Are astronauts, in Earth orbit, without weight?
First, I should mention that there are two types of weight: actual weight and apparent weight. An Earth-bound object's actual weight is the downward force exerted upon it by the Earth's gravity. The object's apparent weight is the upward force, typically transmitted through the ground, that opposes gravity and prevents the object from falling through the floor or ground (towards the center of the Earth). When you stand on your bathroom scale, it's measuring your apparent weight (i.e. how hard it's having to push up on you to prevent you from accelerating downwards through the scale, crushing it). This doesn't necessarily have to be equal to your actual weight. In your bathroom, your apparent weight is equal to your actual weight, but if you carry the scale into an elevator and weigh yourself while accelerating upwards, the scale will register an apparent weight that is greater than your actual weight. Since a cable is pulling the elevator up rapidly, the elevator's floor is pushing up on the scale, and the scale is, in turn, pushing up on your feet. In order to force you upwards, against gravity, the scale is having to push on you harder than if it (and you) were stationary. Your actual weight doesn't change here, because it's dependent on your mass (which isn't changing) and your distance from the center of the Earth (which is changing only a negligible amount). But to the scale, it feels like you're growing heavier, because it's having to not only support you but push you (accelerate you) upwards. So it registers a heavier "weight", which we now know to be your apparent weight. What about when the elevator accelerates downwards? Your apparent weight becomes less than your actual weight. And now, what if the elevator's cable were to break and the elevator, scale, and you were all to freefall towards the ground below? The scale would be falling at the same rate you were falling, and so it wouldn't be supporting any of your weight. It would indicate a weight of zero. That is, your apparent weight would be zero. This is the definition of weightlessness. Weightlessness means without apparent weight; it has nothing to do with your actual weight. So ... an astronaut in orbit, in a constant state of freefall, kept to a near circular path around the Earth by gravity, is weightless, but only in the sense that he or she has no apparent weight. Astronauts most definitely do have actual weight!
Apparent weight can change fairly easily, we see. All you have to do is take a ride on an elevator, or rollercoaster, or some other device that accelerates you in a vertical direction. Does actual weight ever change (ignoring the effects of food)? Yes, it does. The force of gravity on you (which determines your actual weight) is dependent on your mass, the mass of the Earth, the gravitational constant G, and the distance between you and the center of the Earth. Assuming your mass is held constant, you can reduce your actual weight by increasing your distance from the center of the Earth. So astronauts have actual weight, but an actual weight that is slightly less than their actual weight back on Earth. How much less? In orbit around 300 km (185 miles) above the surface of the Earth, astronauts' actual weight will be about 8.8% less than back on Earth.
What makes you feel weightless when you're falling, even though you still have an actual weight? Or one could ask, what makes you feel heavy (or with weight) when you're standing on the ground? It's not gravity. It's the force of the surface you're standing on, pushing against you. If you're standing on a sidewalk, the concrete is pressing against your feet, which are in turn pressing against your ankles, which are pressing against your lower legs, which are pressing against your upper legs, and on and on. The feet are supporting your entire mass. Your chest, for example, only supports the mass of the body above the chest. You don't feel pressure evenly distributed throughout your body. (Well, you're used to the feeling of standing on a surface, and so you may have a hard time sensing this uneven pressure distribution, but it's there.) Your sense of weight also comes from your arms pulling down on your shoulders. When in freefall, this pressure gradient (or change over space) disappears. Each section of your body, each cell, is falling at the same rate. Therefore, your upper body isn't pushing on your lower body. Your ankles aren't pushing on your feet. There is no pushing at all. Neither are your arms pulling down on your shoulders. The absence of these sensations is what one equates to feeling weightless.
How does NASA simulate a weightless environment for astronaut training? They could put their astronauts in an elevator, take it to the top of a tall building, cut the supporting cable, and allow the elevator and its inhabitants to freefall for several seconds. But the impact upon hitting the ground would be extreme and most unpleasant. Instead, NASA sends its astronauts up in an airplane, and the airplane flies in the parabolic trajectories of freely falling objects. Soaring over the Gulf of Mexico, pilots level off at about 26,000 feet. They then shoot the plane upward at about a 45-degree angle. At this point, the apparent weight of the people inside the nearly empty, padded fuselage increases to about 1.8 times their actual weight. Half a minute later, pilots push the aircraft's nose over the top of this "parabola", and the plane falls some 8,000 feet or so until its pointing downward at about 30 degrees. During this freefall, the aircraft's acceleration matches Earth's acceleration of gravity, making everything inside weightless for 17 to 25 seconds. (Parts of the movie Apollo 13 were filmed on this aircraft.) Over a two-hour flight, the aircraft may fly through some 40 of these parabolas. NASA used two KC-135 Stratotanker aircraft for these sessions from 1973 until 2005, when they were retired and replaced with a McDonnell Douglas C-9. The plane, not too surprisingly, earned the nickname "Vomit Comet."
