Question

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.]

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.

Answer 1

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.