1. Why Cloud Droplets Don't FallYou are probably familiar with the
legendary, late-sixteenth-century experiment of Galileo Galilei, who
dropped two objects—a light one and a heavy one—off the Leaning Tower of
Pisa. The objects, being subjected to the same gravitational
acceleration, hit the ground at nearly the same time. Galileo's
demonstration may seem inconsistent with our everyday experience,
because an ant would surely take longer than a golf ball to fall from
the top of a tall building. It is also at odds with our claim that small
droplets fall slowly. The solution must be that a force besides gravity
acts on falling objects; that force is called wind resistance, or drag.
By examining how these two forces work together, we will gain some
insight into why cloud droplets do not fall. To keep the discussion
simple, we will assume spherical droplets throughout—using more
realistic shapes, however, would not change our conclusions.Newton's
Second Law tells us that if a net force is applied to a mass, it will
undergo an acceleration (or change in velocity through time). For a
given mass, the acceleration is directly proportional to the net force.
In equation form, the law is given as...Notice that Newton's Second Law
says that we must consider the net force, the result of all the forces
acting on the object. As far as a falling droplet is concerned, there is
the down-ward gravitational force, which is opposed by the force of
wind resistance (drag). A droplet suddenly released in the atmosphere
falls at increasing speed, but not indefinitely. Eventually the force of
drag (Fd) balances the force of gravity (Fg), resulting in no net
force:...With no net force, there is no acceleration, and the droplet
falls at its terminal velocity. How fast does it fall? To answer that,
we need to know something about the magnitude of the two forces.Force of
GravityThe force of gravity is equal to mass times the acceleration of
gravity, g. Whenever we step on a scale, we measure this force. For a
liquid droplet presumed to be spherical, mass is the density of water,
ρ, times the droplet volume, 4/3/πr3, where r is the droplet radius. We
therefore obtain Fg as follows:...Force of DragDrag between the droplet
and surrounding air depends on the rate of fall and on the size of the
droplet. Just like an automobile on a highway a faster-moving droplet
experiences greater resistance as it moves through the air. In fact, to a
good approximation, the drag force increases with the square of wind
speed, v2. So how does size influence drag?It can be shown that the drag
can be expressed as...where CD is a constant referred to as the drag
coefficient, and ρa is the density of air (about one-thousandth the
density of water). The value of CD is not important here; what matters
is that Fd ts proportional to the square of both the fall rate (v2) and
the radius (r2).Terminal VelocityFor a droplet falling at terminal
velocity, we have said that gravity and drag are equal. If we use vt for
terminal velocity, we get...To find the terminal velocity, we rearrange
and first solve for ...:...By consolidating the numerical values and
constants in the above equation into a single constant, k1, we
get...where .... From this equation, we see that as droplet radius
increases, so does terminal velocity. Large droplets fall faster than
small droplets. What happens physically is that both Fg and Fd increase
with radius, but the gravitational force increases more than drag, and a
higher fall rate is therefore required to cancel Fg. Notice that as far
as the droplet is concerned, falling through a still atmosphere at vt
is the same as remaining stationary in an updraft of speed vt. Thus, the
equation says that a strong updraft is needed to hold a large droplet
aloft, whereas a small droplet is easily suspended.Going back to the
Leaning Tower of Pisa situation described at the beginning of this box,
we can now understand why Galileo's objects fell at nearly the same
speed. With large and therefore heavy objects, the gravitational forces
far exceeded the drag forces throughout their short fall. With
negligible drag, gravity accelerated both at nearly the same rate. If he
had used objects of greatly different size, or if the objects had
fallen far enough to reach their terminal velocities, differences in vt
would have emerged. The old, familiar story would be about wind
resistance, and books like this would have no need for a feature on the
topic.7-1-1 Why Cloud Droplets Don't FallYou are probably familiar with
the legendary, late-sixteenth-century experiment of Galileo Galilei, who
dropped two objects—a light one and a heavy one—off the Leaning Tower
of Pisa. The objects, being subjected to the same gravitational
acceleration, hit the ground at nearly the same time. Galileo's
demonstration may seem inconsistent with our everyday experience,
because an ant would surely take longer than a golf ball to fall from
the top of a tall building. It is also at odds with our claim that small
droplets fall slowly. The solution must be that a force besides gravity
acts on falling objects; that force is called wind resistance, or drag.
By examining how these two forces work together, we will gain some
insight into why cloud droplets do not fall. To keep the discussion
simple, we will assume spherical droplets throughout—using more
realistic shapes, however, would not change our conclusions.Newton's
Second Law tells us that if a net force is applied to a mass, it will
undergo an acceleration (or change in velocity through time). For a
given mass, the acceleration is directly proportional to the net force.
