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

$$\nabla p = \rho {\bf g}.$$ (2.12)

If gravity is directed vertically downwards, then this reduces to

$$\nabla p = - \rho g {\bf k}$$ (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

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

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

$$p = \rho g(z_{0} - z),$$

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$^{-2}$.

$$p = 10^{5} + \rho g d \qquad \Rightarrow \qquad \rho g d = 10^{5},$$

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

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 $p(0) = p_{0}$.

$${dp\over dz} = - \rho g,$$

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

$${dp\over dz} = - p{g\over RT} = - {p\over H},$$

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

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

Hence,

$$p = p_{0}e^{-z/H}.$$