Description of buoyancy in relation to pressure and gravity.

In the first picture we have an arbitrary shape. The mass is pulled down by gravity (m*g), and the force of the surrounding liquid causes a pressure towards the center of the mass, and a normal force pointing up. The normal force can be calculated by simplifying the object like it’s done on the next image.

Below is a cube for simplification purposes. The pressure from the liquid can be divided up by each surface area. The sum of the forces on the left side will be equal the sum of the forces on the right side, and thus cancel each other out.

The sum of forces on the top surface will be smaller in size than the sum of forces on the bottom surface. This is due to the increase in pressure at deeper levels in the liquid. So in total the sum of forces on the top and bottom surface will be a force pointing upwards.

This force minus the mass (mg) is the resulting buoyancy. 

If the density of the object is equal that of the liquid, then the object is truly neutral. With an increase in gravity, the pressure/depth ratio in the liquid will increase, and cause the resulting normal force on the object to equally increase. This will happen at the same time as the mg will increase. The Normal forces are the sum of pressure multiplied with the area, and the pressure is given by the density of the liquid times the depth and gravity vector. This results in an equation where the gravity can be isolated on each side, and thus can be canceled.

Pressure(N/m^2) = Density(kg/m^3)*Depth(m)*g(m/s^2)

mg(N) = Sum[ P(N/m^2)*A(m^2) ]

This proves that the the object’s buoyancy will not be affected by the gravity if it is truly neutral.

The only result on the object is the increase of pressure as the gravity increases!