In equation form, the law is given as...Notice that Newton's Second Law
says that we must consider the net force, the result of all the forces
acting on the object. As far as a falling droplet is concerned, there is
the down-ward gravitational force, which is opposed by the force of
wind resistance (drag). A droplet suddenly released in the atmosphere
falls at increasing speed, but not indefinitely. Eventually the force of
drag (Fd) balances the force of gravity (Fg), resulting in no net
force:...With no net force, there is no acceleration, and the droplet
falls at its terminal velocity. How fast does it fall? To answer that,
we need to know something about the magnitude of the two forces.Force of
GravityThe force of gravity is equal to mass times the acceleration of
gravity, g. Whenever we step on a scale, we measure this force. For a
liquid droplet presumed to be spherical, mass is the density of water,
ρ, times the droplet volume, 4/3/πr3, where r is the droplet radius. We
therefore obtain Fg as follows:...Force of DragDrag between the droplet
and surrounding air depends on the rate of fall and on the size of the
droplet. Just like an automobile on a highway a faster-moving droplet
experiences greater resistance as it moves through the air. In fact, to a
good approximation, the drag force increases with the square of wind
speed, v2. So how does size influence drag?It can be shown that the drag
can be expressed as...where CD is a constant referred to as the drag
coefficient, and ρa is the density of air (about one-thousandth the
density of water). The value of CD is not important here; what matters
is that Fd ts proportional to the square of both the fall rate (v2) and
the radius (r2).Terminal VelocityFor a droplet falling at terminal
velocity, we have said that gravity and drag are equal. If we use vt for
terminal velocity, we get...To find the terminal velocity, we rearrange
and first solve for ...:...By consolidating the numerical values and
constants in the above equation into a single constant, k1, we
get...where .... From this equation, we see that as droplet radius
increases, so does terminal velocity. Large droplets fall faster than
small droplets. What happens physically is that both Fg and Fd increase
with radius, but the gravitational force increases more than drag, and a
higher fall rate is therefore required to cancel Fg. Notice that as far
as the droplet is concerned, falling through a still atmosphere at vt
is the same as remaining stationary in an updraft of speed vt. Thus, the
equation says that a strong updraft is needed to hold a large droplet
aloft, whereas a small droplet is easily suspended.Going back to the
Leaning Tower of Pisa situation described at the beginning of this box,
we can now understand why Galileo's objects fell at nearly the same
speed. With large and therefore heavy objects, the gravitational forces
far exceeded the drag forces throughout their short fall. With
negligible drag, gravity accelerated both at nearly the same rate. If he
had used objects of greatly different size, or if the objects had
fallen far enough to reach their terminal velocities, differences in vt
would have emerged. The old, familiar story would be about wind
resistance, and books like this would have no need for a feature on the
topic.How do the forces of gravity and drag interact to give objects
their terminal velocities? Get solution
2. Why Cloud Droplets Don't FallYou are probably familiar with the legendary, late-sixteenth-century experiment of Galileo Galilei, who dropped two objects—a light one and a heavy one—off the Leaning Tower of Pisa. The objects, being subjected to the same gravitational acceleration, hit the ground at nearly the same time. Galileo's demonstration may seem inconsistent with our everyday experience, because an ant would surely take longer than a golf ball to fall from the top of a tall building. It is also at odds with our claim that small droplets fall slowly. The solution must be that a force besides gravity acts on falling objects; that force is called wind resistance, or drag. By examining how these two forces work together, we will gain some insight into why cloud droplets do not fall. To keep the discussion simple, we will assume spherical droplets throughout—using more realistic shapes, however, would not change our conclusions.Newton's Second Law tells us that if a net force is applied to a mass, it will undergo an acceleration (or change in velocity through time). For a given mass, the acceleration is directly proportional to the net force. In equation form, the law is given as...Notice that Newton's Second Law says that we must consider the net force, the result of all the forces acting on the object. As far as a falling droplet is concerned, there is the down-ward gravitational force, which is opposed by the force of wind resistance (drag). A droplet suddenly released in the atmosphere falls at increasing speed, but not indefinitely. Eventually the force of drag (Fd) balances the force of gravity (Fg), resulting in no net force:...With no net force, there is no acceleration, and the droplet falls at its terminal velocity. How fast does it fall? To answer that, we need to know something about the magnitude of the two forces.Force of GravityThe force of gravity is equal to mass times the acceleration of gravity, g. Whenever we step on a scale, we measure this force. For a liquid droplet presumed to be spherical, mass is the density of water, ρ, times the droplet volume, 4/3/πr3, where r is the droplet radius. We therefore obtain Fg as follows:...Force of DragDrag between the droplet and surrounding air depends on the rate of fall and on the size of the droplet. Just like an automobile on a highway a faster-moving droplet experiences greater resistance as it moves through the air. In fact, to a good approximation, the drag force increases with the square of wind speed, v2. So how does size influence drag?It can be shown that the drag can be expressed as...where CD is a constant referred to as the drag coefficient, and ρa is the density of air (about one-thousandth the density of water). The value of CD is not important here; what matters is that Fd ts proportional to the square of both the fall rate (v2) and the radius (r2).Terminal VelocityFor a droplet falling at terminal velocity, we have said that gravity