What is hydrostatic pressure

Hydrostatic pressure

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The hydrostatic pressure (Greek ὕδωρ hýdor, water), too Gravitational pressure or Gravity pressure, is the pressure that is created within a stationary fluid, that is a liquid or a gas, due to the influence of gravity. Contrary to the meaning of the word “water”, the term is also used for other liquids and even for gases. Dynamic pressure from fluid flows such as B. the dynamic pressure is not recorded by the hydrostatic pressure, it only considers static, static fluids.

Incompressible liquids in a homogeneous gravity field

Pascal’s law

The pressure increases with the depth of the water. In addition to the hydrostatic pressure, there is also the air pressure on the surface of the water. The various scales on the y-axis must be observed: the pressure in the water column increases much faster than in the air column.
The hydrostatic pressure on the bottom is the same in all three vessels despite the different filling quantities.

The hydrostatic pressure for Fluids with constant density in a homogeneous gravitational field (= Incompressible fluids, especially liquids) is calculated according to the Pascal's (or pascals) law (named after Blaise Pascal):

$ p (h) = \ rho g h + p_0 $

Formula symbols:

$ \ rho $ = density [for water: $ \ rho $ ≈ 1,000 kg / m³]
$ g $ = Acceleration due to gravity [for Germany: $ g $ ≈ 9.81 m / s²]
$ h $ = Height of the liquid level above the point in question
$ p_0 $ = air pressure on the surface of the liquid
$ p (h) $ = hydrostatic pressure as a function of the height of the liquid level.[1]


The physical units for hydrostatic pressure are:

  • internationally the SI unit
    Pascal (Pa): 1 Pa = 1 N / m²;
  • also in Germany and Austria the "legal unit"
    bar (bar): 1 bar = 100,000 Pa or N / m² (= 100 kPa)

The non-SI-compliant outdated unit of measurement is sometimes used to describe the hydrostatic pressure Meters of water column (mWS) used.

Example of the hydrostatic paradox

  • Water column, homogeneous water temperature: 3.98 ° C, height: 50 meters:
    1,000 kg / m³ × 9.81 m / s² × 50 m ≈ 490,500 N / m² ≈ 4.90 bar
At a temperature of 20 ° C, water has a density of 998.203 kg / m³. The hydrostatic pressure changes minimally to
998.203 kg / m³ × 9.81 m / s² × 50 m ≈ 489.61857 N / m² ≈ 4.90 bar

The hydrostatic pressure does not depend on the shape a vessel from; The only decisive factor for the pressure on the bottom is that Height of the fluid or liquid level and its density (depending on the temperature), but not the absolute amount of the fluid in the vessel. This phenomenon was called hydrostatic (or also Pascal's) Known paradox.

Total pressure (absolute pressure) at the bottom of the liquid

To full description of the pressure at the bottom of an incompressible fluid at rest however, the ambient pressure has to be added to the hydrostatic pressure. For example, the water pressure acting on a diver in a body of calm water corresponds to the sum

from the air pressure that acts on the surface of the water,

+ the hydrostatic pressure of the water itself.


  • It is important for divers to know what pressure their body gases (nitrogen) are exposed to in order to avoid diving illness.
  • A bathyscaphe has to withstand particularly high hydrostatic pressure.
  • Water towers use the hydrostatic pressure to generate the line pressure necessary to supply the end users.
  • In hydrogeology, according to Darcy's law, a flow between two points can only be established if the pressure difference is different from the difference in the hydrostatic pressures at the two points.
  • A siphon is a device or a device with which a liquid can be transferred from a container over the edge of the container to a lower container or emptied into the open without tipping the container over and without a hole or an outlet under the liquid level.

Gravitational pressure in planets, moons, asteroids and meteorites

Depth dependence of G

With increasing depth, $ g $ can no longer be regarded as constant. If the shape of the celestial body is described by a sphere with radius $ R $ and the density is considered to be constant, the pressure can be calculated as follows:

$ p (h) = \ int_ {0} ^ {h} \ rho \, g (R-r) \, \ mathrm dr $.

The spatial factor $ g (r) $ follows from Newton's law of gravitation:

$ g (r) = G \ frac {M (r)} {r ^ 2} $,

where $ M (r) $ is the mass within a concentric sphere inside the celestial body and $ M = M (R) $ its total mass. With the formula for the spherical volume $ V = \ tfrac {4} {3} \ pi R ^ 3 $ results for the pressure in the center:

$ p_ \ text {Z} = p (R) = \ frac {3} {8} \ frac {G M ^ 2} {\ pi R ^ 4} $.

Limitation of the size of a celestial body due to its compressive strength

Different materials have different compressive strengths. Equating $ p_ \ mathrm Z $ and the maximum pressure $ p_ \ mathrm {max} $ leads to an equation that can be solved for $ R $. The resulting value

$ R_ \ mathrm {max} = \ frac {\ sqrt {6 p_ \ mathrm {max}}} {2 \ rho \ sqrt {\ pi G}}} $

indicates the maximum radius that a homogeneous, spherical celestial body may have in order not to exceed the compressive strength of the material, i.e. not to be crushed by its own mass.

Maximum radii for different materials

For the very hypothetical case of a celestial body consisting entirely of Styrofoam ($ \ rho = 20 \, \ mathrm {\ frac {kg} {m ^ 3}} $ and $ p_ \ mathrm {max} = 150 \, \ mathrm {kPa } $) would result in a radius of around $ 1600 \, \ mathrm {km} $ (for comparison: the radius of the Earth's moon is around $ 1700 \, \ mathrm {km} $). For granite the radius is around $ 380 \, \ mathrm {km} $ and for basalt $ 550 \, \ mathrm {km} $. One conclusion is that celestial bodies with a radius significantly larger than that of the earth cannot consist of a single solid material (diamond: $ R_ \ mathrm {max} = 7600 \, \ mathrm {km} $).

Gravitational pressure in stars

Stars in balance

Gravitational pressure in stars is a special case of hydrostatic pressure. This results from the force of gravity contracting the star. In contrast, z. B. the radiation pressure as the star expanding force. In the case of a stable star, an equilibrium of all forces is established and the star has a stable shape. This is approximately the state of stars on the main sequence of the Hertzsprung-Russell diagram.

Examples of stars in imbalance

In the case of emerging stars that contract, the gravitational pressure outweighs the sum of all forces that build up counter pressure. Examples of back pressure are the kinetic gas pressure of the gas itself and, when the fusion reaction starts, the radiation pressure caused by all types of radiation. This changes the hydrostatic pressure within the emerging star.

In some classes of variable stars, periodic or transient changes in star density occur, which change the star's amount of matter inside or outside a sphere with a fixed radius, and with it the hydrostatic pressure at a given radius from the star's center.

Due to the stellar wind, stars steadily lose mass to their surroundings. This also changes the hydrostatic pressure. However, this change is very slow for main sequence stars.

In the later stages of star life there are also changes in the star structure that affect the hydrostatic pressure in the star.

See also

Individual evidence

  1. ^ Lew Dawidowitsch Landau, Jewgeni Michailowitsch Lifschitz: Statistical Physics. Part I. Akademie Verlag, Berlin 1979/1987, ISBN 3-05-500069-2, p. 70.
en: Pascal's law

fr: Principe de Pascal