Newton’s Laws apply in an inertial reference frame, that is, one that is not accelerating. A point on the rotating Earth is not following a straight line through space, but instead is constantly accelerating by rotating away from a straight line. Therefore, Earth does not provide an inertial reference frame. From the point of view of an astronaut in distant space, the air motions she would observe obey Newton’s Law perfectly, but from the point of view of an Earth-bound observer, Newton’s Laws fail to capture the observed motion. To account for this crazy behavior, the Earth-bound observer needs to add some apparent forces to the real forces for the math to explain the observed motion from the point of view of someone standing on the rotating Earth.

Suppose we have an air parcel moving through space with a velocity${\overrightarrow{U}}_{a}$ , which we will call the absolute velocity. We want to relate this absolute velocity to $\overrightarrow{U}$ , the velocity observed with respect to the Earth reference frame. Let ${\overrightarrow{U}}_{e}$ be the velocity of a point in the Earth reference frame. So ${\overrightarrow{U}}_{e}$ is always eastward, greatest at the equator, and zero at the poles.

Let $\overrightarrow{\Omega}$ be Earth’s velocity vector and $\overrightarrow{r}$be the position vector relative to the Earth’s center, and$\overrightarrow{R}$ be the shortest distance vector from the axis of rotation to the air parcel (as in the figure above). The magnitude of $\overrightarrow{\Omega}$ is$\left|\overrightarrow{\Omega}\right|=\frac{2\pi}{23.934\text{hr}}\frac{1\text{hr}}{3600\text{s}}=7.292x{10}^{-5}{\text{s}}^{\text{-1}}$and the direction of $\overrightarrow{\Omega}$is determined by the right hand rule (the direction of your thumb when you curl the fingers of your right hand in the direction of rotation and point towards the North Star.

The following video (:51) will demonstrate the right hand rule:

Click here for a transcript for UE Right Hand Rule Video

The magnitude of ${\overrightarrow{U}}_{e}$ is *RΩ*, but we need to write ${\overrightarrow{U}}_{e}$ as a vector. Note that${\overrightarrow{U}}_{e}$ is pointed into the page in the figure above, which is a direction perpendicular to both$\overrightarrow{\Omega}$ and$\overrightarrow{R}$. Hence we can use the cross product equation to write an expression for Earth's velocity:${\overrightarrow{U}}_{e}=\overrightarrow{\Omega}\times \overrightarrow{R}=\overrightarrow{\Omega}\times \overrightarrow{r}$, since the component of $\overrightarrow{r}$ perpendicular to $\overrightarrow{\Omega}$ is $\overrightarrow{R}$. So:

$${\overrightarrow{U}}_{a}=\overrightarrow{U}+\overrightarrow{\Omega}\times \overrightarrow{r}$$

We have thus related velocity in the absolute reference frame to velocity in the rotating reference frame.

Now we can consider acceleration. Since $\overrightarrow{U}=\frac{D\overrightarrow{r}}{Dt}$ and ${\overrightarrow{U}}_{a}=\frac{{D}_{a}\overrightarrow{r}}{Dt}$ we can write:

$$\frac{{D}_{a}\overrightarrow{r}}{Dt}=\frac{D\overrightarrow{r}}{Dt}+\overrightarrow{\Omega}\times \overrightarrow{r}$$

This equation describes the change in a position of an air parcel with time observed from an inertial reference frame (the derivative on the left) to the change with time observed from Earth’s reference frame (the derivative on the right). Equation [10.11] is general and applies not only to $\overrightarrow{r}$ but also to any other vector.

