Forces acting in the atmosphere. Forces acting in the atmosphere and their effect on the wind

Air is constantly moving in the atmosphere. Their immediate cause is the uneven distribution of pressure, which in turn is due to the inhomogeneity of the temperature field. What are the forces causing these movements:

3.1 Forces acting in the atmosphere.

The forces acting in the atmosphere can be divided into 2 groups: mass and surface. Mass- these are the forces that act on each element of mass (volume), regardless of whether there are other air particles nearby. These forces are: gravity, deflecting force earth rotation, centrifugal power. Surface forces are the forces of interaction of a certain volume of air and the environment. This is power baric gradient and viscous strength.

In mechanics, it is proved that when any body (including air) moves relative to the rotating Earth, it deviates from its original direction to the right in the northern hemisphere and to the left in the southern hemisphere, the force is directed at an angle of 90 0 to the velocity. It does not change the module but only changes direction. The reason for the emergence of force is that the body retains its direction of motion, and the daily rotation of the Earth changes the direction of the meridians and parallels. Therefore, from the Earth, it seems that the bodies deviate from the direction of the meridians and parallels. The horizontal component of the Coriolis force is A= 2 *v*Sinφ, where v is the speed of the body. Therefore, this force increases towards the poles (due to Sinφ) and with increasing speed v. At the equator it is 0.

3.1.3 The force of the baric gradient.

Horizontal atmospheric pressure gradients are almost always observed in the atmosphere. Air tends to move from places of higher pressure to places of lower pressure. The measure of pressure non-uniformity is the horizontal baric gradient (
. Therefore, the greater the pressure gradient, the more intense the air movement. If the baric gradient is related to a unit mass, i.e.
, then in meaning (and in dimension) this expression is an acceleration or force referred to units. masses. In direction, this force at each point of the baric field coincides with the normal to the isobar in the direction of decreasing pressure. The baric gradient force is the only force that causes air to move. All other forces can only slow down the movement or deflect it from the direction of the gradient.

If only the acceleration that the air receives under the action of the baric gradient acted on the air, then the movement of the air would be constantly accelerated. However, in reality, the wind speed cannot exceed several tens of m/s. It follows from this that, in addition to the force of the baric gradient, other forces act on the air, which balance the force of the gradient.

3.1.4. Friction force

The friction force in the atmosphere occurs when the volumes (layers) of moving air have different speeds. There is a certain viscosity between the layers of air, which prevents them from sliding relative to each other. Therefore, the greater the air speed (their difference), the greater the friction force or R= -kv (where k is the coefficient of friction), the more the movement slows down and its direction changes.

The nature of viscosity between air layers is twofold: it is molecular and turbulent. However, calculations show that the coefficient of turbulent viscosity is several orders of magnitude greater than the molecular one. In this regard, the molecular viscosity can be neglected. Then
, where R is the friction force; p is the air density; τ is the tangential stress of internal friction; z is the direction of air movement (perpendicular to the wall).

With height, the effect of friction in the atmosphere rapidly decreases. And at the level of 1000-1500 m it practically disappears. This height is therefore called the friction level, and the rest of the atmosphere is called the friction layer (boundary layer).

In an unstable atmosphere, the level of friction is higher than in a stable one.

3.1.5. Centrifugal force. It occurs if the movement of air occurs along a curved path. In this case, it is equal to: c = v 2 / r, where v is the speed of movement; r is the radius of curvature of movement. For atmospheric motions, c is usually small, because great value.

3.1.6. Motion equation

Thus, the above-mentioned forces act on the volume of air in the atmosphere. The equation of motion in general view will look like:

3.1.7. Geostrophic wind, its changes with height

Consider one of the special cases of air movement in the atmosphere. Let a particle of air, having a unit of mass, enter the atmosphere. In this case, there is no friction and we consider horizontal motion. Then, under the action of the force of the pressure gradient, the particle will begin to move from high pressure to low pressure along the normal to the isobar. But as soon as it begins to move, the Coriolis force will begin to act on it, which will deflect the movement of the particle to the right from the direction at a right angle. In the end, when these two forces are balanced, the particle will make a rectilinear uniform motion.

This movement is called geostrophic wind.

Mathematically, such a movement can be described as follows.
, where G is the pressure gradient force; A is the Coriolis force. Or
= 2*v g *Sinφ, hence
.

