Deformation of tide

The phenomenon of the tides is produced by the rotation of the Earth in the fields of gravity of the Moon, the Sun and, with a quite less degree, other stars disturbing such as the Venus planets and Jupiter. This rotation imposes periodicities in the potential gravific to which all the points are subjected to surface and inside the Earth. The consequence most immediately visible is consisted the oceanic tides by it , but there exist also tidal impulses in the solid Earth (which one calls, in a rather unsuitable way, the terrestrial tides ) and in the atmosphere ( atmospheric tides ). If the Earth were perfectly rigid, the relative positions of its various material points would not undergo variation; there would be thus no tidal impulses, and thus not of possibility of friction and consequently not of braking of terrestrial rotation due to the tides. The only consequence would be thus a multiperiodic variation of the potential gravific in each point: they is what is called the tides the geoid . The amplitudes of those can be calculated in each point and at every moment by the laws of the celestial mechanics. Their knowledge is important, because it is with them that one reports measurements of the tides of the real Earth, which is deformable.

The equipotential ones are deformed at first approximation so as to constitute ellipsoids centered on the center of mass of the Earth and lengthened in the direction of the disturbing star. By neglecting for the moment the friction, material surfaces are deformed same manner, the pad associated with a particular tide pointing in the direction with the star which causes this tide. The phenomenon of tide observed is produced by the rotation of the Earth in this “deformed envelope”.

Potential of tide

Let us consider the geometry of figure 2 in which one represented the Earth and the Moon orbiting around their common center of mass, which is inside the Earth at a distance B of the center of mass of the Earth alone. We will simplify the theory as it is habit to do it, while starting by considering tides , or tides of balance . One thus admits that the Earth always shows the same one vis-a-vis the Moon, i.e. it has an axial rotation as much as an orbital revolution at the angular velocity ω_L. Thus, the whole figure turns, and the point P is at a constant distance from the axis of rotation.

M_L being the mass of the Moon, in the point P let us define the following scalar function:

Φ (P) = - GM_LR ′ ^-1 - ½ ω_L ² R ².

The rule of the cosine enables us to write

R ′ ² = has ² - 2 has R cos ψ + R ²,

so that

R ′ ^-1 = R^-1 ^-1/2 = R^-1 Σ_n=0→∞ (a/R) ⁿ P_n (cos ψ),

where the P_n symbol represents a Polynôme of Legendre of degree N. Let us recall that

P₀ (cos ψ) = 1, P₁ (cos ψ) = cos ψ, P₂ (cos ψ) = (3/2) cos ² ψ - 1/2,….

The scalar function Φ can thus be still written as follows:

Φ (P) = - GM_LR^-1 Σ_n=0→∞ (a/R) ⁿ P_n (cos ψ) - ½ ω_L ² R ².

In addition, we have

R ² = B ² - 2 B has sin θ cos λ + has ² sin ² θ

and, since

cos ψ = sin θ cos λ,

it comes

R ² = B ² - 2 B has cos ψ + has ² sin ² θ.

The distance B is provided by

B = M_LR/(M+M_L).

As R ≅ 60 has and M ≅ 81 M_L, one finds B ≅ 4700 km. While retaining in Φ only the terms until the order n=2, it comes

Φ ≃ - GM_LR^-1 {1 + (a/R) cos ψ + (a/R) ²} - (1/2) ω_L ² has ² sin ² θ.

By holding account in this last relation of the third law of Kepler, namely

ω_L ² R ³ = G (M+M_L)

and from the expression of B, we obtain

- ½ ω_L ² R ² =

Forces of tide

Families of tide

Tide couples

We already announced that the tides which contribute more to braking of terrestrial rotation are the semi-diurnal tides (in other words, sectoral) and, as we have just seen it, especially the M₂ tide. Considering the symmetry of the phenomenon, it is obvious that it is mainly the East-West component of this tide which is implied in braking.

Deceleration of the rotation of the Earth

Growth of the distance from the Earth to the Moon

Random links: