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.

In other words, a central force must act along the line joining O with the present position of the particle.

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[note 2]

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.

because the partial derivatives are zero for a central force; the magnitude F does not depend on the angular spherical coordinates θ and φ.

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.

In reality, however, the Sun also moves (albeit only slightly) in response to the force applied by the planet Mercury.

The motion of a particle under a central force F always remains in the plane defined by its initial position and velocity.

Consequently, the particle's position r (and hence velocity v) always lies in a plane perpendicular to L.[9] Since the motion is planar and the force radial, it is customary to switch to polar coordinates.

[9] In these coordinates, the position vector r is represented in terms of the radial distance r and the azimuthal angle φ.

[11] 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.

[12] Thus, the areal velocity is constant for a particle acted upon by any type of central force; this is Kepler's second law.

Making the change of variables to the inverse radius u = 1/r[17] yields where C is a constant of integration and the function G(u) is defined by

Take the scalar product of Newton's second law of motion with the particle's velocity where the force is obtained from the potential energy

[24] If there are two turning points such that the radius r is bounded between rmin and rmax, then the motion is contained within an annulus of those radii.

[23] As the radius varies from the one turning point to the other, the change in azimuthal angle φ equals[23]

[25] In general, if the angular momentum L is nonzero, the L2/2mr2 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.

In classical physics, many important forces follow an inverse-square law, such as gravity or electrostatics.

For an attractive force (α < 0), the orbit is an ellipse, a hyperbola or parabola, depending on whether u1 is positive, negative, or zero, respectively; this corresponds to an eccentricity e less than one, greater than one, or equal to one.

[27] Similarly, only six possible linear combinations of power laws give solutions in terms of circular and elliptic functions[28][29]

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.

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.

Newton used an equivalent of leapfrog integration to convert the continuous motion to a discrete one, so that geometrical methods may be applied.

The difference vector Δr = rBC − rAB equals ΔvΔt (green line), where Δv = vBC − vAB is the change in velocity resulting from the force at point B.

The areas of the triangles OAB and OBK are equal, because they share the same base (rAB) and height (r⊥).

Conversely, if the areas of all such triangles are equal, then Δr must be parallel to rB, from which it follows that F is a central force.

Since the azimuthal angle φ does not appear in the Hamiltonian, its conjugate momentum pφ is a constant of the motion.

[31] 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

where S = Sφ(φ) + Sr(r) − Etott is Hamilton's principal function, and Etot and t represent the total energy and time, respectively.

where pφ is a constant of the motion equal to the magnitude of the angular momentum L. Thus, Sφ(φ) = Lφ and the Hamilton–Jacobi equation becomes

A long arrows runs from the lower left to the upper right. At the lower left, the arrow begins with a black point labeled "O"; at the upper right, the arrow ends at a solid red circle labeled "P". Above this arrow is a shorter, thicker arrow labeled "F sub att" that points from the center of P towards O.
An attractive central force acting on a body at position P (shown in red). By definition, a central force must point either towards a fixed point O (if attractive) or away from it (if repulsive).
The positions x 1 and x 2 of two bodies can be expressed in terms of their relative separation r and the position of their center of mass R cm .
Any classical two-body problem to be converted into an equivalent one-body problem. The mass μ of the one equivalent body equals the reduced mass of the two original bodies, and its position r equals the difference of their positions.
The image shows a yellow disc with three vectors. The vector L is perpendicular to the disk, the vector r goes from the center of the disk to a point on its periphery, and the vector v is tangential to the disk, starting from the point where r meets the periphery.
Illustration of planar motion. The angular momentum vector L is constant; therefore, the position vector r and velocity vector v must lie in the yellow plane perpendicular to L .
Two perpendicular lines (Cartesian coordinate axes) are labeled x (horizontal) and y (vertical). They intersect at the lower left in a point labeled O (the origin). An arrow labeled r runs form the origin to the upper right, ending in a point P. The angle between the x-axis and the vector r is labeled with the Greek letter φ. A vertical line is dropped from P to the x-axis, and the horizontal and vertical segments are labeled "r cosine phi" and "r sine phi", respectively.
The position vector r of a point P in the plane can be specified by its distance r from the center (the origin O ) and its azimuthal angle φ . The x and y Cartesian components of the vector are r cos φ and r sin φ , respectively.
The specific angular momentum h equals the speed v times r ⊥ , the component of the position vector r perpendicular to the velocity vector v . h also equals the radial distance r times the azimuthal component v φ of the velocity. Both of these formulae are equal to rv cos β.
Since the area A equals 1 ⁄ 2 r ⊥ vt , the areal velocity dA / dt (the rate at which A is swept out by the particle) equals 1 ⁄ 2 r ⊥ v = 1 ⁄ 2 h .
The area A of a circular sector equals 1 ⁄ 2 r 2 φ = 1 ⁄ 2 r 2 ω t = 1 ⁄ 2 r v φ t . Hence, the areal velocity dA/dt equals 1 ⁄ 2 r v φ = 1 ⁄ 2 h . For uniform circular motion, r and v φ are constant; thus, dA/dt is also constant.
An animation showing a small particle moving on a red ellipse; a large blue mass is located at one focus of the ellipse.
Classical gravity is a central force. Solving that central-force problem shows that a bound particle follows an elliptical orbit in which equal areas are swept out in equal times, as described by Kepler's second law .
Blue ellipse with the two foci indicated as black points. Four line segments go out from the left focus to the ellipse, forming two shaded pseudo-triangles with two straight sides and the third side made from the curved segment of the intervening ellipse.
As for all central forces, the particle in the Kepler problem sweeps out equal areas in equal times, as illustrated by the two blue elliptical sectors. The center of force is located at one of the foci of the elliptical orbit.
Illustration of Newton's theorem of revolving orbits. The green planet completes one (subharmonic) orbit for every three orbits of the blue planet ( k =1/3). A GIF version of this animation is found here .
Figure 10: Newton's geometrical proof that a moving particle sweeps out equal areas in equal times if and only if the force acting on it at the point B is a central force. Here, the triangle OAB has the same area as the triangles OBC and OBK.