Thus, geostrophic wind is proportional to pressure gradient and inversely proportional to latitude. It does not exist at the equator (because = infinity). For standard conditions (t= 0 0 C, P= 1000hPa):
, where ∆P/∆n is in hPa per 100 km, v g is in m/s.

Because If the geostrophic wind does not take into account the friction force, then such a wind can only be observed above the friction layer, i.e. above 1-1.5 km. With height, due to a decrease in ρ, the geostrophic wind intensifies.

A more general case of air motion without friction is gradient in the field of curvilinear isobars (cyclone, anticyclone). In this case, in addition to the baric gradient force and the Coriolis force, the equation of motion includes a third force - centrifugal, i.e.
- 2*v*Sinφ-
; or v gr \u003d - *r*Sinφ+
- for the cyclone.

Gradient wind can be represented graphically as follows:

Here in the cyclone, the pressure gradient force is balanced by 2 forces A and C. The gradient wind is directed to the right at a right angle to the gradient.

In an anticyclone, the Coriolis force is balanced by G and C.

In both cases, the gradient wind is directed tangentially to the isobar to the right of the baric gradient.

Gradient wind calculations (v gr) can be expressed in terms of geostrophic:

V gr.cyclone \u003d v g -
;V gr.anticyclone =v g +
.

At the earth's surface, air experiences friction as it moves relative to the earth. The influence of the surface is especially noticeable up to about 50-100 m above the Earth. This layer is called surface (up to 1-1.5 km - boundary). In this layer, when forming the wind, it is necessary to take into account the friction force, which slows down the movement and changes its direction. Consider the scheme of the balance of forces in the atmosphere in this case. In the case of rectilinear isobars, the pressure gradient is directed perpendicular to the isobars (G); wind v and its direction will no longer blow along the isobars, but under acute angle on the force of the baric gradient α (to the right). The friction force R is directed towards opposite side air movement. And the force of the baric gradient must be balanced by 2 forces: the Coriolis force A and the friction force (A + R). Then, from the construction of a rectangle and taking into account that the force A is directed at a right angle kvi to the right of it, we find the position of the Coriolis force.

To determine the real wind speed, you need to make an equation where the sum of the three forces is zero: G + A + R = 0, substituting the expression for each force, you can come to the expression for v: v = *
, where k is the coefficient of friction. Therefore, the wind speed near the Earth is proportional to the pressure gradient and inversely proportional to the coefficient of friction and latitude. The angle α between the wind and the baric gradient in temperate latitudes is 60-75 0 over the oceans and 40-50 0 over land.

With circular isobars, i.e. in cyclones and anticyclones near the Earth, one should also take into account the centrifugal force C. The scheme of the direction of movement in these cases will be:

With height in the friction layer, the wind speed increases, and the direction approaches the isobar (low pressure on the left). The change in wind with height in the friction layer can be represented by a hodograph, i.e. curve which is also called the Ekman spiral. The wind seems to rotate to the right with height.

In the friction layer near the surface, a daily wind pattern is detected, from max at 2 pm, min at night or in the morning. Starting from a height of about 500 m, the diurnal variation is reversed – max at night, min during the day. Such a diurnal variation is explained by the diurnal variation of turbulent exchange. During the day, turbulence is max, so vortices descend to the surface from above at an increased speed, and from below upwards at a reduced speed. Therefore, during the day, the bottom is max, and the top is min speed. At night, the turbulence intensity is below min, and at the top, therefore, eddies with increased speed remain there and the velocities here reach max.

All forces considered in meteorologists are taken per unit mass. If the pressure in the horizontal plane is not the same, then there is an air flow in the direction of the lowest pressure. In other words, there is a force that makes the air move. It is called the baric gradient salt and per unit mass is equal to:

where ρ is the air density. The pressure gradient dp/dn is directed towards the pressure increase. The movement caused by the pressure difference is directed in the opposite direction. Therefore, in order for the pressure gradient force to be opposite, a minus sign is put in the equation.

In addition, there are other forces that affect the movement of air. These are the forces of Cariolis TO,centrifugal force Z, friction F tr and gravity g.

