(ID 15439)

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)

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)

(ID 15434)

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 that

.

(ID 11643)

Con la ley de la gravitaci n de Newton, con , es:

Se puede, con la definici n de la fuerza, con :

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 la aceleraci n:

(ID 11644)

With acceleration generated by the celestial body, en conjunction $m/s^2$, celestial object planet distance $m$, latitude of the place $rad$, planet radio $m$ and variation of acceleration perpendicular to the direction of the star $m/s^2$, the relationship between the variation of acceleration and acceleration is:

And since the expression for acceleration is with acceleration generated by the celestial body, en conjunction $m/s^2$, celestial object planet distance $m$, latitude of the place $rad$, masa del cuerpo que genera la marea $kg$, planet radio $m$ and universal Gravitation Constant $m^3/kg s^2$:

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 by:

(ID 11645)

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 that

(ID 11647)

With acceleration generated by the celestial body, en conjunction $m/s^2$, celestial object planet distance $m$, latitude of the place $rad$, planet radio $m$ and variation of acceleration in the direction of the star, in conjunction $m/s^2$, the relationship is:

And as for acceleration generated by the celestial body, en conjunction $m/s^2$, celestial object planet distance $m$, latitude of the place $rad$, masa del cuerpo que genera la marea $kg$, planet radio $m$ and universal Gravitation Constant $m^3/kg s^2$,

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 by:

(ID 11650)