# 4.3.5.3: Liquid Under Varying Gravity

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For comparison reason consider the deepest location in the ocean which is about 11,000 [m]. If the liquid "equation of state'' (61) is used with the hydrostatic fluid equation results in

$\dfrac{\partial P}{\partial r} = - {\rho_0} \text{ e}^{\dfrac{P- {P_0}}{ B_T}} \dfrac{G}{r^2} \label{static:eq:liquidGhydro}$
which the solution of equation (118) is

$\text{e}^{\dfrac{P_0-P}{B_T}} =Constant -\dfrac{B_T\,g\;\rho_0}{r} \label{static:eq:liquidGhroS}$

Since this author is not aware to which practical situation this solution should be applied, it is left for the reader to apply according to problem, if applicable.

## Contributors and Attributions

• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.

This page titled 4.3.5.3: Liquid Under Varying Gravity is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

This page titled 4.3.5.3: Liquid Under Varying Gravity is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Genick Bar-Meir via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.