Cariolis Force K or, in other words, the deflecting force of the Earth's rotation, is an inertial apparent force. It arises because the Earth rotates around its axis and per unit mass is equal to:

K = 2ω С sinφ, (14)

where ω is the angular velocity of the Earth's rotation, equal to ω = 2 π /T, where T is the period of the Earth's revolution around its axis, T = 24*60*60s;

C is the speed of air movement;

φ is the latitude of the place.

Thus, the Cariolis force depends on the speed of movement and the latitude of the place. The Cariolis force acts only on moving bodies perpendicular to the direction of motion. It is greatest at the poles, and at the equator it is equal to zero. As a result, the bodies move along the earth's plane, deviate to the right in the northern hemisphere, and to the left in the southern hemisphere from the original direction of their movement.

Centrifugal force Z. The centrifugal force is also an apparent, inertial force that occurs when moving along a curved path. It is directed along the radius from the center and per unit mass is equal to:

Z \u003d C 2 / r,(15)

where r- radius of curvature.

Analytic expression for friction force F tr It has complex view. In navigation, problems are solved in the so-called geostrophic model, without taking into account the friction force, and the friction force is then introduced by a coefficient. And finally, there is the well-known gravity g. It is often viewed as a constant.

Gravity g. Incomparably more than other forces (9.81 ~ 10 m / s 2). It operates along the vertical axis. However, we do not notice noticeable vertical movements in the atmosphere directed towards the Earth's surface (downwards). This is due to the fact that such great strength is balanced by an equally large vertical baric gradient force. It follows from the basic equation of statics:

g = - dp/dz (16)

As you can see, the force of gravity is on the left side of the equation, and the force of the baric gradient along the vertical is written on the right side. The vertical baric gradient is a large value, which means that the force of the baric gradient is also large. Similarly, it can be stated that a very large force of the baric gradient along the vertical does not cause upward movements, as it is balanced by the force of gravity. These forces are on the same axis, directed in different sides and usually balance each other out.

Thus, the wind, by which we mean the horizontal movement of air, is not affected by gravity g. Its projection onto the horizontal plane is zero. Cariolis forces K and centrifugal force Z appear only after the movement has already begun. That is, the only force that causes air movement is the horizontal pressure gradient force G r . The difference in pressure in different places gives rise to the movement of air, seeking to smooth out these differences. The remaining forces reverse the movement relative to the original direction and slow it down.

Forces acting in the atmosphere in a state of equilibrium

STATIC ATMOSPHERE

The system is in equilibrium (at rest) if the resultant of all forces acting on the system is zero.

Forces are divided into mass and surface.

The mass forces acting on the atmosphere as a whole and on its part are gravity and the deflecting force of the Earth's rotation (Coriolis force).

The surface forces acting in the atmosphere are the pressure force and the friction force.

However, the Coriolis force and the force of friction appear only when the atmosphere moves relative to the surface of the Earth or some of its parts relative to others. Therefore, the forces acting in the atmosphere at rest are gravity and pressure.

Let the atmosphere be at rest with respect to the earth's surface. Then the horizontal component of the pressure gradient must vanish (otherwise the air will begin to move). For this, it is necessary and sufficient that the isobaric surfaces coincide with the level ones.

Let us single out two isobaric surfaces in the atmosphere located at heights z and z+dz (Fig.). Between isobaric surfacesp p+dp let's allocate the volume of air with horizontal bases 1 m 2 . The pressure force p directed from bottom to top acts on the lower base; on the top - the pressure force p + dp, directed from top to bottom. The pressure forces acting on the side faces of the selected volume are mutually balanced.

Rice. To the derivation of the equation of statics.

This volume is affected by gravity P, directed vertically downwards and equal in modulus

Let's project all the forces on the z-axis. Since the sum of all forces is zero, then the sum of these projections is also zero:

Substituting the expression for gravity, we get .

Dividing by dz, we define the second form of the main equation of atmospheric statics:

The left side is the vertical component of the pressure gradient, the right side is the force of gravity acting on a unit volume of air. Thus, the equation of statics expresses the balance of two forces - the pressure gradient and gravity.

Three important conclusions can be drawn from the equation of statics:

1. An increase in altitude (dz>0) corresponds to a negative increment in pressure (dp>0), which means that pressure decreases with altitude. The equation of statics is performed with high precision and in the case of atmospheric movement.

