airatmospheric sciencedensitynewtonian-gravity
I have actually two different questions that may result in the same answer.
The first one : if the Earth would have less gravity, would it result in less air density?
The second one : On a different planet, let's say on Uranus which is bigger than Earth, the gravity is less than 1g but does that automatically mean that the density of the air is less? Or in some circumstances it can behave the opposite?
As stated above, your linear calculation is correct and your assumption on compression is correct too. I can try to give a rough answer to the shrinkage. Lets start by looking at pressure.
http://cseligman.com/text/planets/internalpressure.htm
In simple terms, the pressure is the weight above you, over the surface area, which would be linear, but it's weight, not mass and as the gravity doubles, the weight doubles, so, for pressure, you get the square of the effect than you get for gravity. So, doubling the planet's diameter and keeping everything else the same, the pressure goes up 4 times, before we factor in shrinkage.
Heat is also a factor. Heating a metal expands it, not a great deal, but some but lets ignore that for now. Also, different elements and alloys will have different densities and different resistance to pressure. I suspect, pressure is more important than heat, unless it's a gas giant planet and close to it's sun, creating what's known as a "puffy Jupiter", but I think for a rough calculation you can ignore heat. For more accuracy, probably not.
Now, unlike Gas, which has a pressure that's basically molecules bouncing off each other, solids and liquids resist pressure by the electromagnetic force which is why most of them are largely incompressible, but they can still compress. We know for a fact that white dwarf stars for example. The pressure is sufficient, that once the fusion process stops creating heat inside the star, a sun mass star shrinks to about the size of the earth, and a density about a million times as dense. A curious thing about White dwarf stars and Neutron stars is, that as they add matter, they shrink.
This can happen with large planets too. Jupiter is kind of close to the maximum size that a planet can get. A gas giant with 10 times Jupiter's mass wouldn't be much bigger than Jupiter, it might even be a little smaller.
more on maximum planet size here: http://www.universetoday.com/13757/how-big-do-planets-get/
Now, lets look at the density of the core of the earth:
http://hyperphysics.phy-astr.gsu.edu/hbase/geophys/earthstruct.html
The earth's core has a density of 12-13 g/cm^3
Source: http://hyperphysics.phy-astr.gsu.edu/hbase/geophys/earthstruct.html
and Iron has a density of 7.87 g/cm^3 and Nickel about 8.9 and there are trace amounts of heavier elements and the core is actually an alloy so it's density would be slightly different, probably slightly denser, but doubtfully all the way up to 13 g/cm^3, so, it's not unreasonable to say that the Earth's pressure increases the density of the Core, something like 30%-40% as a ballpark estimate. Lets go with 35% increased density in the solid core due to pressure as a ballpark estimate. 35% increased density means about 91% of the diameter, but the solid core is only 1% of the volume of the earth, so that's a small factor.
The liquid outer core is much larger, and it (same link above) has an average density of 9.9 and 12.2, so, I could run more estimates, but it would get very rough. It's safe to say that there is some compression of the Earth going on, how much? Likely less than the 91% of uncompressed diameter at the core, but there might be - and this is a very ballpark guess maybe, 3% - 5% shrinkage of the Earth's diameter due to it's mass and composition, or, perhaps more useful, 95%-97% of it's uncompressed diameter.
This article kind of backs that estimate up. http://www.universetoday.com/26771/density-of-the-earth/
Earth is the most dense planet in the Solar System; however, if gravitational compression where factored out, the second most dense planet, Mercury, would be more dense
So, rough estimates can be made here as well, if we guess the uncompressed density of Mercury and Venus and compare them to earth, but I'm comfortable with the 3% - 5% shrinkage estimate (95%-97%)
If we use Coulomb's law
Source: http://home.fnal.gov/~cheung/rtes/RTESWeb/LQCD_site/pages/calculatingtheforces.htm
,
as the distance between the electron orbitals and the positive atomic nucleus is decreased the resistance increases by the square of the reduced radius. In effect, as pressure goes up 4 times, (and diameter goes up twice), the inverse of the shrinkage also doubles, so there is a corresponding relation between the diameter and the inverse of the shrinkage.
So, 1 earth mass, and lets use 96% of natural diameter
1 earth mass, 1 earth diameter, 1 earth pressure, 96% uncompressed diameter (1/96% = 4.166% compression) - you need to take the inverse, I can explain that if needed.
8 earth mass, 2 earth diameter, 4 earth pressure, 8.33% compression,
64 earth mass, 4 earth diameter, 16 earth pressure, 16.67% compression
512 earth mass (1.7 times the mass of Jupiter), 8 earth diameter, 64 times earth pressure, 33.3% compression. (so, by this calculation, 8 earth diameters, you'd have a diameter 1/1.333 or 75% of the uncompressed diameter).
and a million earths, 100 earth diameters, 10,000 earth pressures, we might expect 400% compression or 20% the uncompressed size. This, is, of-course way way way off, also at 20% size the weight would grow exponentially, so a 2nd round of calculation would be needed as you get too large, But for a very rough estimate, if you set a compression percentage (1 earth, 4% or 5% or whatever you like) each doubling of the diameter should roughly double the compression and you can then go back and re-calculate the gravity using a few percentage points of increased compression.
It's worth noting that too much bigger than 5 or 10 earths and the planet would collect enough hydrogen to effectively be a gas giant, not a giant earth - that's mentioned in the article linked above.
Hope that's not too rough. If anyone wants to correct, feel free.
The object would actually float exactly the same for both values of $g$. Let $V$ be the volume of the body, $d$ its relative density, and $V'$ be the volume inside water. Then for equilibrium of the body,
$V \cdot d \cdot g=V' \cdot 1 \cdot g$
So, $V'/V$ is independent of acceleration due to gravity.
Best Answer
Yes. Less gravity implies less pressure because the gravity "pulls" the air to the surface of the Earth. And less pressure implies more volume. Ergo, the density of the gases on the atmosphere are less with the new atmosphere pressure.
And your second question, the answer is no. It depends of chemical composition (as someone commented above). What we call air, here in the Earth is not the same that the Uranus people call air.