1. The Coriolis ForceIn describing wind and other motions,
convention says we take the surface of Earth as a reference frame, For
example, when we say there is a 10-meter-per-second wind, we mean the
air is moving past the surface at 10 m/sec. Because the surface rotates,
we are describing motions relative to a rotating reference frame. The
result is that an object moving in a straight line with respect to the
stars appears to follow a curved path on Earth’s surface, as shown in
Figure 4–12b. To maintain our Earth-bound reference frame, we need to
introduce a force that produces the apparent deflection. The Coriolis
force is just that force. It can be thought of as an accounting device
that lets us describe winds on a rotating Earth without violating
Newton’s law. Although the Coriolis force technically acts in three
dimensions, we will considerate it to be a horizontal force. This is an
acceptable approximation given that motions on Earth are essentially
horizontal thanks to how thin the atmosphere is compared to the size of
the planet.Rotation Around the Local VerticalThe key to understanding
the Coriolis force is to realize that what matters is rotation around
the local vertical direction. In other words, no matter where we are we
must ask: What direction is “up”, and how much rotation is there around
that direction? Clearly, the vertical direction is a line running from
the center of Earth through the location in question. In other words, it
is a line perpendicular to the surface. At the poles the vertical
direction corresponds to Earth’s axis of rotation, and thus the local
reference frame rotates 360 degrees per day. At 89° latitude, the local
vertical is not perfectly aligned with Earth’s axis, and so the
effective rotation rate is somewhat less. At still lower latitudes the
rotation rate is even less because the misalignment is greater (Figure
4-5-1). At the equator the local vertical is at a right angle to and not
aligned at all with Earth’s axis. Thus, there is no rotation around the
local vertical. As seen in Figure 4-5-1, the local vertical sweeps
around the axis of rotation, but remains perpendicular throughout the
day.To find the rotation at any latitude, we need to determine the angle
between the axis and the local vertical. This amounts to projecting the
axis onto the local vertical, and is given by the sine of latitude. In
other words,...Figure 4-5-1 shows some representative values. Notice
that Southern Hemisphere rotations are given negative values, which
reflects their opposite sense of rotation. Figure 4-5-2 shows the
Coriolis effect on four projectiles moving at 500 km/hr (311 mph). In
three hours Earth rotates by 45 degrees and all projectiles move a total
of 1500 km. However, their paths curve to the right, with more
curvature at higher latitudes than lower latitudes. (If launched at the
same latitude in the Southern Hemisphere, these projectiles would curve
leftward by the same amounts.)Wind Speed and the Coriolis ForceThe other
factor affecting the Coriolis force is the object’s speed. Look at
Figure 4-5-3 and notice the paths for objects moving half as fast, 250
km/hr (155 mph), as those in Figure 4-5-2. in three hours they have
traveled half as far as the objects traveling at 500 km/hr and have
experienced much less total deflection. If the deflection over the same
time period is less, the corresponding deflective “force” must be
smaller In fact the Coriolis force is directly proportional to wind
speed Putting the two factors together, the Coriolis force (FC) is given
by...where Ω is Earth’s rotation rate, φ is latitude, and υ is wind
speed.As written here, the equation gives the Coriolis force per unit
mass; that is, the force per kilogram of moving air.Combining the
information with Newton’s Second Law, which tells us that acceleration
is force per unit mass, we see that FC is an acceleration—specifically,
the Coriolis acceleration. That said, from here forward we won’t draw a
distinction between the Coriolis acceleration and the Coriolis force,
but will instead use the two interchangeably.FIGURE 4-5-1 Rotation
Around the Vertical Direction. Straight arrows show the direction of the
local vertical. Rotation around the local varies from zero at the
equator to once per day at the poles....FIGURE 4-5-2 Travel Paths of
Fast-Moving Objects. Four projectiles are launched north, south, east,
and west from a common point in the Northern Hemisphere. The figures
show their positions after one, two, and three hours of travel. All
cover the same distance and follow paths that curve to the right. Paths
extending farther north have more deflection and a stronger than average
Coriolis force....FIGURE 4-5-3 Travel Paths of Slower Objects. Four
projectiles moving half as fast as in Figure 4-5-2. With less time to
travel there is less deflection. For each unit of the travel time the
Coriolis Force is half as strong as in Figure 4-5-2....Calculate the
local rotation rate for latitude 45°. Is it half of the value at the
poles? Get solution
