Central Force Problems

6.1 Reduction to One Dimension

For a particle of mass $m$ in a central potential $V(r)$ Using polar coordinates $(r, \phi)$ in the Plane of motion:

$$L = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\phi}^2) - V(r)$$

The angular momentum $l = mr^2\dot{\phi}$ is conserved. Substituting $\dot{\phi} = l/(mr^2)$:

$$L = \frac{1}{2}m\dot{r}^2 + \frac{l^2}{2mr^2} - V(r)$$

The effective potential is $V_{\mathrm{eff}(r) = V(r) + \frac{l^2}{2mr^2}}$ Where the second term is the centrifugal barrier.

6.2 Effective Potential Analysis

Definition. The effective one-dimensional energy is:

$$E = \frac{1}{2}m\dot{r}^2 + V_{\mathrm{eff}(r)}$$

Since $E$ and $l$ are conserved, the radial motion is completely determined by $V_{\mathrm{eff}(r)}$.

Circular orbits occur at radii $r_0$ where $V_{\mathrm{eff}'(r_0) = 0}$:

$$V'(r_0) - \frac{l^2}{mr_0^3} = 0$$

The orbit is stable if $V_{\mathrm{eff}''(r_0) \gt 0}$ and unstable if $V_{\mathrm{eff}''(r_0) \lt 0}$.

For the Kepler problem $V(r) = -k/r$:

$$V_{\mathrm{eff}(r) = -\frac{k}{r} + \frac{l^2}{2mr^2}}$$ $$V_{\mathrm{eff}'(r) = \frac{k}{r^2} - \frac{l^2}{mr^3} = 0 \implies r_0 = \frac{l^2}{mk}}$$ $$V_{\mathrm{eff}''(r_0) = -\frac{2k}{r_0^3} + \frac{3l^2}{mr_0^4} = \frac{m^3k^2}{l^4} \gt 0}$$

So the circular orbit is always stable for the Kepler problem.

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The effective potential $V_{\mathrm{eff}}(r) = -k/r + l^2/(2r^2)$ (blue) combines the attractive $-1/r$ well with the centrifugal barrier $\propto 1/r^2$. Adjust sliders k (force constant), l (angular momentum), and E (total energy) to explore bound and unbound orbits.

6.3 The Orbit Equation

Starting from conservation of energy and angular momentum, we derive the orbit equation. Let $u = 1/r$ and use $d/dt = (l/mr^2)\, d/d\phi = (lu^2/m)\, d/d\phi$:

$$\dot{r} = \frac{dr}{dt} = \frac{l}{mr^2}\frac{dr}{d\phi} = -\frac{l}{m}\frac{du}{d\phi}$$

Substituting into the energy equation:

$$E = \frac{l^2}{2m}\left(\frac{du}{d\phi}\right)^2 + \frac{l^2 u^2}{2m} + V\left(\frac{1}{u}\right)$$

Differentiating with respect to $\phi$:

$$\frac{l^2}{m}\frac{du}{d\phi}\frac{d^2u}{d\phi^2} + \frac{l^2 u}{m}\frac{du}{d\phi} - \frac{1}{u^2}V'\left(\frac{1}{u}\right)\frac{du}{d\phi} = 0$$

Dividing by $l^2 u'/(m)$ (assuming $u' \neq 0$):

$$\frac{d^2u}{d\phi^2} + u = -\frac{m}{l^2 u^2}V'\left(\frac{1}{u}\right)$$

This is the Binet equation.

6.4 The Kepler Problem

For $V(r) = -k/r$ (gravitational or Coulomb potential):

$$V'\left(\frac{1}{u}\right) = \frac{dV}{dr} = \frac{k}{r^2} = ku^2$$

The Binet equation becomes:

$$\frac{d^2u}{d\phi^2} + u = \frac{mk}{l^2}$$

This is an inhomogeneous ODE with solution:

$$u(\phi) = \frac{mk}{l^2}(1 + e\cos(\phi - \phi_0))$$

Where $e$ is the eccentricity and $\phi_0$ is a constant. Setting $\phi_0 = 0$:

$$r(\phi) = \frac{l^2/(mk)}{1 + e\cos\phi} = \frac{p}{1 + e\cos\phi}$$

Where $p = l^2/(mk)$ is the semi-latus rectum.

The eccentricity is determined by the energy:

$$e = \sqrt{1 + \frac{2El^2}{mk^2}}$$

Classification of orbits:

Bertrand’s Theorem: The only central potentials that give closed orbits for all bound states are $V(r) \propto 1/r$ (Kepler/Coulomb) and $V(r) \propto r^2$ (harmonic oscillator).

Key results for Kepler orbits:

• Orbits are conic sections (ellipses, parabolas, or hyperbolas).
• The semi-major axis $a$ satisfies $E = -k/(2a)$ for bound orbits.
• The period is $T = 2\pi\sqrt{ma^3/k}$ (Kepler’s third law).

