Re: Boyancy and Gravity?
From: Mark Martin (qed100_at_hotmail.com)
Date: 11/14/04
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Date: 14 Nov 2004 15:14:12 -0800
Gremlin wrote:
> The average atmospheric pressure at Earth's surface is about 14.6
pounds
> per square inch, which is about 100,000 pascals.
>
> What would gravity on the earth be like without this air pressure,
with 0
> pascals of air pressure? If you look at a helium balloon it
obviously has
> mass, yes.. but it has no actual weight.. if you were to measure it
on a
> scale. So obviously air pressure factors in to somethings weight.
My
> question is how much would an average person weigh if it were not for
air
> pressure? Or in general any discussion about the topic.
>
> Apparently the force of gravity keeps air around the earth. Perhaps
though
> the air is just more dense than the near vacuum of space so it
collects
> itself into on place. That is my theory, as what gravity actually
is..
> Just an effect of buoyancy. No need to argue with me about it
because I
> don't have any evidence to support my idea.. yet. Solids also
collect
> themselves because of buoyancy in near vacuum and in air.. As does
water
> etc. This is why a rock will sink in water, and water will sink in
air..
> and helium will rise in air. Everything though will sink in the near
> vacuum of space though because buoyancy is caused by density and
vacuum is
> the least dense of everything. It is just there is no 'down' for
anything
> to sink in, in vacuum so everything just collects together and
creates a
> down consistent with that collection of matter.
>
> What do you think!?
If I stand on a batroom scale at the bottom of a swimming pool, as you
expect the number on the meter will read smaller than if I stand upon
it in open air. But my total weight is the same in either case. How can
this be?
It's because when I stand immersed in the pool, water is supporting
some fraction of my total body weight, depending on the ratio of my
mass density to that of the water. If my density is greater than the
water's, then I'll still fall downward, but will reach terminal
velocity more quickly than by falling through the thinner air. I weigh
the same in the water, but less of my weight is bearing down on the
scale. Some of it is bearing down on the surrounding liquid. If you
measure the change in the depth of the liquid, then in the end you find
my total weight after all.
In the event that my mass density is less than that of the surrounding
fluid, then my mass will be supported to the extent that the pressure
beneath me exceeds the force of gravity, and I'll travel upward. (I
also reach terminal velocity in this scenario as well.)
This is of course how it works at Earth's surface, where things aren't
in free-fall. Suppose I'm way out in space, falling freely. You say
then that I'm weightless. But what does it mean? It means that
everything in my frame of reference is falling, and in a local gravity
field all things fall due to gravity with equal acceleration. I'm in a
spaceship filled with air. (Or, perhaps, even water!) But in this case
I don't float in mid-air because of bouyancy. Both the air & I are
falling with equal acceleration. It doesn't matter what our mass
density ratio is, because the fluid is supporting none of my body mass.
So, returning to Earth, my body mass is supported partially by the
solid ground, partially by the air. My bathroom scale weight is a
little less than if I were standing similarly in a vacuum chamber. But
keep in mind that the difference is all in the mass density ratio, not
in the air pressure itself. If I stand in a sealed chamber with the air
at a certain pressure, the pressure can be raised or lowered by pumping
heat into or out of the system. But the mass density ratio stays the
same, and so does my weight. Sea level air pressure is in the
neighborhood of 15lbs psi, but removing the air to a vacuum won't
increase my weight by 15lbs.
-Mark Martin
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