Consider the International Standard Atmosphere (ISA) with $h$ being the geopotential altitude (in km) and $\dfrac{dT}{dh}$ being the temperature gradient (in K/m).
Which of the following combination(s) of $\Big(h, \dfrac{dT}{dh}\Big)$ is/are correct as per ISA?
Show Hint
- $\,(7, -6.5 \times 10^{-3})$
- $\,(9, 4 \times 10^{-3})$
- $\,(15, 0)$
- $\,(35, 3 \times 10^{-3})$
The Correct Option is A, C
Solution and Explanation
Step 1: ISA temperature profile.
- From sea level to 11 km: troposphere, linear temperature decrease at rate \[ \dfrac{dT}{dh} = -6.5 \,\text{K/km} = -6.5 \times 10^{-3} \,\text{K/m}. \] - From 11 km to 20 km: isothermal region, \[ \dfrac{dT}{dh} = 0. \] - From 20 km to 32 km: temperature increases with positive lapse rate $+1 \,\text{K/km}$. - From 32 km to 47 km: temperature increases further at about $+2.8 \,\text{K/km}$.
Step 2: Check options.
- (A) $h=7$ km: This is in troposphere. Correct gradient $-6.5 \times 10^{-3}$. ✓
- (B) $h=9$ km: Should be $-6.5 \times 10^{-3}$, not $+4 \times 10^{-3}$. ✗
- (C) $h=15$ km: This is in isothermal region. Gradient = 0. ✓
- (D) $h=35$ km: Gradient $\approx +2.8 \times 10^{-3}$, not $+3.0 \times 10^{-3}$. Value mismatch, so ✗. \[ \boxed{\text{Correct statements: (A) and (C)}} \]
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