Central Forces and the Two-Body Problem
~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with vector calculus, specifically cross products and gradients, and have a solid grasp of Newton's second law in vector form.
Why this matters
Central forces govern the motion of planets, satellites, and atomic electrons, making them the cornerstone of classical dynamics.
While general motion in three dimensions is complex, motion under a central force is integrable and reveals deep connections between symmetry and conservation laws. This lecture bridges the gap between simple particle dynamics and the sophisticated treatment of interacting bodies, preparing you for the analytical mechanics that follows.
The Geometry of Central Forces
Imagine a particle moving under the influence of a force that always points directly toward or away from a fixed point, the center of force. Because the force has no component perpendicular to the radius vector, it cannot change the particle's angular momentum. This geometric constraint forces the particle to move in a plane. Furthermore, because the force is conservative, energy is conserved. These two conservation laws—angular momentum and energy—reduce the problem from three spatial dimensions to a one-dimensional problem in the radial coordinate, making the trajectory predictable and solvable.
Conservation Laws in Central Fields
For a particle of mass \(m\) subject to a central force \(\mathbf{F} = f(r)\hat{\mathbf{r}}\), the torque about the center is zero. According to the result from D-027, this implies that the angular momentum \(\mathbf{L} = \mathbf{r} \times \mathbf{p}\) is constant in both magnitude and direction. The constancy of the direction of \(\mathbf{L}\) defines the plane of motion. The magnitude of the angular momentum is given by \(L = m r^2 \dot{\theta}\) in polar coordinates, where \(\theta\) is the angular coordinate.
This equation is known as the conservation of areal velocity. It states that the line connecting the particle to the center sweeps out equal areas in equal times, a result first noted by Kepler for planetary motion.
The Two-Body Problem and Reduced Mass
In reality, forces are interactions between two bodies, such as the Earth and the Moon. The general two-body problem is complex because both bodies accelerate. However, by transforming to the center-of-mass frame, we can decouple the motion. As shown in D-028, the relative motion of the two bodies is mathematically identical to the motion of a single fictitious particle with reduced mass \(\mu\) moving in the potential of the other body.
This reduction is powerful because it allows us to apply the single-particle central force results to any two-body system. The center of mass moves uniformly, and the relative coordinate \(\mathbf{r}\) obeys \(\mu \ddot{\mathbf{r}} = -\nabla V(r)\). This simplification is crucial for analyzing binary stars, planetary orbits, and scattering processes.
Connecting to Circular Motion
A special case of central force motion is uniform circular motion. Here, the radial distance \(r\) is constant, and the force provides exactly the centripetal acceleration required. From D-421, we know that for a particle of mass \(m\) moving with speed \(v\) in a circle of radius \(R\), the required force is \(F_c = m v^2 / R\). In the context of central forces, this means \(f(R) = -m v^2 / R\) (the negative sign indicating attraction). This specific case serves as a benchmark for understanding more general elliptical orbits, where the speed and radius vary but the angular momentum remains constant.
The derivations
The Two-Body Problem and Reduced Mass
For two point masses \(m_1\) and \(m_2\) interacting through a potential \(V\) that depends only on their separation \(\mathbf{r}=\mathbf{r}_1-\mathbf{r}_2\), the change of variables to the center-of-mass position \(\mathbf{R}\) and the relative position \(\mathbf{r}\) decouples the dynamics exactly: the center of mass moves as a free particle of mass \(M=m_1+m_2\), while the relative coordinate obeys the equation of motion of a single fictitious particle of reduced mass \(\mu = \dfrac{m_1 m_2}{m_1+m_2}\) moving in the fixed external potential \(V(\mathbf{r})\).
Angular Momentum, Torque, and Central Forces
For a particle of position \(\vec{r}\) and momentum \(\vec{p}=m\vec{v}\), define the angular momentum about a fixed origin \(O\) as \(\vec{L}=\vec{r}\times\vec{p}\). Newton's second law implies that its time rate of change equals the net torque \(\vec{\tau}=\vec{r}\times\vec{F}\) about the same origin, \(\dfrac{d\vec{L}}{dt}=\vec{\tau}\). Consequently, whenever the net force is central — everywhere parallel or antiparallel to \(\vec{r}\), i.e. \(\vec{F}=f(r)\,\hat{r}\) — the torque vanishes and \(\vec{L}\) is conserved.
Uniform Circular Motion and Centripetal Acceleration
For a particle of mass \(m\) moving in a circle of radius \(R\) with constant angular speed \(\omega\), the acceleration is directed radially inward with magnitude \(a_c = \omega^2 R = v^2/R\), where \(v = \omega R\) is the constant speed; Newton's second law then requires a net inward force \(F_c = m v^2/R = m \omega^2 R\), and the period of revolution is \(T = 2\pi/\omega = 2\pi R/v\).
Students often confuse the reduced mass \(\mu\) with the actual mass of either body. Remember that \(\mu\) is a mathematical construct that appears only in the relative coordinate equation; it does not represent a physical object. The actual masses \(m_1\) and \(m_2\) still determine the center-of-mass motion. C
Worked example
Problem. Two stars of masses \(m_1 = 2.0 \times 10^{30}\ \mathrm{kg}\) and \(m_2 = 1.0 \times 10^{30}\ \mathrm{kg}\) orbit their common center of mass in circular orbits with a separation of \(r = 1.0 \times 10^{11}\ \mathrm{m}\). Calculate the reduced mass \(\mu\) and the centripetal force required to maintain this orbit, assuming the gravitational force provides the centripetal acceleration. Use \(G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}\).
- Reduced mass: \(\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{(2.0)(1.0)}{3.0}\times10^{30}\ \mathrm{kg} = 6.67\times10^{29}\ \mathrm{kg}\).
- Determine the gravitational force: \(F_g = \frac{G m_1 m_2}{r^2} = \frac{(6.67\times10^{-11})(2.0\times10^{30})(1.0\times10^{30})}{(1.0\times10^{11})^2} = 1.33\times10^{28}\ \mathrm{N}\).
- Verify that this force equals the centripetal force \(F_c = \mu \frac{v^2}{r}\) by finding the orbital speed \(v\) from the force balance.
Units check. The reduced mass has units of \(\mathrm{kg}\). The force calculation yields \(\frac{\mathrm{N\,m^2/kg^2 \cdot kg^2}}{\mathrm{m^2}} = \mathrm{N}\), which is consistent.
Three quick tests
- Why does a central force conserve angular momentum?
Answer
The torque \(\mathbf{\tau} = \mathbf{r} \times \mathbf{F}\) is zero because \(\mathbf{F}\) is parallel to \(\mathbf{r}\). - What is the physical significance of the reduced mass \(\mu\)?
Answer
It allows the two-body problem to be treated as a one-body problem in the relative coordinate. - If the force is attractive, what is the sign of \(f(r)\) in \(\mathbf{F} = f(r)\hat{\mathbf{r}}\)?
Answer
Negative, as the force points opposite to the radial unit vector.
Problem set for PU-101 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Derivations cited: D-028 · D-027 · D-421 · Threads: force, energy, matter