solve it again, use g=9.80665 and convert bar to pascal

Let's solve the problem again using \( g = 9.80665 \, \text{m/s}^2 \) and converting the pressure from bar to Pascal.

### Given:
- \( T(h) = 300 - 0.007h \) for \( 0 \leq h < 13000 \) m
- \( T(h) = \text{constant} = 209 \, \text{K} \) for \( 13000 \leq h < 22000 \) m
- \( T(h) = T_{22} + 0.0029h' \) for \( 22000 \leq h \leq 34000 \) m, where \( h' = h - 22000 \)
- \( P(0) = 1 \, \text{bar} = 10^5 \, \text{Pa} \)
- \( R = 287.04 \, \text{J/kg K} \)
- \( g = 9.80665 \, \text{m/s}^2 \)

### Segment 1: From \( h = 0 \) to \( h = 13000 \, \text{m} \)

We have the temperature variation:
\[
T(h) = 300 - 0.007h
\]

The differential equation for pressure is:
\[
\frac{dP}{dh} = -\frac{P \cdot g}{R \cdot T(h)}
\]
Substitute \( T(h) \):
\[
\frac{dP}{P} = -\frac{g}{R} \cdot \frac{dh}{300 - 0.007h}
\]
Integrating both sides:
\[
\ln\left(\frac{P_{13}}{P_0}\right) = -\frac{g}{R} \cdot \int_{0}^{13000} \frac{dh}{300 - 0.007h}
\]
\[
\ln\left(\frac{P_{13}}{10^5}\right) = -\frac{g}{R \times 0.007} \left[\ln(300 - 0.007 \times 13000) - \ln(300)\right]
\]

Substitute the values:
\[
\ln\left(\frac{P_{13}}{10^5}\right) = -\frac{9.80665}{287.04 \times 0.007} \times \ln\left(\frac{209}{300}\right)
\]
\[
\ln\left(\frac{P_{13}}{10^5}\right) = -4.924 \times (-0.3617) = 1.781
\]
\[
P_{13} = 10^5 \times e^{1.781} = 10^5 \times 5.937 \approx 593700 \, \text{Pa}
\]

### Segment 2: From \( h = 13000 \, \text{m} \) to \( h = 22000 \, \text{m} \)

For the second segment, the temperature is constant at \( 209 \, \text{K} \):
\[
\frac{dP}{P} = -\frac{g}{R \cdot T_{13}} dh
\]
Integrating:
\[
\ln\left(\frac{P_{22}}{P_{13}}\right) = -\frac{g \cdot (22000 - 13000)}{R \cdot 209}
\]
\[
P_{22} = P_{13} \times e^{-\frac{9.80665 \times 9000}{287.04 \times 209}} 
\]
\[
P_{22} = 593700 \times e^{-1.484} \approx 593700 \times 0.2273 = 134940 \, \text{Pa}
\]

### Segment 3: From \( h = 22000 \, \text{m} \) to \( h = 30000 \, \text{m} \)

For the final segment:
\[
T(h) = 209 + 0.0029(h - 22000)
\]
At \( h = 30000 \, \text{m} \):
\[
T(30000) = 209 + 0.0029 \times 8000 = 232.2 \, \text{K}
\]
The pressure differential equation:
\[
\frac{dP}{P} = -\frac{g}{R \cdot T(h)} dh
\]
Substitute:
\[
\ln\left(\frac{P_{30}}{P_{22}}\right) = -\frac{g}{R} \int_{22000}^{30000} \frac{dh}{T_{22} + 0.0029(h-22000)}
\]
\[
P_{30} = 134940 \times e^{-\frac{9.80665}{287.04} \times \frac{\ln\left(\frac{232.2}{209}\right)}{0.0029}}
\]
\[
P_{30} = 134940 \times e^{-0.02991 \times 3.16} = 134940 \times 0.1115 \approx 1504.24 \, \text{Pa}
\]

Thus, the ambient pressure at 30 km altitude is:
\[
P_{30} = 1504.2410 \, \text{Pa}
\]

Would you like to explore any further steps in detail? Here are some related questions you might find interesting:

1. How do different altitude ranges influence the pressure gradient?
2. How does constant temperature influence the pressure in the second segment?
3. What role does specific gas constant \( R \) play in the atmospheric pressure calculation?
4. What is the significance of the natural logarithm in this context?
5. How would pressure change if the temperature gradient in the first segment were different?
6. Why is the pressure so much lower at 30 km altitude?
7. How does the result compare with the standard atmosphere model?
8. How would the calculation change if the atmosphere had varying gravitational acceleration?

**Tip:** Converting all units consistently at the beginning can help avoid mistakes later on in atmospheric calculations.

Explore how to solve atmospheric pressure variation using differential equations and integration. Detailed step-by-step solution for calculating pressure at different altitudes with specific temperature gradients and gravitational acceleration. Suitable for advanced learners and atmospheric science enthusiasts.

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