Central force motion

A satellite curving around Earth, a comet swinging past the Sun, an electron looping a nucleus — all obey the same secret bookkeeping rule. Whenever the only force on a particle points straight at (or away from) one fixed point, something about its motion never changes, no matter how the distance or speed shifts. That fixed quantity is angular momentum about the centre of force, and it falls out of Newton's second law the moment you write it in polar coordinates: a force with no sideways (transverse) component can exert no twist about its own centre, so nothing can speed up or slow down the particle's "sweep" around that point. The result is a single conserved number, $mr^2\dot\theta$ , that lets you jump straight from the geometry at one point on the path to the speed at another — without ever solving for the shape of the orbit. This article builds that conservation law from the moment-of-force argument, applies it to two-point speed problems exactly as ME10001 expects, and stops right at the doorway to orbital mechanics — the differential equation of the orbit itself is a later course's business.

The core insight

A central force has zero moment about its own centre, so a particle's angular momentum $mr^2\dot\theta$ about that centre stays constant, which directly links speed and radius at any two points of the path.

The core idea

What makes a force "central". A force $\vec F$ on a particle is central if its line of action always passes through one fixed point $O$ , called the centre of force. In polar coordinates centred on $O$ , this means $\vec F$ has only a radial component:

\vec F = F(r)\,\hat e_r,

where: $F(r)$ is the (signed) magnitude of the force, a function of distance $r$ alone, and $\hat e_r$ is the unit vector pointing from $O$ to the particle. There is no $\hat e_\theta$ (transverse) component — that absence is the entire content of "central."

Why angular momentum survives. Moment of a force about a point equals (perpendicular distance) times (force), and for a central force that perpendicular distance from $O$ to the line of action is zero — the force points straight along the radius through $O$ itself. Zero moment about $O$ means the rate of change of angular momentum about $O$ is zero, which is exactly what the next derivation proves formally.

The conserved quantity in practice. Once $H_O = mr^2\dot\theta$ is constant, it is equivalent to say the particle's areal velocity — the rate at which the radius vector sweeps out area — is constant:

\frac{dA}{dt} = \tfrac12 r^2\dot\theta = \text{constant}.

where: $A$ is the area swept by the position vector $\vec r$ since some reference time. Equal areas in equal times is the geometric face of the same conservation law; the figure below shows why a particle sweeping a thin wedge near $O$ must move much faster there than a particle sweeping an equally sized wedge far from $O$ .

Figure

A particle orbiting a centre of force sweeps two thin wedges of equal area in equal time, a short fast arc near the centre and a long slow arc far from it

What this buys you. Because $H_O$ is constant, the transverse velocity component $v_\theta = r\dot\theta$ at any two points satisfies $r_1 (v_\theta)_1 = r_2 (v_\theta)_2$ , letting you solve two-point speed problems without ever finding $F(r)$ or the trajectory shape. Combine it with the radial equation of motion $\sum F_r = m(\ddot r - r\dot\theta^2)$ when a problem asks for the force itself. Carrying this further into the actual differential equation of the orbit — and from there to ellipses, parabolas and Kepler's laws for gravity — is orbital mechanics and lies outside this course.

Derivation AAngular momentum about the centre of force is conserved

The goal is to show, starting only from Newton's second law written in polar coordinates and the defining property of a central force, that the quantity $H_O = mr^2\dot\theta$ does not change with time, so that its value at any one instant of the motion can be used to relate conditions at any other instant.

Step 1. Write Newton's second law in polar coordinates. For a particle of mass $m$ at position $(r,\theta)$ from $O$ , the equations of motion are $\sum F_r = m(\ddot r - r\dot\theta^2)$ in the radial direction and $\sum F_\theta = m(r\ddot\theta + 2\dot r\dot\theta)$ in the transverse direction. These come from differentiating the position vector twice in polar form and are standard results of particle kinematics.

Step 2. Apply the central-force condition. Since $\vec F = F(r)\hat e_r$ has no transverse component, $\sum F_\theta = 0$ , so the transverse equation becomes $m(r\ddot\theta + 2\dot r\dot\theta) = 0$ , which simplifies (dividing by $m$ and multiplying by $r$ ) to $r^2\ddot\theta + 2r\dot r\dot\theta = 0$ .

Step 3. Recognize this as a derivative. The left side is exactly $\dfrac{d}{dt}\left(r^2\dot\theta\right)$ , obtained by the product rule. So Step 2's result says $\dfrac{d}{dt}\left(r^2\dot\theta\right) = 0$ , meaning $r^2\dot\theta$ is constant throughout the motion.

