Central forces and two-body motion

1. Radial interactions

A planet and a star interact along their line of separation. The force direction therefore changes as they move, but it always passes through the same relative origin. That geometrical restriction has stronger consequences than a particular choice of force magnitude.

Definition. A time-independent central force about an origin has the form

Here $F(r)$ is a signed scalar: negative for attraction, positive for repulsion. The force points along the radius and its magnitude depends only on distance. Examples include gravitational attraction $F=-k/r^2$, electrostatic attraction or repulsion, and an isotropic spring force $F=-\kappa r$. A force that merely points toward a moving unrelated point need not satisfy this fixed-center model.

A radial force with explicit time dependence still has zero torque, but need not conserve energy. A radial force depending on speed can also remove energy. We shall state when an argument uses direction alone and when it needs the stronger time-independent form above.

2. Torque, angular momentum, and planarity

For a particle of mass $\mu$ in an inertial frame, define linear momentum $\mathbf p=\mu\dot{\mathbf r}$ and angular momentum $\mathbf L=\mathbf r\times\mathbf p$. The symbol $\mu$ will soon represent the reduced mass; for a particle about a fixed center, simply put $\mu=m$.

Differentiate the cross product:

The first term is $\mu\dot{\mathbf r}\times\dot{\mathbf r}=0$, because a vector crossed with itself is zero. Newton's law gives $\dot{\mathbf p}=F(r)\hat{\mathbf r}$, so the second term is $F(r)\mathbf r\times\hat{\mathbf r}=0$. Consequently,

The torque about the force center vanishes and both the magnitude and direction of angular momentum remain constant. This is stronger than conservation of only one component.

At every instant $\mathbf r\cdot\mathbf L=0$ by the definition of a cross product. If $\mathbf L\ne0$, a fixed nonzero normal vector defines a fixed plane through the origin. Thus the entire orbit lies in that plane. Its initial position and velocity determine it.

If $\mathbf L=0$, position and velocity are parallel and the motion is radial along a line, away from collision singularities. The statement “the plane is perpendicular to $\mathbf L$” does not select a plane when $\mathbf L=0$. Formulas dividing by its magnitude will exclude this radial case.

3. Areal velocity

During a short displacement $d\mathbf r$, the radius vector sweeps a triangle of area $dS=|\mathbf r\times d\mathbf r|/2$ to first order. Divide by $dt$:

In oriented polar coordinates within the orbital plane, choose increasing $\theta$ in the direction of motion so that $\ell=\mu r^2\dot\theta>0$. The relation becomes $dS/dt=r^2\dot\theta/2$. Equal time intervals sweep equal areas for every central force, not only inverse-square gravity. Kepler's second law is a particular application.

4. Potential and energy conservation

For a time-independent radial function $F(r)$, define a potential on the radial interval of interest by

Because $\nabla r=\hat{\mathbf r}$, $\nabla V=V'(r)\hat{\mathbf r}$ and $-\nabla V=F(r)\hat{\mathbf r}$. Thus the central force is conservative on the region where this potential is defined. For $F=-k/r^2$, integration gives $V=-k/r$ if the zero is chosen at infinity.

Take the dot product of Newton's law with velocity:

The left side is $d(\mu v^2/2)/dt$ and the right side is $-dV/dt$. Hence

The proof uses time independence of $V$. For $V(r,t)$, the same calculation gives $\dot E=\partial V/\partial t$, while angular momentum remains conserved. Linear momentum of a particle about a fixed attracting center is generally not conserved, because its force is nonzero.

