Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 316 / 1

Use the separation vector $\mathbf{r}$ from the star to the comet, and let $m_{\mathrm{red}}=m{M}_{\star }/(m+M_{\star })$. The two-body problem separates into uniform centre of mass motion andTaking a cross product with $\mathbf{r}$ proves conservation of angular momentum; taking the scalar product with $\dot{\mathbf{r}}$ proves conservation of energy. The relative constants areThe total relative angular momentum and energy are $m_{\mathrm{red}}\mathbf{h}$ and $m_{\mathrm{red}}\epsilon$, respectively. The conserved eccentricity vector is $\mathbf{e}=\dot{\mathbf{r}}\times \mathbf{h}/\mu -\hat{\mathbf{r}}$. Its scalar product with $\hat{\mathbf{r}}$ gives $r=h^2/[\mu (1+e\cos f)]$, with $f$ the true anomaly. The periapsis and apoapsis distances are therefore $h^2/[\mu (1+e)]$ and $h^2/[\mu (1-e)]$. Their sum is twice the semi-major axis, soAt either apsis $v=h/r$; substituting an apsidal radius gives $\epsilon =-\mu /(2a)$. Thus the specific orbital energy implies the vis-viva equationThese are exact relative-coordinate formulas for an elliptic orbit, $0\le e<1$. In the barycentric frame the comet's speed is $M_{\star }/(M_{\star }+m)$ times the relative speed; the printed identification with the comet's stellar orbital speed uses $M_{\star }\gg m$.

The transverse velocity is $v_{\theta }=h/r$. Subtracting its square from the vis-viva equation, and putting $u=a/r$, givesConsequentlyThe positive branch is outward motion and the negative branch is inward motion. The PDF's unsigned expression gives the radial speed, or the outward branch, rather than the signed radial velocity everywhere. For example, on the inward part of any noncircular elliptic orbit, $\dot{r}<0$.

The same result follows from $\dot{r}=(\mu e/h)\sin f$, with $f$ the true anomaly. The radicand is nonnegative precisely on $1/(1+e)\le u\le 1/(1-e)$; it vanishes at periapsis and apoapsis. For a circular Kepler orbit, $e=0$, $u=1$ and $\dot{r}=0$.

Assume spherical fragments of the common mass density $\rho$, and let $n(D)=K{D}^{-\alpha }$ count fragments per unit diameter. Mass normalization of a power-law size distribution givesTheir geometric cross-sections are $\pi D^2/4$, hence the cross-section of a power-law fragment population is exactlyFor $3<\alpha <4$, large fragments dominate the mass integral and small fragments dominate the geometric cross-section integral. Neglecting the subdominant endpoints givesThe two actual error parameters are $(D_{\min }/D_{\max }{)}^{\alpha -3}$ and $(D_{\min }/D_{\max }{)}^{4-\alpha }$. They must both be small; the approximation is not uniform as $\alpha$ approaches either endpoint. At $\alpha =3$ or $4$ the corresponding integral is logarithmic, so the endpoint-dominated formula must be replaced by that integral.

The printed line-density statement is incorrect for a phase-mixed orbit. A steady line density on a Kepler orbit is inversely proportional to speed: for a cross-sectional-area current $J_{\sigma }$, the area per unit arc length is $d\sigma /ds=J_{\sigma }/v$. Equivalently, phase mixing gives $d\sigma =({\sigma }_{\mathrm{tot}}/P)dt$, uniform in mean anomaly, where $P=2\pi \sqrt{a^3/\mu }$ is the orbital period.

For an optically thin population of blackbodies in radiative equilibrium, a fragment absorbs $\sigma L_{\star }/(4\pi r^2)$ and reradiates the same luminosity. Thus the fractional luminosity of a phase-mixed eccentric wire isThe specific angular momentum gives $dt/r^2=d\theta /h$, so the integral is $2\pi /h$. Since $Ph=2\pi a^2\sqrt{1-e^2}$,This derives the intended result after explicitly correcting the density to $1/v$.

The error is consequential. If one instead imposes the literal density $d\sigma /ds\propto v$, its time weighting is $v^{2}dt$. Using $⟨v^2⟩=\mu /a$, $⟨r^{-2}⟩=[a^2\sqrt{1-e^2}{]}^{-1}$ and $⟨r^{-3}⟩=[a^3(1-e^2{)}^{3/2}{]}^{-1}$ givesAt $e=1/2$ this is $5/3$ times the printed result. The inverse-radius averages can also be obtained directly with eccentric anomaly $E$, $r=a(1-e\cos E)$ and $dt=(1-e\cos E)dE/n$.

Let $D_c$ be the parent comet's diameter, with the same mass density as the fragments, so $m=\pi \rho D_c^3/6$. The required geometric cross-section is ${\sigma }_{\mathrm{tot}}=4\pi a^2\sqrt{1-e^2}f$, using the corrected phase-mixed orbit interpretation.

Substitute this and the spherical mass into the endpoint-dominated fragment formula. The parent-body size from a fragment cross-section becomesThe common mass density cancels. The comet's radius is $D_c/2$. This expression treats both fragment cutoffs as specified independently and has the same endpoint-approximation restrictions as part (iii).

If a model instead ties the largest fragment to the parent through $D_{\max }=\eta D_c$, with fixed $0<\eta \le 1$, one must solve that dependence too:That extra assumption is not supplied by the PDF; setting $D_{\max }=D_c$ without saying so would change the requested size relation.

On the corrected steady phase-mixed orbit, each fragment crosses a fixed orbital longitude once per orbital period. Hence the cross-sectional-area current through that point isFrom the fractional luminosity of a phase-mixed eccentric wire, ${\sigma }_{\mathrm{tot}}=4\pi a^2\sqrt{1-e^2}f=2Phf$. Therefore the area rate at the foreground crossing of the line of sight isOnly the foreground segment blocks the star; the far-side intersection does not add another occultation current. The orbital area current is constant because the line density varies as $1/v$, even though the local orbital speed varies.

Use an edge-on orbit and an equatorial, effectively central chord. Near the foreground crossing the radial velocity is along the line of sight; the projected transverse speed is $v_{\perp }=h/r$. In the small stellar-angular-radius limit, $r\gg R_{\star }$, the crossing time is $2R_{\star }/v_{\perp }=2R_{\star }r/h$.

The cross-sectional-area current therefore places an areain front of the stellar disk. For an optically thin wire and a uniformly bright stellar disk, transit dimming by an optically thin orbital wire givesThis is a linear occultation estimate, requiring negligible overlap and a fractional dimming much smaller than one. A finite impact parameter shortens the chord; limb darkening, finite wire thickness and variation of $r$ across a large stellar angular extent modify the coefficient. The estimate cannot be extrapolated to dimming greater than unity.