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Ok. Fortified by agreeing on something, I would like to find common ground on your atmospheric theory.
Weightlessness becomes a pointless term. We need a new term. Would low gravity environment (LGE) and ultra low gravity environment (ULGE) be acceptable? If your ship was stationary in deep space it would be in an ULGE. Everything would still have its own gravity but the values would be so low that any effects would be negligible. I agree if you have got mass you wont have zero G..just ultra low G. How does all that sound?
I am less convinced by your atmosphere/gravity argument but would this Earth based experiment help prove anything:-
The atmosphere has weight, so as the object falls the atmospheric weight gets continuously heavier above it, increasing the rush of the object towards the ground
What do you set the scale upon, to weigh anything, when everything floats in the air, including the scale?
I also did say, in the very first message in my theory, “When something falls through the atmosphere, and/or through the water, the fall (or sink) velocity increases at a rate that is globally uniform. It seems like there should be something besides mass/ weight/magnetism that would account for the global uniformity of rates of falling speed. Is it not more logical, that a reliable factor like overhead atmospheric pressure would be the influential factor? Acceleration by overhead and gradually increasing downward (pushing) pressure seems like a far more responsible agent of rising velocity.”
If the ship could stop falling and remain motionless in the vacuum by firing a powerful set of retro-engines, do you suppose that everybody would crash down to the floor? Not a chance.
mass is non-scalable in the vacuum
(1) Why do things accelerate when dropped in a vacuum e.g. in an evacuated tube or on the moon?
(2) Why do things accelerate when dropped in a sealed air-filled tube - because from Pascal's law the force on them is equalised?
If you believe, as Einstein did, in the equivalence of intertial and gravitational mass then the way to 'weigh' something in a weightless environment is simply to apply a force and see how much the object accelerates. Of course what you are measuring is 'mass' - but this is what most people mean when they say 'weight'.
Air pressure at the surface of the earth changes by +/-10% with no measurable change in acceleration due to gravity.How can atmospheric pressure be responsible for the apparent acceleration due to gravity when it is independent of changes in air pressure?
No. The ship, and everything in it, would fall (accelerate) directly to Earth. But the occupants would still feel 'weightless' until the friction with the Earth's atmosphere applied a decelerating (upwards) force. The occupants then found that they were apparently being pushed down into their seats. This is exactly what was reported by the Apollo returnees. But this doesn't support your theory because the deceleration (or 'weight') felt by the astronauts was much greater than their 'weight' at the surface even though they were still in a sealed vessel, and they were high up in the atmosphere.
Not all rocket paths are orbital. For example, the early Mercury flights were sub-orbital. One thing they all had in common though was that the occupants in the ship experienced weightlessness as soon as the engines were shut-off. This was regardless of the height within or above the atmosphere that was achieved.
Don't understand this. The concept of a scalable mass is strange to me. A mass is a mass is a mass unless you do something physical to add or remove that mass (or add/remove a very great deal of energy).
Dark Energry/Negative Pressure - but this is only one explanation and as many sources point-out (e.g. http://physicsweb.org/articles/world/17/5/7) the difference between observation and theory here could be 120 order of magnitude! IF negative pressure is real then the effect is going to be so small on any scale with which we are familiar to be completely invisible.
Acceleration due to “gravity” as it’s called, (which I call, “due to overhead pressure”), is “the same” at sea level around the world. That’s called 1G at sea level.
The ship had to be in at least the “fairly-pure“ vacuum for the people to leave the floor.
The vertical column of air pressure above a falling object of any surface area that is dropping straight down, weighs 14.7 per square inch of area at sea level, so the area of the falling object is under the influence of however much atmospheric weight is above it as it begins to fall, increasing above it until it hits sea level, when the area on top of that falling object is the entire 14.7 PSI that normally “rests” upon the area of a size that matches the area of the falling object.
But as you know, my theory has no belief that gravity accelerates anything, as gravity is a property of matter. Gravity imparts weight to a mass (volume), and it just falls because it has to.
You get exactly the same effect in any freefalling experiment e.g. the German freefall tower experiments a few years ago. Obviously these experiments are short (limited by the height of the Tower) but any of these experiments disprove immediately that there is any connection between weightlessness and vacuum.
But you've already quoted Pascal's law for a sealed vessel as meaning that the pressure exerted on anything within the tube is omni-directional. So the falling of the object clearly has nothing to do with air pressure.
Here's another thought – (etc.) Of course this is all nonsense because in real freefall the effects of air resistance/drag will quickly dominate but it does illustrate why we don't all feel as if we are walking around with (14.7 psi x area) on our heads!
Even without any air thimgs still have weight.
No. While Pascal’s pressure law does state that a pressure EXERTED in a closed vessel is multi-directional, you can’t extend that to mean that if all you do is cap off the two ends of a tube that it will now contain an (exerting) pressure that is any stronger than the atmospheric pressure that exists outside the tube. All you did was put a tube of glass around an existing volume of atmospheric air. You didn’t claim to have “packed in any extra air” that’s “trying to escape”. Effectively, free air in the closed-ended tube has the same interior “pressure” as a double-open-ended tube held in your other hand.
I can ONLY work with known averages i
This general principle works for the practice of my theorizing, and is common across many facets of my theorizing, because “their math” might not be correct if some element in my theory works differently than “theirs”. I am OBLIGATED (by my theoretical variations) to do things this way
No “guessing”, unless you want to promote it as your own theory. I’m theorizing based on logic about attributes of participating factors. We can’t just go making up numbers
A weight on a spring balance in a bell jar - If you pull the air out what happens to the weight? I say it falls slightly because there is no longer any air buoying it up. If air pressure is the cause of weight wouldn't it rise as, without air causing the weight, shouldn't the spring contract and pull the weight up?