6.5 Worked Example: Satellite Orbit

Problem. A satellite of mass $m$ orbits Earth ($M = 5.97 \times 10^{24}\,\mathrm{kg}$) in an elliptical orbit with perigee (closest approach) $r_p = 7000\,\mathrm{km}$ and apogee (farthest point) $r_a = 42000\,\mathrm{km}$. Find the eccentricity, semi-major axis, and orbital period.

Solution

The semi-major axis is $a = (r_p + r_a)/2 = (7000 + 42000)/2 = 24500\,\mathrm{km}$.

From the orbit equation, at perigee ($\phi = 0$): $r_p = p/(1 + e)$ And at apogee ($\phi = \pi$): $r_a = p/(1 - e)$.

Therefore:

$$e = \frac{r_a - r_p}{r_a + r_p} = \frac{42000 - 7000}{42000 + 7000} = \frac{35000}{49000} \approx 0.714$$

The energy is $E = -k/(2a)$ where $k = GMm = 6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times m = 3.986 \times 10^{14} m\,\mathrm m^3/\mathrm s^2$.

The period (independent of mass $m$):

$$T = 2\pi\sqrt{\frac{ma^3}{k}} = 2\pi\sqrt{\frac{a^3}{GM}} = 2\pi\sqrt{\frac{(2.45 \times 10^7)^3}{3.986 \times 10^{14}}}$$$$= 2\pi\sqrt{\frac{1.471 \times 10^{22}}{3.986 \times 10^{14}}} = 2\pi\sqrt{3.691 \times 10^7} \approx 2\pi \times 6075 \approx 38170\,\mathrm s \approx 10.6\,\mathrm{hours}$$

$\blacksquare$

6.6 Worked Example: Rutherford Scattering

Problem. A particle of mass $m$ and energy $E \gt 0$ is scattered by a repulsive Coulomb potential $V(r) = k/r$ ($k \gt 0$). Find the scattering angle $\Theta$ as a function of the impact parameter $b$.

Solution

The angular momentum is $l = mv_\infty b$ where $v_\infty = \sqrt{2E/m}$. The eccentricity is:

$$e = \sqrt{1 + \frac{2El^2}{mk^2}} = \sqrt{1 + \left(\frac{2Eb}{k/m}\right)^2}$$

The orbit is a hyperbola. The asymptotic angles satisfy $r \to \infty$ I.e., $1 + e\cos\phi = 0$ Giving $\phi_{\pm} = \pi \pm \arccos(1/e)$. The scattering angle is:

$$\Theta = \pi - 2\arccos(1/e) = 2\arcsin(1/e)$$

Using $\sin(\Theta/2) = 1/e$:

$$\cot\frac{\Theta}{2} = \frac{2Eb}{k/m} = \frac{mv_\infty^2 b}{k/m}$$

This is the Rutherford scattering formula relating the scattering angle to the impact parameter. $\blacksquare$

Intuition

Central force problems describe motion under forces that point toward or away from a fixed center. The key insight is that angular momentum is conserved, confining motion to a plane. The effective potential combines the real potential with the centrifugal barrier, creating an energy landscape that determines the orbit shape. For attractive inverse-square forces, the orbits are conic sections: circles, ellipses, parabolas, or hyperbolas. Rutherford scattering shows how alpha particles deflected by the Coulomb force reveal the atomic nucleus. The orbit equation transforms the problem from tracking position versus time to tracking the trajectory shape directly, eliminating time as a variable.

Common Mistakes

Mistake 1: Confusing the effective potential with the actual potential The effective potential $V_{\mathrm{eff}}(r) = V(r) + l^2/(2mr^2)$ includes the centrifugal barrier term $l^2/(2mr^2)$, which is not a real force but a consequence of angular momentum conservation. Students often forget this term and attempt to find circular orbits from $V'(r_0) = 0$ instead of the correct $V_{\mathrm{eff}}'(r_0) = 0$.

Mistake 2: Assuming all central force orbits are closed Bertrand’s theorem states that only the Kepler potential ($V \propto 1/r$) and the harmonic oscillator potential ($V \propto r^2$) produce closed orbits for all bound states. For other central potentials, orbits generally precess and are not closed. The orbit equation still applies, but the trajectory does not repeat.

Mistake 3: Mixing up the semi-latus rectum with the semi-major axis The semi-latus rectum $p = l^2/(mk)$ determines the shape of the orbit, while the semi-major axis $a$ determines the energy via $E = -k/(2a)$. For a circular orbit $p = a$, but for eccentric orbits $p < a$. Using $p$ in place of $a$ in energy formulas gives incorrect results.

Cross-References

• Lagrangian Mechanics: The Lagrangian formulation provides the effective potential analysis for central force problems.
• Hamiltonian Mechanics: The Hamiltonian formalism gives the total energy and phase space representation of central force orbits.