2. The Coriolis ForceIn describing wind and other motions, convention says we take the surface of Earth as a reference frame, For example, when we say there is a 10-meter-per-second wind, we mean the air is moving past the surface at 10 m/sec. Because the surface rotates, we are describing motions relative to a rotating reference frame. The result is that an object moving in a straight line with respect to the stars appears to follow a curved path on Earth’s surface, as shown in Figure 4–12b. To maintain our Earth-bound reference frame, we need to introduce a force that produces the apparent deflection. The Coriolis force is just that force. It can be thought of as an accounting device that lets us describe winds on a rotating Earth without violating Newton’s law. Although the Coriolis force technically acts in three dimensions, we will considerate it to be a horizontal force. This is an acceptable approximation given that motions on Earth are essentially horizontal thanks to how thin the atmosphere is compared to the size of the planet.Rotation Around the Local VerticalThe key to understanding the Coriolis force is to realize that what matters is rotation around the local vertical direction. In other words, no matter where we are we must ask: What direction is “up”, and how much rotation is there around that direction? Clearly, the vertical direction is a line running from the center of Earth through the location in question. In other words, it is a line perpendicular to the surface. At the poles the vertical direction corresponds to Earth’s axis of rotation, and thus the local reference frame rotates 360 degrees per day. At 89° latitude, the local vertical is not perfectly aligned with Earth’s axis, and so the effective rotation rate is somewhat less. At still lower latitudes the rotation rate is even less because the misalignment is greater (Figure 4-5-1). At the equator the local vertical is at a right angle to and not aligned at all with Earth’s axis. Thus, there is no rotation around the local vertical. As seen in Figure 4-5-1, the local vertical sweeps around the axis of rotation, but remains perpendicular throughout the day.To find the rotation at any latitude, we need to determine the angle between the axis and the local vertical. This amounts to projecting the axis onto the local vertical, and is given by the sine of latitude. In other words,...Figure 4-5-1 shows some representative values. Notice that Southern Hemisphere rotations are given negative values, which reflects their opposite sense of rotation. Figure 4-5-2 shows the Coriolis effect on four projectiles moving at 500 km/hr (311 mph). In three hours Earth rotates by 45 degrees and all projectiles move a total of 1500 km. However, their paths curve to the right, with more curvature at higher latitudes than lower latitudes. (If launched at the same latitude in the Southern Hemisphere, these projectiles would curve leftward by the same amounts.)Wind Speed and the Coriolis ForceThe other factor affecting the Coriolis force is the object’s speed. Look at Figure 4-5-3 and notice the paths for objects moving half as fast, 250 km/hr (155 mph), as those in Figure 4-5-2. in three hours they have traveled half as far as the objects traveling at 500 km/hr and have experienced much less total deflection. If the deflection over the same time period is less, the corresponding deflective “force” must be smaller In fact the Coriolis force is directly proportional to wind speed Putting the two factors together, the Coriolis force (FC) is given by...where Ω is Earth’s rotation rate, φ is latitude, and υ is wind speed.As written here, the equation gives the Coriolis force per unit mass; that is, the force per kilogram of moving air.Combining the information with Newton’s Second Law, which tells us that acceleration is force per unit mass, we see that FC is an acceleration—specifically, the Coriolis acceleration. That said, from here forward we won’t draw a distinction between the Coriolis acceleration and the Coriolis force, but will instead use the two interchangeably.FIGURE 4-5-1 Rotation Around the Vertical Direction. Straight arrows show the direction of the local vertical. Rotation around the local varies from zero at the equator to once per day at the poles....FIGURE 4-5-2 Travel Paths of Fast-Moving Objects. Four projectiles are launched north, south, east, and west from a common point in the Northern Hemisphere. The figures show their positions after one, two, and three hours of travel. All cover the same distance and follow paths that curve to the right. Paths extending farther north have more deflection and a stronger than average Coriolis force....FIGURE 4-5-3 Travel Paths of Slower Objects. Four projectiles moving half as fast as in Figure 4-5-2. With less time to travel there is less deflection. For each unit of the travel time the Coriolis Force is half as strong as in Figure 4-5-2....Suppose you wanted to calculate the Coriolis force on Jupiter. What factors would you have to consider? Get solution
