[math-fun] (Tides and) Public education (more a soliloquy than a rant)
I wish someone would explain tides to me. It makes sense that the side of the earth facing the moon would have a water bulge, but I've never really understood why the opposite side of the earth has one, too. The antipodal bulges are usually depicted as about the same size. Is this close to accurate? If so, can someone tell me why? * * * As for public education -- I'm a liberal myself. Anyone who argues that compulsory public education is the answer to the U.S.'s education problems -- when public education is failing so miserably right now -- is out of their mind. I would argue, however, that the minuscule money that's allocated for teachers' salaries (attracting only those who can't get a more lucrative job elsewhere) and educational materials, and the massively unpleasant discipline problems that public school teachers have to often deal with, are significant reasons that public education is failing so egregiously. I really don't know what the root of the problem is. I went to a lower-middle-class elementary school with 40+ students per class and mostly dedicated teachers, and most kids there got a reasonable education from that school. Later I went to a middle-middle-class high school and most kids got a reasonable education that way, too (class of '64). My feeling is that my experience was typical for that era; what's changed in the last 40 years to vastly lower the quality of public education, I really can't say. I *do* agree that home schooling makes more and more sense these days -- to avoid the colossally nasty behavior of the bullies and the cliquies, the mind-numbing boredom that smart kids (and probably most others) have to endure in the classroom, and the poorly trained teachers who barely know their subject matter, not to mention how to teach. (Note the huge resistance teachers' unions have mounted against proposals that all public school teachers pass a minimum competency exam in many states.) I would love to run a school for gifted kids, giving them all the things I wish I'd had in school but never came close to until going to college. (My model is the most memorable book I read as a child, "Children of the Atom" by Wilmar Shiras. Anyone else read that?) --Dan
Consider the earth's orbit relative to the moon as determined by the moon's gravity. Regard the earth as sufficiently approximately spherical. The moon's gravity varies between parts of the earth that are closer or farther from the moon. The orbit of the earth as a whole is calculated correctly (as Newton proved) as if the entire mass of the earth were concentrated at the center. A point on the earth's surface facing the moon is closer to the moon than the center and so feels additional force. This force tends to create a bulge facing the moon. The size of the bulge is an equilibrium between the increased attraction of the moon there and the gravity of the earth pulling on the bulge. On the side of the earth away from the moon the moon's gravity is less and centrifugal acceleration creates a bulge. According to the mathematics the two bulges are of the same size. Because the moon always keeps the same face to the earth, the bulges on the near and far sides of the moon stay in place. Since the earth rotates in 24 hours the bulges on the earth move.
Because the Earth is free falling in the Moon's gravitational field, Earth does not experience a force due to the gravitational acceleration at its center. But the Moon's force is not constant over the Earth's surface, and this is responsible for the tides. The gravitational potential at Earth due to the Moon, expressed in coordinates centered at Earth's center, and expanded to second order, is Vm = (G Mm / L^3)(y^2/2 - x^2), L = distance to Moon, x points toward Moon, in which the constant and linear terms have been removed, the constant because it is unimportant, the linear because we're in free fall. The gravitational potential due to Earth at height h above its surface is Ve = - G Me / (Re + h), which becomes, if the value at h = 0 is removed, Ve = (G Me / Re^2) h. The ocean surface is an equipotential of V = Ve + Vm. In the absence of the Moon, the geoid h = 0 corresponds to V = 0. In the presence of the Moon the V = 0 geoid is perturbed to h = (Mm / Me) (Re^2 / L^3) (x^2 - y^2/2) = (Mm / Me) (Re^4 / L^3) (cos(theta)^2 - (1/2)sin(theta)^2). The bulge at theta = 0, on the side facing the Moon, equals the bulge at theta = pi, on the side opposite the Moon, while at theta = pi/2, the ocean drops by one half the amount. Re = 6.40e6 m L = 3.85e8 m Me = 5.97e24 kg Mm = 7.35e22 kg The predicted height of the Moon's contribution to the tidal bulge is 0.36 m or 1.2 ft., and the high-to-low tide difference is 1.8 ft. The Sun's effect has been ignored, as well as the interaction between the rotating tidal bulge and the bathymetric terrain. __________________________________ Do you Yahoo!? The New Yahoo! Search - Faster. Easier. Bingo. http://search.yahoo.com
Imagine yourself weightless inside a spherical space station, orbiting a planet. Yhere are no windows. you have four marbles. Find out where the planet is that you are orbiting. (stolen from some old SciFi book). hint: think of orbiting marbles at slightly different altitudes. W. ----- Original Message ----- From: "Eugene Salamin" <gene_salamin@yahoo.com> To: <math-fun@mailman.xmission.com> Sent: Sunday, May 11, 2003 3:47 AM Subject: [math-fun] Re: Tides
Because the Earth is free falling in the Moon's gravitational field, Earth does not experience a force due to the gravitational acceleration at its center. But the Moon's force is not constant over the Earth's surface, and this is responsible for the tides.
