Coriolis force on a particle moving on sphere

Question

Consider a particle moving on a rotating disc, as shown below. The direction of the Coriolis force acting on the particle can be found using the cross-product of the angular velocity vector and the velocity vector (right-hand rule).

Now, consider a particle moving on a rotating sphere (say, Earth) as shown below. The Earth is rotating about the Z-axis. The particle is located at the equator (A) and moving towards the north pole along the meridian. Thus, the particle's velocity would be along the Z-axis.

My question is: Will the Coriolis force on the particle be zero since Earth's angular velocity and the particle's velocity are in the same direction (Z)? However, this goes against what is written in Geography textbooks describing the deflection of winds due to Coriolis force:

Where am I making a mistake?

Answer 1

Considering only Earth rotation around its axis with angular velocity $\boldsymbol{\Omega}_E = \Omega_E \mathbf{\hat{z}}$, you're right that at the equator there is no Coriolis force on a point that starts moving along a meridian line with velocity $\mathbf{v}(0) = v(0) \mathbf{\hat{z}}$, since $\mathbf{F}^{Coriolis} = 2 \boldsymbol{\Omega}_{E} \times \mathbf{v}$. But, as it moves away from the equator, if the point keeps moving along the surface, Coriolis force appears since $\boldsymbol{\Omega}_{E}$, $\mathbf{v}$ are not aligned anymore.

In general, for the description of the dynamics of mechanical systems in a non-inertial reference frame, you need to rely on relative kinematics. Just as an example, the acceleration w.r.t. to an inertial reference frame (where Newton's equation holds in their "standard form", e.g. for a point mass $m \mathbf{a} = \mathbf{F}$) with origin $O$ can be written w.r.t. a moving reference frame with origin $Q$ and angular velocity $\boldsymbol{\omega}_{Q/O}$ (and angular acceleration $\dot{\boldsymbol{\omega}}_{Q/O} = \boldsymbol{\alpha}_{Q/O}$) as

$$\mathbf{a}_{P/O} = \mathbf{a}_{Q/O} + \mathbf{a}_{P/Q} + \boldsymbol{\alpha}_{Q/O} \times \mathbf{r}_{P/Q} + \boldsymbol{\omega}_{Q/O} \times (\boldsymbol{\omega}_{Q/O} \times \mathbf{r}_{P/Q}) + 2 \boldsymbol{\omega}_{Q/O} \times \mathbf{v}_{P/Q}$$

It's not possible to predict the dynamics of global winds only with Coriolis force, since many different process governs it. In order to describe the dynamics of global winds, you need to solve complex equations governing the motion of fluids (a sort of hard-core version of Navier$-$Stokes equations) in the Earth non-inertial reference frame. If you were to solve Navier$-$Stokes equations, you would need to replace the expression of acceleration $$\mathbf{a}_O = \frac{\partial \mathbf{u}}{\partial t}+ \left( \mathbf{u} \cdot \nabla \right) \mathbf{u}$$ with it's non-inertial counterpart

$$\mathbf{a}_{/Q} = \mathbf{a}_{/O} - \boldsymbol{\Omega}_{E} \times (\boldsymbol{\Omega}_{E} \times \mathbf{r}_{/Q}) - 2 \boldsymbol{\Omega}_{E} \times \mathbf{u}_{/Q}$$

having assumed $\boldsymbol{\Omega}_E$ constant and $\mathbf{a}_{Q/O} = \mathbf{0}$ for the reference frame rotating with the Earth. Momentum equation can be recast as

$$\rho \underbrace{\left[ \frac{\partial \mathbf{u}}{\partial t}\bigg|_{Q} + (\mathbf{u} \cdot \nabla) \mathbf{u} \right]}_{= \mathbf{a}_{/Q}} = \underbrace{\rho \mathbf{g}}_{\text{volume force}} + \underbrace{\nabla \cdot \mathbb{T}}_{\text{stress: pressure + viscosity}} \underbrace{- \rho \boldsymbol{\Omega}_{E} \times (\boldsymbol{\Omega}_{E} \times \mathbf{r}_{/Q})}_{\text{centrifugal force}} \underbrace{- 2 \boldsymbol{\Omega}_{E} \times \mathbf{u}_{/Q}}_{\text{Coriolis force}} \ .$$

While it's not easy to solve these equations, prevailing winds and global wind cells are a result of these equations. As an example, see the attached picture and the links to wiki pages about:

• prevailing winds
• Hardley cells