Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 4 / 9B (Dynamics and Relativity)

The central force has no transverse component. Its angular equation is thereforeThus the specific angular momentum is the constantThe radial equation isSubstitute $\dot{\theta }=l/r^2$:Consequently the radial motion for $r>0$ is exactly a unit-mass one-dimensional motion in the effective potentialIts conserved energy is $\frac{1}{2}{\dot{r}}^2+V_{\mathrm{eff}}(r)$; the second term in the effective potential accounts for the angular kinetic energy.

For circular orbits in an exponential central potential, a constant positive radius requires $V_{\mathrm{eff}}^{\prime }(r)=0$, orLet $h(r)=r^3{e}^{-r}$. Its derivative is $h^{\prime }(r)=r^2{e}^{-r}(3-r)$: it rises from zero to $h(3)=27/e^3$, then falls back to zero. HenceThe strict lower bound matters: the printed upper bound also allows $l=0$, but then there is no finite-radius circular orbit. Nonzero angular momentum is an implicit hypothesis of the two-orbit claim.

At a circular orbit, use $l^2=r^3{e}^{-r}$ to obtainA small radial displacement $\eta$ satisfies $\ddot{\eta }=-V_{\mathrm{eff}}^{\prime \prime }(r)\eta$. Therefore the inner circular orbit is radially stable at fixed angular momentum, while the outer orbit is unstable. The inner perturbation oscillates with squared frequency $e^{-r_1}(3/r_1-1)$; the outer perturbation has an exponentially growing solution. At the limiting value $l^2=27/e^3$, the radii coalesce at $r=3$ and the linear restoring coefficient vanishes.

For inferring an inverse-square force from Kepler laws, first use the area swept by an infinitesimal radius sector:Kepler's second law makes the areal velocity constant, so $l=r^2\dot{\theta }$ is constant. Then $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$, meaning that the acceleration has no transverse component. Newton's second law therefore gives a central force directed along the Sun–planet line.

For the radial dependence, set $u=1/r$. Since $\dot{\theta }=l{u}^2$,The radial acceleration is the Binet equation expressionThe focus-based ellipse has $u=(1+\epsilon \cos \theta )/A$, hence $u_{\theta \theta }+u=1/A$. ThusFor a planet of mass $m$, the force is $F_r=-m{l}^2/(A{r}^2)$: it is attractive and obeys the inverse-square law. This deduction uses both the shape of the orbit and its constant areal velocity; the orbit shape alone would not determine its time-dependent acceleration.