Sunday, November 22, 2009
Monday, November 2, 2009
Why Do Golf Balls Have Dimples?
Why do golf balls have dimples? The dimples enable the ball to fly much farther through the air. A swing, driving a smooth golf ball 70 yards, could drive a dimpled ball perhaps 250 yards. Why?
First, air pressure is the force exerted by air molecules divided by the area on which the force is exerted. That is, force per unit area. The force comes from the countless collisions of the air molecules (i.e. nitrogen and oxygen molecules, as well as a very small number of carbon dioxide molecules and argon atoms) against the surface in question. Keep in mind that a net force on an object causes that object to accelerate (or decelerate). If the net force acts in the direction of the object's motion, it accelerates the object; acting in the direction opposite the object's motion, it decelerates the object.
Daniel Bernoulli was born in the Dutch Republic (now known as The Netherlands) in the year 1700. He's perhaps best known for discovering a relationship between the pressure, velocity (speed in a certain direction), and height (above some arbitrary reference level) of an incompressible fluid in perfect steady-state flow. Water being pumped through a pipe can fit this description. It's virtually incompressible, and the pump can keep it moving at a steady rate through the pipe. Air, while not incompressible, is close enough to an incompressible fluid in steady-state flow under certain conditions (velocity less than 300 km/h and no pressure differences of more than one tenth of an atmosphere) that we can use Bernoulli's equation to understand its behavior. So what's the relationship? For a fluid as described above, the pressure, plus one-half times the density times the velocity squared, plus the density times the acceleration due to gravity times the height above some arbitrary reference level, is constant. In equation form, P + 1/2 dv2 + dgh = constant. So what happens if I increase the velocity (v) of the fluid? Either the pressure (P) must decrease or the height (h) must decrease, so that the left side of the equation remains equal to the constant. You should take from this equation the following: for an incompressible, steady-state flow liquid, of a particular density (d), and at a set height (that doesn't change), pressure and velocity always move in opposite directions. If pressure decreases, velocity increases. If pressure increases, velocity decreases.
When the path of a fluid in steady-state flow bends, the pressure on the outside of the bend is always higher than the pressure on the inside of the bend. It's this pressure imbalance that causes the fluid to bend. This pressure change indicates a change in the fluid's velocity. So does the fluid on the outside of the bend speed up or slow down? It slows down. And the fluid on the inside of the bend? It speeds up, of course.
When a ball is hurtling through the air, the air it encounters is forced to flow around it. Some of the air flows over the top of the ball, some flows beneath the ball, and some air flows around each side. Air pressure above, beneath, and aside the ball is not everywhere the same. As the air encounters the front of the ball, it bends away from the ball, moving out of the way. (The ball is on the outside of the bend.) This creates a high-pressure zone in front of the ball. And the air here slows down. The air then curves back towards the ball, on all sides of the ball, hugging its surface as it moves towards the back of the ball. This puts the ball on the inside of many curved paths (or bends). Therefore, the air around the ball's middle is at low pressure and high speed. As the air reaches the back of the ball, it peels away from the ball and straightens back out. This bending of the air away from the ball creates a high-pressure zone behind the ball. Low speed air. Now you ask, how can the low-pressure air along the sides of the ball move into the high-pressure zone behind the ball? Doesn't air always move from a high-pressure zone into a low-pressure zone? Normally, yes. Here, the low-pressure air is definitely moving against the tide, so to speak. It's fighting its way into the high-pressure zone, slowing down (decelerating) as the high-pressure air pushes on it. But it has enough energy to successfully make the trip. It does reach the back of the ball. Now, these pressure imbalances are symmetric about the ball; they balance one another and produce no net force on the ball. They don't accelerate or decelerate the ball itself. Air resistance does exist, but it's a result of air near the ball's surface rubbing against the surface, producing a type of friction. Viscous drag, it's called. The air resistance is not a result of the pressure variations just described. Okay, now for a qualifier! The behavior of the air about the ball, as described in this paragraph, applies to balls traveling at slow speeds. This is important. The air behaves differently when it encounters a ball moving at high speed.
