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 rousejohnny</i>
<br />Will a stationary atomic clock and a fast spinning atomic clock keep the same time?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That would depend on construction. There is no essential reason why spin should affect a clock. But to the extent that the spin affects speeds in the cesium/rubidium atomic transitions, then that rotational speed will affect the clock's output too.

For example, an atomic clock on Earth's equator shares Earth's spin in a way that is just like translational speed through the local gravity field. By contrast, an atomic clock at the north pole also shares Earth's spin, but not in a way that simulates translational motion. -|Tom|-

[Jim] "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."

It has been done. You will probably find one or more homework problems covering this in most second year physics text books.

For the simplest case (spherical planet with uniform density) the drop in gravitational acceleration with depth as you descend below the surface is linear. IOW, full surface gravity at r = R (at the surface), half of surface gravity at r = R/2, 10% of surface gravity at r = R/10, and zero at r = 0 (at the center).

This would appear to be a "velocity retardation" compared to an object falling in a constant g field.

LB

LB, The estimate of the rate of acceleration is easy to make but how much is the period retarded due to this happening? I'm sure you will agree the period would be as per Kepler's 3rd law if the orbit was calculated with the mass focused at a center point. But with this example the mass is not at a central point. The motion would be a round orbit if angular momentum was factored in and the period would be the same whatever the amount of angular momentum.

perhaps i'm missing or overlooked something here but where is it that the diameter of Jim's hole has been defined? no matter its location!!

i find this detail significant.

Hi North. The hole is through a sphere and is big enough to allow an object to pass through without friction. This is not something that can exist in the real world-it is just a model of ideal properties.

Jim

good , lets go from here

since the ability of the gravitons to affect mass, is by the density of a mass, in the end , since from what i understand , gravitons have less effect in the center of this sphere of mass than at its surface , then what would happen if this hole was say , a 0.25 miles across and that a ton of lead was dropped from 10 miles above this hole. does the density of the lead(Pb) have the density to make the gravitons actions non-significant,or for that matter , gravity in any form? is the density of lead(Pb)enough to not allow any gravitational action upon it's self because of its density?

or does the density of the mass , matter? if not why not?

i think that the consentration or density of mass is important.

since it it is easier to stop a 50lb balloon(air spreads the weight) than a 50lb sphere of lead! no