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05-11-2003, 11:18 AM | #81 |
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my posts
I seem to have been right. Care to address the problem I submitted.
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05-12-2003, 06:22 AM | #82 | |
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05-12-2003, 06:42 AM | #83 |
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Pi
I assume that was a sarcastic comment.
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05-12-2003, 03:27 PM | #84 | |
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The single most important thing that I have "learned" in my life is that I know nothing, but I can't be sure about that, because there is always the possiblity that I actually know something. If you know some physics you should already know how primitive much (if not all) of it is. It still uses point masses, and dozens of approximate constants. There isn't even a good formula for the approximate air-resistance of a object at any velocity. (Not that I know of, at least.) Ok, I know that these systems and formulae are the best things that humans have, but that does not mean that they are anywhere near accurate or true. |
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05-13-2003, 07:29 AM | #85 |
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points
What is your solution to the problem?
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05-13-2003, 09:57 AM | #86 | |
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Therefore, I assume you would be heavier on the pole than the equator (even if the earth were not spinning). crc |
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05-13-2003, 10:09 AM | #87 |
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solution
Where is your proof. I have just had a look at other ideas for measuments involving gaussening solutions apart from electromagntic fields.
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05-13-2003, 11:55 AM | #88 | |
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But the real issue is what do you need to know. Somehow, using our "primitive" understanding of gravity and the motions of the planets, we have been able to fly several tiny space probes from our planet to other planets. We can even choose the landing locations on Mars and rendesvous with comets! Using GPS, we can pinpoint your location on the planet to a couple of meters. I don't know what your threshold for the burden of proof is for you to "know" something, but it seems like you've set it to an unreasonable value. |
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05-13-2003, 12:27 PM | #89 | |
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Re: solution
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crc |
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05-13-2003, 03:58 PM | #90 | |
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Let's say the Earth is a perfect sphere of radius R. For 0 < r < R/2, the mass density is p1. For R/2 < r < R, the mass density is p2. The mass of the core would be M1 = 4/3 pi (R/2)³ p1 = 1/6 pi R³ p1 The mass of the mantle (or crust, or whatever you want to call it) would be M2 = 4/3 pi (R³ - (R/2)³) p2 = 7/6 pi R³ p2 The mass of the entire planet would be M = M1 + M2 = pi/6 R³ (p1 + 7 p2) For r > R, gravity would be g = GM/r² For R/2 < r < R, gravity would be g = G(M1 - (1/7)M2)/r² + (8/(7R³)) GM2 r For r < R/2, gravity would be g = (8/R³)GM1 Note that all three of these equations are continuous at the boundaries r = R and r = R/2, as they must be. So, outside you get the traditional 1/r² dependance one would expect--given only the total mass we know the gravity. Inside the mantle, we have two competing components: a 1/r² dependence along with an r dependence. Inside the core we have only an r dependence. So it seems the interesting stuff happens in the mantle. Try graphing y = A/x² + Bx for arbitrary constants A and B and you'll see that y goes to infinity as x -> 0 and as x -> infinity, with a minimum somewhere inbetween. Depending on the radius at which this minimum occurs (which in itself is a function of R, M1, and M2), we see that gravity within the mantle can conform to the following three scenarios: 1) As you dig down into the mantle towards the core, gravity is always decreasing. 2) As you dig down into the mantle, gravity is initially decreasing but then at some critical depth it begins increasing 3) As you dig down into the mantle, gravity is always increasing |
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