# 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.

• Have you tried to write down the force balance to see why bodies on earth can float? What would happen if the magnitude of the gravitational constant changed? – Sanya Aug 6 '16 at 17:21
• An object displaces an equivalent mass of water. Water would be more dense under higher gravity (all other conditions being equal), so it depends on whether your object has the same density. If it does, then it would displace less water and float better. – Jason Goemaat Aug 7 '16 at 4:17
• @JasonGoemaat The question also talks about objects rising through the body of the liquid. A submerged object displaces an equal volume of water; an object floating at the surface displaces an equal mass. – David Richerby Aug 8 '16 at 9:02
• Ah, that was added after my comment I think. The addition makes the problem much more complicated. I think they buoyant force would be more because the water is denser, but I'm not sure if the density of the water would affect the speed of its ascent when submerged, which would also depend a lot on the shape of the object. – Jason Goemaat Aug 8 '16 at 9:18
• @JasonGoemaat Good point -- that part was added after your comment. – David Richerby Aug 9 '16 at 0:07

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.

• This assumes the density of water is not a function of gravity: or, rather, that water has a uniform density. Higher pressure water tends to be denser. This effect is likely to be tiny, and possibly overwealmed by the compression of the object itself under pressure! – Yakk Aug 6 '16 at 18:50
• @Yakk Increased density of higher pressure water is not a function of g. – Jason C Aug 6 '16 at 21:12
• @JasonC Pressure per inch down in water is a function of g. Pressure is, to a first approximation, the weight of the water above you (plus air etc). – Yakk Aug 6 '16 at 21:19
• This is good. Except please one that if g=0 it does not apply and won't float – Bob Bee Aug 7 '16 at 4:01
• I'm not sure how this addresses the OP's question. You are pointing out essentially that it will, when floating on top of water, displace the same amount of water as on Earth, but OP's clarification states: "By "float more," I mean it rises to the surface of the water faster, and it takes more force to push it down." - by that definition it is "floating more". It displaces the same amount of water, but all forces involved are larger [with higher gravity], as is the buoyant acceleration if it starts below the water level. – Random832 Aug 8 '16 at 6:47

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.

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:-)

• Correct, though I suppose it's debatable whether this can be called floating in the first place. – leftaroundabout Aug 7 '16 at 1:28
• @leftaroundabout: It may be debatable that this can be called "floating" but it can be debatable that this cannot be called "floating":-). For example, one of definitions of "floating" in Merriam-Webster is "buoyed on or in a fluid", and one of the definitions of "afloat" there is "borne on or as if on the water". – akhmeteli Aug 7 '16 at 3:55

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.

• Might this mean you need more freeboard to avoid being swamped? Maybe not, since the waves will also be smaller? Or maybe so, since other disturbances (moving loads etc) will remain the same? – Doug McClean Aug 8 '16 at 2:59

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.

## protected by Qmechanic♦Aug 7 '16 at 12:46

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