Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics IV / 4

Divide Newton's second law by the mass and resolve along the radial and transverse unit vectors in polar coordinates. A central force has no transverse component, soMultiplication of the second equation by $r$ shows that $d(r^2\dot{\theta })/dt=0$. Thuswhere $h$ is the signed specific angular momentum perpendicular to the orbital plane.

If $h\ne 0$ and $r>0$, the polar angle is locally a valid independent variable. For $u=1/r$, use primes for derivatives with respect to $\theta$. The chain rule givesSubstitution into the radial equation gives the Binet equationIf $h=0$, the orbit is radial and division by $h$ is unavailable; the first radial equation still applies with $\dot{\theta }=0$ away from the origin.

The initial transverse velocity gives $h=a(L/a)=L$. From $\dot{r}=-h{u}^{\prime }$, the initial conditions are $u(0)=1/a$ and $u^{\prime }(0)=1/a$. DefineThe Binet equation becomes the constant-coefficient linear ordinary differential equationIts particular solution is $({\nu }^2-1)/(a{\nu }^2)$. Applying the two initial conditions to the complementary solution givesThe physical branch is the connected range containing $\theta =0$ on which the right-hand side is positive. A zero corresponds to escape to infinite radius, rather than continuation to negative $r$.

The source calls $F$ an inward force magnitude, but here $F(r)=(r-a)/(a{r}^3)$ changes sign at $r=a$. The stated initial condition has inward radial velocity, so the particle immediately enters $r<a$, where the force is outward. The consistent interpretation is that $F$ is the signed inward radial component of the force per unit mass.

The oscillatory numerator $\cos (\nu \theta )+\nu \sin (\nu \theta )$ has amplitude $\sqrt{1+{\nu }^2}$. Therefore the reciprocal radius stays bounded away from zero exactly whenIn this case the radial bounds areFor completeness, the potential energy per unit mass is $\Phi =-1/(ar)+1/(2r^2)$, and conservation of mechanical energy givesThis effective potential diverges positively at zero, has one minimum at $r=a(1+L^2)$, and tends to zero from below at infinity. Since the initial radial velocity is negative, the particle first reaches an inner turning point. If $L^2\ge 1/2$, the subsequent outward motion has no finite outer turning point and escapes. At equality the escape speed tends to zero, but the radius is still unbounded; infinity is approached only as time tends to infinity. HenceThe minimum of $U$ permits a circular orbit for other initial data, but the nonzero initial radial velocity here excludes that exceptional solution.

With $L=1/\sqrt{n^2-1}$, one has $\nu =n$. Since $n\ge 3>\sqrt{3}$, ii proves boundedness. Write the positive rational number $n=p/q$ in lowest terms. The radius and its angular derivative repeat after $\Delta \theta =2\pi /n$. After $p$ such radial oscillations,so both the position and the radial and transverse velocities return to their initial values. The elapsed time iswhich is finite and positive because $r$ is bounded above and below by positive constants. Thus the motion is periodic. The orbit is a closed, generally self-intersecting rosette orbit, with $p$ radial oscillations during $q$ complete revolutions; for integral $n$ it has $n$ evenly spaced lobes around the origin. It never reaches the origin and is not a circular orbit.