Question

Two parallel isobars are separated by 250 km at 30°N with air density = 0.70 kg/m$^3$, Coriolis parameter $f = 2\Omega \sin \phi$ (where $\Omega = 14.6 \times 10^{-5}$ rad/s), and pressure gradient = 5 hPa. Calculate the geostrophic velocity in m/s.

Hint:

Geostrophic velocity depends inversely on density and Coriolis parameter. Lower density or latitude → stronger winds.

Answer 1

Correct Answer: 39.06

Step 1: Geostrophic balance.
\[ V_g = \frac{1}{\rho f} . \frac{\Delta p}{\Delta x} \]
Step 2: Convert units.
Pressure difference = 5 hPa = 500 Pa.
Distance = 250 km = $2.5 \times 10^5$ m.
\[ \frac{\Delta p}{\Delta x} = \frac{500}{2.5 \times 10^5} = 0.002 \, \text{Pa/m} \]
Step 3: Calculate Coriolis parameter.
\[ f = 2\Omega \sin \phi = 2 \times (14.6 \times 10^{-5}) \times \sin 30^\circ \] \[ = 2.92 \times 10^{-4} \times 0.5 = 1.46 \times 10^{-4} \, s^{-1} \]
Step 4: Calculate $V_g$.
\[ V_g = \frac{0.002}{0.70 \times 1.46 \times 10^{-4}} \] \[ = \frac{0.002}{1.022 \times 10^{-4}} = 19.57 \, m/s \] Final Answer: \[ \boxed{19.57 \, m/s} \]