Difficulty: Introductory review | Prerequisites: P211 circular motion, centripetal acceleration, Newton's law of gravitation
This topic derives Kepler's third law from Newton's second law applied to circular orbits, then generalises the result to any central force that may depend on both velocity and radius. The punchline is the cyclotron result: when the force is proportional to velocity and independent of radius, the orbital frequency is constant regardless of the orbit size. In P212 this is exactly what happens to a charged particle in a uniform magnetic field.
Setting the centripetal force equal to the gravitational force for a circular orbit gives Kepler's third law: T^2 is proportional to R^3. Generalising to a force of the form F = D v^a R^b yields a power-law relationship between angular frequency and radius. The special case where the force is proportional to velocity and independent of radius produces a constant angular frequency, which is the operating principle of the cyclotron.
Kepler's third law (K-III)
For circular planetary orbits: T^2 = C R^3, where T is the orbital period, R is the orbital radius, and C is a constant that depends on the central mass. In simple terms, larger orbits take disproportionately longer to complete.
Angular frequency (omega)
The rate of rotation in radians per unit time: omega = 2 pi / T. Think of it as how quickly the particle sweeps around the circle, measured in angle rather than distance.
Centripetal acceleration
The inward acceleration required to keep an object moving in a circle: a = v^2 / R = omega^2 R. In simple terms, the object is constantly being pulled toward the centre, which bends its path into a curve.
Cyclotron frequency
The angular frequency of a charged particle moving in a uniform magnetic field. It is independent of the orbital radius, meaning all particles of the same charge-to-mass ratio orbit at the same rate regardless of how fast they are going. This is the key insight behind the cyclotron particle accelerator.
For a planet of mass m in a circular orbit of radius R around a star of mass Ms:
Gravitational force provides the centripetal force: G m Ms / R^2 = m v^2 / R.
Cancel m and solve for v^2: v^2 = G Ms / R.
The period is T = 2 pi R / v. Squaring: T^2 = 4 pi^2 R^2 / v^2.
Substituting v^2: T^2 = 4 pi^2 R^3 / (G Ms).
This is K-III with C = 4 pi^2 / (G Ms).
omega = 2 pi / T, so omega^2 = 4 pi^2 / T^2 = G Ms / R^3.
This tells you that for gravitational orbits, omega decreases as R increases: planets further from the sun orbit more slowly.
For a central force of this form, set F equal to the centripetal force: D v^a R^b = m v^2 / R.
Substitute v = omega R throughout and solve for omega.
Result: omega^(2-a) = D R^(a+b-1) / m.
This is the generalised K-III for any power-law central force.
This recovers the standard gravitational case: F = D / R^2, with D playing the role of G m Ms.
Substituting: omega^2 = D R^(-3) / m = D / (m R^3).
With D = G m Ms: omega^2 = G Ms / R^3. This matches the result from part (b), confirming consistency.
The force is proportional to velocity and independent of radius: F = D v.
Substituting into the generalised result: omega^(2-1) = D R^(1+0-1) / m = D / m.
Therefore omega = D / m.
The angular frequency is independent of R. Every orbit, regardless of size, has the same period.
This is the cyclotron frequency. In P212 you will see that the magnetic force on a moving charge is F = qvB, which has exactly this form (D = qB). The cyclotron frequency is then omega = qB / m.
Quantity | Formula | Notes |
|---|---|---|
Kepler's third law | T^2 = 4 pi^2 R^3 / (G Ms) | For gravity; C = 4 pi^2 / (G Ms) |
Angular frequency (gravity) | omega^2 = G Ms / R^3 | omega decreases with R |
Generalised K-III | omega^(2-a) = D R^(a+b-1) / m | For F = D v^a R^b |
Cyclotron frequency | omega = D / m = qB / m | Independent of R; the secret of the cyclotron |
Students often confuse angular frequency (omega, in rad/s) with ordinary frequency (f, in Hz). They are related by omega = 2 pi f. Make sure you know which one the question asks for.
