Treat the morning and evening halves of the <day-night terminator> as independent, isothermal, hydrostatic atmospheres with the same reference radius $R_0$, reference pressure $P_0$, composition, gravity $g$, and extinction coefficient $\kappa_\lambda$. Their <atmospheric scale heights> are
$$
H_m=\frac{k_BT_m}{\mu m_Hg},
\qquad
H_e=\frac{k_BT_e}{\mu m_Hg}.
$$
For $H_i\ll R_0$, the slant <optical depth> of half $i$ at tangent altitude $z$ is approximately
$$
\tau_{\lambda,i}(z)
=\frac{\kappa_\lambda P_0}{g}
\sqrt{\frac{2\pi R_0}{H_i}}e^{-z/H_i}.
$$
The standard isothermal effective altitude is therefore
$$
z_{\lambda,i}=H_i\left[
\gamma_E+log\left(
\frac{\kappa_\lambda P_0}{g}
\sqrt{\frac{2\pi R_0}{H_i}}
\right)
\right],
$$
up to a wavelength-independent choice of reference radius. The two semicircular limbs add in projected area, so the <exoplanet transmission spectrum> is
$$
\boxed{
D_\lambda
=\frac{(R_0+z_{\lambda,m})^2+(R_0+z_{\lambda,e})^2}
{2R_*^2}}.
$$
Equivalently, $R_{\rm tr}^2=[(R_0+z_m)^2+(R_0+z_e)^2]/2$. Differentiating with respect to $\log\kappa_\lambda$ gives
$$
\frac{dR_{\rm tr}}{d\log\kappa_\lambda}
=\frac{(R_0+z_m)H_m+(R_0+z_e)H_e}{2R_{\rm tr}}
\simeq\frac{H_m+H_e}{2}.
$$
Thus a homogeneous retrieval measures, to leading order,
$$
\boxed{H_{\rm av}=\frac{k_B(T_m+T_e)}{2\mu m_Hg}},
$$
provided the opacity and composition do not themselves differ between the two limbs.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 315 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-315/1/a/solution