5. Reduction of the isolated two-body problem

An actual star also moves under the planet's force. Let two masses $m_1,m_2$ interact through $V(|\mathbf r_1-\mathbf r_2|)$ with no external forces. Define

Solving the two definitions gives $\mathbf r_1=\mathbf R+(m_2/M)\mathbf r$ and $\mathbf r_2=\mathbf R-(m_1/M)\mathbf r$. Expand their velocity squares in $T=m_1\dot{\mathbf r}_1^2/2+m_2\dot{\mathbf r}_2^2/2$. The cross coefficient is $m_1m_2/M-m_2m_1/M=0$. The relative-square coefficient is

Therefore

The center coordinate is cyclic, so $M\ddot{\mathbf R}=0$. The relative coordinate obeys

The isolated two-body problem separates into uniform center-of-mass motion and an equivalent one-body central-force problem of mass $\mu$. In the center-of-mass frame the total energy and angular momentum are precisely the relative $E$ and $\mathbf L$. In a general inertial frame, add $M\dot{\mathbf R}^2/2$ to energy and $\mathbf R\times M\dot{\mathbf R}$ to angular momentum about the same origin.

6. Applications

Example 1 · Torque at a specified position

A particle is at $\mathbf r=(3,4,0)\,\mathrm m$ and experiences an attractive force of magnitude $10\,\mathrm N$. Since $r=5\,\mathrm m$, $\mathbf F=-10(3,4,0)/5=(-6,-8,0)\,\mathrm N$. The only possible torque component is $\tau_z=3(-8)-4(-6)=0$. The force is nonzero, so the linear momentum changes even though angular momentum does not.

Example 2 · The orbital plane from initial data

Let $\mu=2\,\mathrm{kg}$, $\mathbf r_0=(2,0,0)\,\mathrm m$, and $\mathbf v_0=(1,3,0)\,\mathrm{m\,s^{-1}}$. Then $\mathbf L=\mu\mathbf r_0\times\mathbf v_0=(0,0,12)\,\mathrm{kg\,m^2\,s^{-1}}$. The motion stays in the $xy$ plane and $dS/dt=12/(2\times2)=3\,\mathrm{m^2\,s^{-1}}$. The initial radial velocity $1\,\mathrm{m\,s^{-1}}$ contributes no angular momentum.

Example 3 · Comparable masses

For $m_1=3m$ and $m_2=m$, $\mu=3m/4$. In the center-of-mass frame, $\mathbf r_1=\mathbf r/4$ and $\mathbf r_2=-3\mathbf r/4$. Each body follows a scaled version of the relative trajectory, on opposite sides of the common center. Replacing $\mu$ by the lighter mass would overestimate the inertial coefficient of relative motion by a factor $4/3$.

Example 4 · A radial force with changing strength

Take $\mathbf F=-k(t)\hat{\mathbf r}/r^2$ and $V=-k(t)/r$. Torque is still zero, so the orbital plane and areal velocity remain fixed. But $\dot E=\partial_tV=-\dot k/r$. If $k$ increases, the instantaneous mechanical energy decreases. A conservation proof based only on radial direction would therefore be incomplete.

7. What has been reduced?

Planarity removes one spatial direction. Angular-momentum conservation relates angular velocity to radius. Energy conservation then reduces radial motion to one first-order equation. None of these steps yet chooses the force to be inverse-square. The next chapters use this general structure before specializing to planetary orbits.

Exercises

Hints and solutions are collected separately.

1. 16.1 What two restrictions define the time-independent central force used here?

2. 16.2 Prove angular-momentum conservation for a central force.

3. 16.3 Prove planarity and explain the zero-angular-momentum exception.

4. 16.4 For $\mu=3$, $\mathbf r=(2,0,0)$, $\mathbf v=(1,2,0)$ in SI units, find $\mathbf L$ and areal rate.

5. 16.5 Find reduced mass for masses $2m$ and $3m$.

6. 16.6 Derive the kinetic-energy separation for an isolated two-body system.

7. 16.7 For $F=-Cr^2$ with $C>0$, find a potential and prove energy conservation.

8. 16.8 If the force is radial but explicitly time-dependent, which central-force properties survive?

9. 16.9 Why is the reduced mass approximately the lighter mass for a very heavy central body?

10. 16.10 Does constant angular momentum imply constant speed?

11. 16.11 Show that the center of mass moves uniformly in the isolated two-body problem.

12. 16.12 Why is conservation of relative energy not the same numerical statement as total laboratory energy?