Force of gravity and tides in conjunction

Storyboard

Gravity and centrifugal acceleration generate tides, the movement of oceans that raises and lowers their level with a frequency of 12 hours. Their origin can be generated by both the moon and the sun.

>Model

ID:(1523, 0)



Force of gravity and tides in conjunction

Storyboard

Gravity and centrifugal acceleration generate tides, the movement of oceans that raises and lowers their level with a frequency of 12 hours. Their origin can be generated by both the moon and the sun.

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$a_c$
a_c
Acceleration generated by the celestial body, en conjunction
m/s^2
$d$
d
Celestial object planet distance
m
$\theta$
theta
Latitude of the place
rad
$M$
M
Masa del cuerpo que genera la marea
kg
$R$
R
Planet radio
m
$G$
G
Universal Gravitation Constant
m^3/kg s^2
$\Delta a_{cx}$
Da_cx
Variation of acceleration in the direction of the star, in conjunction
m/s^2
$\Delta a_{cy}$
Da_cy
Variation of acceleration perpendicular to the direction of the star
m/s^2

Calculations


First, select the equation:   to ,  then, select the variable:   to 

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

 Variable   Given   Calculate   Target :   Equation   To be used



Equations


Examples


mechanisms

There is a contribution from the gravitational attraction of the celestial body that pulls water towards the equatorial region:

image

The hypotenuse of the triangle is related to the vertical leg by:

$R\sin\theta$



and the horizontal leg by:

$d - R\cos\theta$



Using the Pythagorean theorem, we have:

$R^2\sin^2\theta+(d-R\cos\theta)^2=d^2+R^2-2Rd\cos\theta$

There is a contribution from the gravitational attraction of the celestial body that pulls water towards the radius, which tends to displace the water towards the equatorial zone:

image

The hypotenuse of the triangle is formed by the vertical leg:

$R\sin\theta$



and the horizontal leg:

$d - R\cos\theta$



According to the Pythagorean theorem, we have:

$R^2\sin^2\theta+(d-R\cos\theta)^2=d^2+R^2-2Rd\cos\theta$


model

To determine the variation of the acceleration perpendicular to the radius, we can use triangle similarity to equate the relation

$\displaystyle\frac{\Delta a_{cy}}{a_c}$



with the length

$d-R\cos\theta$



and the hypotenuse

$\sqrt{d^2+R^2-2dR\cos\theta}$

.

By triangle similarity, we have with the list that

kyon.

Con la ley de la gravitaci n de Newton, con list=9238, es:

equation=9238



Se puede, con la definici n de la fuerza, con list=10975:

equation=10975

Y el radio al cuadrado:

$r^2=d^2+R^2-2dR\cos\theta$



Calcular la aceleraci n reemplazando el radio en la fuerza y despejando la aceleraci n. Esto da con list la aceleraci n:

kyon

With list=11643, the relationship between the variation of acceleration and acceleration is:

equation=11643



And since the expression for acceleration is with list=11644:

equation=11644

It follows that:

$\Delta a_{cy} = GM\displaystyle\frac{R\sin\theta}{(d^2 + R^2 - 2dR\cos\theta)^{3/2}}\sim \displaystyle\frac{GM}{d^2}\displaystyle\frac{R\sin\theta}{d}$



Therefore, in the approximation d\gg R, we can approximate with the list by:

kyon

To determine the variation of the acceleration parallel to the radius, we can use triangle similarity to equate the relation

$\displaystyle\frac{\Delta a_{cx}}{a_c}$



with the length

$d+R\cos\theta$



and the hypotenuse

$\sqrt{d^2+R^2-2dR\cos\theta}$



By triangle similarity, we have with the list that

kyon

With list=11647, the relationship is:

equation=11647



And as for list=11644,

equation=11644

Thus, we have:

$\Delta a_{cx} =GM\displaystyle\frac{d - R\cos\theta}{(d^2 + R^2 - 2dR\cos\theta)^{3/2}}\sim \displaystyle\frac{GM}{d^2}\left(1+\displaystyle\frac{2R\cos\theta}{d}\right)$



Therefore, in the approximation d\gg R, we can approximate with the list by:

kyon


>Model

ID:(1523, 0)



Mechanisms

Definition


ID:(15439, 0)



Variation of gravity perpendicular to the radius, in conjunction

Image

There is a contribution from the gravitational attraction of the celestial body that pulls water towards the equatorial region:



The hypotenuse of the triangle is related to the vertical leg by:

$R\sin\theta$



and the horizontal leg by:

$d - R\cos\theta$



Using the Pythagorean theorem, we have:

$R^2\sin^2\theta+(d-R\cos\theta)^2=d^2+R^2-2Rd\cos\theta$

ID:(11635, 0)



Variation of gravity parallel to the radius, in conjunction

Note

There is a contribution from the gravitational attraction of the celestial body that pulls water towards the radius, which tends to displace the water towards the equatorial zone:



The hypotenuse of the triangle is formed by the vertical leg:

$R\sin\theta$



and the horizontal leg:

$d - R\cos\theta$



According to the Pythagorean theorem, we have:

$R^2\sin^2\theta+(d-R\cos\theta)^2=d^2+R^2-2Rd\cos\theta$

ID:(11658, 0)



Model

Quote


ID:(15434, 0)