Apollo also used pure oxygen, this is why the Apollo fire happened, and
the fix was to remove as much combustible material as possible from the
cabin, not to change the atmospheric composition.

The lungs respond to the partial pressure of oxygen and ignore
(to a first approximation) everything else.  The "normal"
partial pressure of oxygen in the atmosphere is about 20%.
You can thrive then in an atmosphere that is pure oxygen at
0.20 atm, a 50/50 oxygen-nitrogen mixture at 0.40 atm, or
whatever.

However, many other processes run more as we expect at 1 atm.
Thus NASA's current cabin usage.  For a space suit, there's
no point in lugging extra nitrogen up into outer space.

As for outgassing, that depends on the composition of the
plastic and the method of manufacture.  Modern plastics
are made with very low or no outgassing problems.  Chemistry
has moved forward a bit since 1960.

By the way, the "normal" human can't take an oxygen pressure
of more than about 0.3 atm.  This becomes a problem with
people in severe heart failure.  

Plonk me too please.

Actually the force is not equivalent.  The principle of equivalence
applies to uniform gravitational fields.  The Earth's gravitational field
is not uniform.

But had it really been uniform (globally), looking out could tell you
that you're accelerating downwards in a field, or that something else
is accelerating upwards. Of course, if the something else is a planet,
it's a good bet that you're the one being accelerated :) And even for a
nonuniform field, you're using non-gravitational information to
conclude that you're being accelerated in a field rather than being in
a field-free region.

No Roger, they are locally completely equivalent in all observable
properties. Looking out the window isn't a local experiment. Also, this
is like saying that Dutch and English are the same language because you
saw drawings by Dutchmen and Englishmen and understood what they were
showing :) [I know it is hard to see the connection between this and
what we're discussing, but it's there:) It involves using one type of
data to draw conclusions about another form of communication: visual
data-verbal communication, and  visual data+context--gravitational
interaction as a means of communicating, in the sense of detecting a
grav field by measurements of quantities related to gravity].

And talking about force in this context is simply bad terminology,
because then you need to make a distinction between a real force (I
push you) and a fictitious one (due to using an accelerated frame of
reference). Extremely inelegant (and should give a hint that we're
missing something; in this case, the equivalence between gravity and
acceleration). This is ignoring GR; in that case, it's not bad
terminology but simply wrong (well, assuming that GR, or in any case
some geometrical description of gravity is correct, which does seem to
be the case as far as anybody can tell).

Anyway, their local equivalence, in the sense of being unable to
determine whether you're in a field or being accelerated by doing local
experiments, is indisputable, and all I was saying.

Merry Christmas!

Achilleas

Yes, exactly: all the earth's mass is uniformly distributed around you
and the pull cancels out. imagine splitting up the earth into tiny
little cubes; each pulls you in one direction with some force. But,
since it's a sphere and you're in the centre, there exists another,
identical little cube diametrically opposed to this one; their pulls
cancel. The same argument holds for all cubes, so the force cancels!

Actually, for the hell of it, I worked out the following. A
gravitational force F [vector] can be expressed as the gradient of a
scalar potential: grad V. This V satisfied
div grad V=rho(r)
(rho is the local density; I ignore a constant of proportionality as I
don't know it, although it would be trivial to work out by compaing to
the more familiar F=G m1 m2/r^2; but I don't have paper here and anyway
am too lazy). This div grad V basically means the second derivative wrt
x+second derivative wrt y+second derivative wrt z of V. That is, it is
the laplacian.

Now, consider any hollow object; that is, rho(r) is zero everywhere
except on some surface (I take it to be infinitely thin, although this
is not important), on which it is a constant.  That means that
everywhere except on that surface, the laplacian is zero; thus, it is
not possible for all three derivs to vanish at the same time, that is,
there is no maximum or minimum at any point (except perhaps on the
surace, where the laplacian doesn't vanish).

Outside the object, it is obvious (and may also be calculated) that the
field simply monotonically decays to zero. Thus, there's a force
towards the object outside it (gravitational attraction).

Now specialise to a sphere.

Inside, think of this: First, assume V isn't constant inside (ie there
is a force). there can be no max or min, as we said; so only saddle
points.. But, by symmetry, this will not occur. Thus, V is a constant
inside (and the force is zero everywhere inside a hollow sphere)!

A sphere of finite thickness can be constructed by a limiting procedure
involving spheres of infinitesimal thickness (everything is linear, so
this works).

Similar arguments can be constructed for cylinders, ellipsoids etc.

Of course, what you're talking about (ie the earth) isn't a hollow
object, so what I said is irrelevant. Sorry. I lose control of myself
easily.

To get back to the topic, I believe that most plastics do not outgas
(but, frankly, I have no clue about this).

Merry Christmas!