Isn't Pascal's Law valid even if we DON'T ignore gravity?

Pascal's Law can be read anywhere online.

Now suppose I want to apply on earth taking into account the variation of pressure with depth. So I don't want to ignore the effect of gravity.

So I have colum like this, filled with a fluid, open at top, and due to gravity, pressure is continuously increasing with depth.

|        |
|        | --> P
|        |
|        |
|        |
|        |
|        | --> 2P
|        |
|        |
|        |
|        |
|        | --> 3P
|        |
|        |


The pressure let us say is incerasing as P, then 2P, then 3P with the depth.

Suppose I push a cylinder from the top and hence put an external additional pressure of 10P.

Wouldn't this external pessure of 10P be equally transmitted throughout the rod to increase the pressure at each point by 10P ?? Like this:

   |  |
PISTON
|        |
|        | --> P + 10P = 11P
|        |
|        |
|        |
|        |
|        | --> 2P + 10P = 12P
|        |
|        |
|        |
|        |
|        | --> 3P + 10P = 13P
|        |
|        |


Or will the additional pressure of 10P also be variedly added to the pressure at each point like this??:

   |  |
PISTON
|        |
|        | --> P + 5P (Let's say)
|        |
|        |
|        |
|        |
|        | --> 2P + 10P (Let's say)
|        |
|        |
|        |
|        |
|        | --> 3P + 15P (Let's say)
|        |
|        |


That is, will the extra pressure which is transmitted due to piston also vary?? Or will it remain constant ?? WHICH CASE IS TRUE ??

• – Phoenix87 Feb 9 '15 at 10:46

The fact that $g$ appears explicitly in Pascal's law tells you the answer. Setting $g \rightarrow 0$ in a limiting procedure is identical to taking $\Delta h \rightarrow 0$ (or $\rho \rightarrow 0$).
So Pascal's law reduces in these (equivalent) limits as the rather trivial statement that $\Delta P =0$.