2. Let's single out a vertical column of air in the atmosphere with a base of 1m2 and a height from the level z to the upper boundary of the atmosphere. The weight of this column is . Integrating both parts () in the range from z, where the pressure p, to , the pressure is 0 (by definition of the upper bound), we get: , or .

Thus, we come to the second definition of the concept of pressure. Atmospheric pressure at each level is equal to the weight of a column of air of a single cross section and the height from this level to the upper boundary of the atmosphere. This explains the physical meaning of the decrease in pressure with height.

3. Equations of statics make it possible to draw a conclusion about the rate of decrease in pressure with height. The decrease in pressure is the greater, the greater the density of air and the acceleration of free fall. Density plays a major role. Air density decreases with increasing altitude. The higher the level, the lower the pressure drop.

If the points are located on the same isobaric surface, then the air density will depend only on the temperature at these points. At a point with a lower temperature, the density is higher. This means that when ascending to the same height, the decrease in pressure at a point with a higher temperature is less than at a point with a lower temperature.

In a cold air mass, the pressure decreases with height faster than in a warm one. This conclusion is confirmed by the fact that at heights (in the middle and upper troposphere) low pressure prevails in cold air masses, and high pressure in warm air masses.

Let's estimate the value of the vertical gradient. At normal conditions near sea level r=1.29 kg/m3, g=9.81 m/s2. Substituting these values ​​into (), we find: G = 1205 hPa / 100m.

The law of conservation of mass, from which the equation of continuity follows, is the first of the fundamental laws of mechanics. The second basic law is the law of change of momentum or Newton's second law, according to which the change in momentum (momentum) per unit time is equal to the sum of the forces applied to the body in question. In hydromechanics, Newton's second law is used in the form of the d'Alembert principle, according to which, when a control volume moves, all forces applied to it balance each other. In order to find out how the forces acting on a particle are described mathematically atmospheric air, we should consider an important special case - the state of rest.

Forces acting on air particles

Volume and surface forces

Volume (mass) forces: the magnitude of these forces is proportional to the volume (mass) of the fluid on which they act. The body force acting in the control volume is expressed by the formula , in which the characteristic of the volume (mass) force at each point is the distribution density of this force in space, vector quantity, equal to strength acting per unit volume (mass)
. An example of a body force is gravity. In this case, the distribution density is the force per unit masses continuum.

surface forces, act between parts of a given volume of liquid. They cannot change the momentum of this volume, since inside it each internal force is balanced by an internal force equal to it in absolute value and having the opposite direction. At the same time, the work of internal forces can change the kinetic and (or) potential energy of the volume of liquid under consideration. The magnitude of these forces is proportional to the surface area on which they act. The characteristic of the surface force on a given surface is the density of its distribution, which is called tension. This is a vector quantity. Its direction, in the general case, does not coincide with the direction of the normal to the given surface. The projection of the stress on this normal is called the normal stress, and the projection of the stress on the tangent plane to the given surface is called the shear stress.

Below are the basic information about the volumetric and surface forces acting in the atmosphere.

Gravity - body force

The gravitational force vector according to Newton's law can be written as

F = f m 1 m 2 / r 2 i F

, where f = 6.673 10 -11 [n m 2 /kg 2 or m 3 /With 2 ] – gravitational constant, i F – ort direction of force from a smaller mass ( m 2 ) to the larger ( m 1 ). In what follows, it is assumed that m 1 = m ( for the earth M) , m 2 = 1 kg (single mass). Choosing the unit mass of the attracted body, the force field of the massMbegin to describe using the acceleration of gravity. (In what follows, the geocentric gravitational constant will also be used fM\u003d 3.086 10 14 [m 3 / s 2]).

If, as shown in the figure, if the mass M is located at the point {ξ, η, ζ ), and the unit mass is located at the point ( x, y, z), then the force direction vector is opposite to the distance vector r 2 = (x 2 - x 1 ) 2 + (y 2 - y 1 ) 2 +(z 2 - z 1 ) 2 to the point of attraction.

