Classical central-force problem

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 $${\displaystyle \mathbf{L}=\mathbf{r}\times \mathbf{p}=\mathbf{r}\times m\mathbf{v}}$$ where m is the mass of the particle and p is its linear momentum. In this equation the times symbol × indicates the vector cross product, not multiplication. 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 $${\displaystyle \begin{array}{cc}\frac{d\mathbf{L}}{dt}& =\dot{\mathbf{r}}\times m\mathbf{v}+\mathbf{r}\times m\dot{\mathbf{v}}\\ & =\mathbf{v}\times m\mathbf{v}+\mathbf{r}\times \mathbf{F}\\ & =\mathbf{r}\times \mathbf{F},\end{array}}$$

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 $${\displaystyle \mathbf{r}\cdot \mathbf{L}=\mathbf{r}\cdot (\mathbf{r}\times \mathbf{p})=\mathbf{p}\cdot (\mathbf{r}\times \mathbf{r})=0}$$

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.

$${\displaystyle \mathbf{r}=(x,y)=(r\cos \phi ,r\sin \phi )}$$

Taking the first derivative with respect to time yields the particle's velocity vector v $${\displaystyle \mathbf{v}=\frac{d\mathbf{r}}{dt}=\dot{r}(\cos \phi ,\sin \phi )+r\dot{\phi }(-\sin \phi ,\cos \phi )}$$

Similarly, the second derivative of the particle's position r equals its acceleration a $${\displaystyle \mathbf{a}=\ddot{r}(\cos \phi ,\sin \phi )+2\dot{r}\dot{\phi }(-\sin \phi ,\cos \phi )+r\ddot{\phi }(-\sin \phi ,\cos \phi )-r{\dot{\phi }}^2(\cos \phi ,\sin \phi )}$$

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 $${\displaystyle \hat{\mathbf{r}}=(\cos \phi ,\sin \phi )}$$

The azimuthal unit vector is given by $${\displaystyle \hat{\mathit{\phi }}=(-\sin \phi ,\cos \phi )}$$

Thus, the velocity can be written as $${\displaystyle \mathbf{v}=v_r\hat{\mathbf{r}}+v_{\phi }\hat{\mathit{\phi }}=\dot{r}\hat{\mathbf{r}}+r\dot{\phi }\hat{\mathit{\phi }}}$$ whereas the acceleration equals $${\displaystyle \begin{array}{cc}\mathbf{a}& =a_r\hat{\mathbf{r}}+a_{\phi }\hat{\mathit{\phi }}\\ & =\left(\ddot{r}-r{\dot{\phi }}^2\right)\hat{\mathbf{r}}+\left(2\dot{r}\dot{\phi }+r\ddot{\phi }\right)\hat{\mathit{\phi }}\end{array}}$$

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_φ must be zero $${\displaystyle a_{\phi }=2\dot{r}\dot{\phi }+r\ddot{\phi }=0}$$

Therefore, $${\displaystyle \begin{array}{cc}\frac{d}{dt}\left(r^2\dot{\phi }\right)& =r\left(2\dot{r}\dot{\phi }+r\ddot{\phi }\right)\\ & =r{a}_{\phi }=0\end{array}}$$

This expression in parentheses is usually denoted h $${\displaystyle \begin{array}{cc}h=r^2\dot{\phi }& =r{v}_{\phi }=\left|\mathbf{r}\times \mathbf{v}\right|\\ & =v{r}_{\perp }=\frac{L}{m}\end{array}}$$

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 $${\displaystyle \omega =\dot{\phi }=\frac{d\phi }{dt}}$$

However, it should not be assumed that ω is constant. Since h is constant, ω varies with the radius r according to the formula $${\displaystyle \omega =\frac{h}{r^2}}$$

Since h is constant and r² 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_φ is always zero. Hence, by Newton's second law, F = ma, 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 Δt equals ω r²Δt; hence, equal areas are swept out in equal times Δt. 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 Δt, the particle sweeps out an area 1⁄2vΔtr_⊥ (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 1⁄2vr_⊥ 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 $${\displaystyle \tau =\int \frac{dt}{y^2}}$$