2. The Coriolis ForceIn describing wind and other motions, convention says we take the surface of Earth as a reference frame, For example, when we say there is a 10-meter-per-second wind, we mean the air is moving past the surface at 10 m/sec. Because the surface rotates, we are describing motions relative to a rotating reference frame. The result is that an object moving in a straight line with respect to the stars appears to follow a curved path on Earth’s surface, as shown in Figure 4–12b. To maintain our Earth-bound reference frame, we need to introduce a force that produces the apparent deflection. The Coriolis force is just that force. It can be thought of as an accounting device that lets us describe winds on a rotating Earth without violating Newton’s law. Although the Coriolis force technically acts in three dimensions, we will considerate it to be a horizontal force. This is an acceptable approximation given that motions on Earth are essentially horizontal thanks to how thin the atmosphere is compared to the size of the planet.Rotation Around the Local VerticalThe key to understanding the Coriolis force is to realize that what matters is rotation around the local vertical direction. In other words, no matter where we are we must ask: What direction is “up”, and how much rotation is there around that direction? Clearly, the vertical direction is a line running from the center of Earth through the location in question. In other words, it is a line perpendicular to the surface. At the poles the vertical direction corresponds to Earth’s axis of rotation, and thus the local reference frame rotates 360 degrees per day. At 89° latitude, the local vertical is not perfectly aligned with Earth’s axis, and so the effective rotation rate is somewhat less. At still lower latitudes the rotation rate is even less because the misalignment is greater (Figure 4-5-1). At the equator the local vertical is at a right angle to and not aligned at all with Earth’s axis. Thus, there is no rotation around the local vertical. As seen in Figure 4-5-1, the local vertical sweeps around the axis of rotation, but remains perpendicular throughout the day.To find the rotation at any latitude, we need to determine the angle between the axis and the local vertical. This amounts to projecting the axis onto the local vertical, and is given by the sine of latitude. In other words,...Figure 4-5-1 shows some representative values. Notice that Southern Hemisphere rotations are given negative values, which reflects their opposite sense of rotation. Figure 4-5-2 shows the Coriolis effect on four projectiles moving at 500 km/hr (311 mph). In three hours Earth rotates by 45 degrees and all projectiles move a total of 1500 km. However, their paths curve to the right, with more curvature at higher latitudes than lower latitudes. (If launched at the same latitude in the Southern Hemisphere, these projectiles would curve leftward by the same amounts.)Wind Speed and the Coriolis ForceThe other factor affecting the Coriolis force is the object’s speed. Look at Figure 4-5-3 and notice the paths for objects moving half as fast, 250 km/hr (155 mph), as those in Figure 4-5-2. in three hours they have traveled half as far as the objects traveling at 500 km/hr and have experienced much less total deflection. If the deflection over the same time period is less, the corresponding deflective “force” must be smaller In fact the Coriolis force is directly proportional to wind speed Putting the two factors together, the Coriolis force (FC) is given by...where Ω is Earth’s rotation rate, φ is latitude, and υ is wind speed.As written here, the equation gives the Coriolis force per unit mass; that is, the force per kilogram of moving air.Combining the information with Newton’s Second Law, which tells us that acceleration is force per unit mass, we see that FC is an acceleration—specifically, the Coriolis acceleration. That said, from here forward we won’t draw a distinction between the Coriolis acceleration and the Coriolis force, but will instead use the two interchangeably.FIGURE 4-5-1 Rotation Around the Vertical Direction. Straight arrows show the direction of the local vertical. Rotation around the local varies from zero at the equator to once per day at the poles....FIGURE 4-5-2 Travel Paths of Fast-Moving Objects. Four projectiles are launched north, south, east, and west from a common point in the Northern Hemisphere. The figures show their positions after one, two, and three hours of travel. All cover the same distance and follow paths that curve to the right. Paths extending farther north have more deflection and a stronger than average Coriolis force....FIGURE 4-5-3 Travel Paths of Slower Objects. Four projectiles moving half as fast as in Figure 4-5-2. With less time to travel there is less deflection. For each unit of the travel time the Coriolis Force is half as strong as in Figure 4-5-2....Suppose you wanted to calculate the Coriolis force on Jupiter. What factors would you have to consider? Get solution
