The purpose of this somewhat lengthy discussion on apparent weightlessness and a bit of centripetal acceleration
is to determine if astronauts
fill their spacesuits
, or what. Reading this will make you extremely popular
, as you will be able to weed out
at dinner parties
who talk about everything but make it all up
. Your hat size
should also increase by several measures.
First of all, we should distinguish between two things: weightlessness and apparent weightlessness.
If you are in the universe with another object (and nothing else), and that object has mass M kg, you have mass m kg and the distance between your centres is r, then the force exerted by this body on you (and vice versa) is F, where:
F = (GMm)/(r^2), where G is the gravitational constant.
This force is actually your weight. If you put in the values for M as the mass of the Earth and m as your mass, then F is what you weigh. If you consider more than two bodies, there is no exact solution and you need to use an approximation. In our case, though, since there is nothing of considerable mass near the Earth (to the extent that it can appreciably affect our calculation), we can ignore other bodies.
So we know how to calculate our weight. Now, a gravitational field is infinite in range, so you will always have some weight, hence you can never be truly weightless. However, we have all seen astronauts floating merrily in space, and even (a rarer sight) in aircraft flying within the atmosphere, to train astronauts for the effects of apparent weightlessness, endearingly termed the Vomit Comet (thanks to m_turner for that information).
You can see, therefore, that these astronauts are not truly weightless. In fact, their weight is close to what it would be on Earth. This is because on Earth, r was the same as the radius of the Earth, and in orbit, the additional distance from the centre of the Earth is relatively small.
Now we finally begin to answer the question: how do they do it? Why do astronauts appear weightless, when in fact they have considerable weight? Well, first of all, the astronauts are not just floating in one place. They are actually revolving around the Earth like the dickens. Let's consider an analogy for a moment.
Say I have a stone on a string (this is actually quite hard to do, especially if the stone is small). Now, I start swinging this stone round in a horizontal circle, and I keep it going at a steady speed. The only thing that is keeping that stone moving in a circle is the tension in the string: there is a force on the stone inwards. If I suddenly let go of the string, ignoring the effects of gravity for now, the stone will travel in a straight line (at a tangent to the circle at the point where I let go). So you can see that in order to move something in a circle at a steady speed, you need a force on it that is always towards the centre of the circle.
There is a more mathematically rigorous way to arrive at the same conclusion; it shows that if something is moving in a circle, then between two nearby points in that circle the change in velocity will always be towards the centre of the circle. As acceleration the change in velocity divided by the change in time (and acceleration is, like velocity, a vector quantity), then the thing is accelerating towards the centre of the circle. Consequently, as force is mass times acceleration (and, you guessed it, force is a vector quantity), there will always be a force towards the centre of the circle. It has something to do with triangles. No, really, I know what I'm doing, but I would need a diagram or two. If you are interested, it turns out that the force F needed to keep you, mass m, going in a circle of radius r at a speed v is:
F = m(v^2)/r
Now, that was all to show that the astronaut:
has to be going in a circle (otherwise he'd just crash into the earth), and
needs his weight to keep him going in that circle.
But if he definitely has a force on him, why does he seem weightless? (The astronaut does not feel any force, by the way.) The reason is that his acceleration is equal to the acceleration due to gravity. So actually, if you jump, you are apparently weightless until you hit the ground. This can be seen if you take a glass of water (try this outside) and splash the water upwards. If you keep your frame of reference constant relative to the water, that is, you jump with the water, it will look more or less as it would in space, except for air resistance. Or, another example: get into a lift (elevator) and go to the top floor of a skyscraper. Once there, cut the cord (yes I know they have safety systems; frankly, I don't care). Your final moments will be spent in a completely weightless state, as far as you know. This again will not be perfect as the lift will experience a force due to air resistance, so its acceleration will be less than that due to gravity. In the same way, if a spring weight (a weighing device) is attached to an object that is in orbit (or falling freely), the weight measured will be zero. To summarize, then: in apparent weightlessness, you still experience a force (your weight), but since that is the only force acting on you, you will be apparently weightless to anything in your frame of reference. (N.B. This concept of apparent weightlessness is related to the equivalence principle, read up.)
If all has been understood, your hat should now be of comic proportion to your head.