To describe the path of air flowing around a fast-moving ball, I must introduce the term boundary layer. A thin layer of air moving very close to the surface of the ball is called the boundary layer, and it behaves differently from air farther from the surface. It moves more slowly and has less total energy than the freely flowing air farther out. Why? Because friction with the ball's surface (i.e. viscous drag) slows it down and robs it of energy.
Hmmm. So you're thinking, it's hard for the air along the sides of the ball to push into the high-pressure zone behind the ball. Okay. But it sounds like it can do it anyways. Guess it has enough energy to do so. And that boundary layer. It has less energy than the air just a bit farther out. But, well, it seems that it, too, is able to push into the high-pressure zone. At least when the ball is moving slowly. (Good. You're right so far.) And so does this change when the ball is moving rapidly? Yes. When the ball is moving rapidly, this lower-energy boundary layer of air is no longer able to push into the high-pressure zone behind the ball. In fact, it is pushed back towards the sides of the ball by the adverse pressure gradient, cutting like a wedge between the ball and the freely flowing air outside this boundary layer. No longer does the air curve around behind the ball. This leaves us with an air pocket behind the ball; a turbulent wake, in other words. In this wake, the air pressure is roughly atmospheric. There goes the symmetry of pressure forces on the ball. Now there is no high-pressure zone behind the ball to cancel the high-pressure zone in front of the ball. There is a large pressure drag, a force on the ball in the direction of downwind, slowing the ball down. Decelerating it. This pressure drag is what limits the range of a smooth golf ball. Yes, there is also viscous drag, but it's not nearly as significant as the large pressure drag caused by the turbulent wake.
(turbulent wake behind ball, which is moving to the left)
So dimpled golf balls travel farther than smooth golf balls. Do the dimples somehow reduce the size (and severity) of this turbulent wake, reducing the pressure drag on the ball, preventing the ball from slowing so much as it arcs through the air? Yes, indeed. The dimples, or surface irregularities, cause the air in the boundary layer to tumble about. This tumbling about gives the boundary-layer air more energy, and more forward momentum. It now has a much better chance of pushing around to the back side of the ball, into the high-pressure zone. Alas, it still doesn't make it, but it comes much closer. It travels partially around the back of the ball before its progress is stopped and it separates from the surface. The air outside the boundary layer, following along, hugs the ball for a longer time, as well. It separates from the ball at the same spot where the boundary layer separates, this being a fair ways down the backside of the ball. The result is a smaller air pocket. A small turbulent wake. A less dramatic variation in air pressure between the front of the ball and the back of the ball. A more modest force of pressure drag. And this reduction in pressure drag is what enables the dimpled ball to soar some 200 yards farther than a smooth ball.
First, air pressure is the force exerted by air molecules divided by the area on which the force is exerted. That is, force per unit area. The force comes from the countless collisions of the air molecules (i.e. nitrogen and oxygen molecules, as well as a very small number of carbon dioxide molecules and argon atoms) against the surface in question. Keep in mind that a net force on an object causes that object to accelerate (or decelerate). If the net force acts in the direction of the object's motion, it accelerates the object; acting in the direction opposite the object's motion, it decelerates the object.
Daniel Bernoulli was born in the Dutch Republic (now known as The Netherlands) in the year 1700. He's perhaps best known for discovering a relationship between the pressure, velocity (speed in a certain direction), and height (above some arbitrary reference level) of an incompressible fluid in perfect steady-state flow. Water being pumped through a pipe can fit this description. It's virtually incompressible, and the pump can keep it moving at a steady rate through the pipe. Air, while not incompressible, is close enough to an incompressible fluid in steady-state flow under certain conditions (velocity less than 300 km/h and no pressure differences of more than one tenth of an atmosphere) that we can use Bernoulli's equation to understand its behavior. So what's the relationship? For a fluid as described above, the pressure, plus one-half times the density times the velocity squared, plus the density times the acceleration due to gravity times the height above some arbitrary reference level, is constant. In equation form, P + 1/2 dv2 + dgh = constant. So what happens if I increase the velocity (v) of the fluid? Either the pressure (P) must decrease or the height (h) must decrease, so that the left side of the equation remains equal to the constant. You should take from this equation the following: for an incompressible, steady-state flow liquid, of a particular density (d), and at a set height (that doesn't change), pressure and velocity always move in opposite directions. If pressure decreases, velocity increases. If pressure increases, velocity decreases.
When the path of a fluid in steady-state flow bends, the pressure on the outside of the bend is always higher than the pressure on the inside of the bend. It's this pressure imbalance that causes the fluid to bend. This pressure change indicates a change in the fluid's velocity. So does the fluid on the outside of the bend speed up or slow down? It slows down. And the fluid on the inside of the bend? It speeds up, of course.