The corresponding equations of motion for ξ and η are given by $${\displaystyle \begin{array}{cc}\frac{d\xi }{d\tau }& =\frac{d}{dt}\left(\frac{x}{y}\right)\frac{dt}{d\tau }\\ & =\left(\frac{\dot{x}y-\dot{y}x}{y^2}\right)y^2=-h\\ \frac{d\eta }{d\tau }& =\frac{d}{dt}\left(\frac{1}{y}\right)\frac{dt}{d\tau }\\ & =-\frac{\dot{y}}{y^2}{y}^2=-\dot{y}\end{array}}$$

Since the rate of change of ξ is constant, its second derivative is zero $${\displaystyle \frac{d^2\xi }{d{\tau }^2}=0}$$

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 $${\displaystyle \begin{array}{cc}\frac{d^2\eta }{d{\tau }^2}& =\frac{dt}{d\tau }\frac{d}{dt}\left(\frac{d\eta }{d\tau }\right)\\ & =-y^2\ddot{y}=-\frac{y^3}{mr}F(r)\end{array}}$$ because the acceleration in the y-direction equals $${\displaystyle \ddot{y}=\frac{1}{m}{F}_y=\frac{1}{m}F(r)\frac{y}{r}}$$

Here, F_y 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 $${\displaystyle F(r)=m\ddot{r}-mr{\omega }^2=m\frac{d^{2}r}{d{t}^2}-\frac{m{h}^2}{r^3}}$$

This equation has the integration factor ${\displaystyle \dot{r}=\frac{dr}{dt}}$ $${\displaystyle \begin{array}{cc}F(r)dr& =F(r)\dot{r}dt\\ & =m\left(\dot{r}\ddot{r}-\frac{h^2}{r^3}\dot{r}\right)dt\\ & =\frac{m}{2}d\left({\dot{r}}^2+h^2{r}^{-2}\right)\end{array}}$$

Integrating yields $${\displaystyle {\int }^{r}F(r)dr=\frac{m}{2}\left[{\left(\frac{dr}{dt}\right)}^2+{\left(\frac{h}{r}\right)}^2\right]}$$

If h is not zero, the independent variable can be changed from t to ϕ $${\displaystyle \frac{d}{dt}=\omega \frac{d}{d\phi }=\frac{h}{r^2}\frac{d}{d\phi }}$$ giving the new equation of motion $${\displaystyle {\int }^{r}F(r)dr=\frac{m{h}^2}{2}\left[{\left(-\frac{1}{r^2}\frac{dr}{d\phi }\right)}^2+{\left(\frac{1}{r}\right)}^2\right]}$$

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 $${\displaystyle G(u)=-\frac{2}{m{h}^2}{\int }^{\frac{1}{u}}F(r)dr}$$

This equation becomes quasilinear on differentiating by ϕ $${\displaystyle \frac{d^{2}u}{d{\phi }^2}+u=-\frac{1}{m{h}^2{u}^2}F(1/u)}$$

This is known as the Binet equation. Integrating (1) yields the solution for ϕ $${\displaystyle \phi ={\phi }_0+{\int }^{\frac{1}{r}}\frac{du}{\sqrt{C-u^2-G(u)}}}$$ where ϕ₀ 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

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 $${\displaystyle m\ddot{\mathbf{r}}=\mathbf{F}=-\mathbf{\nabla }U(\mathbf{r})/\cdot \dot{\mathbf{r}}}$$ gives $${\displaystyle \begin{array}{cc}m\dot{\mathbf{r}}\cdot \ddot{\mathbf{r}}=\frac{1}{2}m\frac{d}{dt}\left(\dot{\mathbf{r}}\cdot \dot{\mathbf{r}}\right)& =-\dot{\mathbf{r}}\cdot \mathbf{\nabla }U(\mathbf{r})\\ & =-\sum _{i=1}^3{\dot{x}}_i\frac{\mathrm{\partial }U}{\mathrm{\partial }{x}_i}=-\frac{dU}{dt}\end{array}}$$ where summation is assumed over the spatial Cartesian index ${\displaystyle i=1,2,3}$ and we have used the fact that $\frac{d}{dt}\left(\dot{\mathbf{r}}\cdot \dot{\mathbf{r}}\right)=\ddot{\mathbf{r}}\cdot \dot{\mathbf{r}}+\dot{\mathbf{r}}\cdot \ddot{\mathbf{r}}=2\dot{\mathbf{r}}\cdot \ddot{\mathbf{r}}$ and used the chain rule $\frac{d}{dt}U(x,y,z)=\sum _i\frac{\mathrm{\partial }U}{\mathrm{\partial }{x}_i}\frac{d{x}_i}{dt}$. Rearranging $${\displaystyle \frac{d}{dt}\left(\frac{1}{2}m{\dot{\mathbf{r}}}^2+U(\mathbf{r})\right)=0}$$ The term in parentheses on the left hand side is a constant, label this with ${\displaystyle E_{\mathrm{t}\mathrm{o}\mathrm{t}}}$, the total mechanical energy. Clearly, this is the sum of the kinetic energy and the potential energy.

