present position of the particle is known as the position vector. Therefore, a central force must have the mathematical form formula_1 where "r" is the vector magnitude |r| (the distance to the center of force) and r̂ = r/r is the corresponding  unit vector. According to Newton's second law of motion, the central force F generates a parallel acceleration a scaled by the mass "m" of the particle formula_2 For attractive forces, "F"("r") is negative, because it works to reduce the distance "r" to the center. Conversely, for repulsive forces, "F"("r") is positive. Potential energy. If the central force is a conservative force, then the magnitude "F"("r") of a central force can always be expressed as the derivative of a time-independent potential energy function "U"("r") formula_3 Thus, the total energy of the particle—the sum of its kinetic energy and  its potential energy "U"—is a constant; energy is said to be conserved. To show this, it suffices that the work "W" done by the force depends only on initial and final positions, not on the path taken between them. formula_4 Equivalently, it suffices that the curl of the force field F is zero; using the formula for the curl in spherical coordinates, formula_5 because the partial derivatives are zero for a central force; the magnitude "F" does not depend on the angular spherical coordinates θ and φ. Since the scalar potential "V"("r") depends only on the distance "r" to the  origin, it has spherical symmetry. In this respect, the central-force problem is analogous to the Schwarzschild geodesics in general relativity and to the quantum mechanical treatments of particles in potentials of spherical symmetry. One-dimensional problem. If the initial velocity v of the particle is aligned with position vector r, then the motion remains forever on the line defined by r. This follows because the force—and by Newton's second law, also the acceleration a—is also aligned with r. To determine this motion, it suffices to solve the equation formula_6 One solution method is to use the  conservation of total energy formula_7 Taking the reciprocal and integrating we get: formula_8 For the remainder of the article, it is assumed that the initial velocity v of the particle is not aligned with position vector r, i.e., that the angular momentum vector L = r × "m" v is not zero. Uniform circular motion. Every central force can produce uniform circular motion, provided that the initial radius "r" and speed "v" satisfy the equation for the centripetal force formula_9 If this equation is satisfied at the initial moments, it will be satisfied at all later times; the particle will  continue to move in a circle of radius "r" at speed "v" forever. Relation to the classical two-body problem. The central-force problem concerns an ideal situation (a "one-body problem") in which a single particle is attracted or repelled from an immovable point O, the center of force. However, physical forces are generally between two bodies; and by Newton's third law, if the first body applies a force on the second, the second body applies an equal and opposite force on the first. Therefore, both bodies are accelerated if a force is present between them; there is  no perfectly immovable center of force. However, if one body is overwhelmingly more massive than the other, its acceleration relative to the other may be neglected; the center of the more massive body may be treated as approximately fixed. For example, the Sun is overwhelmingly more massive than the planet Mercury; hence, the Sun may be approximated as an immovable center of force, reducing the problem to the motion of Mercury in response to the force applied by the Sun. In reality, however, the Sun also moves (albeit only slightly) in response to the force applied by the planet Mercury . Such approximations are unnecessary, however. Newton's laws of motion allow any classical two-body problem to be converted into a corresponding exact one-body problem. To demonstrate this, let x and x be the positions of the two particles, and let r = x − x be their relative position. Then, by Newton's second law, formula_10 The final equation derives from Newton's third law; the force of the second body on the first body (F) is equal and opposite to the force of the first body on the second (F). Thus, the equation of motion for r  can be written in the form formula_11 where formula_12 is the reduced mass formula_13 As a special case, the problem of two bodies interacting by a central force can be reduced to a central-force problem of one body. Qualitative properties. Planar motion. The motion of a particle under a central force F always remains in the plane defined by its initial position and velocity. This may be seen by symmetry. Since the position r, velocity v and force F all lie in the same plane, there is never an acceleration perpendicular to that plane, because that would break the  symmetry between "above" the plane and "below" the plane. To demonstrate this mathematically, it suffices to show that the angular momentum of the particle is constant. This angular momentum L is defined by the equation formula_14 where "m" is the mass of the particle and p is its linear momentum. Therefore, the angular momentum vector L is always perpendicular to the plane defined by the particle's position vector r and velocity vector v. In general, the rate of change of the angular momentum L equals the net torque r × F formula_15 The first term "m" v × v  is always zero, because the vector cross product is always zero for any two vectors pointing in the same or opposite directions. However, when F is a central force, the remaining term r × F is also zero because the vectors r and F point in the