Maximum gravity

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[Rudolf]: Given the gravitational shielding and other effects (based on a Le Sage type model) massive objects should start reaching a level where gravity does not increase based on the 'mass' of the object - assumption. Now, I was wondering, have anyone tried to do calculations based on available data (or assumptions) what and how big this limit is?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

From p. 119 of <i>Pushing Gravity</i>, it is 5x10^18 kg/m^2.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>If more mass is added to the body, would the gravity increase? My quess would be yes but more like linearly or less than the normal gravitational equation would predict.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

No, not unless the surface area of the body increased too. -|Tom|-

Thanks for the response,

But surely if mass is added to an object then the surface area should increase as well - given that enough material is added of course. The material would be compressed given the high force of gravity in those conditions but nevertheless it should contribute a little, IMO. The change, if any would probably so small that it could be neglected.

Then there is the issue of its affects on light, either coming of the object (some might think of it as a black hole) or photons passing close to it.

I did not do the math - guilty I'm afraid, but would the maximum force you described have a great enough effect on photons to (1) suck then in - aka 'black hole' or (2) just red shift them? I suspect the latter. I believe it is then called a michaelson (name might be wrong) star and not a 'black hole'. Given these assumptions it should then be possible to detect objects like these if you know where and what to look for.

Rudolf

A question, maybe a dumb one but makes sense to me:
If gravity was a field that moves with the Earth, a person dropped from an airplane in any hemisphere should experience the same terminal momentum correct? OTOH if gravity was separate from the Earth and rather caused by transfer of momentum by gravitons, wouldn't there be a difference in the terminal momentum between a freefaller falling toward the motion of the Earth from one falling away from it?

I might not be an expert on the subject but compared to the speed of 'C gravitons' falling in any direction would not cause any significant change.
However, As I have it there is more to the story. The local 'elysium' field moves with the earth making any such measurement impossible.

Let me try to give my reasoning. (as a layman)
The reason is to do with the fact that 'elysium' is more dense closer to a massive object like the earth. Additionally as the earth move through the elysium it will cause a kind of a wake (like a boat in water). On average, incomming gravitons from the leading edge of the earth will find it more difficult to get 'through' meaning less will get through - even so the difference would be very small. The trailing edge will be less dense allowing more gravitons to get through. The result is simply that they balance out. Thus, no difference in the rate which the earth would pull you closer. The full explanation might be much more complex than this.

That is probably why the Michaelson & Morgan type experiments failed in detecting any ether.

I stand to be corrected. Lets get the 'experts' to comment on this.

Rudolf

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[123...]: if gravity was separate from the Earth and rather caused by transfer of momentum by gravitons, wouldn't there be a difference in the terminal momentum between a freefaller falling toward the motion of the Earth from one falling away from it?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

Yes, such a theoretical asymmetry must exist, and is described by the "graviton drag" equations in Slabinski's paper in <i>Pushing Gravity</i>. The absence of any observed asymmetry or drag sets a constraint on graviton parameters (mass, speed, and cross-sectional area for both absorption and scattering). The known constraints are summarized in a table at the end of Slabinski's article. None of the constraints contradict one another or anything known. -|Tom|-

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[Rudolf]: would the maximum force you described have a great enough effect on photons to (1) suck then in - aka 'black hole' or (2) just red shift them?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

The latter. Remember, light-bending and redshift are caused by refraction, which is determined by the density of elysium. The limiting mass per unit area (note that this is a limiting pressure) is the pressure of gravitons if all of them come from one direction and none from the opposite direction. That maximum would produce a maximum compaction and density of elysium, which in turn produces a maximum refraction. This would be seen as a very high redshift of emerging light.

Keep in mind that redshift is not produced directly by the force of gravity (by gravitons), but rather by refraction in the elysium, whose density is determined by the force of gravity. So light does not have to fight gravity to get away from a huge, dense mass. Rather, it must work its way through a very dense medium.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>I believe it is then called a michaelson (name might be wrong) star and not a 'black hole'.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

That would be "Mitchell star".

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Given these assumptions it should then be possible to detect objects like these if you know where and what to look for.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

We would look for compact objects of very high redshift that have properties suggesting they are relatively nearby. Many such exist, and are called "quasars". -|Tom|-