Furthermore if the potential is central, ${\displaystyle U(x,y,z)=U(r)}$ and so the force is along the radial direction. In this case, the cross product of Newton's second law of motion with the particle's position vector ${\displaystyle \mathbf{r}}$ must vanish since the cross product of two parallel vectors is zero: $${\displaystyle \begin{array}{cc}\mathbf{r}\times m\ddot{\mathbf{r}}& =\frac{d}{dt}\left(\mathbf{r}\times m\dot{\mathbf{r}}\right)-\left(\dot{\mathbf{r}}\times m\dot{\mathbf{r}}\right)\\ & =\mathbf{r}\times \mathbf{F}=0\end{array}}$$ but ${\displaystyle \dot{\mathbf{r}}\times m\dot{\mathbf{r}}=0}$ (cross product of parallel vectors), so $${\displaystyle \frac{d}{dt}\left(\mathbf{r}\times m\dot{\mathbf{r}}\right)=0}$$ The term in parentheses on the left hand side is a constant, label this with ${\displaystyle \mathbf{L}}$ the angular momentum, In particular, in polar coordinates, ${\displaystyle L=mr{v}_{\phi }=m{r}^2\dot{\phi }}$ or $${\displaystyle \dot{\phi }=\frac{L}{m{r}^2}}$$ Further, ${\displaystyle {\dot{\mathbf{r}}}^2=v_r^2+v_{\phi }^2={\dot{r}}^2+(r\dot{\phi }{)}^2}$, so the energy equation may be simplified with the angular momentum as $${\displaystyle \begin{array}{cc}{E}_{\mathrm{t}\mathrm{o}\mathrm{t}}& =\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\dot{\phi }}^2+U(r)\\ & =\frac{1}{2}m{\dot{r}}^2+\frac{L^2}{2m{r}^2}+U(r)\\ & =\frac{1}{2}m{\dot{r}}^2+U_{\mathrm{e}\mathrm{f}\mathrm{f}}(r)\end{array}}$$ This indicates that the angular momentum contributes an effective potential energy $${\displaystyle U_{\mathrm{e}\mathrm{f}\mathrm{f}}(r)=U(r)+\frac{L^2}{2m{r}^2}}$$ Solve this equation for ${\displaystyle \dot{r}}$ $${\displaystyle \dot{r}=\frac{dr}{dt}=\sqrt{\frac{2}{m}}\sqrt{E_{\mathrm{t}\mathrm{o}\mathrm{t}}-U_{\mathrm{e}\mathrm{f}\mathrm{f}}(r)}}$$ which may be converted to the derivative of ${\displaystyle r}$ with respect to the azimuthal angle ${\displaystyle \phi }$ as $${\displaystyle \frac{dr}{d\phi }=\frac{dr/dt}{d\phi /dt}=\frac{\dot{r}}{\dot{\phi }}=\frac{m{r}^2}{L}\sqrt{\frac{2}{m}}\sqrt{E_{\mathrm{t}\mathrm{o}\mathrm{t}}-U_{\mathrm{e}\mathrm{f}\mathrm{f}}(r)}}$$ This is a separable first order differential equation. Integrating and yields the formula $${\displaystyle \phi ={\phi }_0+\frac{L}{\sqrt{2m}}{\int }^r\frac{dr}{r^2\sqrt{E_{\mathrm{t}\mathrm{o}\mathrm{t}}-U_{\mathrm{e}\mathrm{f}\mathrm{f}}(r)}}}$$

Changing the variable of integration to the inverse radius yields the integral $${\displaystyle \phi ={\phi }_0+{\int }^u\frac{du}{\sqrt{\frac{2m}{L^2}{E}_{\mathrm{t}\mathrm{o}\mathrm{t}}-\frac{2m}{L^2}U(1/u)-u^2}}}$$ which expresses the above constants C = 2mE_tot/L² and G(u) = 2mU(1/u)/L² above in terms of the total energy E_tot 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 $${\displaystyle E_{\mathrm{t}\mathrm{o}\mathrm{t}}=U(r)+\frac{L^2}{2m{r}^2}}$$