same or opposite directions. Therefore, the angular momentum vector L is constant. Then formula_16 Consequently, the particle's position r (and hence velocity v) always lies in a plane perpendicular to L. Polar coordinates. Since the motion is planar and the force radial, it is customary to switch to polar coordinates. In these  coordinates, the position vector r is represented in terms of the radial distance "r" and the azimuthal angle "φ". formula_17 Taking the first derivative with respect to time yields the particle's velocity vector v formula_18 Similarly, the second derivative of the particle's position r equals its acceleration a formula_19 The velocity v and acceleration a can be expressed in terms of the radial and azimuthal unit vectors. The radial unit vector is obtained by dividing the position vector r by its magnitude "r", as described above formula_20 The azimuthal unit vector is given by formula_21 Thus, the velocity  can be written as formula_22 whereas the acceleration equals formula_23 Specific angular momentum. Since F = ma by Newton's second law of motion and since F is a central force, then only the radial component of the acceleration a can be non-zero; the angular component "a" must be zero formula_24 Therefore, formula_25 This expression in parentheses is usually denoted "h" formula_26 which equals the speed "v" times "r", the component of the radius vector perpendicular to the velocity. "h" is the magnitude of the specific angular momentum because it equals the magnitude "L" of the angular momentum divided  by the mass "m" of the particle. For brevity, the angular speed is sometimes written "ω" formula_27 However, it should not be assumed that ω is constant. Since "h" is constant, ω varies with the radius "r" according to the formula formula_28 Since "h" is constant and "r" is positive, the angle "φ" changes monotonically in any central-force problem, either continuously increasing ("h" positive) or continuously decreasing ("h" negative). Constant areal velocity. The magnitude of "h" also equals twice the areal velocity, which is the rate at which area is being swept out by the particle relative to the  center. Thus, the areal velocity is constant for a particle acted upon by any type of central force; this is Kepler's second law. Conversely, if the motion under a conservative force F is planar and has constant areal velocity for all initial conditions of the radius r and velocity v, then the azimuthal acceleration "a" is always zero. Hence, by Newton's second law, F = "m"a, the force is a central force. The constancy of areal velocity may be illustrated by uniform circular and linear motion. In uniform circular motion, the particle moves with constant speed "v" around  the circumference of a circle of radius "r". Since the angular velocity ω = "v"/"r" is constant, the area swept out in a time Δ"t" equals ω "r"Δ"t"; hence, equal areas are swept out in equal times Δ"t". In uniform linear motion (i.e., motion in the absence of a force, by Newton's first law of motion), the particle moves with constant velocity, that is, with constant speed "v" along a line. In a time Δ"t", the particle sweeps out an area "v"Δ"tr" (the impact parameter). The distance "r" does not change as the particle moves along the line; it  represents the distance of closest approach of the line to the center O (the impact parameter). Since the speed "v" is likewise unchanging, the areal velocity "vr" is a constant of motion; the particle sweeps out equal areas in equal times. Equivalent parallel force field. By a transformation of variables, any central-force problem can be converted into an equivalent parallel-force problem. In place of the ordinary "x" and "y" Cartesian coordinates, two new position variables ξ = "x"/"y" and η = 1/"y" are defined, as is a new time coordinate "τ" formula_29 The corresponding equations of motion for  ξ and "η" are given by formula_30 formula_31 Since the rate of change of "ξ" is constant, its second derivative is zero formula_32 Since this is the acceleration in the "ξ" direction and since "F"="ma" by Newton's second law, it follows that the force in the "ξ" direction is zero. Hence the force is only along the "η" direction, which is the criterion for a parallel-force problem. Explicitly, the acceleration in the "η" direction equals formula_33 because the acceleration in the "y"-direction equals formula_34 Here, "F" denotes the "y"-component of the central force, and "y"/"r" equals the cosine  of the angle between the "y"-axis and the radial vector r. General solution. Binet equation. Since a central force F acts only along the radius, only the radial component of the acceleration is nonzero. By Newton's second law of motion, the magnitude of F equals the mass "m" of the particle times the magnitude of its radial acceleration formula_35 This equation has integration factor formula_36 formula_37 Integrating yields formula_38 If "h" is not zero, the independent variable can be changed from "t" to "ϕ" formula_39 giving the new equation of motion formula_40 Making the change of variables to the  inverse radius "u" = 1/"r" yields where "C" is a constant of integration and the function "G"("u") is defined by formula_41 This equation becomes quasilinear on differentiating by "ϕ" formula_42 This is known as the Binet equation. Integrating yields the solution for "ϕ" formula_43 where "ϕ" is another constant of integration. A central-force problem is said to be "integrable" if this final integration can be solved in terms of known functions. Orbit of the particle. The total energy of the system "E" equals the sum of the potential energy and the kinetic energy formula_44 Since the total energy is  constant, the