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Solar and lunar tides
Storyboard 
The second type of tides that are recorded on land are solar tides. Its size is less than that of the moon.
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Parallel increase in acceleration generated, as opposed to
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The change in gravitational acceleration leads to a flow of water that tends to alter the height of the water column (sea depth) in order to compensate for pressure:
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Representation as ellipse
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Variations in acceleration lead to changes in water pressure around the planet, allowing water columns to differ in heights.
In particular, the deviations caused are as follows:
For the sun's case: 8.14 cm, 16.28 cm
For the moon's case: 17.9 cm, 35.6 cm
This situation can be represented as a deformation of a circle, corresponding to an ellipse.
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Sun case parameters
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In the case of the sun,
the following parameters are considered:
Mass: 1.987e+30 kg
Sun-Earth distance: 1.50e+11 m
The tidal heights can be calculated using the following relationships:
For the x-direction, with angle from the planet line - celestial object $rad$, celestial object planet distance $m$, gravitational Acceleration $m/s^2$, masa del cuerpo que genera la marea $kg$, planet radio $m$, tidal height parallel to the ecliptic $m$ and universal Gravitation Constant $m^3/kg s^2$, we have:
| $h_x = \displaystyle\frac{2 G M }{ g }\displaystyle\frac{ R ^2}{ d ^3}\cos\theta $ |
And for the y-direction, with angle from the planet line - celestial object $rad$, celestial object planet distance $m$, gravitational Acceleration $m/s^2$, masa del cuerpo que genera la marea $kg$, planet radio $m$, tidal height perpendicular to the ecliptic $m$ and universal Gravitation Constant $m^3/kg s^2$, we obtain:
| $h_y = \displaystyle\frac{ G M }{ g }\displaystyle\frac{ R ^2}{ d ^3}\sin\theta$ |
With the Earth's radius of 6371 km, at the point of minimum tide ($\theta = \pi/2$), we have:
$h_y = 8.14 cm$
And at the point of maximum tide ($\theta = 0$), it is:
$h_x = 16.28 cm$
Thus, the fluctuations due to the sun amount to $h_x + h_y = 24.42 cm$.
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Moon case parameters
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In the case of the Moon,
we have the following parameters:
Mass: 7.349e+22 kg
Distance Earth-Moon: 3.84e+8 m
For the x-direction, with angle from the planet line - celestial object $rad$, celestial object planet distance $m$, gravitational Acceleration $m/s^2$, masa del cuerpo que genera la marea $kg$, planet radio $m$, tidal height parallel to the ecliptic $m$ and universal Gravitation Constant $m^3/kg s^2$, we have:
| $h_x = \displaystyle\frac{2 G M }{ g }\displaystyle\frac{ R ^2}{ d ^3}\cos\theta $ |
And for the y-direction, with angle from the planet line - celestial object $rad$, celestial object planet distance $m$, gravitational Acceleration $m/s^2$, masa del cuerpo que genera la marea $kg$, planet radio $m$, tidal height perpendicular to the ecliptic $m$ and universal Gravitation Constant $m^3/kg s^2$, we have:
| $h_y = \displaystyle\frac{ G M }{ g }\displaystyle\frac{ R ^2}{ d ^3}\sin\theta$ |
With the Earth's radius of 6371 km, at the point of lowest tide ($\theta = \pi/2$), we obtain:
$h_y = 17.9 cm$
And at the point of highest tide ($\theta = 0$), we have:
$h_x = 35.6 cm$
So, the fluctuations due to the Moon amount to $h_x + h_y = 53.5 cm$.
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Model
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Parameters
Variables
Calculations
to 
, then, select the variable: 
to 
Calculations
Calculations
Variable 
Given Calculate 
Target : Equation To be used 
Equations
$ g h_x =\displaystyle\frac{1}{2}( \Delta a_{cx} - \Delta a_{ox} ) R $
g * h_x = ( Da_cx - Da_ox )* R / 2
$ g h_y = \Delta a_{cy} R $
g * h_y = Da_cy * R
$h_x = \displaystyle\frac{2 G M }{ g }\displaystyle\frac{ R ^2}{ d ^3}\cos\theta $
h_x = 2* G * M * R ^2* cos( theta )/( g * d ^3)
$h_y = \displaystyle\frac{ G M }{ g }\displaystyle\frac{ R ^2}{ d ^3}\sin\theta$
h_y = G * M * R ^2* sin( theta )/( g * d ^3)
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Depth relationship and acceleration variation in x
Equation
The change in acceleration means that the water column experiences a different pressure unless the depth adjusts. To achieve a steady state, this is precisely what happens. The modification of gravitational acceleration is compensated by a change in depth corresponding to the tide:
$p_x=\rho g h_x=\rho\displaystyle\frac{1}{2} (\Delta a_{cx} - \Delta a_{ox}) R$
Therefore,
| $ g h_x =\displaystyle\frac{1}{2}( \Delta a_{cx} - \Delta a_{ox} ) R $ |
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Depth variation in the x direction
Equation
The change in acceleration implies that the column of water experiences a different pressure unless the depth adjusts. Achieving a steady state involves precisely this. The modification of gravitational acceleration is compensated by a change in depth corresponding to the tide:
| $ g h_x =\displaystyle\frac{1}{2}( \Delta a_{cx} - \Delta a_{ox} ) R $ |
With the variation on the conjunction side with
| $ \Delta a_{cx} = \displaystyle\frac{ G M }{ d ^2}\left(1+\displaystyle\frac{2 R \cos \theta }{ d }\right)$ |
and with
| $ \Delta a_{ox} =\displaystyle\frac{ G M }{ d ^2}\left(1-\displaystyle\frac{2 R \cos \theta }{ d }\right)$ |
It follows that the surface rises with in
| $h_x = \displaystyle\frac{2 G M }{ g }\displaystyle\frac{ R ^2}{ d ^3}\cos\theta $ |
where only the variable part of the variation was taken into account, since the term $GM/d^2$ acts on the entire system and does not create differences.
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Depth relationship and acceleration variation in y
Equation
The change in acceleration implies that the column of water experiences a different pressure unless the depth adjusts. Achieving a steady state involves precisely this. The modification of gravitational acceleration is compensated by a change in depth corresponding to the tide:
$p_y=\rho g h_y=\rho\Delta a_{cy} R$
Therefore, it follows that:
| $ g h_y = \Delta a_{cy} R $ |
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Depth variation in the y direction
Equation
The change in acceleration means that the column of water experiences a different pressure unless the depth adjusts. To achieve a steady state, this is precisely what happens. The modification of gravitational acceleration is compensated by a change in depth corresponding to the tide:
| $ g h_y = \Delta a_{cy} R $ |
With the variation on the side of conjunction with
| $ \Delta a_{cy} = \displaystyle\frac{ G M }{ d ^2 }\displaystyle\frac{ R \sin \theta }{ d }$ |
As a result, the surface rises with at
| $h_y = \displaystyle\frac{ G M }{ g }\displaystyle\frac{ R ^2}{ d ^3}\sin\theta$ |
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