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In the midlatitudes, both Coriolis and pressure gradient forces are present in the atmosphere. Can a statement be made about which one of the two forces is bigger in this region?

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According to these lecture notes, the Coriolis parameter at mid-latitudes is on the order of $f_c = 1\times10^-4 \text{s}^{-1}$ and this needs to be multiplied by a wind speed to get a force. This is the first important note -- Coriolis forces do not create wind/motion, they merely change the direction of it.

For a pressure force, let's look at a worst-case scenario. The greatest pressure drop in a tornado is on the order of 100 millibars. If we assume an F5 tornado, we can say that the radius is on the order of 1000 meters. So the pressure force is $F_p = \frac{1}{\rho}\frac{\partial p}{\partial r} \approx \frac{1}{1.10}\frac{10000}{1000} \approx 9$. Such a tornado probably has wind speeds of approximately 90 m/s, giving a Coriolis force of $F_c = 1\times10^{-4} \times 90 = 0.009$, or roughly 1000 times smaller than the pressure force.

Even if we consider some more reasonable numbers, a pressure drop may be 10 millibars and wind speeds may be 10 m/s. The thickness of the front may be, I don't know, 10 kilometers. This would give something like $F_p = 0.08$ and a Coriolis force of $F_c = 0.001$.

Another way to look at this is that wind speed (required to find a Coriolis force) is proportional to the pressure gradient. So you could say $F_c \propto 1\times10^{-4} \times \frac{\partial p}{\partial x}$ where any constant of proportionality is going to be order one. Doing this, it should be obvious that the pressure gradient force driving the wind itself is going to be larger than the Coriolis force on that wind.

There are bound to be exceptions to this analysis, but in general because wind speed requires a pressure gradient to exist, and because the Coriolis force is proportional to the wind speed by a small number, it is unlikely for the Coriolis force to exceed the pressure gradient force.

There is a nondimensional number, the Rossby number, that indicates the relative strength of the Coriolis term and the inertial terms. Consistent with the analysis above, in a tornado the Rossby number is large and Coriolis forces may be neglected. For low pressure systems, the Rossby number may be order one and therefore the Coriolis forces are important -- this is not different from the analysis above, except to indicate that 10 kilometers is probably too small of a length scale to pick for a low-pressure system pressure gradient. It's probably more on the order of hundreds of kilometers.

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  • $\begingroup$ According to the notes in the link below (page 6), the relative magnitudes of the two forces determine whether the flow curvature is cyclonic or anticyclonic. On weather maps, you see both kinds of flows at midlatitudes. This seems to contradict your answer. Any thoughts? snowball.millersville.edu/~adecaria/ESCI241/… $\endgroup$
    – user37222
    Nov 15, 2015 at 23:26
  • $\begingroup$ @user37222 Is there anything in particular you would like to point out in that link? $\endgroup$
    – tpg2114
    Nov 15, 2015 at 23:27
  • $\begingroup$ I pressed enter before I finished my comment. Please read the full comment. $\endgroup$
    – user37222
    Nov 15, 2015 at 23:31
  • $\begingroup$ @user37222 So my analysis is looking at the total pressure gradient while the notes you linked to are looking at the normal pressure gradient. What I said applies to the sum of the normal and tangential gradients (remember, velocity cannot exist without pressure gradients). The normal pressure gradient component can have any value from zero to huge and so nothing, in general, can be said about it. $\endgroup$
    – tpg2114
    Nov 15, 2015 at 23:38

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