Central-force motion and the Kepler problem

Start with a force, a position and a velocity

Place a 2 kg point particle at (1, 0) m with velocity (0, 3) m/s. The attracting center stays fixed at the origin of an inertial Cartesian frame: right is positive x and up is positive y. Supply just one force, directed toward that center, with magnitude κ/r², where r is the distance from the origin and κ = 12 N m². At this starting point the force is (−12, 0) N and the acceleration is (−6, 0) m/s². These are exact stipulated model values.

A central force points along the line joining the particle and the chosen center. Here F = −(κ/r²)e_r, where e_r is the outward radial unit vector. The minus sign makes the force attractive. The force magnitude changes with r; it is not a constant acceleration model. We use Newtonian motion, constant mass, a fixed center, no other forces and r > 0.

This is the fixed-center Kepler problem for an attractive inverse-square force. We will obtain an ellipse whose nearest point is the starting position. It is a mathematical teaching model, not a measured planetary orbit. The center is fixed by assumption; an isolated two-body system needs a separate relative-motion treatment.

Keep energy and angular momentum distinct

Momentum is p = m v. Angular momentum about the fixed origin is r × p. A central force has zero torque about that origin because r and F lie on the same line. Newton’s equation therefore keeps this angular-momentum vector constant. Our initial value is ℓ = +6 kg m²/s along the positive z direction, perpendicular to the drawing. The positive sign means counterclockwise motion. Nonzero angular momentum keeps this central-force motion in one plane.

For the supplied time-independent force, choose potential energy V = −κ/r with its zero at infinite separation. Kinetic energy is K = m|v|²/2, and E = K + V is constant along a solution of this conservative model. Initially K = 9 J and V = −12 J, so E = −3 J. Negative potential energy reflects the stated reference, not negative speed or kinetic energy.

The two conservation statements have different conditions. A force that always remains radial has zero torque about the fixed origin, but a time-dependent force strength need not conserve this E. Our force and potential are explicitly time independent. We use the same origin, force law and potential reference throughout.

Separate outward motion from motion around the center

Let the angle from positive x be θ = φ rad. Here φ is its dimensionless numerical radian value; counterclockwise is positive. The outward radial component is v_r = dr/dt, and the transverse component is v_φ = r dφ/dt. The two local directions are perpendicular, so |v|² = v_r² + v_φ². Positive v_r means moving away from the center, even on the left side of the plot.

The angular momentum is ℓ = m r v_φ = m r² dφ/dt. Thus v_φ = ℓ/(m r) = (3 m²/s)/r in this model. At the nearest radius 1 m it is 3 m/s; at the farthest radius 3 m it is 1 m/s. Angular momentum is constant while transverse speed and angular rate change.

At a radial turning point v_r is zero: the distance from the center momentarily stops increasing or decreasing. The particle still moves transversely when ℓ is nonzero. A turning radius is therefore not a point of rest. Also, the local radial direction turns as the particle moves; the radial component of acceleration is not just the derivative of v_r.

Use the effective potential to find allowed radii

Insert v_φ = ℓ/(m r) into the energy. It gives E = m v_r²/2 + U_eff(r), where U_eff = ℓ²/(2 m r²) − κ/r. At fixed ℓ = 6 kg m²/s, U_eff = (9 J m²)/r² − (12 J m)/r. The first term accounts for transverse kinetic energy. It is not another applied outward force on the particle.

Radial kinetic energy cannot be negative, so a possible radius must satisfy E ≥ U_eff. Equality gives a radial turning point. With E = −3 J, this condition restricts the radius to 1 m ≤ r ≤ 3 m. Values outside that interval do not belong to this supplied energy and angular momentum, although the potential graph can still be drawn there.

A different energy at the same angular momentum gives a useful comparison. U_eff has its minimum −4 J at r = 1.5 m. Setting E = −4 J and v_r = 0 gives a circular solution with transverse speed 2 m/s. The inward physical force supplies the needed circular acceleration. This comparison is a different orbit; the E = −3 J ellipse passes through r = 1.5 m with nonzero v_r.

Read the orbit and the energy graph together

The upper panel is position in the fixed x–y plane, with the same metre scale on both axes. The attracting center at (0, 0) m is one focus of the ellipse, not its geometric center at (−1, 0) m. The lower panel is energy versus distance from that focus. Its vertical axis is joules, not physical height.

The four circle markers select angles 0°, 90°, 180° and 270°. They are not equal-time samples. The table gives the signed local velocity components at those angles. At 90° the radius is increasing; at 270° it is decreasing. Both states have the same K, V and E.

The horizontal dashed energy line intersects U_eff at the two square turning markers. The triangle is the separate circular minimum. The drawn U_eff curve joins 131 supplied radii on 0.75 m ≤ r ≤ 4 m; its interpolation error is bounded by 2/225 J, less than 0.009 J. That is a drawing approximation bound, not measurement uncertainty. The upper ellipse is a geometric orbit, not a plot of position against time.

State the ellipse and check its scope

For this supplied state, the orbit is r(φ) = (1.5 m)/(1 + 0.5 cos φ), with 0 ≤ φ ≤ 2π. Its eccentricity is e = 0.5, its semi-major axis is a = 2 m and its semi-minor axis is b = √3 m. Equivalently, its Cartesian equation is (x + 1 m)²/(4 m²) + y²/(3 m²) = 1. A semi-axis is half an ellipse’s full width or height. The nearest and farthest radii are 1 m and 3 m.

The inverse-square force is essential to this conic form. With nonzero angular momentum, inverse-square solutions include ellipses for negative E, a parabolic case at E = 0 and hyperbolic cases for positive E. We are using the bounded ellipse only. A generic central force need not produce an ellipse or a closed bound orbit.