Let’s replace$\overrightarrow{r}$ on the left hand side with${\overrightarrow{U}}_{\mathrm{a,}}$ and $\overrightarrow{r}$ on the right hand side with $\overrightarrow{U}+\overrightarrow{\Omega}\times \overrightarrow{r}$ since these two expressions equal each other in Equation [10.10]. By making these substitutions, we can relate acceleration in the absolute frame to acceleration in the rotating frame:

We can simplify this equation and then we can make sense of it physically. First,$\overrightarrow{\Omega}$ is not changing significantly with time, so $\frac{D\overrightarrow{\Omega}}{Dt}$ can be set to zero. Second,$(\overrightarrow{\Omega}\times \overrightarrow{r})$ has the magnitude of $\Omega R$ and points to the east (by the right hand rule) and thus $\overrightarrow{\Omega}\times (\overrightarrow{\Omega}\times \overrightarrow{r})$ has the magnitude ${\Omega}^{2}R$ and points toward -$\overrightarrow{R}$ . Finally, noting that $\overrightarrow{U}=\frac{D\overrightarrow{r}}{Dt}$, we end up with the equation:

$$\frac{{D}_{a}{\overrightarrow{U}}_{a}}{Dt}=\frac{D\overrightarrow{U}}{Dt}+2\overrightarrow{\Omega}\times \overrightarrow{U}-{\Omega}^{2}\overrightarrow{R}$$

The term on the left is the acceleration in the absolute inertial reference frame. The first term on the right is the acceleration in the Earth reference frame. The next terms are the apparent accelerations. The first one is the Coriolis acceleration and the second one is the centrifugal acceleration.

We can now substitute the real accelerations in Equation [10.9] for $\frac{{D}_{a}{\overrightarrow{U}}_{a}}{Dt}$ to get:

$-\frac{1}{\rho}\overrightarrow{\nabla}p+\overrightarrow{g}*-\frac{{C}_{d}}{h}\left|\overrightarrow{V}\right|\overrightarrow{V}=\frac{D\overrightarrow{U}}{Dt}+2\overrightarrow{\Omega}\times \overrightarrow{U}-{\Omega}^{2}\overrightarrow{R}$

and then rearrange this equation to get the equation:

$$\begin{array}{l}\frac{D\overrightarrow{U}}{Dt}=-\frac{1}{\rho}\overrightarrow{\nabla}p+\overrightarrow{g}*-\frac{{C}_{d}}{h}\left|\overrightarrow{V}\right|\overrightarrow{V}-2\overrightarrow{\Omega}\times \overrightarrow{U}+{\Omega}^{2}\overrightarrow{R}\\ \text{|------realforces------||-apparentforces-|}\end{array}$$

### Centrifugal Force

The centrifugal force pulls away from the Earth's axis of rotation and is the same type of force that you feel when you are in a car going around a sharp curve. Over its long history, all the material that makes up the Earth has adjusted to the real gravitational force, $\overrightarrow{g}*$, which is directed to Earth’s center, and the apparent centrifugal force that goes directly outward from Earth’s rotation axis (see figure below).

The resulting gravity that the Earth and everything on it feels is the vector sum of this real and this apparent force:

$$\overrightarrow{g}=\overrightarrow{g*}+{\Omega}^{2}\overrightarrow{R}$$

Since the centrifugal force depends on $\overrightarrow{R}$ , it is greatest at the equator and zero at the poles. As a result, the Earth has become oblate, with the least gravity at the equator and the greatest gravity at the poles. Note that $\overrightarrow{g}$ is always perpendicular to Earth’s surface, which is very useful because the vertical coordinate is always chosen to be perpendicular to Earth's surface, so that$\overrightarrow{g}$ is only in the z direction. Earth’s shape is consistent with effective gravity (Earth's radius, a, at the equator equals 6378.1 km and at the poles equals 6356.8 km). The difference at the equator for *Ω ^{2}R_{A}* = (7.27x10

^{-5})

^{2}6.378x10

^{6}m = 0.033 m s

^{-2}. If we assume a constant value for$\overrightarrow{g}$, the maximum error in our calculations will be ~0.3% error and is usually neglected.

We can combine the two terms on the right hand side of Equation [10.14] with $\overrightarrow{g}$ to get the equations of motion:

$$\frac{D\overrightarrow{U}}{Dt}=-\frac{1}{\rho}\overrightarrow{\nabla}p+\overrightarrow{g}-\frac{{C}_{d}}{h}\left|\overrightarrow{V}\right|\overrightarrow{V}-2\overrightarrow{\Omega}\times \overrightarrow{U}$$

The equations of motion in the Cartesian coordinates x, y, and z are:

