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 a few 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
F = F ( r ) r ^ {\displaystyle \mathbf {F} =F(r){\hat {\mathbf {r} }}}
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
F = F ( r ) r ^ = m a = m r ¨ {\displaystyle \mathbf {F} =F(r){\hat {\mathbf {r} }}=m\mathbf {a} =m{\ddot {\mathbf {r} }}}
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)
F ( r ) = − d U d r {\displaystyle F(r)=-{\frac {dU}{dr}}}
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.
W = ∫ r 1 r 2 F ⋅ d r = ∫ r 1 r 2 F ( r ) r ^ ⋅ d r = ∫ r 1 r 2 F d r = U ( r 1 ) − U ( r 2 ) {\displaystyle {\begin{aligned}W=\int _{\mathbf {r} _{1}}^{\mathbf {r} _{2}}\mathbf {F} \cdot d\mathbf {r} &=\int _{\mathbf {r} _{1}}^{\mathbf {r} _{2}}F(r){\hat {\mathbf {r} }}\cdot d\mathbf {r} \\[1ex]&=\int _{r_{1}}^{r_{2}}Fdr=U(r_{1})-U(r_{2})\end{aligned}}}
Equivalently, it suffices that the curl of the force field F is zero; using the formula for the curl in spherical coordinates,
∇ × F = 1 r sin θ ( ∂ F ∂ φ ) θ ^ − 1 r ( ∂ F ∂ θ ) φ ^ = 0 {\displaystyle \nabla \times \mathbf {F} ={\frac {1}{r\sin \theta }}\left({\frac {\partial F}{\partial \varphi }}\right){\hat {\boldsymbol {\theta }}}-{\frac {1}{r}}\left({\frac {\partial F}{\partial \theta }}\right){\hat {\boldsymbol {\varphi }}}=0}
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
m r ¨ = F ( r ) {\displaystyle m{\ddot {r}}=F(r)}
One solution method is to use the conservation of total energy
| r ˙ | = | d r d t | = 2 m E tot − U ( r ) {\displaystyle \left|{\dot {r}}\right|=\left|{\frac {dr}{dt}}\right|={\sqrt {\frac {2}{m}}}{\sqrt {E_{\text{tot}}-U(r)}}}
Taking the reciprocal and integrating we get:
| t − t 0 | = m 2 ∫ | d r | E tot − U ( r ) {\displaystyle |t-t_{0}|={\sqrt {\frac {m}{2}}}\int {\frac {|dr|}{\sqrt {E_{\text{tot}}-U(r)}}}}
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
m v 2 r = F ( r ) {\displaystyle {\frac {mv^{2}}{r}}=F(r)}
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 x1 and x2 be the positions of the two particles, and let r = x1 − x2 be their relative position. Then, by Newton's second law,
r ¨ = x ¨ 1 − x ¨ 2 = ( F 21 m 1 − F 12 m 2 ) = ( 1 m 1 + 1 m 2 ) F 21 {\displaystyle {\begin{aligned}{\ddot {\mathbf {r} }}={\ddot {\mathbf {x} }}_{1}-{\ddot {\mathbf {x} }}_{2}&=\left({\frac {\mathbf {F} _{21}}{m_{1}}}-{\frac {\mathbf {F} _{12}}{m_{2}}}\right)\\[1ex]&=\left({\frac {1}{m_{1}}}+{\frac {1}{m_{2}}}\right)\mathbf {F} _{21}\end{aligned}}}
The final equation derives from Newton's third law; the force of the second body on the first body (F21) is equal and opposite to the force of the first body on the second (F12). Thus, the equation of motion for r can be written in the form
μ r ¨ = F {\displaystyle \mu {\ddot {\mathbf {r} }}=\mathbf {F} }
where μ {\displaystyle \mu } is the reduced mass
μ = 1 1 m 1 + 1 m 2 = m 1 m 2 m 1 + m 2 {\displaystyle \mu ={\frac {1}{{\frac {1}{m_{1}}}+{\frac {1}{m_{2}}}}}={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}}
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
L = r × p = r × m v {\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
d L d t = r ˙ × m v + r × m v ˙ = v × m v + r × F = r × F , {\displaystyle {\begin{aligned}{\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} \\[1ex]&=\mathbf {r} \times \mathbf {F} \ ,\end{aligned}}}
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
r ⋅ L = r ⋅ ( r × p ) = p ⋅ ( r × r ) = 0 {\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.
