A column of air mass extending from surface to a height of 10 km moving eastward along 30°N strikes a north-south oriented mountain range. While crossing the mountain range, the air mass acquires a relative vorticity of \( -3.65 \times 10^{-5} \, {s}^{-1} \) at the top. If the air mass maintains the same latitude and conserves potential vorticity, the height of the mountain range is ........ km. (Round off to the nearest integer.)[Assume the angular velocity of the Earth is \( 7.3 \times 10^{-5} \, {s}^{-1} \) and initial relative vorticity is zero.]
Show Hint
In problems involving potential vorticity, remember that the product of relative vorticity and height is conserved if the fluid maintains the same latitude and undergoes no external forces.
In this problem, we are asked to calculate the height of the mountain range while maintaining the potential vorticity. The potential vorticity \( \Pi \) for a rotating fluid is given by the relation:
\[
\Pi = \frac{\zeta}{h} = {constant}
\]
Where:
- \( \zeta \) is the relative vorticity of the fluid,
- \( h \) is the height of the mountain range,
- The constant represents the conservation of potential vorticity.
Given:
- Initial relative vorticity \( \zeta_1 = 0 \),
- Final relative vorticity \( \zeta_2 = -3.65 \times 10^{-5} \, {s}^{-1} \),
- Angular velocity of the Earth \( \omega = 7.3 \times 10^{-5} \, {s}^{-1} \),
- The air mass moves eastward along \( 30^\circ N \), so the latitude and distance traveled affect the vorticity.
We can solve for \( h \), the height of the mountain range using the given values:
\[
h = \frac{\zeta_1}{\zeta_2} \times 1000 \, {km}
\]
This results in:
\[
h \approx 5 \, {km}
\]
Thus, the height of the mountain range is 5 km.
