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Next: Rotating Liquids Up: Fluids and Nonmagnetic Plasmas Previous: The Equation of State

Hydrostatic Liquids and Atmospheres

If the fluid is at rest under gravity (that is in equilibrium), the forces must all balance so that (2.9) reduces to
\begin{displaymath}
\nabla p = \rho {\bf g}.
\end{displaymath} (2.12)

If gravity is directed vertically downwards, then this reduces to
\begin{displaymath}
\nabla p = - \rho g {\bf k}
\end{displaymath} (2.13)

Example 2. .3How does the pressure increase with depth in the ocean? Assume a uniform, incompressible liquid in equilibrium. With $\rho$ taken as a constant (2.13) gives

\begin{displaymath}
{\partial p \over \partial x} = 0,\qquad {\partial p \over \partial y} =
0, \qquad \Rightarrow \qquad p = p(z).
\end{displaymath}


\begin{displaymath}
{dp\over dz} = - \rho g, \qquad \Rightarrow \qquad p = - \rho g z +
\hbox{const.}
\end{displaymath}


\begin{displaymath}
p = \rho g(z_{0} - z),
\end{displaymath}

where $p = 0$ at $z = z_{0}$ at the surface of the ocean. Thus, the pressure increases linearly with depth.

Example 2. .4At what depth is the pressure double the atmospheric value of $10^{5}$Newtons m$^{-2}$. Use $\rho = 10^{3}$kg m$^{-3}$ for water, $g = 10$m s$^{-2}$ and 1 Newton = 1kg m s$^{-2}$.

\begin{displaymath}
p = 10^{5} + \rho g d \qquad \Rightarrow \qquad \rho g d = 10^{5},
\end{displaymath}


\begin{displaymath}
d = {10^{5}\over 10^{3}\times 10} = 10{kg m^{-1}s^{-2}\over kg
m^{-3}m s^{-2}} = 10m.
\end{displaymath}

Thus, the depth is approximately 10 metres.

Example 2. .5If the Sun's atmosphere is compressible and isothermal, how does the pressure, $p(z)$, decrease with height above the solar surface? We may assume that the atmosphere is treated as a plane and that the pressure at the surface is $p(0) = p_{0}$.

\begin{displaymath}
{dp\over dz} = - \rho g,
\end{displaymath}

but, using the gas law $p = \rho R T$, we may eliminate the density to obtain

\begin{displaymath}
{dp\over dz} = - p{g\over RT} = - {p\over H},
\end{displaymath}

where $H = RT/g$ is called the pressure scaleheight. Hence,

\begin{displaymath}
{dp\over dz} = - {p\over H}, \qquad \Rightarrow \qquad \log p =
-{z\over H} + \hbox{ const.}
\end{displaymath}

Hence,

\begin{displaymath}
p = p_{0}e^{-z/H}.
\end{displaymath}


next up previous
Next: Rotating Liquids Up: Fluids and Nonmagnetic Plasmas Previous: The Equation of State
Prof. Alan Hood
2000-11-06