The wind speed in colatitude \(\theta\) is \(v\). By considering the forces on a small volume of air, show that the pressure gradient required to balance the horizontal component of the Coriolis force, and thus to maintain a constant wind direction, is \(\mathrm{d} p / \mathrm{d} x=2 \omega \rho v \cos \theta\), where \(\rho\) is the density of the air. Evaluate this gradient in mbar \(\mathrm{km}^{-1}\) for a wind speed of \(50 \mathrm{~km} \mathrm{~h}^{-1}\) in latitude \(30^{\circ} \mathrm{N} .\left(1\right.\) bar \(=10^{5} \mathrm{~Pa}\); density of air \(=\) \(\left.1.3 \mathrm{~kg} \mathrm{~m}^{-3} .\right)\)

The Coriolis force (\(F_{cor}\)) acting on a small volume of air is given by the following equation: \(F_{cor}=2m \omega v \cos \theta\) where \(m\) is the mass of the small volume of air, \(\omega\) is the angular velocity of Earth, \(v\) is the wind speed and \(\theta\) is the colatitude. Here, we are interested in the horizontal component of this force, which acts in the pressure gradient direction, i.e., in the x-direction. Since the force is already in the x-direction, we don't need to modify the expression for the Coriolis force.

To maintain a constant wind direction, the horizontal component of the Coriolis force must be balanced by the pressure force acting on the small volume of air. The pressure force can be written as: \(F_{pressure} = \rho \frac{\mathrm{d} p}{\mathrm{d} x}\) Where \(\rho\) is the air density and \(\frac{\mathrm{d} p}{\mathrm{d} x}\) is the pressure gradient in the x-direction. Since the pressure force must balance the horizontal component of the Coriolis force, we can write the equation as: \(\rho \frac{\mathrm{d} p}{\mathrm{d} x} = 2m \omega v \cos \theta\) As mass \(m\) can be expressed as density \(\rho\) times volume \(V\), we have: \(\rho \frac{\mathrm{d} p}{\mathrm{d} x} = 2(\rho V) \omega v \cos \theta\) Dividing both sides by \(\rho V\), we obtain the pressure gradient: \(\frac{\mathrm{d} p}{\mathrm{d} x} = 2 \omega \rho v \cos \theta\)

Now we need to plug in the given values of wind speed, latitude, and the constants required to convert the units to evaluate the pressure gradient in mbar/km. We are given: - Wind speed (\(v\)): \(50 \mathrm{~km} \mathrm{~h}^{-1}\) - Latitude (\(\phi\)): \(30^{\circ} \mathrm{N}\) - Earth's angular velocity (\(\omega\)): \(7.292 \times 10^{-5}\mathrm{~rad/s}\) - Air density (\(\rho\)): \(1.3 \mathrm{~kg} \mathrm{~m}^{-3}\) First, we need to convert the wind speed from km/h to m/s: \(v = 50 \times \frac{1000}{3600} = 13.89\,\mathrm{m/s}\) Next, we need to convert latitude to colatitude: \(\theta = 90^{\circ} - \phi = 90^{\circ} - 30^{\circ} = 60^{\circ}\) Now we can find the pressure gradient: \(\frac{\mathrm{d} p}{\mathrm{d} x} = 2 \omega \rho v \cos \theta = 2 (7.292 \times 10^{-5}\,\mathrm{rad/s}) (1.3\,\mathrm{kg/m^3}) (13.89\,\mathrm{m/s}) \cos (60^{\circ})\) \(\frac{\mathrm{d} p}{\mathrm{d} x}= 2.28 \times10^{3} \,\mathrm{Pa/m}\) Finally, we need to convert the units of the pressure gradient to mbar/km: \(1\,\mathrm{Pa} = 1\times 10^{-2}\,\mathrm{mbar}\) \(1\,\mathrm{m} = 1\times 10^{-3}\,\mathrm{km}\) \(\frac{\mathrm{d} p}{\mathrm{d} x}= 2.28 \times10^{3} \times(10^{-2}\,\mathrm{mbar/Pa})(10^{-3}\,\mathrm{km/m}) = 22.8\,\mathrm{mbar/km}\) The pressure gradient required to maintain a constant wind direction is 22.8 mbar/km.