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Why Symmetry of Tides on Earth (influenced by moon, sun)

  1. Dec 15, 2006 #1

    Ouabache

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    I have been trying to understand how our ocean tides work and read about the effect of our moon's and sun's gravitation.
    I have learned there are diurnal, semi-durnal and mixed semi-diurnal tidal patterns

    When the moon is directly overhead equatorially (near side), I've seen illustrations showing how the oceans bulge outward on both the sides of the earth, that facing the moon and the opposite side of the earth (180degr), such that both sides experience a high tide simultaneously. So the coasts facing the atlantic and pacific oceans, experience 2 high and 2 low tides per day, a semi-diurnal or mixed patterns. I also realize that the sun and moon create 'spring' and 'neap' tides.

    My question is, when the moon is adjacent to one side, why is the bulge "equally large" on the far side of the earth? What physical forces cause this to happen?
    (If I had not seen the illustrations, my intuition leads me to think the ocean would bulge to a much greater degree on the side facing the moon).

    Once I understand how this happens, it will make more sense how the sun causes a "spring" tide when the earth, moon and sun line up during "both" a full and new moon.
     
    Last edited: Dec 15, 2006
  2. jcsd
  3. Dec 15, 2006 #2
    You're actually quite right that the bulge should be smaller on the far side. It's just that that is a very small effect compared to the effect that creates tidal bulges in the first place. The relevant effect is the difference in the moon's gravitational pull between the center of the earth and a point at the edge. To a leading order approximation, this difference is given by [tex]\frac{2GMr}{R^3}[/tex], where M is the moon's mass, r is Earth's radius and R is the distance from the center of the moon to the center of the Earth. This lowest order formula holds for both tidal bulges. However, if we want a more precise approximation, we need the next term, which is [tex]\pm \frac{3GMr^2}{R^4}[/tex]. This term is positive for the closer edge and negative for the farther edge. But, you'll notice that this term has an extra factor of r/R, which means that this effect has only about 2% of the strength of the leading term.
     
  4. Dec 21, 2006 #3

    Ouabache

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    Thanks for your reply..
    Now you have me curious, how may I derive these and higher order terms?

    Plugging in the numbers, I found the values and orders of magnitude, for both your terms. Dimensional analysis indicates they are both accelerations [itex]m/s^2[/itex]

    After a little reading, I gain some feel your 1st term. It close to this term for tidal acceleration.

    You have an additional r term in the numerator, which appears to be, from comparing acceleration at the center of the earth, to a point at its edge. But for the initial term, wouldn't there actually be two situations along the moon-earth axis yielding +/- terms with [itex] (r - r_o) = r [/itex] on the side nearest the moon and [itex](r_o - r) = -r [/itex] on the far side?

    Your 2nd term seems like a differentiation of the 1st term but not quite.
     
    Last edited: Dec 21, 2006
  5. Dec 22, 2006 #4
    Tides arise from the difference between the gravitational acceleration (I actually prefer the term gravitational field, but that's just me) at two different points. We always take one of the points to be the Earth's center of mass. At this point, the gravitational acceleration is [tex]\frac{GM}{R^2}[/tex], where again M is the moon's mass and R is the distance from Earth's COM to the moon's. We also consider the acceleration at two points along the Earth/moon axis, each a distance r from the center. The acceleration at the point closer to the moon is [tex]\frac{GM}{(R-r)^2}[/tex] and at the farther point it's [tex]\frac{GM}{(R+r)^2}[/tex].

    From these we find that the near-side tidal acceleration is [tex]a_{near} = \frac{GM}{(R-r)^2} - \frac{GM}{R^2}[/tex]. This simplifies to [tex]a_{near} = \frac{GM}{R^2} \left (\frac{1}{\left (1-\frac{r}{R} \right )^2} - 1 \right )[/tex]. By a similar argument, the far-side acceleration is [tex]a_{far} = \frac{GM}{R^2} \left (1 - \frac{1}{\left (1+\frac{r}{R} \right )^2} \right )[/tex].

    To generate the terms I posted above (and as many more as you'd like), assume r is much smaller than R and expand the terms involving their ratio in a binomial series.
     
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