Find the altitude of a passenger plane if the outside air temperature is $$ -51^{\circ} \mathrm{C} $$

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We start with the standard atmosphere relation for temperature, which can be written as:
$$
T = T_0 - L \cdot h
$$
where:
- $$T_0 = 15^{\circ} \mathrm{C}$$ is the sea level temperature,
- $$L = 6.5^{\circ} \mathrm{C}/\mathrm{km}$$ is the lapse rate,
- $$h$$ is the altitude in kilometers.

Given the outside temperature $$T = -51^{\circ} \mathrm{C}$$, we substitute into the formula:
$$
-51 = 15 - 6.5 \cdot h
$$

Now, solve for $$h$$:

1. Add $$6.5 \cdot h$$ to both sides and add $$51$$ to both sides:
   $$
   6.5 \cdot h = 15 + 51
   $$

2. Simplify the right-hand side:
   $$
   6.5 \cdot h = 66
   $$

3. Divide both sides by $$6.5$$ to isolate $$h$$:
   $$
   h = \frac{66}{6.5}
   $$

4. Calculate the value:
   $$
   h \approx 10.15 \text{ km}
   $$

Thus, the altitude of the passenger plane is approximately $$10.15$$ kilometers.
The altitude of the passenger plane is approximately 10.15 kilometers.

Find the altitude of a passenger plane if the outside air temperature is

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