Whats the difference between apparent weightlessness and having no weight?

Question

Whats the difference between apparent weightlessness and having no weight?

Does an athlete clearing the bar in a high jump experience apparent weightlessness?

I feel that a parachutist descending at terminal velocity with the parachute open experiences apparent weightlessness but the book says that he doesn't. Why?

How do you find the change in momentum of an object in circular motion? I just sort of remembered it as 2mv but I don't know how to calculate since I think it is zero because
change in p= m(v-u) since v and u are in the same direction after completing the circle shouldn't change in p be zero???

Try to explain as clearly as possible!

or by the way is mass deficit or mass excess the correct term to use in nuclear physics?

jenny this is not a place for you to show off your vocabularly

UNIFORM circular motion

Answer 1

Good questions!

To "have weight" means that you are close enough to a massive object (such as a planet) to experience the effect of its gravitational pull. The strength of the pull depends on the planet's mass, as well as how far away you are from the (center of) the planet. The pull gets weaker and weaker the farther away you get; but it never actually drops to zero. So technically, even if you are billions of miles from the earth, you will still feel an (extremely tiny) gravitational pull from it. However, for all practical purposes, you are "weightless" if you are very, very far away from all massive bodies.

Notice that just being in outer space doesn't make you weightless. The space shuttle typically orbits at an altitude of 200 miles or so. At that altitude, the earth's gravitational pull is still very strong--about 90% as strong as if you were standing on the surface. So how come the astronauts can float around? That's because of "apparent" weightlessness.

"Apparent weightlessness" means that you are still near a planet (or other massive body), but that the _only_ force acting on you is the planet's gravity. We don't normally experience this condition. If you are standing on a sidewalk, gravity pulls down on you, but in addition, the sidewalk pushes _up_ on you (this is consistent with Newton's laws which say: if the sidewalk were _not_ pushing up on you, you would suddenly accelerate downwards). The result is that you feel a pressure between your feet and the sidewalk.

But now say you're in an elevator, and suddenly the cable snaps, and the elevator plummets earthward. The floor drops away from you, and now the only force acting on you is gravity. You are falling downward, but the elevator is falling downward just as fast. Every time you descend an inch, the elevator floor also descends an inch, so you never get any closer to the elevator floor. You are essentially "floating" in the elevator. This is "apparent weightlessness." (Lots of fun until you hit the ground!)

The situation is similar with astronauts in the space shuttle. The only force acting on the shuttle, and on the astronauts, is the earth's gravity (which bends the shuttle's path into a circle, rather than letting it fly off in a straight line). Since the shuttle and the astronauts are both "falling" at the same rate, the astronauts never move any closer to the shuttle's floor (unless they actually push themselves toward it).

Both of these situations are sometimes called "freefall." Freefall means that the only force acting on you is gravity. When you're in freefall, you "feel" weightless.

Here's an interesting situation: Suppose you stood on top of a tower that was 200 miles high. There'd be no air, so you'd have to wear a space suit--but would you feel weighless? The answer is "no". You'd feel 90% as heavy as when you're on the ground. You could wave at the space shuttle as it zooms past right in front of your eyes--the astronauts inside would feel weightless (because they're in freefall), but you would not.

> Does an athlete clearing the bar in a high jump experience apparent weightlessness?

Sort of. Technically, there is a second force--air resistance--that slightly spoils the freefall effect. But in the case of the high jump, the air resistance is fairly negligible, so the athlete will feel weightless for a short time.

> I feel that a parachutist descending at terminal velocity with the parachute open experiences apparent weightlessness...

I've done a few parachute jumps myself, so I can tell you this is definitely _not_ the case. When your parachute's deployed, the upward force of air resistence is definitely _not_ negligible (thank goodness, else you'd hit the ground pretty hard). This means that you are _not_ in freefall. Indeed, any time that you are descending at a fixed speed (not accelerating), it implies that their must be an upward force on you (in this case air resistance) that exactly balances the gravitational force. In such a case, you feel your full weight.

Another way to look at it: Say you're under your parachute, you take a marble out of your pocket, then let go of the marble. Does the marble float next to you? No--it doesn't have the advantage of all that air resistance provided by your parachute, so the marble immediately drops away from you. Likewise, while the marble's in your pocket, it sticks to the bottom of your pocket rather than floating around within it. You (and your clothes, your marbles, etc.) do not feel weightless.