rate of change of "r" can be calculated formula_45 which may be converted (as before) to the derivative of "r" with respect to the azimuthal angle φ formula_46 Integrating and using the angular-momentum formula "L"="mh" yields the formula formula_47 which indicates that the angular momentum contributes an effective potential energy formula_48 Changing the variable of integration to the inverse radius yields the integral formula_49 which expresses the above constants "C" = 2"mE"/"L" and "G"("u") = 2"mU"(1/"u")/"L" above in terms of the total energy "E" and the potential energy "U"("r"). Turning points and closed orbits. The rate of  change of "r" is zero whenever the effective potential energy equals the total energy formula_50 The points where this equation is satisfied are known as "turning points". The orbit on either side of a turning point is symmetrical; in other words, if the azimuthal angle is defined such that "φ" = 0 at the turning point, then the orbit is the same in opposite directions, "r"("φ") = "r"(−"φ"). If there are two turning points such that the radius "r" is bounded between "r" and "r", then the motion is contained within an annulus of those radii. As the radius varies  from the one turning point to the other, the change in azimuthal angle "φ" equals formula_51 The orbit will close upon itself provided that Δφ equals a rational fraction of 2"π", i.e., formula_52 where "m" and "n" are integers. In that case, the radius oscillates exactly "m" times while the azimuthal angle φ makes exactly "n" revolutions. In general, however, Δφ/2π will not be such a rational number, and thus the orbit will not be closed. In that case, the particle will eventually pass arbitrarily close to every point within the annulus. Two types of central force always produce closed  orbits: "F"("r") = α"r" (a linear force) and "F"("r") = α/"r" (an inverse-square law). As shown by Bertrand, these two central forces are the only ones that guarantee closed orbits. In general, if the angular momentum "L" is nonzero, the "L"/2"mr" term prevents the particle from falling into the origin, unless the effective potential energy goes to negative infinity in the limit of "r" going to zero. Therefore, if there is a single turning point, the orbit generally goes to infinity; the turning point corresponds to a point of minimum radius. Specific solutions. Kepler problem. In classical physics, many  important forces follow an inverse-square law, such as gravity or electrostatics. The general mathematical form of such inverse-square central forces is formula_53 for a constant formula_54, which is negative for an attractive force and positive for a repulsive one. This special case of the classical central-force problem is called the Kepler problem. For an inverse-square force, the Binet equation derived above is linear formula_55 The solution of this equation is formula_56 which shows that the orbit is a conic section of eccentricity "e"; here, "φ" is the initial angle, and the center of force is at  the focus of the conic section. Using the half-angle formula for sine, this solution can also be written as formula_57 where "u" and "u" are constants, with "u" larger than "u". The two versions of the solution are related by the equations formula_58 and formula_59 Since the sin function is always greater than zero, "u" is the largest possible value of "u" and the inverse of the smallest possible value of "r", i.e., the distance of closest approach (periapsis). Since the radial distance "r" cannot be a negative number, neither can its inverse "u"; therefore, "u" must be  a positive number. If "u" is also positive, it is the smallest possible value of "u", which corresponds to the largest possible value of "r", the distance of furthest approach (apoapsis). If "u" is zero or negative, then the smallest possible value of "u" is zero (the orbit goes to infinity); in this case, the only relevant values of φ are those that make "u" positive. For an attractive force (α < 0), the orbit is an ellipse, a hyperbola or parabola, depending on whether "u" is positive, negative, or zero, respectively; this corresponds to an eccentricity "e" less than  one, greater than one, or equal to one. For a repulsive force (α > 0), "u" must be negative, since "u" is positive by definition and their sum is negative; hence, the orbit is a hyperbola. Naturally, if no force is present (α=0), the orbit is a straight line. Central forces with exact solutions. The Binet equation for "u"("φ") can be solved numerically for nearly any central force "F"(1/"u"). However, only a handful of forces result in formulae for "u" in terms of known functions. As derived above, the solution for "φ" can be expressed as an integral over "u  formula_60 A central-force problem is said to be integrable" if this integration can be solved in terms of known functions. If the force is a power law, i.e., if "F"("r") = α "r", then "u" can be expressed in terms of circular functions and/or elliptic functions if "n" equals 1, -2, -3 (circular functions) and -7, -5, -4, 0, 3, 5, -3/2, -5/2, -1/3, -5/3 and -7/3 (elliptic functions). Similarly, only six possible linear combinations of power laws give solutions in terms of circular and elliptic functions formula_61 formula_62 formula_63 formula_64 formula_65 formula_66 The following special cases of the  first two force types always result in circular functions. formula_67 formula_68 The special case formula_69 was mentioned by Newton, in corollary 1 to proposition VII of the principia, as the force implied by circular orbits passing through the point of attraction. Revolving orbits. The term "r" occurs in all the force laws above, indicating that the addition of the