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_min and r_max, 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 $${\displaystyle \mathrm{\Delta }\phi =\frac{L}{\sqrt{2m}}{\int }_{r_{min}}^{r_{max}}\frac{dr}{r^2\sqrt{E-U(r)-\frac{L^2}{2m{r}^2}}}}$$

The orbit will close upon itself provided that Δφ equals a rational fraction of 2π, i.e., $${\displaystyle \mathrm{\Delta }\phi =2\pi \frac{m}{n}}$$ 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) = αr (a linear force) and F(r) = α/r² (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 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. 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 $${\displaystyle F=\frac{\alpha }{r^2}=\alpha u^2}$$ for a constant ${\displaystyle \alpha }$, 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 $${\displaystyle \frac{d^{2}u}{d{\phi }^2}+u=-\frac{\alpha }{m{h}^2}.}$$

The solution of this equation is $${\displaystyle u(\phi )=-\frac{\alpha }{m{h}^2}\left[1+e\cos \left(\phi -{\phi }_0\right)\right]}$$ which shows that the orbit is a conic section of eccentricity e; here, φ₀ 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 $${\displaystyle u(\phi )=u_2+(u_2-u_1){\sin }^2\left(\frac{\phi -{\phi }_0}{2}\right)}$$

where u₁ and u₂ are constants, with u₂ larger than u₁. The two versions of the solution are related by the equations $${\displaystyle u_1+u_2=-\frac{2\alpha }{m{h}^2}}$$ and $${\displaystyle e=\frac{u_2-u_1}{u_2+u_1}}$$

Since the sin² function is always greater than zero, u₂ 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, u₂ must be a positive number. If u₁ 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 u₁ 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 u₁ 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), u₁ must be negative, since u₂ 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 $${\displaystyle \phi ={\phi }_0+\frac{L}{\sqrt{2m}}{\int }^u\frac{du}{\sqrt{E_{\mathrm{t}\mathrm{o}\mathrm{t}}-U(1/u)-\frac{L^2{u}^2}{2m}}}}$$

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) = αrⁿ, 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 $${\displaystyle F(r)=A{r}^{-3}+Br+C{r}^3+D{r}^5}$$ $${\displaystyle F(r)=A{r}^{-3}+Br+C{r}^{-5}+D{r}^{-7}}$$ $${\displaystyle F(r)=A{r}^{-3}+B{r}^{-2}+Cr+D}$$ $${\displaystyle F(r)=A{r}^{-3}+B{r}^{-2}+C{r}^{-4}+D{r}^{-5}}$$ $${\displaystyle F(r)=A{r}^{-3}+B{r}^{-2}+C{r}^{-3/2}+D{r}^{-5/2}}$$ $${\displaystyle F(r)=A{r}^{-3}+B{r}^{-1/3}+C{r}^{-5/3}+D{r}^{-7/3}}$$

The following special cases of the first two force types always result in circular functions. $${\displaystyle F(r)=A{r}^{-3}+Br}$$ $${\displaystyle F(r)=A{r}^{-3}+B{r}^{-2}}$$

The special case $${\displaystyle F(r)=A{r}^{-5}}$$ 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⁻³ 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 F₁(r), and let its radius r and azimuthal angle φ be denoted as r(t) and φ₁(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 φ₂(t) = k φ₁(t). Newton showed that the force acting on the second particle equals the force F₁(r) acting on the first particle, plus an inverse-cube central force $${\displaystyle F_2(r)=F_1(r)+\frac{L_1^2}{m{r}^3}\left(1-k^2\right)}$$ where L₁ is the magnitude of the first particle's angular momentum.

If k² is greater than one, F₂−F₁ is a negative number; thus, the added inverse-cube force is attractive. Conversely, if k² is less than one, F₂−F₁ 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 1⁄3, 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 = Δt, at point C at time t = 2Δt, and so on for all times t = nΔt, where n is an integer. The velocity is assumed to be constant between these time points. Thus, the vector r_AB = r_B − r_A equals Δt times the velocity vector v_AB (red line), whereas r_BC = r_C − r_B equals v_BCΔt (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_AB to v_BC. The difference vector Δr = r_BC − r_AB equals ΔvΔt (green line), where Δv = v_BC − v_AB is the change in velocity resulting from the force at point B. Since the acceleration a is parallel to Δv and since F = ma, the force F must be parallel to Δv and Δr. If F is a central force, it must be parallel to the vector r_B from the center O to the point B (dashed green line); in that case, Δr is also parallel to r_B.

