Does an object float more or less with more or less gravity? This might be a stupid question, but I'm a newbie to physics.
An object less dense than water (or any other fluid, but I'm going to use water for this example) floats normally on Earth when placed in water. But if the object was placed in a hypothetical place where there is no gravity and there is air, it would not float on water. So if the object was placed in water on a planet with more gravity than Earth, would it float more or would it float less, or float the same as on Earth?
Would it float more because it doesn't float without gravity, but it does float with Earth gravity, therefore it'd float even more with more gravity.
Or would it float less because more gravity would pull the object down, so it won't float as much.
Or would it'd float the exact same as on Earth because the above two scenarios cancel each other out.

EDIT: By "float more," I mean it rises to the surface of the water faster, and it takes more force to push it down. By "float less," I mean it rises to the surface of the water slower, and it takes less force to push it down.
 A: Assuming both the water and the object are rigid and incompressible (pretty good approximation for water, may or may not be so good for the object) and that we can ignore surface tension (good approximation for large objects, not so good for tiny ones) then in equalibrium the same proportion of the object will be above the water regardless of the strength of gravity.
However stronger gravity will mean that the forces involved in the non-equalibrim state will be larger. Those larger forces will lead to faster movement. 
A: If you submerge an object that floats underwater, it will rise to the top more slowly with smaller $g$ and more quickly with higher $g$. Likewise, objects that sink will sink faster with higher $g$, etc.
The buoyant force is $\rho_L V g$ where $\rho_L$ is the density of the liquid (assumed to be independent of $g$) and $V$ is the volume of the object. The net acceleration of the submerged object is 
$$a = g\left(\frac{\rho V}{m} - 1\right)$$
Everything in the parentheses on the right-hand side is independent of $g$. So the acceleration is just proportional to $g$, whether the object is floating or sinking.
As others have noted, $\rho_L$ could in theory could depend on $g$, but this is a small effect. Most likely higher $g$ leads to higher $\rho$, making buoyant objects rise faster and dense objects sink slower.
A: 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.
A: If your object is compressible, like wood, it definitely might not float at higher gravity. The higher pressure in both the water and the air might compress the object to the point that its density exceeds the density of the water (which is much less compressible than spongy things like wood).  This is an important plot point in the classic science fiction novel Mission of Gravity by Hal Clement. 
A: I generally agree with Amritansh Singhal's answer and Yakk's comment, but I would like to add that in some situations there is another mechanism of floating that significantly depends on the value of g. For example, water striders (https://en.wikipedia.org/wiki/Gerridae) walk on water using surface tension to prevent sinking. In this case, higher g would make their life harder:-)
