Solving barometric formulae for height

Question

I am wanting to find the height in the barometric equations below. Could anyone solve this?

Equation 1: $$ {P}=P_b \cdot \left[\frac{T_b}{T_b + L_b\cdot(h-h_b)}\right]^{\textstyle \frac{g_0 \cdot M}{R^* \cdot L_b}} $$

Equation 2: $$ P=P_b \cdot \exp \left[\frac{-g_0 \cdot M \cdot (h-h_b)}{R^* \cdot T_b}\right] $$

where

• $P_b =$ Static pressure (pascals)
• $T_b =$ Standard temperature ([[kelvin|K]])
• $L_b =$ Standard temperature lapse rate -0.0065 (K/m) in [[International Standard Atmosphere|ISA]]
• $h =$ Height above sea level (meters)
• $h_b =$ Height at bottom of layer b (meters; e.g., h_1 = 11,000 meters)
• $R^* =$ [[Universal gas constant]] for air: 8.31432 N·m /(mol·K)
• $g_0 =$ Gravitational acceleration (9.80665 m/s²)
• $M =$ Molar mass of Earth's air (0.0289644 kg/mol)

Answer 1

Equation $1$ :

$$\left[\frac{T_b}{T_b + L_b\cdot(h-h_b)}\right]^{\textstyle \frac{g_0 \cdot M}{R^* \cdot L_b}} = \frac {P}{P_b} \Rightarrow \left[\frac{T_b}{T_b + L_b\cdot(h-h_b)}\right] =\left(\frac {P}{P_b}\right)^{\textstyle \frac{R^* \cdot L_b}{g_0 \cdot M}} \Rightarrow$$

$$\Rightarrow T_b+L_b(h-h_b)=T_b\cdot \left(\frac {P_b}{P}\right)^{\textstyle \frac{R^* \cdot L_b}{g_0 \cdot M}} \Rightarrow L_b(h-h_b)=T_b\left(\left(\frac {P_b}{P}\right)^{\textstyle \frac{R^* \cdot L_b}{g_0 \cdot M}}-1\right)\Rightarrow$$

$$\Rightarrow h = h_b+ \frac{T_b}{L_b}\left(\left(\frac {P_b}{P}\right)^{\textstyle \frac{R^* \cdot L_b}{g_0 \cdot M}}-1\right)$$

Equation $2$ :

$$\left[\frac{-g_0 \cdot M \cdot (h-h_b)}{R^* \cdot T_b}\right] = \ln \left(\frac {P}{P_b}\right) \Rightarrow -g_0 \cdot M \cdot(h-h_b)=R^* \cdot T_b \cdot \ln \left(\frac {P}{P_b}\right) \Rightarrow$$

$$\Rightarrow h = h_b - \frac {R^* \cdot T_b}{g_0 \cdot M} \cdot \ln \left(\frac {P}{P_b}\right)$$

Answer 2

In the 1st one: start with $P$, divide by $P_b$, raise to the power $R^*L_b/g_oM$, take the reciprocal, multiply by $T_b$, subtract $T_b$, divide by $L_b$, and add $h_b$.