Gravitational acceleration

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18 years 9 months ago #14703 by Cindy
Replied by Cindy on topic Reply from
Hi Uncle Tom,

I thank you for your giving me two velocity equations:

1. v = [(1-2GM/rc^2)/(1+2GM/rc^2)^3] sqrt(2GM/r)
2. v = (1-2GM/rc^2) sqrt(2GM/r)

<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>
Naturally, that is merely the appearance when coordinate time is used. In proper time, things remain Newtonian-|Tom|-
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Do you mean that:
In case I stand on a surface of a neutron star to measure velocity of a neutron (which came from infinity ), I should use the Newtonian formula v = sqrt(2GM/r) ?


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18 years 9 months ago #17310 by jrich
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Tom,

I wasn't actually thinking of anything as large as a spaceship, more like protons and neutrons. The electron drag explanation wouldn't seem to apply to neutrons, but if elysium limits a neutron's speed then elysium must exert some form of drag on the particle sufficient to prevent FTL speeds. Even a skydiver is capable of breaking the sound barrier from very high altitudes if his drag coefficient is low enough either because he assumes an aerodymanic shape or because the air density is low. If a freefalling neutron cannot attain FTL speeds, then a neutron's drag coefficient must be too high. This would either be due to an inefficient particle geometry or insufficiently low elysium density or both. It seems to me that the possibility exists that the elysium may be sufficiently dense everywhere so as to prevent FTL speeds of anything larger than some multiple of the size of an elysium particle. The problem seems more similar to that of a submarine attempting to exceed the speed of sound in water. On the other hand one would expect an elysium of that density to exert a drag on planets and stars which doesn't seem to be the case. Have you been able to calculate any limits on elysium particle size and medium density given these types of considerations?

JR

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18 years 9 months ago #14704 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 Cindy</i>
<br />Do you mean that: in case I stand on a surface of a neutron star to measure velocity of a neutron (which came from infinity ), I should use the Newtonian formula v = sqrt(2GM/r)?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Right. On a neutron star, your own clocks are slowed greatly, but light and test bodies are slowed greatly too. So it appears to you as if nothing has changed. However, to outside observers, everything has slowed. -|Tom|-

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18 years 9 months ago #14707 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 jrich</i>
<br />The electron drag explanation wouldn't seem to apply to neutrons, but if elysium limits a neutron's speed then elysium must exert some form of drag on the particle sufficient to prevent FTL speeds.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Free neutrons are too unstable and short-lived to conduct relevant experiments. For our purposes here, we should probably think of a neutron as a combination proton-electron bound in a nucleus with other protons, so that the same reasoning about electric "drag" forces still applies.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Even a skydiver is capable of breaking the sound barrier from very high altitudes if his drag coefficient is low enough either because he assumes an aerodymanic shape or because the air density is low.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Of course. He is being propelled by a force (gravity) that propagates much faster than the speed of sound.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">It seems to me that the possibility exists that the elysium may be sufficiently dense everywhere so as to prevent FTL speeds of anything larger than some multiple of the size of an elysium particle.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">You seem to be thinking here of classical drag caused by friction in a dense medium. However, the "drag" (in quotes) that I described was caused by negative electric charges attracting positive electric charges. The effect is therefore independent of the body size and shape, and therefore not at all like frictional drag.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Have you been able to calculate any limits on elysium particle size and medium density given these types of considerations?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Yes. Slabinski's chapter in <i>Pushing Gravity</i> (also available on our "Gravity" CD) derived the drag formula and uses it to help set constraints. Some of these relate to elysium, although Slabinksi was thinking only of gravitons in his chapter. -|Tom|-

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18 years 9 months ago #14708 by jrich
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<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>
You seem to be thinking here of classical drag caused by friction in a dense medium. However, the "drag" (in quotes) that I described was caused by negative electric charges attracting positive electric charges. The effect is therefore independent of the body size and shape, and therefore not at all like frictional drag.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
True, except MM is a mechanistic theory so all forces, including those associated with charge, result from collisions of particles at some scale tranferring kinetic energy. So, it seems to me that the transfer of kinetic energy from neutrons, protons or whatever to the elysium medium can properly be called <i>friction</i>. And since it is acting in opposition to the movement of the object through the medium it can properly be called <i>drag</i>.

JR

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18 years 9 months ago #14710 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 jrich</i>
<br />it seems to me that the transfer of kinetic energy from neutrons, protons or whatever to the elysium medium can properly be called <i>friction</i>. And since it is acting in opposition to the movement of the object through the medium it can properly be called <i>drag</i>.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I'd have to agree that the words fit. However, my point was that we must not carry the analogy with friction too far or we would be led to conclude, as you seemed to do, that larger masses will experience more of this "drag". It doesn't work that way, which is why drag on planets is still undetectably small. Moreover, in principle we apparently can build containers not subject to this kind of drag. -|Tom|-

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