User talk:82.15.53.173

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Hi Someguy,thanks for the effort with the gravity problem but the whole thing was missed. the point I am puzzled about is the mass of earth that is now above the subject and very much closer to it than the centre of the earth.This mass is related by the inverse square and could be more significant.

This was actually covered, but I'm sorry we didn't make it clear enough. Finding the gravitational field caused by a sphere of mass is a somewhat difficult mathematical problem to solve. Physicists have found it easiest to think not in terms of was mass is behind you and what mass is in front, but rather to calculate the field of spherical shells of mass. Thus, the Earth is modelled as very many concentric shells. To calculate the gravitational field at any point, a physicist would sum the fields of each individual shell. So what happens under this way of thinking is that as you drill into the Earth, you are entering the outer shells, and getting closer to all those you haven't entered. The gravitational field of a shell of mass is shown and derived within our shell theorem article. It works out that a spherical shell of mass produces within itself no gravitational field. Thus, as you go toward the center of the Earth, you have two counteracting sources of gravitational change: The mass of shells outside of you is growing larger, and thus the mass of the earth with a net pull on you is decreasing; and simultaneously you're getting closer to the remaining shells, and thus their pull is getting stronger. Calculus can easily derive the net effect of these counteracting influences, and the result is that once you start going beneath the surface of the earth, gravity weakens linearly until reaching zero at the very center. This is shown in shell theorem#Solid spheres. I hope this answers your question. Someguy1221 (talk) 04:23, 16 December 2007 (UTC)

[edit] Atoms

I'm sorry this took a while, but I struggled to find an explanation that would work for someone who isn't taking a quantum mechanics corse. In a single atom, you have two ranges of size: the electron and the nucleus. The nucleus is far smaller than the orbits of the electrons about it, so if you're looking at the electrons, the nucleus is but a tiny and immobile spec that we can look at later. So then, what of the electrons? In quantum mechanics, particles are typically described as a wave function. This wave function gives the probability of finding the particle at a given point in space. For an atom, the wave function takes the form of a very pretty atomic orbital. However, the wave function is not what you would literally see. Electrons don't emit light on their own, so you have to shine light on them to see them. I'll now give you the explanation typically given in popular literature or general chemistry texts. When you do anything to a wave that can reveal the position of the particle it describes, the wave changes; it "collapses." The simple reason is that, if you've just located the particle, then it doesn't have a probability of being found anywhere else. So let's say you shine a great big spotlight on that atom you're holding in your hand. The atomic orbitals will all collapse, and you'll see in your hand a field of tiny specs (one spec for each electron). Unfortunately, you will only see them for a moment. The energy you just pumped them with by shinging that light on them will cause them to fly off faster than you can track them, and now you don't have an atom anymore! (just a nucleus.) If you tried to look at the constituent particles of the nucleus, the exact same thing would happen. Someguy1221 (talk) 17:52, 5 January 2008 (UTC)