Step 4. Multiply through by the (constant) mass $m$ to write the result in the standard form for angular momentum about $O$ : $H_O = mr^2\dot\theta = \text{constant}$ . This constant is fixed by the initial conditions of the particular problem and can be evaluated at any single instant — often the simplest one, such as a point of closest or farthest approach where the radial velocity is momentarily zero — then reused at any other point on the path.

Worked example

A particle of mass $m = 0.5\ \text{kg}$ moves under a central force directed toward a fixed point $O$ . At one instant it is at $r_1 = 2\ \text{m}$ with transverse velocity $(v_\theta)_1 = 6\ \text{m/s}$ .

Step 1. Identify the conserved quantity. Because the force is central, angular momentum about $O$ is conserved: $mr_1(v_\theta)_1 = mr_2(v_\theta)_2$ , and the mass cancels to give $r_1(v_\theta)_1 = r_2(v_\theta)_2$ .

Step 2. Find the transverse speed at $r_2 = 3\ \text{m}$ . $(v_\theta)_2 = \dfrac{r_1(v_\theta)_1}{r_2} = \dfrac{2\times 6}{3} = 4\ \text{m/s}$ . This matches the reference calculation for this problem — moving farther from $O$ , the transverse speed drops proportionally.

Step 3. Extend to a third point using the same constant. Suppose the particle later reaches $r_3 = 1\ \text{m}$ (swinging back closer to $O$ ). The same conserved quantity gives $(v_\theta)_3 = \dfrac{r_1(v_\theta)_1}{r_3} = \dfrac{2\times 6}{1} = 12\ \text{m/s}$ .

Step 4. Sanity-check the trend. As $r$ decreases from $2\ \text{m}$ to $1\ \text{m}$ , the transverse speed rises from $6\ \text{m/s}$ to $12\ \text{m/s}$ — doubling as $r$ halves, exactly consistent with $r\,v_\theta$ staying fixed at $mr v_\theta / m = 2\times 6 = 12\ \text{m}^2/\text{s}$ throughout. At $r_2 = 3\ \text{m}$ , $(v_\theta)_2 = 4\ \text{m/s}$ ; at $r_3 = 1\ \text{m}$ , $(v_\theta)_3 = 12\ \text{m/s}$ .

Common traps

Applying angular momentum conservation about a point other than the actual centre of force — the conservation law only holds about $O$ , not about an arbitrary reference point.
Assuming the particle's total speed is conserved. It is not, in general; only the product $r v_\theta$ (equivalently $H_O$ ) stays constant, and radial speed can still change.
Forgetting that at a point of closest or farthest approach the velocity is purely transverse ( $v_r = 0$ momentarily), which is often the easiest place to evaluate $H_O$ .
Reaching for the full orbit-shape equation (ellipse, parabola, hyperbola) when the problem only needs the two-point conservation relation — that derivation is beyond this course's scope.

Ask the tutor: Want to see how combining the radial equation of motion with angular momentum conservation lets you find the force magnitude at a specific point, or how this conservation law becomes the starting point for deriving orbit shapes in a later course?

Why does a central force have zero moment about its own centre

Moment of a force about a point equals the perpendicular distance from that point to the force's line of action, multiplied by the force magnitude. For a central force, the line of action passes directly through the centre of force $O$ itself, so the perpendicular distance from $O$ to that line is zero. Zero perpendicular distance means zero moment, and since angular momentum's rate of change about a point equals the moment of the net force about that same point, angular momentum about $O$ cannot change — it is conserved for the entire motion.

Does this conservation law work for any force, or only gravity

It works for any force whose line of action always passes through one fixed point, regardless of how the force magnitude depends on distance, $F(r)$ . Gravity is the most common example in this course because planetary and satellite problems are central-force problems, but the same $mr^2\dot\theta = \text{constant}$ result would hold equally for a spring-like attractive force toward $O$ or a repulsive force away from $O$ — the derivation in the box above never assumed a specific $F(r)$ .

If speed is not conserved, what physical quantity is actually constant here

Two equivalent things are constant: the angular momentum $H_O = mr^2\dot\theta$ , and the areal velocity $dA/dt = \tfrac12 r^2\dot\theta$ , the rate at which the radius vector sweeps out area. Neither the particle's speed nor its transverse velocity alone stays fixed — as $r$ shrinks, $v_\theta$ must grow to keep the product $r v_\theta$ constant, and vice versa. Total speed only stays constant in the special case of a circular orbit, where $r$ itself never changes.

Central force motion

The core idea

Worked example

Common traps

Before this

Where this leads