inverse-cube force does not influence the solubility of the problem in terms of known functions. Newton showed that, with adjustments in the initial conditions, the addition of such a force does not affect the radial motion of the particle, but  multiplies its angular motion by a constant factor "k". An extension of Newton's theorem was discovered in 2000 by Mahomed and Vawda. Assume that a particle is moving under an arbitrary central force "F"("r"), and let its radius "r" and azimuthal angle φ be denoted as "r"("t") and "φ"("t") as a function of time "t". Now consider a second particle with the same mass "m" that shares the same radial motion "r"("t"), but one whose angular speed is "k" times faster than that of the first particle. In other words, the azimuthal angles of the two particles are related  by the equation φ("t") = "k" "φ"("t"). Newton showed that the force acting on the second particle equals the force "F"("r") acting on the first particle, plus an inverse-cube central force formula_70 where "L" is the magnitude of the first particle's angular momentum. If "k" is greater than one, "F"−"F" is a negative number; thus, the added inverse-cube force is "attractive". Conversely, if "k" is less than one, "F"−"F" is a positive number; the added inverse-cube force is "repulsive". If "k" is an integer such as 3, the orbit of the second particle is said to  be a "harmonic" of the first particle's orbit; by contrast, if "k" is the inverse of an integer, such as , the second orbit is said to be a "subharmonic" of the first orbit. Historical development. Newton's derivation. The classical central-force problem was solved geometrically by Isaac Newton in his "Philosophiæ Naturalis Principia Mathematica", in which Newton introduced his laws of motion. Newton used an equivalent of leapfrog integration to convert the continuous motion to a discrete one, so that geometrical methods may be applied. In this approach, the position of the particle is considered only at evenly  spaced time points. For illustration, the particle in Figure 10 is located at point A at time "t" = 0, at point B at time "t" = Δ"t", at point C at time "t" = 2Δ"t", and so on for all times "t" = "n"Δ"t", where "n" is an integer. The velocity is assumed to be constant between these time points. Thus, the vector r = r − r equals Δ"t" times the velocity vector v (red line), whereas r = r − r equals vΔ"t" (blue line). Since the velocity is constant between points, the force is assumed to  act instantaneously at each new position; for example, the force acting on the particle at point B instantly changes the velocity from v to v. The difference vector Δr = r − r equals ΔvΔ"t" (green line), where Δv = v − v is the change in velocity resulting from the force at point B. Since the acceleration a is parallel to Δv and since F = "m"a, the force F must be parallel to Δv and Δr. If F is a central force, it must be parallel to the vector r from the center O to the point B  (dashed green line); in that case, Δr is also parallel to r. If no force acts at point B, the velocity is unchanged, and the particle arrives at point K at time "t" = 2Δ"t". The areas of the triangles OAB and OBK are equal, because they share the same base ("r") and height ("r"). If Δr is parallel to r, the triangles OBK and OBC are likewise equal, because they share the same base ("r") and the height is unchanged. In that case, the areas of the triangles OAB and OBC are the same, and the particle sweeps out  equal areas in equal time. Conversely, if the areas of all such triangles are equal, then Δr must be parallel to r, from which it follows that F is a central force. Thus, a particle sweeps out equal areas in equal times if and only if F is a central force. Alternative derivations of the equations of motion. Lagrangian mechanics. The formula for the radial force may also be obtained using Lagrangian mechanics. In polar coordinates, the Lagrangian "L" of a single particle in a potential energy field "U"("r") is given by formula_71 Then Lagrange's equations of motion formula_72  take the form formula_73 since the magnitude "F"("r") of the radial force equals the negative derivative of the potential energy "U"("r") in the radial direction. Hamiltonian mechanics. The radial force formula may also be derived using Hamiltonian mechanics. In polar coordinates, the Hamiltonian can be written as formula_74 Since the azimuthal angle φ does not appear in the Hamiltonian, its conjugate momentum "p" is a constant of the motion. This conjugate momentum is the magnitude "L" of the angular momentum, as shown by the Hamiltonian equation of motion for "φ" formula_75 The corresponding equation of motion for "r" is formula_76  Taking the second derivative of "r" with respect to time and using Hamilton's equation of motion for "p" yields the radial-force equation formula_77 Hamilton-Jacobi equation. The orbital equation can be derived directly from the Hamilton–Jacobi equation. Adopting the radial distance "r" and the azimuthal angle "φ" as the coordinates, the Hamilton-Jacobi equation for a central-force problem can be written formula_78 where "S" = "S"("φ") + "S"("r") − "E""t" is Hamilton's principal function, and "E" and "t" represent the total energy and time, respectively. This equation may be solved by successive integrations of ordinary  differential equations, beginning with the "φ" equation formula_79 where "p" is a constant of the motion equal to the magnitude of the angular momentum "L". Thus, "S"(φ) = "L"φ and the Hamilton–Jacobi equation becomes formula_80 Integrating this equation for "S" yields formula_81 Taking the derivative of "S" with respect to "L" yields the orbital equation derived above formula_82