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Shadow Mint

No Gravity -- Falling?

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Can someone explain the physics of this to me please? The shuttle / space station has no effective gravity. However, thats not because its way up; its really just because its falling toward the earth, and the earths surface is curving away from it as it falls. Similarly, I''ve heard that astronauts are trained for no gravity by being put in big empty cargo planes that fly through many giant parabolic arcs to simulate the same effect. What is actually happening in these cases? As you fall (unrestricted by air or other things), you effectively experience no gravity in your local intertial frame. Thats what I understand it to be because of. But I dont understand the details of why. In theory, since the entire frame of reference is falling at the same rate, relative to each other, there is no detectable gravity, right; and a pound of lead and a feather (ignoring resistance) fall at exactly the same rate. Which is cool... in a vacume. But what about in air? Imagine you have a room full of people, dropped out of the sky from 10km up. Ignoring the fact its going to fall apart, consider the whole thing as a frame of reference. The whole thing starts falling; acceleration is -9.8ms^-2, right? Relative to each other, no decernable gravity. Now after a while, the thing hits terminal velocity; it cant go any faster because of the air resistance (imagine the atmosphere is totally uniform in density for this. =p). The acceleration is still -9.8 ms^-2 due to gravity, but you have F=ma=9.8 * mass of room upwards totally cancelling out the force of acceleration. What now? The thing is still dropping out of the sky at a pretty damn quick rate...whats the effect on the people inside? Still weightless? I would imagine so; even though there is now no effective acceleration on the frame, from inside that frame of reference they''re all still at the same speed relative to each other; no decernable gravity. Am I wrong? I think I must be. I would actually expect them either to be mushed against the roof (not quite sure why, but there you go), and the acceleration on the frame is changing, and the object in the frame have different intertia, so there must be -some- change in the way they behave as the acceleration changes... Anyway, any comments on this quite welcome. I need to be a simulation of a bunch of objects in a box when the box gets dropped (out of a plane). Ie. How they fly around in the box. So anyone bold enough to suggest some differntial equations or something on the behaviour, much appreciated also.

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Guest Anonymous Poster
Haha, you''re silly . Ok. What would happen once the room hit terminal velocity and stopped accelerating (or slowed down accelerating) would be the people inside would start dropping to the floor. As the acceleration slowed down, the people inside would still be accelerating at 9.8ms^2, while the room (due to friction) is not falling quite as fast. The people in refrence to each other would be falling at the same rate in relation to each other, but they will start sinking to the floor as the rooms acceleration slows down (aka, feeling gravity again, but slowly). I hope you understand what I said, but that''s the best I could describe it :D.

Billy

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Your problem comes from these two paragraphs:


The whole thing starts falling; acceleration is -9.8ms^-2, right? Relative to each other, no decernable gravity. Now after a while, the thing hits terminal velocity; it cant go any faster because of the air resistance (imagine the atmosphere is totally uniform in density for this. =p). The acceleration is still -9.8 ms^-2
due to gravity, but you have F=ma=9.8 * mass of room upwards totally cancelling out the force of acceleration.

What now? The thing is still dropping out of the sky at a pretty damn quick rate...whats the effect on the people inside? Still weightless? I would imagine so; even though there is now no effective acceleration on the frame, from inside that frame of reference they''re all still at the same speed relative to each other; no decernable gravity.


The empty cargo plane, frequently called the "Vomit Comet" simulates weightlessness for short periods of time by starting at a high altitude and moving towards the earth (the negative Z direction) at 9.8m/(s^2) with the X and Y velocities constant. Thus the people in the plane have no acceleration relative to the plane, and are weightless.

f and a below are for the passengers relative to the plane.

g=-9.8m/(s^2)
a=9.8m/(s^2)

f=(a+g)*m=0

The plane can only do this for a short period of time. Once the plane reaches terminal velocity, the passengers feel the normal effects of gravity. The plane must then accelerate upward, or crash into the ground. If it accelerates upward,

a=-something
f=(a+g)*m =-(9.8+something) * m

Hope that helps.

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Terminal velocity is caused by an additional force of air resistance ''pushing'' against the surface you are exposing to the ''moving'' air. You downward side during a fall. In the case of a closed room, this means that the room will begin to be braked, while the people inside will not (as they are sheltered from the air by the floor.) So the people will settle to the floor, since they are moving somewhat faster than the room.

I''ve been skydiving. When I fall belly to earth, it feels as if I am lying on a giant air mattress when I reach terminal velocity. My parachute (worn like a backpack) doesn''t have any surface into the air, but it doesn''t pull away form my back. It pressed against my back, because it is trying to fall faster than my body, which is braked by air resistance.

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Cool. Gotcha. I see what you guys mean. So to model it all I need to do is come up with an air resistance, air density, etc. model. From that generate air resistance force due to the areodynamics of the object falling, related to time, and run it as effective acceleration = (gravitational force - resistive force) / mass of frame.

Anyone see anything else I''m going to have to worry about with the gravity effects when I model it?

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Guest Anonymous Poster
Depending on the scale and desired accuracy of your model, you might want to take into account that the acceleration due to gravity decreases as you move away from the earth.

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