Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S30 / 1

Use the relative star–planet separation and specific angular momentum per unit reduced mass,For $r=l/(1+e\cos \theta )$, differentiation gives $\dot{r}=Je\sin \theta /l$, while the transverse speed is $r\dot{\theta }=J/r=J(1+e\cos \theta )/l$. Adding the squared orthogonal components yieldsThe Binet equation for the relative two-body inverse-square force is $u^{\prime \prime }+u=G(M+m)/J^2$, with $u=1/r$. The conic has $u^{\prime \prime }+u=1/l$, soThe printed division by $l$ is an error: $J^2$ has dimensions $\mathrm{lengt}{\mathrm{h}}^4/tim{\mathrm{e}}^2$, while $G(M+m)/l$ has dimensions of speed squared. A circular orbit independently checks the correction: $J^2=r^2{v}^2=G(M+m)r$, with $l=r$.

For a closed ellipse, $l=a(1-e^2)$ and $1+2e\cos \theta +e^2=2l/r-(1-e^2)$. Substitution now gives the vis-viva equation,This is relative orbital speed; either body's speed about the barycentre follows by the corresponding mass-ratio factor.

A circular orbit has uniform angular motion, so its Keplerian radial-velocity curve is sinusoidal. Choose phase $\varphi =(t-t_{\mathrm{tr}})/P$, positive stellar velocity away from the observer, and an edge-on circular orbit. One consistent convention isTransit occurs at the decreasing systemic-velocity crossing $\varphi =0$; secondary eclipse is at the increasing crossing $\varphi =1/2$. Extrema occur one quarter of an orbit away. Reversing the velocity sign convention reverses the crossing directions but not the conjunction phases.

For an eccentric orbit, $\nu$ is not proportional to time. Solve $\mathcal{M}=E-e\sin E$, then use $\tan (\nu /2)=\sqrt{(1+e)/(1-e)}\tan (E/2)$ in the stated velocity relation. Motion is fastest near periastron, so the curve has an abrupt feature there and a more prolonged change near apastron, with its shape also depending on $\varpi$. It is not merely a shifted sinusoid. An eccentric transit need not coincide with $V=V_0$, because the $e\cos \varpi$ offset remains at conjunction.

Assume the companion is much less massive than the star, the sinusoid represents a circular Keplerian orbit, and the transit implies $\sin i\simeq 1$. Ignore dilution, limb darkening and grazing geometry when estimating the radius. With $P=5\times 86400\mathrm{s}$,The stellar reflex radial-velocity semi-amplitude gives $K\simeq (m/M_{\odot })(2\pi a/P)$, while transit depth gives $\delta \simeq (R_p/R_{\odot }{)}^2$. ThereforeThe mean density relative to the supplied Jupiter values isThis is a low-density hot gas giant. Strong irradiation can slow contraction or help sustain an inflated envelope; internal heating and a different heavy-element fraction are other possibilities. The inferred low mass also reduces gravitational compression. Mass and radius alone do not determine which inflation mechanism operates.