02 Jun 2025
Central-Force Motion and Angular Momentum
Planar motion under a central force, angular-momentum conservation, areal velocity, and centrifugal force.
Let the fixed force centre be the origin and let $r>0$. A central force has the form
\[\mathbf F(\mathbf r)=F(r)\,\hat{\mathbf r},\]where the signed scalar $F(r)$ is positive for an outward force and negative for an inward force. Its torque about the centre is
\[\boldsymbol\tau =\mathbf r\times\mathbf F =r\hat{\mathbf r}\times F(r)\hat{\mathbf r} =\mathbf 0.\]Therefore
\[\frac{d\mathbf L}{dt}=\boldsymbol\tau=\mathbf 0, \qquad \boxed{\mathbf L=\text{constant}}.\]If $\mathbf L\ne\mathbf 0$, both $\mathbf r$ and the velocity remain in the fixed plane perpendicular to $\mathbf L$. If $\mathbf L=\mathbf 0$, the motion is purely radial along a fixed line. Thus a central-force orbit can always be treated as a plane problem.
Polar equations of motion
In the orbital plane,
\[\frac{d\hat{\mathbf r}}{dt}=\dot\theta\,\hat{\boldsymbol\theta}, \qquad \frac{d\hat{\boldsymbol\theta}}{dt}=-\dot\theta\,\hat{\mathbf r}.\]Differentiating $\mathbf r=r\hat{\mathbf r}$ once gives
\[\dot{\mathbf r}=\dot r\,\hat{\mathbf r} +r\dot\theta\,\hat{\boldsymbol\theta}.\]Differentiating again and collecting radial and transverse components gives
\[\ddot{\mathbf r} =(\ddot r-r\dot\theta^2)\hat{\mathbf r} +(r\ddot\theta+2\dot r\dot\theta)\hat{\boldsymbol\theta}.\]Since a central force has no transverse component, Newton’s law yields
\[m(\ddot r-r\dot\theta^2)=F(r),\] \[m(r\ddot\theta+2\dot r\dot\theta)=0.\]Multiplication of the transverse equation by $r$ gives
\[m\left(r^2\ddot\theta+2r\dot r\dot\theta\right) =\frac{d}{dt}\left(mr^2\dot\theta\right)=0.\]Hence the constant angular-momentum magnitude is
\[\boxed{L=mr^2\dot\theta}.\]Areal velocity
During an infinitesimal angular displacement $d\theta$, the radius vector sweeps the sector area
\[dA=\frac12r(r\,d\theta)=\frac12r^2d\theta.\]Consequently,
\[\boxed{\frac{dA}{dt}=\frac12r^2\dot\theta =\frac{L}{2m}=\text{constant}}.\]This result follows from centrality alone; it does not require an inverse-square force.
Radial energy and the centrifugal term
For a conservative central force, define the potential by
\[F(r)=-\frac{dV}{dr}.\]The mechanical energy is
\[E=\frac12m\dot r^2+\frac12mr^2\dot\theta^2+V(r).\]Using $\dot\theta=L/(mr^2)$ in the angular kinetic energy,
\[\frac12mr^2\dot\theta^2 =\frac12mr^2\left(\frac{L}{mr^2}\right)^2 =\frac{L^2}{2mr^2}.\]Thus
\[\boxed{E=\frac12m\dot r^2+V_{\rm eff}(r)}, \qquad \boxed{V_{\rm eff}(r)=V(r)+\frac{L^2}{2mr^2}}.\]The radial equation may also be rearranged as
\[m\ddot r=F(r)+mr\dot\theta^2 =F(r)+\frac{L^2}{mr^3}.\]In a frame rotating with angular velocity $\boldsymbol\Omega$, the centrifugal pseudo-force is
\[\mathbf F_{\rm cf} =-m\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf r).\]When $\boldsymbol\Omega$ is perpendicular to the orbital plane, this becomes
\[\boxed{\mathbf F_{\rm cf}=m\Omega^2r\,\hat{\mathbf r}},\]directed outward. For a circular orbit, $\Omega=\dot\theta$ and $\ddot r=0$, so the inward central force and the outward centrifugal pseudo-force balance in the rotating description:
\[F(r)+m\Omega^2r=0.\]The centrifugal force is not a new physical interaction; it appears because the chosen frame rotates. In the inertial description, the same term is part of the radial acceleration.
The polar, angular-momentum, effective-potential, and centrifugal identities are verified in the Maxima worksheet; every printed residual is zero.
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