If no force acts at point B, the velocity is unchanged, and the particle arrives at point K at time t = 2Δt. The areas of the triangles OAB and OBK are equal, because they share the same base (r_AB) and height (r_⊥). If Δr is parallel to r_B, the triangles OBK and OBC are likewise equal, because they share the same base (r_B) 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_B, 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 $${\displaystyle L=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\dot{\phi }}^2-U(r)}$$

Then Lagrange's equations of motion $${\displaystyle \frac{d}{dt}\left(\frac{\mathrm{\partial }L}{\mathrm{\partial }\dot{r}}\right)=\frac{\mathrm{\partial }L}{\mathrm{\partial }r}}$$ take the form $${\displaystyle m\ddot{r}=mr{\dot{\phi }}^2-\frac{dU}{dr}=\frac{m{h}^2}{r^3}+F(r)}$$ 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 $${\displaystyle H=\frac{p_r^2}{2m}+\frac{p_{\phi }^2}{2m{r}^2}+U(r)}$$

Since the azimuthal angle φ does not appear in the Hamiltonian, its conjugate momentum p_φ 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 $${\displaystyle \dot{\phi }=\frac{\mathrm{\partial }H}{\mathrm{\partial }{p}_{\phi }}=\frac{p_{\phi }}{m{r}^2}=\frac{L}{m{r}^2}}$$

The corresponding equation of motion for r is $${\displaystyle \dot{r}=\frac{\mathrm{\partial }H}{\mathrm{\partial }{p}_r}=\frac{p_r}{m}}$$

Taking the second derivative of r with respect to time and using Hamilton's equation of motion for p_r yields the radial-force equation $${\displaystyle \begin{array}{cc}\ddot{r}=\frac{{\dot{p}}_r}{m}& =-\frac{1}{m}\left(\frac{\mathrm{\partial }H}{\mathrm{\partial }r}\right)\\ & =\frac{p_{\phi }^2}{m^2{r}^3}-\frac{1}{m}\frac{dU}{dr}\\ & =\frac{L^2}{m^2{r}^3}+\frac{1}{m}F(r)\end{array}}$$

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 $${\displaystyle \frac{1}{2m}{\left(\frac{d{S}_r}{dr}\right)}^2+\frac{1}{2m{r}^2}{\left(\frac{d{S}_{\phi }}{d\phi }\right)}^2+U(r)=E_{\mathrm{t}\mathrm{o}\mathrm{t}}}$$ where S = S_φ(φ) + S_r(r) − E_tott is Hamilton's principal function, and E_tot 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 $${\displaystyle \frac{d{S}_{\phi }}{d\phi }=p_{\phi }=L}$$ where p_φ is a constant of the motion equal to the magnitude of the angular momentum L. Thus, S_φ(φ) = Lφ and the Hamilton–Jacobi equation becomes $${\displaystyle \frac{1}{2m}{\left(\frac{d{S}_r}{dr}\right)}^2+\frac{L^2}{2m{r}^2}+U(r)=E_{\mathrm{t}\mathrm{o}\mathrm{t}}}$$

Integrating this equation for S_r yields $${\displaystyle S_r(r)=\sqrt{2m}\int dr\sqrt{E_{\mathrm{t}\mathrm{o}\mathrm{t}}-U(r)-\frac{L^2}{2m{r}^2}}}$$

Taking the derivative of S with respect to L yields the orbital equation derived above $${\displaystyle \begin{array}{cc}{\phi }_0& =\frac{\mathrm{\partial }S}{\mathrm{\partial }L}=\frac{\mathrm{\partial }{S}_{\phi }}{\mathrm{\partial }L}+\frac{\mathrm{\partial }{S}_r}{\mathrm{\partial }L}\\ & =\phi -\frac{L}{\sqrt{2m}}{\int }^r\frac{dr}{r^2\sqrt{E_{\mathrm{t}\mathrm{o}\mathrm{t}}-U(r)-\frac{L^2}{2m{r}^2}}}\end{array}}$$