If dF = dFx i + dFy j + dFz k – vector attraction force element dm masses M, unit mass in projections on the axes of the Cartesian coordinate system with the center at the center of gravity of the body M, then the calculation of the force of attraction by a body of finite volume can be performed using the volume integral.

dFx = dF cos(Fx)= - dF cos(rx ) = - (f dm/r 2 ) (x 2 -x 1 )/r Fx = -f cos(rx ) /r 2 dm

dFy = dF cos(F y)= - dF cos(r y ) = - (f dm/r 2 ) (y 2 -y 1 )/r Fy = -f cos(r y ) /r 2 dm

dFz = dF cos(Fz)= - dF cos(rz ) = - (f dm/r 2 ) (z 2 -z 1 )/r Fz = -f cos(rz ) /r 2 dm

If the Z axis is aligned with the direction operating force, then fx= fy= 0. Then

The force of attraction of a unit mass from the side of the mass M, is expressed by the formula

(6.1)

Attraction of a uniform ball

Let the center of the attracted mass be at a distance ρ from the center of the sphere. Arbitrary point A on the attracting sphere is at a distance r from the attracted point, and r 2 = R 2 + ρ 2 –2 R ρ cos whence it follows that R/ ρ dr= R 2 sin d / r

An element of an attracting mass located on a surface area dσ R 2 sin( ) d d can be found using the formula

dm= dσ R 2 sin( ) d d   (6.2)

where  W (R) dR surface density (bulk density denoted  W (R)). The force of attraction of the mass element of the surface area dm, is calculated by the formula

dF= - f μ cos(r, z) dσ =- f μ (ρ - Rcosθ)/ r dσ = f μ (ρ 2 - R 2 + r 2 )/2 pr dσ , (6.3)

wherein Rcosθ expressed in terms of distances.

The attractive force of the entire spherical surface can be calculated by integrating dF over all surfaces of the sphere

F = f
(6.4)

The force of attraction of a ball can be calculated by expressing the surface density in terms of a constant bulk density  W =  dR, summing up the effect of all internal infinitely thin layers dR and considering that within the atmosphere of altitude z(0-50 km) almost a thousand times smaller than the radius globe Rsh(6400 km), according to the formula

F = =9.8 m/s 2 = g (6.5)

Thus, it is shown that when estimating the force of gravity, we can assume that the force of attraction of the Earth is concentrated in its center and is calculated according to the law of universal gravitation for material points. This means that a force acts on each particle of air. P , directed towards the center of the Earth, called the weight of this particle and calculated by the formula

(6.6)

Gravitational force potential and geopotential

If  V/  x = fx,  V/  y = fy ,  V/  z = fz , then the scalar field V(x, y, z) – vector field potential F (x, y, z). For the gravity field of the Earth in meteorology, one can restrict oneself only to an approximate estimate of its vertical component using the formula

dV=  V/  x dx+  V/  y dy +  V/  z dz = fx dx + fy dy + fz dz = g dz

Given that the potential is a total differential, it is determined by integrating over an arbitrary contour between two points of the field

V(B) – V(A) = A  B dV= A  B Fx dx + Fy dy + Fz dz =

Physically, the potential - this is the work of the earth's gravitational force to move a unit mass between points A B. With great accuracy, we can assume that it depends only on the height difference between the points. In meteorology, it is commonly called the geopotential. It is useful to remember that for central vector fields, which include the gravity field, for the force vector F (x, y, z) the potential is inversely proportional to the distance to the point (V = f M/ r). There is no inconsistency between these definitions, since the latter goes over to the former when using the assumption 1/ r =1/(R sh + z)≈ - z/ R sh 2 .

Stress tensor - a form of writing surface forces

D In order to show why there are surface forces, we divide, as is customary in continuum mechanics, an arbitrary part of the control volume of a continuous medium by the surface AB into two parts (see figure). In this case, part 1 will act on part 2 with the force ΔF AB . Denoting the part of the surface area AB, located at the point M through ΔА AB , we can write the formula for the stress vector P AB acting on this site, in the form

It should be noted that the part of the area ΔA DE of the surface DE, located at the same point M, is affected by a different stress vector

It means that the vector representation of surface forces at the same point in the atmosphere is ambiguous, it depends on the orientation of the elementary area. In order to separate the unambiguous description of the stress state at a point from the influence of the site orientation, it is necessary to take into account that for any site whose orientation is given by the normal vector, the stress vector P decomposes in three non-coplanar vectors, according to the chosen coordinate system. (see picture). Each of the vectors P X , P Y , P Z represents the stress acting at a point on the coordinate planes. In general, these vectors may not be perpendicular to the coordinate planes. Therefore, each of them has a three-component representation.

