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发表于 2013-1-23 10:08:33 | 显示全部楼层 |阅读模式
Weightlessness idea
The purpose of this relatively lengthy discussion on visible weightlessness and a bit of centripetal acceleration will be to determine if astronauts fill his or her's spacesuits with helium, or what. Encountered this will make you extremely popular, as you will be prepared to weed out those frauds while dining parties who talk about all things but make it all way up. Your hat size should also increase by several measures.
To begin with, we should distinguish between two things: weightlessness as well as apparent weightlessness.
If you are in the whole world with another object (there exists nothing that else), and that object offers mass M kg, you've gotten mass m kg and then the distance between your centres is r, then the force exerted by this body on you (plus vice versa) is F, at which:
F = (GMm)/(r^2), where by G is the gravitational constant.
The following force is actually your weight. If you happen to put in the values for M as the mass of the Earth not to mention m as your mass, afterward F is what you look at. If you consider more than 2 bodies, there is no exact answer and you need to use an approximation. Throughout our case, though, since there is practically nothing of considerable mass close to the Earth (to the extent it may appreciably affect our formula), we can ignore other bodies.
So we know how to calculate our own weight. Now, a gravitational market is infinite in spectrum, so you will always have some bodyweight, hence you can never be certainly weightless. However, we have all seen jet pilots floating merrily in space, and even (any rarer sight) in aeroplane flying within the atmosphere, to exercise astronauts for the effects of obvious weightlessness, endearingly termed the Vomit Comet (caused by m_turner for that information).
You can see, consequently, that these astronauts are not definitely weightless. In fact, their weight is approximately what it would be on Earth. The reason being on Earth, r was identical to the radius of the Earth, and in orbit, the extra distance from the centre of the world is relatively small.
Now many of us finally begin to answer the issue: how do they do it? Counseling ? astronauts appear weightless, when in realization they have considerable weight? Most certainly, first of all, the astronauts are not just floating in one place. They are the truth is revolving around the Earth enjoy the dickens. Let's consider an analogy if you will.
Say I have a stone over a string (this is actually quite hard to do, especially if the stone is smallish). Now, I start swinging this stone round in any horizontal circle, and I it going at a steady pace. The only thing that is keeping the fact that stone moving in a cir is the tension in the sequence: there is a force on the gemstone inwards. If I suddenly let go of all the string, ignoring the effects regarding gravity for now, the natural stone will travel in a in a straight line line (at a tangent to the cir at the point where Simply put i let go). So you can see that to be able to move something in a ring at a steady speed, you want a force on it that is often towards the centre of the cir.
There is a more mathematically rigorous way to arrive at the same summary; it shows that if a specific thing is moving in a eliptical, then between two near by points in that circle the change in velocity will always be to your centre of the circle. As acceleration the change in pace divided by the change in occasion (and acceleration is, enjoy velocity, a vector quantity), the thing is accelerating towards the centre of the circle. For this reason, as force is mass times acceleration (and, you guessed it, force is a vector quantity), you can always find a force towards the center of the circle. It has something related to triangles sac longchamp pliage. No, really, I know exactly what I'm doing, but I would wish a diagram or a couple. If you are interested, it turns out that that force F needed to keep, mass m, going in the circle of radius l at a speed v is definitely:
Now, that was all in order to show that the astronaut: has to be going in some circle (otherwise he'd just crash into the earth), and requirements his weight to keep your man going in that circle.
When he definitely has a power on him, why does he or she seem weightless? (The astronaut does not believe any force, by the way.) The reason is that his acceleration is equal to this acceleration due to gravity. Therefore actually, if you jump, you are apparently weightless until you hit the bottom. This can be seen if you take a fabulous glass of water (try this out-of-doors) and splash the water max. If you keep your frame of reference constant relative to the stream, that is, you jump using the water, it will look very as it would in space, with the air resistance. Or, a second example: get into a lift (lift) and go to the top floors of a skyscraper. Once there, cut the cord (yes I know they've safety systems pas cher longchamps; frankly, I wouldn't care). Your final moments can be spent in a completely weightless state, in terms of you know. This again will not be perfect as the lift will experience the force due to air level of resistance sac longchamp cuir, so its acceleration can be less than that due to gravitational pressure. In the same way, if a spring bodyweight (a weighing device) is definitely attached to an object that is found in orbit (or falling freely), the weight measured will be zero. In summary, then: in apparent weightlessness, still you experience a force (your weight), question that is the only force functioning on you solde sac longchamp, you will be apparently weightless towards anything in your frame with reference. This concept of apparent weightlessness is related to the equivalence principle, educate yourself.)
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