Optional derivative check: write u = 1/r = (2 + cos φ)/(3 m). The inverse-square orbit equation is d²u/dφ² + u = mκ/ℓ² = 2/(3 m). Differentiating this u twice makes the cosine terms cancel. Together with dφ/dt = ℓ/(m r²), this checks the whole model orbit, not only its four marked states. The shape equation alone would not specify the timing; the angular rate supplies that part.

Use swept area to connect shape and time

The radius line from the attracting focus to the particle sweeps out area as the particle moves. Its area rate is dA/dt = r²(dφ/dt)/2 = ℓ/(2m) = 1.5 m²/s. Equal time intervals therefore sweep equal areas in this model. They do not give equal angle increments, equal path lengths or constant speed.

One full ellipse encloses area πab = 2π√3 m². Dividing by 1.5 m²/s gives the modeled period T = 4π/√3 s. We take the starting nearest point at t = 0 and one complete orbit on 0 ≤ t ≤ T. The diagram labels angles rather than assigning equally spaced times to them.

For fixed-center inverse-square ellipses with the same μ = κ/m, the general relation is T² = 4π²a³/μ. Our μ is 6 m³/s², and T²/a³ = 2π²/3 s²/m³. The common force parameter is part of the comparison. This is not a period rule for arbitrary central forces.

Keep the model’s boundaries visible

The model is singular at r = 0, and no formula here is applied there. Our nonzero ℓ and the supplied negative energy keep this orbit between 1 m and 3 m. Zero-angular-momentum radial motion, moving attracting bodies, added forces and relativistic corrections require separate treatments.

The exact numbers test internal consistency of the stipulated model; they do not show how accurately it describes any actual astronomical system. Changing the force law changes the orbit equation and effective potential. Changing the potential reference shifts E and U_eff together without changing the force or the condition E − U_eff ≥ 0.

A circular comparison at an effective-potential minimum is one way to organize a local question. Small oscillations and normal modes develops nearby motion further. For a first reading here, the force, component table, radial bounds and swept-area calculation already give a complete example; the derivative check is an optional return.

Worked examples

Try it yourself

1. At a state with m = 2 kg, r = 1.5 m and transverse velocity v_φ = 2 m/s, calculate signed ℓ and dφ/dt. Use the same positive transverse direction as the lesson.

   Show the answer and reasoning

   ℓ = +6 kg m²/s and dφ/dt = 4/3 s⁻¹.

   ℓ = m r v_φ = 2 × 1.5 × 2 = 6 kg m²/s. The numerical radian rate is v_φ/r = 2/1.5 s⁻¹; the physical angular rate is (4/3) rad/s. The sign is positive because the transverse direction is positive.

2. At the farthest radius 3 m in the supplied ellipse, v_r = 0. Is the particle at rest? Find its speed and kinetic energy.

   Show the answer and reasoning

   It is not at rest. Its speed is 1 m/s and K = 1 J.

   v_φ = ℓ/(m r) = 6/(2 × 3) m/s = 1 m/s. With v_r = 0 the speed is |v_φ|. K = m v_φ²/2 = 1 J; only the radial distance is at a turn.

3. For the same E = −3 J and ℓ = 6 kg m²/s, could the particle reach r = 4 m? Evaluate E − U_eff rather than just reading the plot.

   Show the answer and reasoning

   No. E − U_eff = −9/16 J is negative.

   U_eff(4 m) = 9/16 J − 3 J = −39/16 J. Thus E − U_eff = −48/16 J + 39/16 J = −9/16 J, which cannot be radial kinetic energy. This rejection concerns the supplied E and ℓ, not every possible state at 4 m.

4. At fixed ℓ = 6 kg m²/s, the effective minimum is at r = 1.5 m. Find the circular state’s transverse speed, K, V and E. Is it the ellipse’s 90° state?

   Show the answer and reasoning

   v_φ = 2 m/s, K = 4 J, V = −8 J and E = −4 J. It is a different state from the ellipse’s 90° state.

   The circular state has v_r = 0 and K = (2 kg)(2 m/s)²/2 = 4 J. The ellipse at 90° has v_r = +1 m/s, so its K is 5 J and E is −3 J. Same radius and transverse component do not mean the same full velocity or energy.

5. A reader calls the four angle markers “four equal-time snapshots” and says angular momentum conservation makes speed constant. Are either claims justified?

   Show the answer and reasoning

   Neither claim is justified.

   The markers select angles, not times. Since dφ/dt = ℓ/(m r²), the angular rate is greater nearer the center. Transverse speed is ℓ/(m r), and the full speed also includes v_r. Equal area per unit time does not make speed constant.

6. Find the ratio of total speed at the nearest point to total speed at the farthest point. Explain why the simple transverse formula suffices at those two points.

   Show the answer and reasoning

   The ratio is 3.

   Both are radial turning points, so v_r = 0. Their speeds are |v_φ| = 3 m/s and 1 m/s. With the same ℓ and mass, v_φ is inversely proportional to r; the smaller radius gives the greater speed.

What to carry forward

• A central force has zero torque about its fixed center; the time-independent potential of the supplied inverse-square force also conserves energy.
• Constant angular momentum allows changing transverse speed and angular rate; a radial turn need not be rest.
• At fixed angular momentum, effective potential combines potential energy with transverse kinetic accounting and identifies allowed radii.
• This geometric ellipse is an inverse-square solution with the attracting focus at the origin; its angle markers are not time samples.
• Equal swept areas connect shape to time within the stated model; generic central forces and actual orbit accuracy need separate checks.