BTW, another way to feel weighless, even when there's a ton of air resistance, is to introduce a _third_ force that will counteract the upward air resistance that's counteracting gravity. In that case, you move exactly as if gravity were the only effective force, so you're effectively in freefall again. For example, airplanes normally experience a lot of upward air resistance; for example on the bottom of their wings. But by effectively using their engines and control surfaces, they can produce opposing forces so that they move exactly as if their were no air resistance at all (which means: in a nice big parabolic arc). And during that time, the people inside the airplane experience apparent weightlessness.

Okay, switch topics--about momentum of an object in circular motion:

...after completing the circle shouldn't change in p be zero???

Yes. If you are remembering "2mv", you're thinking of the change in momentum after completing _half_ a circle. The change in momentum will in general be different depending on which points you make the beginning and ending measurements at.

Answer 2

When a body reaches terminal velocity v ~ constant, the net weight w = W - D ~ 0 is almost zero. But the actual weight W = mg remains the same when g ~ constant near Earth's surface. The air drag D ~ rho V^2 depends on the shape and aerodynamics of the falling body, plus the air mass density rho and the velocity V of the fall relative to the air. Thus, as the body picks up velocity V and gets closer to Earth where rho is larger, the drag D gets larger and offsets more of the weight W. Finally W <= D and the net weight w ~ 0 so the body no longer accelerates to a greater V. In fact, the body will actually decelerate a bit due to the increasing rho as the body approaches the ground. So there really isn't a terminal velocity where there is absolutely no change in velocity, but the change is very small during the latter part of the fall. For all intents and purposes the velocity is constant when W = D so that w = 0.

Answer 3

When a person is present in a gravitational field, he is pulled by it. The force that gives this pull is wat we call as weight. If you say that ur weight is 500 N (actual unit of weight is N .. but we us kg.wt .. just multiply by 9.8 to get weight in N), it means that u r having a force of 500 N acting on u due to gravity of earth.

Now, suppose you are not in any gravitational field ... no gravity acts on u .. and ur weight is zero ... this is wat we mean by no weight.

Now, how do u feel this weight? when you are standing on a floor, the floor opposes the gravitational force by applying 500 N to ur feet. This is why u feel ur own weight. If there were no floor, you will be undergoing freefall and u wont be able to feel ur own weight ... this is called apparant weightlessness.

The case of an athelete ... the only force acting on him is gravity .. and nothing to oppose it ... i.e., he is undergoing freefall (he will take a trajectory path) ... and hence this is apparant weightlessness.

For a parachute diver, there is air resistance that pulls the parachute up and hence the parachute pulls the person upwards ... he will feel the same way a person will feel when he wears a backpack and someone lifts him by lifting the backpack. In this case, he doesnt feel weightlessness.

You can take it this way ... if a person is having an acceleration equal to that of acceleration due to gravity (9.8 m/s^2), he will feel weightlessness .. if he is not accelerating, he doesnt feel weightlessness. If he is accelerating at an acceleration less than g, he feels reduced weight (eg., when lift starts moving downwards).

Hope this helps

Edit:
btw, for finding change in momentum, you must consider velocity .. not speed ... i.e., change in direction of movement will also give a change in velocity even if speed doesnt change .. and hence, circular motion change in momentum will not be 0. To find the change in momentum, u must do the vector subtraction of v and u and then multiply it with m .. it wont be 0 unless both the points coincide.

Answer 4

According to general relativity: not a god d@mned thing. The two conditions (being in a gravitational field and being in an accelerating reference fram) are identical.

Yes, the athlete feels weightless in mid-air.

The parachute doesn't feel weightless because his velocity is constant. He has a force down (gravity) and a force up (air drag). To him, it feels like he's resting on a bed of air. He doesn't have the queasy in your stomach freefall feeling at this point.

For UNIFORM circular motion (is that what you're talking about), the change in momentum (a vector quantity) comes from the centripetal force, which makes the momentum change directions all the time.

Centripetal force = magnitude (dp (vector) / dt) = mv^2 / r = p^2 / mr

edit: the most general term is mass defect. That could describe a mass deficit (you end up with less mass than you started with, so the reaction is exothermic) or a mass excess (you end up with more mass than you started with, so the reaction is endothermic)

Answer 5

True Weightlessness

Answer 6

If you have mass you have weight, weight is what you weigh when at rest in a gravitational field. A parachutist is weightless before his chute opens, prior to that he is in free fall. It is impossible to weigh yourself in free fall.

Answer 7

oh sorry. I think the answer is 43