Gravitational acceleration

<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>
It's always best to talk to the source -- especially since you did not ask any specific question here, so it's hard to know how to help. -|Tom|-
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Oh! My specific question was that the velocity equation (27) make any sense to you ?

v = c.sqrt[1-exp(-2GM/rc^2)]

( I have never talked with Mr. Hatch yet. I just also find out his web. But I thought you knew this equation and this was the reason I asked you )

<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 />Oh! My specific question was that the velocity equation (27) make any sense to you?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Um, yes, it does. In celestian mechanics, 2GM/r is the escape velocity, so this equation is very similar to the one in the standard Lorentz transformation.

Propagation velocities slow in a denser medium, which is what is happening physically. Either going closer to a source mass or moving faster relative to it exposes the test body to more (or denser) elysium per second, which makes it take longer for electrons in atoms to complete their circuits. This is like a form of drag on the test body, which ultimately limits its speed as long as only electromagnetic-type forces are involved. -|Tom|-

Hi Uncle Tom,

I thank for your comment about the velocity equation. I have other two small questions about the equation v = c.sqrt[1-exp(-2GM/rc^2)]

1. Velocity v in the equation is escape velocity ?
2. Because the equation make sense, so the "Exponential scale factor s = exp(-GM/rc^2)" also make sense ?

<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 />Velocity v in the equation is escape velocity?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">No, v is presumably the velocity of a photon or a test body. (But its your equation, not mine.) 2GM/r is equal to escape velocity for a test body at a distance r from a source mass.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Because the equation make sense, so the "Exponential scale factor s = exp(-GM/rc^2)" also make sense?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">As r tends toward infinity (away from the source mass), the exponential tends toward unity (which should mean no effect on velocity). As r tends toward zero, the exponential tends toward zero, which should drive the speed of light toward zero. That's the part that makes sense.

But because your equation does not work that way, I'd say your equation must be incorrect. -|Tom|-

<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>
No, v is presumably the velocity of a photon or a test body. (But its your equation, not mine.) 2GM/r is equal to escape velocity for a test body at a distance r from a source mass.

-|Tom|-
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Oh! I thought you knew the equation! It is the equation (27) which is in the article of Mr. Hatch: v = c.sqrt[1-exp(-2GM/rc^2)]

If you don't like to hit the link of this short article (which I think is much clearer), I will copy and paste it into here.

Through his explanation about the equation, the equation seem to be an escape velocity to me ?!!

<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 />Oh! I thought you knew the equation! It is the equation (27) which is in the article of Mr. Hatch: v = c.sqrt[1-exp(-2GM/rc^2)]<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That is the equation for the velocity of a test particle (the second possibility I mentioned) falling in from infinity, not that of a photon. So the test body's speed at every point is also its escape velocity. Then as I described, the exponential drives its velocity to zero at infinity and to c at r = 0. This is just Hatch's way to limit the speed of the test particle to c, instead of speed going to infinity as Newtonian gravity would imply. -|Tom|-