Gravitons and Push Gravity question.

<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 />How do you get the increase in velocity that is indicated by the Kepler law if the acceleration is reduced to zero at the center of mass?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">There is no velocity in the law you refer to. The correct form is n^2 a^3 = GM, where G = gravitational constant, M = central mass, a = semi-major axis, and n = angular velocity. You probably saw a simplified form, valid only for circular orbits, such as v^2 r = GM. But a tube through the Earth is not even approximately a circular orbit. -|Tom|-

A falling object will obay the R^3/T^2 law and that defines an orbit when angular momentum is factored in. So, the model would need to assume the mass is centered at the center of mass which is not the case here. Mass is assembled as you say in the shell model when the view is internal. As you say the acceleration does go to zero when it is viewed internally and increases when viewed externally. Therefore it seems to me the Kepler law wouldn't work with this model. So, I wonder what would be the time for an object to make the round trip in this model?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
<br /><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by north</i>
<br />i just can not picture a mass slowing down and then bobbing at the earths center, for me it is one of those things i have to see with my own eyes.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Your words here are rather different from what I explained will happen. This "bobbing" is from surface to surface through the entire shaft, not something that happens "at Earth's center" (your words). And the body speeds up while it is falling, and only starts to slow down again while it is rising up the other side of the shaft. How is that "slowing down" any different than the slowing down of a ball thrown upward from the surface?

I'm starting to suspect that your skepticism exists because we failed to communicate. If I can explain the picture correctly, you will see that it is the most natural and obvious way for the body to act, not something in any way odd or unexpected.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">my thinking is that more the mass the more momentum therefore harder to slow down. i have always thought that Galileo's experiment was not high enough to prove that no matter the mass they fall at the same rate. for instance i have a hard time thinking that an object with the weight of a feather but not aerodynamicly disadvantaged and no atmospheric turbulence, would drop from 50,000 ft at the same rate as an plane. therefore the more mass the faster it would fall depending on the height, masses of objects to be compared and turbulance of air.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Okay, you have the same intuition that everyone had before Galileo. But his experiment was plenty high for the weight difference involved. And the feather experiment you wanted to see has been done from a height of four feet by astronaut Scott of the Moon. You can find video on the internet of him simultaneously dropping a hammer and a feather in the 1/6 gravity environment of the Moon, and see with your own eyes that they fall at the same rate.

More directly to the point, we no longer depend on Galileo or lab experiments for that conclusion. Modern experiments are called "Eotvos experiments", and now show that different masses fall continuously at trhe same rate even from orbital heights, with a precision of better than 15 significant digits. Moreover, our orbit calculations would be drastically wrong if Jupiter and a small asteroid fell toward the Sun at different rates. But our highest precision observations show that all bodies of whatever mass fall toward any source of gravity at <i>exactly</i> the same rate. We don't even need to know the mass of the orbiting body (for example, a spacecraft) to determine its orbit.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">what i mean here is that the 10 tons of water would be harder to slow down than say an object of the weight of a feather and therefore would have enough momentum to counter the effects of the force of gravity on the other side of this hole. and therefore continue on through slowing down but not coming to a stop and falling back.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Part of the beauty of the Le Sage model of "pushing gravity" is that we can understand intuitively why bodies of any mass are affected the same by gravity. It is because gravitons are so small that they have access to each matter ingredient (like an atom, but much smaller) in the body. Because each matter ingredient is affected (pushed) by gravitons by an exactly equal amount, it just doesn't matter how many of them are put together to form the body.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">for me in the end i would have to see this "bobbing" at the center of the earth with my own eyes, you could do all the math you want but i would still have to see it physically happen.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That "insight" comes with understanding. I recommend again the vastly improved understanding of gravitation available from the unique perspectives you will find in Meta Science and in "pushing gravity". Being able to see the how and why with the mind's eye can be more satisfying than seeing it happen with one's own eyes in cases when the latter seems inexplicable. -|Tom|-
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

interseting, it seems i may have to adjust my thinking!

by the way i have still not received your CD.

<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 />what would be the time for an object to make the round trip in this model?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I'd estimate a few hours. -|Tom|-

The law of orbiting bodies would apply if the object had enough angular momentum to orbit at a radius equal to the height at the start of the fall through the sphere. And the period of that orbit would be the same as the time it took the falling object to return to the starting point if the mass was centered so the acceleration increased all the way to the center. But, in this case the object is slowed in a way not expected in the law of gravity as it is applied at this time. So, it would be a good thing if a way could be found to calculate how much the object velocity is retarded due to the fact the mass is not centered.

Tom,

Will a stationary atomic clock and a fast spinning atomic clock keep the same time?