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Hydrostatic of oceans

  1. Apr 19, 2012 #1
    Hello,

    1. The problem statement, all variables and given/known data

    We consider the Earth as a homogenous sphere or radius [itex]R_0[/itex], rotating with angular speed of [itex]\theta' = \omega[/itex].
    We work in a terrestrial reference frame (geographical reference of the terrestrial sphere whose center O is the center of the earth and whose axes are fixed to the planet), in rotation relative to the Galilean reference frame centered at the center of the earth and fixed axes.
    We'll uuse spherical coordinates in the terrestrial reference frame.
    The gravity field is considered as a centripetal field whose intensity is described by law :

    [itex]g(R) = g_0\frac{R^2_0}{R^2}[/itex]

    The pressure is equal to the atmospheric pressure Pa at the free surface of the oceans

    1) We set [itex]h = R_S - R_0[/itex] the radial distance between the free surface of oceans and the surfaceland. We set [itex]h_P[/itex] the value at the poles we'll assume known. Show that the equation of the free surface of the oceans is

    [itex]g_0\frac{R_0^2}{R_0 + h} + \frac{\omega^2(R_0 + h)^2\sin^2\,\theta}{2} - g_0\frac{R_0^2}{R_0 + h_P} = 0[/itex]

    2) After making a first-order Taylor expansion of the equation of the free surface, estimate [itex]\Delta h = h_E - h_P[/itex] radial distances between the equator and the poles.

    2. The attempt at a solution

    1) My first problem lies in the establishment of the equation. We're faced to 2 forces : g and A_e, the drive acceleration of a point on the surface of the water relative to the Galilean report. According to the fundamental hydrostatic's law :

    [itex]\overrightarrow{grad}\, P=\rho\overrightarrow{g}-\rho \overrightarrow{A_{e}} \Longleftrightarrow \overrightarrow{grad}\, P=- \rho g_{0}\frac{R_{0}^{2}}{R^{2}} \overrightarrow{e_{x}}+\rho\omega^{2}R \overrightarrow{e_{r}} \Longleftrightarrow \overrightarrow{grad} \, P=-\rho g_{0}\frac{R_{0}^{2}}{R^{2}}\left(\cos\,\theta \overrightarrow{e_{r}}-\sin\,\theta \overrightarrow{e_{\theta}}\right)+ \rho \omega^{2}R\overrightarrow{e_{r}}[/itex]

    Finally,

    [itex]\overrightarrow{grad}\, P=\left(\rho\omega^{2}R-\rho g_{0}\frac{R_{0}^{2}}{R^{2}}\cos\,\theta\right) \overrightarrow{e_{r}}+\rho g_{0}\frac{R_{0}^{2}}{R^{2}}\sin\,\theta \overrightarrow{e_{\theta}}[/itex]

    We get this system to solve :

    [itex]\frac{ \partial P}{\partial R} = R\omega^2\rho - \rho \cos\,\theta g_0 \frac{R^2_0}{R^2}[/itex]
    [itex]\frac{ \partial P}{\partial \theta} = \rho \sin\, \theta g_0\frac{R^2_0}{R}[/itex]

    The first equation gives [itex]P(R,\theta) = \frac{R^2\omega^2\rho}{2} + \rho g_0\cos\,\theta \frac{R^2_0}{R} + C(\theta)[/itex]. Differentiating this function with respect to theta and using the second relation, I get

    [itex]C'(\theta) = 2\rho g_0\frac{R^2_0}{R}(\sin\, \theta)[/itex]

    Hence,

    [itex]C(\theta) = -2\rho g_0\frac{R^2_0}{R}(\cos\, \theta ) + C[/itex]

    Finally,

    [itex]P(R,\theta) = \frac{R^2\omega^2\rho}{2} - \rho g_0\cos\,\theta \frac{R^2_0}{R} + C[/itex]

    The equation of the free surface is the set of (R, theta) satisfying P = P_a? But I can not determine the constant C. More ... I find the equation of the surface quite different compared to the one given, for example, I have no sin ² with the term w ² ...

    Can someone help me ?

    2) I make a Taylor expansion to order 1 :

    [itex]g_0R_0(1-\frac{h}{R_0}) + \frac{\omega^2R^2_0\sin^2\, \theta}{2}(1+\frac{2h}{R_0}) - g_0R_0(1-\frac{h_P}{R_0}) = 0[/itex]

    After simplifications,

    [itex]-g_0h + \frac{\omega^2R^2_0\sin^2\, \theta}{2} + \omega^2R_0h\sin^2\, \theta + g_0h_P = 0[/itex]

    For [itex]\theta = pi/2[/itex], we get : [itex]-g_0h_E + \frac{\omega^2R^2_0}{2} + \omega^2R_0h_E + g_0h_P = 0[/itex]

    Except that I have a [itex] h_E [/itex] with a product, not possible to isolate [itex] \Delta h [/itex] ... Where is my mistake?

    Thank you.
     
  2. jcsd
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