Central-force motion
Objective. Use angular momentum and effective potentials.
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- 2. Worked example
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Learn: the key idea
Central-force motion describes particles moving under a force always directed toward or away from a fixed centre, conserving angular momentum L = mr²θ̇ and reducing the problem to one radial variable via an effective potential. The energy equation becomes ½mṙ² + V_eff(r) = E, where V_eff(r) = V(r) + L²/(2mr²), the second term acting as a repulsive centrifugal barrier. Example: for gravity V(r)=−GMm/r, circular orbits occur where V_eff'(r)=0, giving the balance between gravitational pull and centrifugal term. This effective-potential picture explains bound, unbound and circular orbits without solving the full two-dimensional motion directly.
Worked example
A particle undergoes central-force motion with potential V(r) = −k/r and angular momentum L. Find the radius of the circular orbit in terms of k, L and m.
- 1Write effective potential: V_eff(r) = −k/r + L²/(2mr²)
- 2Circular orbit requires V_eff'(r) = 0: k/r² − L²/(mr³) = 0
- 3Multiply by r³: kr = L²/m
- 4Solve for r: r = L²/(mk)
Answer: r = L²/(mk)
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Start practisingCommon mistake. Students drop the centrifugal term L²/(2mr²) from the effective potential, treating radial motion as if under V(r) alone; this term is essential and prevents particles with angular momentum from reaching r=0.
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