= Solution
In a thin atmosphere in <hydrostatic equilibrium>, take gravity $g$ and mean molecular mass $\mu m_H$ as constant. The <ideal gas> equation of state and hydrostatic balance give
$$
\frac{dP}{dz}=-\rho g
=-\frac{\mu m_Hg}{k_BT}P.
$$
For $P>1\ {\rm mbar}$,
$$
\frac{dT}{dP}=\frac{Ax}{P\ln10},
$$
so the <atmospheric lapse rate> follows from the chain rule:
$$
\boxed{\frac{dT}{dz}
=-\frac{A\mu m_Hg}{k_BT\ln10}
\log_{10}\frac{P}{1\ {\rm mbar}}}.
$$
It vanishes in the assumed isothermal region above $1\ {\rm mbar}$. If natural logarithms are used instead, the factor $\ln10$ is absent and the fitted numerical value of $A$ changes accordingly.
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