Components P XX , P YY , P ZZ are normal stresses, and the remaining components are shear stresses.

If we consider the equilibrium of the control volume in the form of a pyramid with a vertex at point M (see figure), t
about the projection of face ABC having area A n, on the coordinate planes are expressed by the formulas
. The stress vector acting on this face is represented as
, and the stress vectors acting parallel to the coordinate axes have components
,
,

In order for the pyramid to be in balance, the projections of all forces on the coordinate axes must be balanced. This implies the equalities

If cut A n and represent these equalities in matrix form, then these equalities can be rewritten as

(6.7)

It becomes clear that the effect of the orientation of face ABC, expressed by the normal vector to this face n and the effect of the stresses acting at point M, expressed in the table P (3x3) are separated.

table
is called the stress tensor.

Properties of stress tensors in any continuous medium

1. P is a matrix. All properties of matrices are valid.

2. If from the system (x, y, z) go to (x", y", z"), then P" \u003d A P , П" - tensor in new system, A - transition matrix (known). This means that P" is predictable and does not depend on the orientation of the site, the stress tensor definitely determines the surface forces acting at a point in a continuous medium.

3. When changing coordinates, the INVARIANTS of the tensor П are preserved:

a) trace ( p xx + p yy + p zz ), b) Minors; c) Determinant.

4. Since the vector n is dimensionless, then the dimension [ p ij ] = N/m 2

Properties of fluid stress tensors.

Fluidity is the ability of liquid particles to move under any, even infinitely small, shear stress. It follows that at rest, when there is no movement, there are no tangential stresses, that is, the stress tensor in a liquid (and gas) is a diagonal matrix, that is

Since for an arbitrarily oriented area the stress vector in the liquid is perpendicular to it, then P N = n | P N | . In tensor representation P N = n P. Comparing these two definitions, we get that

n | P N | = { n x | P N |; n y | P N |; n z | P N |} = n P = ( n x p xx +0+0; 0+ n y p yy +0; 0+0+ n z p zz }.

Whence it follows that

| P N |= p xx = p yy = p zz = - p and

In a fluid (and gas) at rest, the stress tensor is completely determined by a single scalar quantity p, which is called hydrostatic pressure

Pascal's law: In a fluid at rest, the stresses in any direction are the same and are directed along the normal to the site

The definition of the pad pressure force ∆A coincides with the thermodynamic ∆ F = - p n ∆ A Determination of the baric gradient force generated by the pressure difference and acting on
and the volume element ∆ V = dx dy dz illustrates the figure. On him p - pressure force on the platform dydz located at the point ( x , y , z .), -( p +∂ p /∂ xdx ) - pressure force on the platform dydz located at the point ( x + dx , y , z .). Per volume element in the direction x the component pressure forces p dydz -( p + ∂ p /∂ xdx ) dydz = - ∂ p /∂ x dx dydz

per element ∆ V the pressure force vector acts, which in meteorology is usually called the baric gradient force. He is equal - grad p dx dydz , where grad p = { - ∂ p /∂ x , - ∂ p /∂ y , - ∂ p /∂ z } .

The law of hydrostatics. Atmospheric statics

In a fluid at rest, the vector of gravity acting on the element is balanced by the pressure gradient:

( ρ f - grad p) dx dy dz = 0

In projections on the axis:

{ ρ f x - ∂ p /∂ x =0, ρ f y - ∂ p /∂ y =0, ρ f z - ∂ p /∂ z =0}

It is customary to direct the z-axis to the zenith, then f = { 0, 0, - g } and the balance of gravity and baric gradient reduces to the equalities

∂p /∂ x =0, ∂ p /∂ y =0, ∂ p /∂ z = - ρ g

In a resting atmosphere, the isobars are parallel to the geosphere. The last of the equalities is called the law of hydrostatics.

Atmospheric statics.

In the atmosphere, the law of hydrostatics acts together with the equation of state

O
It follows that the vertical pressure distribution in the atmosphere is completely determined if the vertical temperature profile and the pressure at any one level are known. It would be physically correct to use the pressure value at the highest levels, but due to the low accuracy of observations, the pressure at the level of the underlying surface is used.

