A

To calculate the variation of pressure with height in the upper atmosphere where the temperature remains constant, we start from the hydrostatic equilibrium condition.
Step 1: The hydrostatic equilibrium equation is given by:
$$\frac{dP}{dz} = -\rho g$$
where \( P \) is the pressure, \( z \) is the height, \( \rho \) is the density, and \( g \) is the acceleration due to gravity.
Step 2: The density \( \rho \) can be related to pressure using the ideal gas law:
$$\rho = \frac{P}{RT}$$
where \( R \) is the ideal gas constant and \( T \) is the absolute temperature.
Step 3: Substituting \( \rho \) back into the hydrostatic equilibrium equation gives:
$$\frac{dP}{dz} = -\frac{P}{RT} g$$
Step 4: This can be rearranged to form:
$$\frac{dP}{P} = -\frac{g}{RT} dz$$
Step 5: Integrating both sides, we have:
$$\int \frac{dP}{P} = -\frac{g}{RT} \int dz$$
Step 6: Integrating the left side yields \( \ln P \), and integrating the right side gives \( -\frac{g}{RT} z + C \):
$$\ln P = -\frac{g}{RT} z + C$$
Step 7: Exponentiating both sides gives us:
$$P = e^C e^{-\frac{g}{RT} z}$$
Letting \( P_0 = e^C \) (the pressure at height \( z = 0 \)), we get:
$$P(z) = P_0 e^{-\frac{g}{RT} z}$$
Step 8: Since the mean molecular weight of air is \( M \), we can express \( R \) in terms of \( M \):
$$R = \frac{R_u}{M}$$
where \( R_u \) is the universal gas constant.
Therefore, the expression for pressure variation with height in the upper atmosphere is given by:
$$P(z) = P_0 e^{-\frac{gM}{R_u T} z}$$
This shows how pressure decreases exponentially with height in the atmosphere at constant temperature.