# 4.3.5.3: Liquid Under Varying Gravity

- Page ID
- 675

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.