Gravitons and Push Gravity question.

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19 years 9 months ago #12481 by Jim
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Say the center of mass is half way between two equal masses-do we say the there are two mass centers or three or one? It is clear a center of mass is empty if there is one center of mass in this example. It is in the middle of the two masses and no mass is at that point. This is always true and so why not say there is no mass at a mass center?

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19 years 9 months ago #12482 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />Say the center of mass is half way between two equal masses-do we say the there are two mass centers or three or one?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That's your choice. Centers of mass are for computational convenience. Nature doesn't know anything about them.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">It is clear a center of mass is empty if there is one center of mass in this example. It is in the middle of the two masses and no mass is at that point. This is always true and so why not say there is no mass at a mass center?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">What if the two equal masses are in contact? What if they are unequal, such as Earth and Moon? What if you have a single sphere that is hollow? Or not?

Sometimes there is mass at the center of mass and sometimes there is not. But it doesn't matter. A center of mass is a mathematical device that serves human convenience by simplifying equations. It is not something in nature or physical reality. -|Tom|-

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19 years 9 months ago #12103 by Jim
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It doesn't matter if there is mass or not mass at a point that is a focus of force? Why would that be? Wouldn't there be a different effect if mass was there than the effect would be if no mass was there?

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19 years 9 months ago #12104 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />It doesn't matter if there is mass or not mass at a point that is a focus of force?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The whole point is that no force comes from the barycenter or the center of mass (except a tiny one if there happens to be mass right at that spot). For example, no force comes from the solar system barycenter when it is outside the Sun.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Why would that be?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The concept of a center of mass is just a mathematical device to simplify equations. Consider the Earth's force on a nearby satellite. The real gravitational forces acting on the satellite come from every individual atom in the Earth, whether in the core, the mantle, or the crust. But Newton showed that, instead of having to add up all those countless individual forces, we could simplify the force <i>equation</i> by pretending that the mass is all concentrated at the center of mass, and that usually works pretty well.

For example, suppose the satellite is over the equator at 90 degrees longitude. Now examine the force on the satellite from an atom on the equator at 0 degrees longitude. Part of that force pulls the satellite downward, and part of it pulls the satellite sideways. Now examine the force on the satellite from an atom on the equator at 180 degrees longitude. It also has two components, downward and sideways. But its sideways force is exactly equal and opposite to the sideways force of the first atom, so those two force components cancel and become zero net force. Meanwhile. the two components straight down add together. So the net the the two forces is the same as a single force straight down, which is toward the center of mass. For every point in the Earth there exists a counterpoint that allows the same thing to be true -- the net force is toward the body center.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Wouldn't there be a different effect if mass was there than the effect would be if no mass was there?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">No, no difference at all. The real forces are atom to atom, and our simplified formulas still work pretty well whether there is mass at the center of mass or not. Think again of the hollow spherical shell. There is still a counterpoint for every point in the shell, making all forces operate toward the center of the shell in the balance, even though there is no mass located at that center.

For the solar system barycenter, that is the spot the net force from the Sun and all its planets would appear to come from if we were outside the solar system. But it is just a mathematical device because real forces are operating from many different distances and directions. But the outside test body doesn't feel all the individual forces. It just feels the net force, the sum of all the others acting. That simplification by nature is what allows us to make a very useful model that simplifies our equations for describing nature. -|Tom|-

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19 years 9 months ago #12327 by Jim
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This is a very good lesson in gravity effects. What I am attempting to find out is how a body reacts inside a many point system. You are describing a single point system that has worked and does work very well. Anyway, if a model is made with two equal masses at some distance apart(say two solar masses) and a third small mass(say zero for example) is introduced at some distance from the two large mass structures. The exact location where the third body entered would determine the path or orbit it would follow. Every entry location would produce a different orbit. Is this correct? I'm trying to find a simple way to calculate how an object reacts inside the mass rather than how it reacts outside if that makes any sense.

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19 years 9 months ago #12106 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />if a model is made with two equal masses at some distance apart(say two solar masses) and a third small mass (say zero for example) is introduced at some distance from the two large mass structures. The exact location where the third body entered would determine the path or orbit it would follow. Every entry location would produce a different orbit. Is this correct?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Yes. This is known as "the 3-body problem" in celestial mechanics. There are only a few shortcuts known to exist. Mostly, solving for the orbit requires some nasty-looking differential equations that can only be solved numerically.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I'm trying to find a simple way to calculate how an object reacts inside the mass rather than how it reacts outside if that makes any sense.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The connection with the previous is difficult to see. For most purposes, the force at a point inside a single mass is determined by dividing the mass into two parts: (1) concentric spherical shells outside the point that contribute zero net force; and (2) a spherical mass inside the point that contributes an inverse square force acting from the center. -|Tom|-

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