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.

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.

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 $b$.