r = ( x , y ) = ( r cos φ , r sin φ ) {\displaystyle \mathbf {r} =(x,\ y)=(r\cos \varphi ,\ r\sin \varphi )}
Taking the first derivative with respect to time yields the particle's velocity vector v
v = d r d t = r ˙ ( cos φ , sin φ ) + r φ ˙ ( − sin φ , cos φ ) {\displaystyle \mathbf {v} ={\frac {d\mathbf {r} }{dt}}={\dot {r}}(\cos \varphi ,\ \sin \varphi )+r{\dot {\varphi }}(-\sin \varphi ,\cos \varphi )}
Similarly, the second derivative of the particle's position r equals its acceleration a
a = r ¨ ( cos φ , sin φ ) + 2 r ˙ φ ˙ ( − sin φ , cos φ ) + r φ ¨ ( − sin φ , cos φ ) − r φ ˙ 2 ( cos φ , sin φ ) {\displaystyle \mathbf {a} ={\ddot {r}}(\cos \varphi ,\ \sin \varphi )+2{\dot {r}}{\dot {\varphi }}(-\sin \varphi ,\ \cos \varphi )+r{\ddot {\varphi }}(-\sin \varphi ,\cos \varphi )-r{\dot {\varphi }}^{2}(\cos \varphi ,\sin \varphi )}
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
r ^ = ( cos φ , sin φ ) {\displaystyle \mathbf {\hat {r}} =(\cos \varphi ,\ \sin \varphi )}
The azimuthal unit vector is given by
φ ^ = ( − sin φ , cos φ ) {\displaystyle {\hat {\boldsymbol {\varphi }}}=(-\sin \varphi ,\ \cos \varphi )}
Thus, the velocity can be written as
v = v r r ^ + v φ φ ^ = r ˙ r ^ + r φ ˙ φ ^ {\displaystyle \mathbf {v} =v_{r}\mathbf {\hat {r}} +v_{\varphi }{\hat {\boldsymbol {\varphi }}}={\dot {r}}\mathbf {\hat {r}} +r{\dot {\varphi }}{\hat {\boldsymbol {\varphi }}}}
whereas the acceleration equals
a = a r r ^ + a φ φ ^ = ( r ¨ − r φ ˙ 2 ) r ^ + ( 2 r ˙ φ ˙ + r φ ¨ ) φ ^ {\displaystyle {\begin{aligned}\mathbf {a} &=a_{r}\mathbf {\hat {r}} +a_{\varphi }{\hat {\boldsymbol {\varphi }}}\\[1ex]&=\left({\ddot {r}}-r{\dot {\varphi }}^{2}\right)\mathbf {\hat {r}} +\left(2{\dot {r}}{\dot {\varphi }}+r{\ddot {\varphi }}\right){\hat {\boldsymbol {\varphi }}}\end{aligned}}}
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 aφ must be zero
a φ = 2 r ˙ φ ˙ + r φ ¨ = 0 {\displaystyle a_{\varphi }=2{\dot {r}}{\dot {\varphi }}+r{\ddot {\varphi }}=0}
Therefore,
d d t ( r 2 φ ˙ ) = r ( 2 r ˙ φ ˙ + r φ ¨ ) = r a φ = 0 {\displaystyle {\begin{aligned}{\frac {d}{dt}}\left(r^{2}{\dot {\varphi }}\right)&=r\left(2{\dot {r}}{\dot {\varphi }}+r{\ddot {\varphi }}\right)\\&=ra_{\varphi }=0\end{aligned}}}
This expression in parentheses is usually denoted h
h = r 2 φ ˙ = r v φ = | r × v | = v r ⊥ = L m {\displaystyle {\begin{aligned}h=r^{2}{\dot {\varphi }}&=rv_{\varphi }=\left|\mathbf {r} \times \mathbf {v} \right|\\[1ex]&=vr_{\perp }={\frac {L}{m}}\end{aligned}}}
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
ω = φ ˙ = d φ d t {\displaystyle \omega ={\dot {\varphi }}={\frac {d\varphi }{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
ω = h r 2 {\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 aφ 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
τ = ∫ d t y 2 {\displaystyle \tau =\int {\frac {dt}{y^{2}}}}
The corresponding equations of motion for ξ and η are given by
d ξ d τ = d d t ( x y ) d t d τ