# 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
$\dfrac{\partial P}{\partial r} = - {\rho_0} \text{ e}^{\dfrac{P- {P_0}}{ B_T}} \dfrac{G}{r^2} \label{static:eq:liquidGhydro} \tag{118}$
which the solution of equation (118) is
$\text{e}^{\dfrac{P_0-P}{B_T}} =Constant -\dfrac{B_T\,g\;\rho_0}{r} \label{static:eq:liquidGhroS} \tag{119}$
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

• 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.