When a ball is hurtling through the air, the air it encounters is forced to flow around it. Some of the air flows over the top of the ball, some flows beneath the ball, and some air flows around each side. Air pressure above, beneath, and aside the ball is not everywhere the same. As the air encounters the front of the ball, it bends away from the ball, moving out of the way. (The ball is on the outside of the bend.) This creates a high-pressure zone in front of the ball. And the air here slows down. The air then curves back towards the ball, on all sides of the ball, hugging its surface as it moves towards the back of the ball. This puts the ball on the inside of many curved paths (or bends). Therefore, the air around the ball's middle is at low pressure and high speed. As the air reaches the back of the ball, it peels away from the ball and straightens back out. This bending of the air away from the ball creates a high-pressure zone behind the ball. Low speed air. Now you ask, how can the low-pressure air along the sides of the ball move into the high-pressure zone behind the ball? Doesn't air always move from a high-pressure zone into a low-pressure zone? Normally, yes. Here, the low-pressure air is definitely moving against the tide, so to speak. It's fighting its way into the high-pressure zone, slowing down (decelerating) as the high-pressure air pushes on it. But it has enough energy to successfully make the trip. It does reach the back of the ball. Now, these pressure imbalances are symmetric about the ball; they balance one another and produce no net force on the ball. They don't accelerate or decelerate the ball itself. Air resistance does exist, but it's a result of air near the ball's surface rubbing against the surface, producing a type of friction. Viscous drag, it's called. The air resistance is not a result of the pressure variations just described. Okay, now for a qualifier! The behavior of the air about the ball, as described in this paragraph, applies to balls traveling at slow speeds. This is important. The air behaves differently when it encounters a ball moving at high speed.
To describe the path of air flowing around a fast-moving ball, I must introduce the term boundary layer. A thin layer of air moving very close to the surface of the ball is called the boundary layer, and it behaves differently from air farther from the surface. It moves more slowly and has less total energy than the freely flowing air farther out. Why? Because friction with the ball's surface (i.e. viscous drag) slows it down and robs it of energy.
Hmmm. So you're thinking, it's hard for the air along the sides of the ball to push into the high-pressure zone behind the ball. Okay. But it sounds like it can do it anyways. Guess it has enough energy to do so. And that boundary layer. It has less energy than the air just a bit farther out. But, well, it seems that it, too, is able to push into the high-pressure zone. At least when the ball is moving slowly. (Good. You're right so far.) And so does this change when the ball is moving rapidly? Yes. When the ball is moving rapidly, this lower-energy boundary layer of air is no longer able to push into the high-pressure zone behind the ball. In fact, it is pushed back towards the sides of the ball by the adverse pressure gradient, cutting like a wedge between the ball and the freely flowing air outside this boundary layer. No longer does the air curve around behind the ball. This leaves us with an air pocket behind the ball; a turbulent wake, in other words. In this wake, the air pressure is roughly atmospheric. There goes the symmetry of pressure forces on the ball. Now there is no high-pressure zone behind the ball to cancel the high-pressure zone in front of the ball. There is a large pressure drag, a force on the ball in the direction of downwind, slowing the ball down. Decelerating it. This pressure drag is what limits the range of a smooth golf ball. Yes, there is also viscous drag, but it's not nearly as significant as the large pressure drag caused by the turbulent wake.
(turbulent wake behind ball, which is moving to the left)So dimpled golf balls travel farther than smooth golf balls. Do the dimples somehow reduce the size (and severity) of this turbulent wake, reducing the pressure drag on the ball, preventing the ball from slowing so much as it arcs through the air? Yes, indeed. The dimples, or surface irregularities, cause the air in the boundary layer to tumble about. This tumbling about gives the boundary-layer air more energy, and more forward momentum. It now has a much better chance of pushing around to the back side of the ball, into the high-pressure zone. Alas, it still doesn't make it, but it comes much closer. It travels partially around the back of the ball before its progress is stopped and it separates from the surface. The air outside the boundary layer, following along, hugs the ball for a longer time, as well. It separates from the ball at the same spot where the boundary layer separates, this being a fair ways down the backside of the ball. The result is a smaller air pocket. A small turbulent wake. A less dramatic variation in air pressure between the front of the ball and the back of the ball. A more modest force of pressure drag. And this reduction in pressure drag is what enables the dimpled ball to soar some 200 yards farther than a smooth ball.
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