Considering a gyre system (shown in the figure) where ‘L’ and ‘H’ represent low- and high-pressure regions along 45°N, the average slope between points ‘A’ and ‘B’ is 2.1 cm km$^{-1}$ under no-wind condition. For a steady southerly wind of 10 m s$^{-1}$ over these regions, find out the magnitude of the current velocity in m s$^{-1}$ (rounded off to two decimal places).
Show Hint
Sea surface slope sets up the geostrophic current, while wind stress slightly modifies it. Remember $V_g = (g/f)(\Delta \eta/\Delta x)$.
Step 1: Recall geostrophic current relation.
The geostrophic velocity is determined by the slope of the sea surface:
\[
V_g = \frac{g}{f} . \frac{\Delta \eta}{\Delta x}
\]
where:
- $g = 9.8 \, \text{m/s}^2$ (gravity),
- $f = 2 \Omega \sin \phi$ (Coriolis parameter),
- $\Delta \eta / \Delta x$ = slope of sea surface.
Step 2: Convert given slope.
Slope = $2.1 \, \text{cm per km} = \frac{0.021}{1000} = 2.1 \times 10^{-5}$.
Step 3: Calculate Coriolis parameter.
At $\phi = 45^\circ$, $\Omega = 7.29 \times 10^{-5} \, \text{rad/s}$.
\[
f = 2 \Omega \sin \phi = 2 \times 7.29 \times 10^{-5} \times \frac{\sqrt{2}}{2}
\]
\[
f = 1.03 \times 10^{-4} \, s^{-1}
\]
Step 4: Compute geostrophic velocity (no wind).
\[
V_g = \frac{9.8}{1.03 \times 10^{-4}} \times (2.1 \times 10^{-5})
\]
\[
= 9.51 \times 10^4 \times 2.1 \times 10^{-5} = 2.00 \, m/s
\]
Step 5: Include Ekman transport (wind effect).
For a steady southerly wind ($U = 10 \, m/s$), the surface stress adds to the current. Approximate relation:
\[
V = V_g + U . \alpha
\]
where $\alpha \approx 0.01$ (wind-current coupling efficiency).
Thus additional speed $\approx 0.1 \, m/s$.
Step 6: Total current velocity.
\[
V \approx 2.00 + 0.10 = 2.10 \, m/s
\]
Final Answer:
\[
\boxed{2.10 \, m/s}
\]
