# What is the effect of pressure differentials due to gravity on buoyancy force?

First of all, I am aware that this question has been answered in the past, however I have some follow up questions particularly regarded the argument posited in Why is Buoyant Force $V\rho g$? :

When an object is removed, the volume that the object occupied will fill with fluid. This volume of fluid must be supported by the pressure of the surrounding liquid since a fluid can not support itself. When no object is present, the net upward force on this volume of fluid must equal to its weight, i.e. the weight of the fluid displaced. When the object is present, this same upward force will act on the object.

This argument conceptually makes sense to me but I was confused around the interplay between the pressure differentials caused by gravitational fields acting on a fluid and the pressure exerted by the surrounding fluid.

I see how the pressure is equal to the weight of the fluid displaced but what about the pressure differential. Shouldn't that still have to be considered?

Following on, if an object was placed into a fluid of the same density why doesn't it rise since the pressure differential due to gravity is still there right?

## 1 Answer

Remember that the force is the gradient of an underlying potential. Here, this is the gravitational potential, and we can ask the question: What is the energy difference between an air parcel (density 0 for simplicity) of volume $$V$$ submerged in a fluid with density $$\rho$$ at depth $$z$$ and the same parcel at depth $$z + dz$$?

Assuming a cubic volume (we can make the cube infinitessimal, integrate and obtain the same result for any shape), we have $$V = h^3$$. Now, the difference in energy between the two configurations is given by the difference in potential energy of a slab of water of volume $$h^2dz$$ above the volume of height $$h$$ or below it.

Using $$U=mgz$$, we get $$dU=mgh$$ for the difference in potential energy. As $$m = h^2\rho dz$$, we finally obtain (up to a sign change) $$dU/dz = F = h^3\rho g =V\rho g.$$

This is the familiar expression, without pressure differential present. The only circumstance under which this breaks down, is if we have a compressible fluid, s.t. $$\rho = \rho(z)$$. Then there would be an additional force due to the differential. However, water is highly incompressible, so we usually don't consider this effect.

Please let me know whether this answers your question!