Differentiate the result of part i and use hydrostatic equilibrium:
$$
\frac{dT}{dz}
=aT\frac{d\log P}{dz}
=-\frac{a\mu m_Hg}{k_B}.
$$
Thus the constant <atmospheric lapse rate> is
$$
\boxed{
\Gamma=-\frac{dT}{dz}
=\frac{\mu m_Hg}{k_B}
\frac{\log(T_b/T_t)}{\log(P_b/P_t)}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 315 / 2 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-315/2/a/ii/solution