and drag are equal. If we use vt for terminal velocity, we get...To find the terminal velocity, we rearrange and first solve for ...:...By consolidating the numerical values and constants in the above equation into a single constant, k1, we get...where .... From this equation, we see that as droplet radius increases, so does terminal velocity. Large droplets fall faster than small droplets. What happens physically is that both Fg and Fd increase with radius, but the gravitational force increases more than drag, and a higher fall rate is therefore required to cancel Fg. Notice that as far as the droplet is concerned, falling through a still atmosphere at vt is the same as remaining stationary in an updraft of speed vt. Thus, the equation says that a strong updraft is needed to hold a large droplet aloft, whereas a small droplet is easily suspended.Going back to the Leaning Tower of Pisa situation described at the beginning of this box, we can now understand why Galileo's objects fell at nearly the same speed. With large and therefore heavy objects, the gravitational forces far exceeded the drag forces throughout their short fall. With negligible drag, gravity accelerated both at nearly the same rate. If he had used objects of greatly different size, or if the objects had fallen far enough to reach their terminal velocities, differences in vt would have emerged. The old, familiar story would be about wind resistance, and books like this would have no need for a feature on the topic.What is the formula for estimating the terminal velocities of falling raindrops? What does that formula tell us about how dramatically a change in radius affects the terminal velocity? Get solution
2. Why Cloud Droplets Don't FallYou are probably familiar with the legendary, late-sixteenth-century experiment of Galileo Galilei, who dropped two objects—a light one and a heavy one—off the Leaning Tower of Pisa. The objects, being subjected to the same gravitational acceleration, hit the ground at nearly the same time. Galileo's demonstration may seem inconsistent with our everyday experience, because an ant would surely take longer than a golf ball to fall from the top of a tall building. It is also at odds with our claim that small droplets fall slowly. The solution must be that a force besides gravity acts on falling objects; that force is called wind resistance, or drag. By examining how these two forces work together, we will gain some insight into why cloud droplets do not fall. To keep the discussion simple, we will assume spherical droplets throughout—using more realistic shapes, however, would not change our conclusions.Newton's Second Law tells us that if a net force is applied to a mass, it will undergo an acceleration (or change in velocity through time). For a given mass, the acceleration is directly proportional to the net force. In equation form, the law is given as...Notice that Newton's Second Law says that we must consider the net force, the result of all the forces acting on the object. As far as a falling droplet is concerned, there is the down-ward gravitational force, which is opposed by the force of wind resistance (drag). A droplet suddenly released in the atmosphere falls at increasing speed, but not indefinitely. Eventually the force of drag (Fd) balances the force of gravity (Fg), resulting in no net force:...With no net force, there is no acceleration, and the droplet falls at its terminal velocity. How fast does it fall? To answer that, we need to know something about the magnitude of the two forces.Force of GravityThe force of gravity is equal to mass times the acceleration of gravity, g. Whenever we step on a scale, we measure this force. For a liquid droplet presumed to be spherical, mass is the density of water, ρ, times the droplet volume, 4/3/πr3, where r is the droplet radius. We therefore obtain Fg as follows:...Force of DragDrag between the droplet and surrounding air depends on the rate of fall and on the size of the droplet. Just like an automobile on a highway a faster-moving droplet experiences greater resistance as it moves through the air. In fact, to a good approximation, the drag force increases with the square of wind speed, v2. So how does size influence drag?It can be shown that the drag can be expressed as...where CD is a constant referred to as the drag coefficient, and ρa is the density of air (about one-thousandth the density of water). The value of CD is not important here; what matters is that Fd ts proportional to the square of both the fall rate (v2) and the radius (r2).Terminal VelocityFor a droplet falling at terminal velocity, we have said that gravity and drag are equal. If we use vt for terminal velocity, we get...To find the terminal velocity, we rearrange and first solve for ...:...By consolidating the numerical values and constants in the above equation into a single constant, k1, we get...where .... From this equation, we see that as droplet radius increases, so does terminal velocity. Large droplets fall faster than small droplets. What happens physically is that both Fg and Fd increase with radius, but the gravitational force increases more than drag, and a higher fall rate is therefore required to cancel Fg. Notice that as far as the droplet is concerned, falling through a still atmosphere at vt is the same as remaining stationary in an updraft of speed vt. Thus, the equation says that a strong updraft is needed to hold a large droplet aloft, whereas a small droplet is easily suspended.Going back to the Leaning Tower of Pisa situation described at the beginning of this box, we can now understand why Galileo's objects fell at nearly the same speed. With large and therefore heavy objects, the gravitational forces far exceeded the drag forces throughout their short fall. With negligible drag, gravity accelerated both at nearly the same rate. If he had used objects of greatly different size, or if the objects had fallen far enough to reach their terminal velocities, differences in vt would have emerged. The old, familiar story would be about wind resistance, and books like this would have no need for a feature on the topic.What is the formula for estimating the terminal velocities of falling raindrops? What does that formula tell us about how dramatically a change in radius affects the terminal velocity? Get solution