A common error is thinking that a faster-moving charged particle in a magnetic field orbits at a higher frequency. It does not. It orbits in a larger circle, but at the same frequency. That is the cyclotron result.
Students sometimes forget to eliminate v from the generalised expression. The final answer for omega must not contain v, only omega, R, D, and m.
When checking limiting behaviour by setting a = 0 and b = -2, students sometimes substitute before simplifying. Simplify the general expression first, then substitute.
The cyclotron, one of the earliest particle accelerators, works because the orbital frequency of a charged particle in a magnetic field does not depend on its speed or radius. An alternating electric field can be tuned to a single frequency and will accelerate the particle on every pass, no matter how large its orbit grows. This principle is also central to how mass spectrometers separate ions by charge-to-mass ratio.
The derivation of K-III from Newton's second law is a standard exam question. Know every step.
The generalised result omega^(2-a) = D R^(a+b-1) / m is a powerful tool for any central-force problem. It covers gravity, magnetic forces, and anything in between.
The cyclotron frequency omega = qB/m is one of the most important results in P212. You will use it repeatedly in the magnetism unit.
Always check your generalised result by plugging in a = 0, b = -2 to recover the gravitational case.
Fill in the blank: Kepler's third law states T^2 = C R^3 where C = 4 pi^2 / (___). (G Ms)
True or false: for gravitational orbits, angular frequency increases with orbital radius. (False, it decreases.)
Fill in the blank: for a force F = D v^a R^b, the generalised K-III is omega^(2-a) = D R^___ / m. (a + b - 1)
True or false: the cyclotron frequency depends on the radius of the orbit. (False)
Fill in the blank: when a = 1 and b = 0, omega = ___ / ___. (D / m)
Q: Derive Kepler's third law for a circular orbit. Express the constant C in terms of the sun's mass Ms and the gravitational constant G.
A: Set the gravitational force equal to the centripetal force: G m Ms / R^2 = m v^2 / R. Solve for v^2 = G Ms / R. Substitute into T = 2 pi R / v and square: T^2 = 4 pi^2 R^3 / (G Ms). Therefore C = 4 pi^2 / (G Ms).
Q: Re-express K-III as a relationship between angular frequency omega and radius R.
A: omega = 2 pi / T, so omega^2 = 4 pi^2 / T^2 = G Ms / R^3.
Q: For a general central force F = D v^a R^b, derive the relationship between omega and R.
A: Set D v^a R^b = m v^2 / R. Replace v with omega R: D (omega R)^a R^b = m omega^2 R^2 / R. Simplify: D omega^a R^(a+b) = m omega^2 R. Divide: omega^(2-a) = D R^(a+b-1) / m.
Q: For the case a = 1, b = 0, what happens to the angular frequency? Why is this result physically significant?
A: omega = D/m, independent of R. This means every orbit has the same period. In P212, the magnetic force F = qvB has this form (a = 1, b = 0, D = qB), giving the cyclotron frequency omega = qB/m. A particle accelerator can use a fixed-frequency oscillating field to accelerate particles repeatedly.
Q: Verify that setting a = 0 and b = -2 in the generalised K-III recovers the gravitational result.
A: omega^(2-0) = D R^(0+(-2)-1) / m = D R^(-3) / m = D / (m R^3). With D = G m Ms (since F = G m Ms / R^2 = D R^b with b = -2, so D = G m Ms), omega^2 = G Ms / R^3, which matches.
This connects directly to the P212 magnetism unit, where the magnetic force F = qvB on a moving charge produces circular motion at the cyclotron frequency omega = qB/m. It also connects to Discussion 1A (uniform force producing parabolic deflection, the linear analogue of this circular problem) and to any P212 problem involving charged particles in combined electric and magnetic fields.
Uniform circular motion, Kepler's third law, K-III, angular frequency, centripetal acceleration, centripetal force, cyclotron, cyclotron frequency, magnetic force, Lorentz force, charged particle in magnetic field, mass spectrometer, orbital period, P212 Week 1, P211 review, PHYS 212 UIUC, velocity-dependent force, power-law force