The gravitational potential at Earth due to the Moon, expressed in coordinates centered at Earth's center, and expanded to second order, is
Vm = (G Mm / L^3)(y^2/2 - x^2), L = distance to Moon, x points toward Moon,
in which the constant and linear terms have been removed, the constant because it is unimportant, the linear because we're in free fall. The gravitational potential due to Earth at height h above its surface is
Ve = - G Me / (Re + h),
which becomes, if the value at h = 0 is removed,
Ve = (G Me / Re^2) h.
The ocean surface is an equipotential of V = Ve + Vm. In the absence of the Moon, the geoid h = 0 corresponds to V = 0. In the presence of the Moon the V = 0 geoid is perturbed to
h = (Mm / Me) (Re^2 / L^3) (x^2 - y^2/2)
= (Mm / Me) (Re^4 / L^3) (cos(theta)^2 - (1/2)sin(theta)^2).
The bulge at theta = 0, on the side facing the Moon, equals the bulge at theta = pi, on the side opposite the Moon, while at theta = pi/2, the ocean drops by one half the amount.
Re = 6.40e6 m L = 3.85e8 m Me = 5.97e24 kg Mm = 7.35e22 kg
The predicted height of the Moon's contribution to the tidal bulge is 0.36 m or 1.2 ft., and the high-to-low tide difference is 1.8 ft. The Sun's effect has been ignored, as well as the interaction between the rotating tidal bulge and the bathymetric terrain.
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At 03:20 AM 5/10/2003, asimovd@aol.com wrote:
I wish someone would explain tides to me. It makes sense that the side of the earth facing the moon would have a water bulge, but I've never really understood why the opposite side of the earth has one, too.
I saw the calculations for this about 25 years ago, but I don't remember them. The astro book "Horizons" gives this description "There is also a bulge on the side away from the Moon, which develops because the moon pulls more strongly on the Earth's center than the far side. Thus the moon pulls the Earth away from the oceans, which flow into a bulge on the far side."
The antipodal bulges are usually depicted as about the same size. Is this close to accurate? If so, can someone tell me why?
I think so, to be in equilibrium.
Arthur Robinson faced the problem of home schooling his 6 children after his wife died. He and the older children developed a curriculum, and are selling it on CD-ROM. It was designed to place the responsibility for learning mainly on the student, with a minimal involvement of the parents. The material covers up to high school, and includes mathematics up to calculus. The English and history reading material is for the most part based on books from the old days, before public education became corrupted, and includes the 1911 Britannica. http://www.robinsoncurriculum.com/ The one most important thing we can do for the educational system is to encourage "vouchers". Public education is too far gone to recover by itself, and we need to turn over school operation to the competition of private enterprise, while continuing to provide public financing for the benefit of children who cannot afford full private school tuition. Public schools could continue to exist for the benefit of dumb kids; that is their niche already. Also, vote NO on any tax proposals for public schools. __________________________________ Do you Yahoo!? The New Yahoo! Search - Faster. Easier. Bingo. http://search.yahoo.com
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asimovd＠aol.com -
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wouter meeussen