Constants or Proportionality?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by xenocidere</i>
<br />What does the constant G units represent? Is it just a proportionality?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

See p. 126 of <i>Pushing Gravity</i> for a formula for G in terms of more fundamental parameters, such as graviton number density, mass, speed, and absorption/scattering coefficients. This shows where it units come from. -|Tom|-

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by xenocidere</i>
<br />What does the constant G units represent? Is it just a proportionality?

any opinions[?]:

[?]E=mc^2
c^2=E/m where E=W=Fr (r stands for radius or distance or ‘s’)
therefore c^2= Fr/m
therefore c^2= c^2----when forces interact the forces hold true the constant c^2 in order to remain stable

therefore Fr/m= Fr/m

Fr/m=mar/m ----where F=ma
Fr/mm=ar/m ----where a=F/m

therefore Fr/mm=Fr/mm

Fr/mm=Fvt/mm ----v=r/t where r=vt
Fr/mm=Frt/tmm ----where v=r/t

therefore Frt/mm=Frt/mm

Frt/mm=Frr/vmm ----where t=r/v
Frtv/mm=Frr/mm ----where v=r/t

therefore Frr/mm=Frr/mm

where F=(Frr/mm)mm/r^2 ----(Frr/mm) is constant for example: F=Kq1q2/r^2-electromagnetic force

The constant G(=Frr/mm) is a way of expressing how the speed c is kept true mathematically when forces interact just the way c^2 is a mathematical expression expressing that c should always remain true in order for the matter to remain stable.


xeno
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Pretty neat, you derived Newton's Universal Law for gravitation
from Einstein's matter energy equivalency. I think you gave a perfect
answer to your own question really. There's a problem though-
since you are
equating the Force from the mass energy equivalence to the Force
of gravitational attraction- you end up taking all the energy
available from a mass and assuming it can be used as a force
of attraction between other masses- the value you would get for
the gravitational attraction would consequently be too large.

1234567890
whenever one compares(calculates) gravity force within objects, one calculates using the whole mass of both objects in order to find precise answer. simple example is calculating the gravity force between the sun and earth. we compare the whole masses of both objects not partial mass. therefore deriving the constant G from E=mc2 should be alright since the equation includes the whole mass of an object.

xeno

I plugged in the numbers and got a much larger value for G using your derivation
than the standard value. The final equation I got from your derivation was
F1rr/m1m2 = F2rr/m2m2. So, if F2rr/m2m2 is G and we used c^2= F2r/m2 , letting
m2=1 and r =1, you get F2 = c^2 (kg/m) and G would equal to c^2 (m/kg) and the
dimensions are not equal to G as defined by Newton's gravitational equation.

Maybe I made a mistake somewhere but I think just from looking at the energy aspect,
the forces are different for the gravitational force than that in E = mc^2. When
you use E = W = Fr = mc^2, that's the force that can be gotten if all the mass was
turned into energy. In F of gravity, not only is the force distributed differently
(it falls off as r^2), all the mass of the system has not been converted into energy
to be used as a force of attraction (since the matter still exists as matter and
not pure energy).

Please see:

www.space-talk.com/ForumE/showthread.php3?threadid=1797

Thanks

xeno

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by xenocidere</i>
<br />Please see:

www.space-talk.com/ForumE/showthread.php3?threadid=1797

Thanks

xeno
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Before I make a comment, let's go over the derivation:

Starting from E = mc^2 and F = ma. To derive, it's critical
to label each variable so we can keep track of its physical meaning
throughout the algebra. So, I used 1 and 2 to distinguish
between the masses, forces, accelerations and energies. Those
letters without a number attached are assumed to be constant.
Also keep in mind that the equations on each line is an algebraic consequence of the previous line.

Hence:

E1 = m1*c^2 E2 = m2* c^2
E1 = W1 = F1*r E2 = W2 = F2*r
m1*c^2 = E1 = F1*r
m2*c^2 = E2 = F2*r

hence,

F1*r/m1 = F2*r/m2

since F2 = m2a2
we obtain

F1*r/m1 = m2*a2*r/m2

dividing both sides by m2,
we get
F1*r/(m1*m2) = a2*r/m2

Finally, substituting a2= F2/m2, we get

F1*r/(m1*m2) = F2*r/(m2*m2)... we can stop here
because the rest of the derivation leads to F1 rr/m1m2 = F2rr/m2m2,
so that G = F2rr/m2m2.

I don't see how you obtained F1rr/m1m2 = F2rr/m1m2 unless
you assumed m1 = m2.

AS for your question about the physical significance of G,
I hesitate to speculate until I can understand the physical
meaning of multiplying a mass by another mass. Or maybe this
is my answer to your question- once you figure out the physical
meaning of the multiplication between masses, you get an idea
of what G is.

I think you equated G to the graviton and went further to describe
how it ends up producing the gravitational attractive force.
I can't see the connection you are making and am unclear with
your description of the mechanism. I think the idea is innovative,
just as your derivation was but the details are a little foggy
for me...