$\begin{array}{l}\frac{Du}{Dt}=-\frac{1}{\rho}\frac{\partial p}{\partial x}-\frac{{C}_{d}}{h}\left|\overrightarrow{V}\right|u+2\Omega v\mathrm{sin}\varphi -2\Omega w\mathrm{cos}\varphi \\ \frac{Dv}{Dt}=-\frac{1}{\rho}\frac{\partial p}{\partial y}-\frac{{C}_{d}}{h}\left|\overrightarrow{V}\right|v-2\Omega u\mathrm{sin}\varphi \\ \frac{Dw}{Dt}=-\frac{1}{\rho}\frac{\partial p}{\partial z}-g+2\Omega u\mathrm{cos}\varphi \end{array}$

We will now move on to a discussion of the Coriolis Force. The following video (3:05) gives a basic introduction.

### Coriolis Force

The Coriolis force, $-2\overrightarrow{\Omega}\times \overrightarrow{U}$ , acts on an air parcel (or any other object) only when it is moving with respect to the Earth. It acts perpendicular to Earth's angular velocity vector and the air parcel's velocity vector. The explanation for the Coriolis force is usually broken into an explanation in the **zonal (constant latitude) direction** and the **meridional (constant longitude) direction**.

#### Zonal Flow (East-West Wind Velocity)

Consider an air parcel that is initially at rest and in hydrostatic balance but is impulsively accelerated to a velocity *u* to the east (see left side of figure below). Initially when it was at rest, it had the same acceleration as the Earth below it. However, after it accelerated to velocity *u*, it suddenly had more acceleration than it had before, throwing it out of hydrostatic balance. Look at the change in acceleration that comes from the air parcel suddenly acquiring a velocity to the east, which is just the acceleration after the velocity changesminus the acceleration before the velocity changes and equals *(Ω+u/R) ^{2}R-Ω^{2}R*. To a very good approximation, this change equals the Coriolis force,

*2Ωu*. There is a vertical component that points up, but there is also a horizontal component of force that points to the right of the motion in the Northern Hemisphere and to the left in the Southern Hemisphere.

Now consider an air parcel that is initially at rest but is impulsively accelerated to a velocity *u* to the west (see right side of figure below). The air parcel suddenly has less angular momentum than it had before and experiences a decreased centrifugal force. This decrease in angular momentum, to a very good approximation, equals the Coriolis force, *2Ωu,* but is pointed toward Earth's axis of rotation. There is a vertical component that points down, but the horizontal component of force that points to the right of the motion in the Northern Hemisphere and to the left in the Southern Hemisphere.

We can write down the accelerations in the y and z directions due to the air moving to the east with speed u:

$\text{Coriolisaccelerationintheydirection}=\frac{Dv}{Dt}=-2\Omega \mathrm{sin}\left(\varphi \right)\xb7u$

#### Meridional Flow (North-South Velocity)

What about an air parcel traveling to the north at a constant altitude? Note that the air parcel moving north starts at a greater distance from Earth’s axis and comes closer to Earth’s axis if it moves at the same height above the surface. Its angular momentum is conserved, so it is moving faster to the east than the Earth beneath it. As a result, it appears to move to the right or the east.

If the same air parcel moves to the south at the same height above Earth’s surface, then it moves to a greater distance from Earth’s rotation axis. Its angular momentum becomes less than that of Earth, it slows down relative to Earth, and it veers to the right of south or to the west.

In both the zonal and meridional flow cases, the air parcel's velocity with respect to the Earth causes the air parcel to have a different angular momentum from the Earth below it. Conservation of angular momentum during that motion requires that the apparent Coriolis force be added in order to describe the observed motion. See the video below (2:11) for further explanation:

Click here for a transcript for Coriolis Explanation Video

#### Finding the Magnitude and Direction of the Coriolis Force

The magnitude of the horizontal Coriolis force is simply^{o} or velocity - π/2 in radians) in the Northern Hemisphere and is to the left of the velocity vector (velocity direction + 90^{o} or velocity + π/2 in radians) in the Southern Hemisphere.

#### Quiz 10-1: All about forces.

- Find
**Practice Quiz 10-1**in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz. - When you feel you are ready, take
**Quiz 10-1**. You will be allowed to take this quiz only**once**. Good luck!