Pushing Gravity Question

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17 years 3 months ago #19615 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 KoenigMKII</i>
<br />The question is, what happens to the lead soccer ball?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Newton's law gives an unambiguous answer: the lead soccer ball oscillates from one pole to the other and back repeatedly (assuming no air drag). Are you thinking there is some reason to doubt Newton's law?

Since you ask this in a PG context, note that for any two particles in the universe, the forces applied and the resulting motions are identical, whether that force originates inside each body and pulls on the other, or whether it originates outside each body and pushes the two bodies toward each other. Hence, pushing and pulling gravity have no observable differences insofar as Newton's law is concerned. -|Tom|-

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17 years 3 months ago #19616 by KoenigMKII
Replied by KoenigMKII on topic Reply from Neil Laverty
Tom, what about the graviton flux that goes straight through the bored out cylinder and hits the "underside" of the soccer ball in my "cored apple" Earth scenario?

Surely the 3D geometry of the situation will weaken the "graviton" shadow in the "cored apple" Earth scenario as compared to the real Earth comparison??

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17 years 3 months ago #18008 by Larry Burford

Tom may not be able to respond immediately. Meanwhile, here is my take on an answer to your question.

<b>[KoenigMKII] “Surely the 3D geometry of the situation will weaken the "graviton" shadow in the "cored apple" Earth scenario as compared to the real Earth comparison??”</b>

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Yes, it will. Actually it will weaken the graviton shadow everywhere, but slightly more so along the polar axis. Note that this also includes the portions of the shadow that are exterior to Earth.

Since the removed material is less symmetric than Earth, the details of the changes depend on where exactly you do a comparison. But the amount of material removed is small (in a relative sense) so the differences are small. The lead ball still orbits (oscillates ), but the orbital period is a little longer.

The larger the hole diameter, the larger the differences. If the hole was 12,000 km in diameter all that would be left is a small ring of material near the equator, about 1 or 2 thousand km wide and having a maximum thickness of about 250 km , dropping to zero as you move away from the equator. Releasing the test mass from the original location of the North Pole would still result in an oscillation, but it would take a lot longer to complete one orbit because the mass driving it is so much less.

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17 years 3 months ago #19650 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 KoenigMKII</i>
<br />what about the graviton flux that goes straight through the bored out cylinder and hits the "underside" of the soccer ball in my "cored apple" Earth scenario? Surely the 3D geometry of the situation will weaken the "graviton" shadow in the "cored apple" Earth scenario as compared to the real Earth comparison??<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The entire Earth is almost perfectly transparent to gravitons. Only roughly one graviton in every 100 million hits something and is absorbed. So there is only the tiniest, virtually negligible difference between a cored Earth and a solid Earth.

To express this differently, if we replaced the lead soccer ball with a single neutrino, which is another particle able to pass through the "solid" Earth without hitting anything, the motion of the neutrino and the soccer ball are visuially indistinguishable. -|Tom|-

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