Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 4 / 10A (Dynamics and Relativity)

A central force has no component in the angular direction. The angular component of the acceleration in polar coordinates therefore gives $2\dot{r}\dot{\theta }+r\ddot{\theta }=0$, orFor unit mass, $h=r^2\dot{\theta }$ is the signed angular momentum perpendicular to the orbital plane, so its magnitude $|h|$ is conserved. Equivalently, the torque $\mathbf{r}\times \mathbf{F}$ vanishes, which also fixes that plane.

The given potential energy is $k/r$, whose radial force is $k/r^2$: positive $k$ is repulsive and negative $k$ attractive. The kinetic energy in polar coordinates is $({\dot{r}}^2+r^2{\dot{\theta }}^2)/2$, so conservation of energy givesFor $h\ne 0$ and $k>0$, the effective potential is positive, strictly decreasing from $+\infty$ to $0$, and has no stationary point. For $k=-\kappa <0$, it still tends to $+\infty$ at zero, crosses zero at $r=h^2/(2\kappa )$, and has its unique minimum atIt then approaches zero from below. The minimum corresponds to a stable circular orbit. If $h=0$, the centrifugal barrier disappears; the attractive effective potential is simply $-\kappa /r$ and has no minimum.

Interpret projection from very far away as the asymptotic incoming state at $r\to \infty$. For $v>0$ and impact parameter $b>0$, conservation of energy and conservation of angular momentum give $E=v^2/2$ and $|h|=bv$. At the closest approach the radial velocity vanishes. With $k=-\kappa <0$, the central-force radial turning point equation isOnly the positive root is physical. Put $q=\kappa /v^2$. The pericentre distance and the purely tangential speed there areThus $p<b$ and $u>v$: attraction bends the path inward and increases the speed. With positive energy, the Kepler orbit is a hyperbola. Measured from the pericentre direction, its polar coordinates satisfy $r=h^2/[\kappa (1+e\cos \varphi )]$ with $e=\sqrt{1+b^2{v}^4/{\kappa }^2}>1$.

The hyperbola shown has a nonzero impact parameter. For $b=0$, the attractive radial orbit instead reaches the singular origin; there is no regular turning point with a finite closest-approach speed.

For $k=\kappa >0$, the same asymptotic data give $E=v^2/2$ and $|h|=bv$. At the central-force radial turning point,The positive root and the tangential speed, again with $q=\kappa /v^2$, areFor $b>0$, $\tilde{p}>b$ and $\tilde{u}<v$. The repulsive inverse-square force bends the trajectory away from the origin, as in the right-hand sketch. Its polar equation of a repulsive inverse-square orbit is $r=h^2/[\kappa (e\cos \varphi -1)]$, with $e=\sqrt{1+b^2{v}^4/{\kappa }^2}>1$ and $\varphi$ measured from closest approach.

For $b=0$, the regular radial turning point is $\tilde{p}=2\kappa /v^2$ and its speed is zero; the particle reverses direction there. The formula $bv/\tilde{p}$ still gives zero, while its rationalized form containing $1/b$ should be interpreted only by a limit.

For equal $|k|=\kappa$ in the two cases, their closest distances are conjugate roots expressed with the same $q=\kappa /v^2$. ConsequentlyAt either central-force radial turning point, the velocity is tangential and conservation of angular momentum gives $pu=\tilde{p}\tilde{u}=bv$. ThusThese attractive and repulsive inverse-square closest approaches pair a shorter, faster attractive passage with a longer, slower repulsive one. The speed-product conclusion assumes $b>0$; at $b=0$, the attractive orbit hits the singular centre with unbounded speed, so the formal product with the repulsive zero speed is not defined.