Hanak, have you EVER studied any physics? Basic mechanics? Newtonian dynamics? If so, why is your analysis of ocean tides so ridiculously wrong?
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On 2002-08-23 20:51, Richard J. Hanak wrote:
The ocean tides on Earth are caused by the gravitational effects of the Moon, Sun, and the planets. The Moon, averaging about 230,000 miles between its surface and the surface of the Earth has the strongest effect on the oceans. The Sun, at about 93 million miles, has a lesser tidal effect. The other planets have much smaller effects.
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This is fine, but let's evaluate the statement. The earth is a non-inertial frame of reference, so we need to evaluate the acceleration (call it "a") starting at the center of the Earth. Taking into account the Earth's gravity and the moon's gravity, we get the following equation (r being the radius of the Earth, M_e being the mass of the Earth, G being the gravitational constant, M_m being the mass of the moon,
e_r being the unit vector in the direction from the surface of the Earth to the center of the earth,
e_R being the unit vector in the direction of a line drawn between the Earth's surface and the center of the moon,
e_d being the unit vector in the direction of the line connecting the center of the moon to the center of the Earth, R being the distance from the center of the moon to the surface of Earth, and D being the distance from the moon to the center of the Earth):
a=[(G*M_e*
e_r)/r^2]-
G*M_m*{(
e_R/R^2)-(
e_D/D^2)}
The first term which is due to the Earth and the second partis due to the difference bewteen the moon's gravitational pull at the center of the Earth and on the Earth's surface.
We then can evaluate the tida force on a given particle with mass "m" at the Earth's surface which will be simply the second term of the above equation multiplied by "m" (according to Newton's Second Law).
Are you following?
Good, now we ask the question, what is the force along the axis of the tidal bulge at high tide? To do that we use the fact that R^2 is basically (D+r)^2 and combine the two terms that are associated with the two different unit vectors. This means that the F_x force in the direction we're interested in,
F_x = -(G*m*M_m/D^2)*{(1/(1+r/D)^2)-1}
or
F_x = -(G*m*M_m/D^3)[1-2*(r/D)+3(r/D)^2-...-1]
which taking the first term (since r/D is small) to
F_x = +2G*m*M_m*r/D^3
Then we wish to calculate it for the direction at low tide (that's going to be 90 degrees away, the y-axis), and that analysis is much simpler, namely, if you're following me...
F_y= -(G*m*M_m)*{(1/D^2)(r/D)}=-G*m*M_m*r/D^3
so for any point on the Earth's surface you have a linear combination of the vectors...
(let G*m*M_m = N)
F_x=2N*r*cos(theta)/D^3
F_y=-N*r*sin(theta)/D^3
Which will give you the correct heights for the tides. Notice it is a 1/D^3 force. This is true because r/D is small. As you approach Earth r/D is no longer small, so you can't take this approximation, but have to remain in the regime for high tide where
F_x=-(N/D^2)*(1/(1+r/D)^2-1)
the height, then, will be determined by the definition of work, and since we can take any path we choose, we will go the path moving the water from low tide to high tide... or h being the height, g being the acceleartion due to gravity...
h=[Integral[r to 0](F_y*dy)+Integral[0 to r](F_x*dx)]/(g*m)
So you see, Hanak, your analysis of the tides is dead wrong.
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If the moon were at half the distance, 115,000 miles, a place that might otherwise experience a 10-foot tide would experience a 40-foot tide. The moon at 57,500 miles would create a 160-foot tide, and at 28,750 miles a 640-foot tide, at 14,375 miles a 2,560-foot tide. Keep going with the series and by the time the Moon’s surface is 457 miles from Earth’s surface, the tide will be 496 miles high. Not only would the Moon get a bath, but the Moon’s gravitational force would make lots of water leave the Earth and go to the Moon.
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So you see why this is wrong, right?
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The Earth is larger and heavier than the moon. The bucket, being very small and insignificant compared to the Earth, would not have enough gravitational force to keep any of its water. Perhaps you can now understand that if the Earth orbited even a couple of miles away from (let alone right next to) a stationary bucket containing water, the Earth would suck that bucket dry to the last drop.
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and the tides have nothing to do with "Earth sucking a bucket dry. Think of it this way: If I hold a bucket of water upsidedown, what happens to the water? It is NOT subject to tidal forces. It is subject to the gravitational acceleration that everything else is subject to.
(Thanks to Marion and Thorton's
Classical Dynamics of Particles and Systems for helping me to present the origin of the tides more elegantly than I would have been able to do.)