For various assessments, it is useful to know how approximately the pressure changes with height in the standard atmosphere, that is, with a linear temperature drop (polytropic atmosphere) up to 11 km, characteristic of the troposphere, and at a constant temperature (isothermal atmosphere), which is a simplified description of the stratosphere (see Fig. drawing).

In a polytropic atmosphere (troposphere)

At the top of the troposphere z= z 11 = 11000 m T= T 11 =217 o K, p= p 11 =225 hPa

In an isothermal atmosphere (stratosphere)

V
the vertical pressure distribution obtained from these dependences is shown in the figure

Consequences of the equations of statics and state

Mass of a single column of atmosphere

Internal energy of a single column of the atmosphere

Potential Energy and the Dines Theorem

Writing the Dynes theorem in terms of the height of the center of gravity and the average temperature

Satisfaction of the Dynes theorem at the maximum levelψ

Proof of isopycnicityaverage energy level

Approximate values ​​of variables for the average energy level

Forces acting in the atmosphere in a state of equilibrium

STATIC ATMOSPHERE

The system is in equilibrium (at rest) if the resultant of all forces acting on the system is zero.

Forces are divided into mass and surface.

The mass forces acting on the atmosphere as a whole and on its part are gravity and the deflecting force of the Earth's rotation (Coriolis force).

The surface forces acting in the atmosphere are the pressure force and the friction force.

However, the Coriolis force and the force of friction appear only when the atmosphere moves relative to the surface of the Earth or some of its parts relative to others. Therefore, the forces acting in the atmosphere at rest are gravity and pressure.

Let the atmosphere be at rest with respect to the earth's surface. Then the horizontal component of the pressure gradient must vanish (otherwise the air will begin to move). For this, it is necessary and sufficient that the isobaric surfaces coincide with the level ones.

Let us single out two isobaric surfaces in the atmosphere located at heights z and z+dz (Fig.). Between isobaric surfacesp p+dp let's allocate the volume of air with horizontal bases 1 m 2 . The pressure force p directed from bottom to top acts on the lower base; on the top - the pressure force p + dp, directed from top to bottom. The pressure forces acting on the side faces of the selected volume are mutually balanced.

Rice. To the derivation of the equation of statics.

This volume is affected by gravity P, directed vertically downwards and equal in modulus

Let's project all the forces on the z-axis. Since the sum of all forces is zero, then the sum of these projections is also zero:

Substituting the expression for gravity, we get .

Dividing by dz, we define the second form of the main equation of atmospheric statics:

The left side is the vertical component of the pressure gradient, the right side is the force of gravity acting on a unit volume of air. Thus, the equation of statics expresses the balance of two forces - the pressure gradient and gravity.

Three important conclusions can be drawn from the equation of statics:

1. An increase in altitude (dz>0) corresponds to a negative increment in pressure (dp>0), which means that pressure decreases with altitude. The equation of statics is also fulfilled with high accuracy in the case of atmospheric motion.

2. Let's single out a vertical column of air in the atmosphere with a base of 1m2 and a height from the level z to the upper boundary of the atmosphere. The weight of this column is . Integrating both parts () in the range from z, where the pressure p, to , the pressure is 0 (by definition of the upper bound), we get: , or .

Thus, we come to the second definition of the concept of pressure. Atmospheric pressure at each level is equal to the weight of a column of air of a single cross section and the height from this level to the upper boundary of the atmosphere. This explains the physical meaning of the decrease in pressure with height.

3. Equations of statics make it possible to draw a conclusion about the rate of decrease in pressure with height. The decrease in pressure is the greater, the greater the density of air and the acceleration of free fall. Density plays a major role. Air density decreases with increasing altitude. The higher the level, the lower the pressure drop.

If the points are located on the same isobaric surface, then the air density will depend only on the temperature at these points. At a point with a lower temperature, the density is higher. This means that when ascending to the same height, the decrease in pressure at a point with a higher temperature is less than at a point with a lower temperature.

In a cold air mass, the pressure decreases with height faster than in a warm one. This conclusion is confirmed by the fact that at heights (in the middle and upper troposphere) low pressure prevails in cold air masses, and high pressure in warm air masses.

Let's estimate the value of the vertical gradient. Under normal conditions near sea level r=1.29 kg/m3, g=9.81 m/s2. Substituting these values ​​into (), we find: G = 1205 hPa / 100m.