On 12/21/2018 7:29 AM, John Clark wrote:
On Thu, Dec 20, 2018 at 11:57 PM Brent Meeker <[email protected]
<mailto:[email protected]>> wrote:
>> The mass of the Earth played no part in Cavendish's
determination of G because he was measuring gravitational
attraction in a direction that was parallel to the Earth's
surface.
/> But in comparing it to the clock precision you have to consider
that the clock is measuring the change over a cm of a very much
greater potential, /
Yes, that makes it even more valuable.
> /while Cavendish is measuring a much smaller change in a much
smaller potential.
/
No, the change Cavendish measured was much larger not smaller.
>> the new clock can detect the difference in time dilation
between 1g and 1.000000003g, so I'm sure it could detect the
time dilation caused by a 348 pound mass a foot or so away.
/> Could it? /
Yes indeed.
> /The gravitational time dilation factor is tau = sqrt[1 -
2GM/rc^2]. So a clock that can detect a one cm change in height at
the Earth's surface is in fact not accurate enough to to detect
the difference between being adjacent to a 348lbm 18" ball and
being arbitrarily far away from it./
Cavendish could not detect the gravitational field arbitrarily far
away from an object
I clearly wrote "...to detect /*the difference*/ between being adjacent
to a 348lbm 18" ball and being arbitrarily far away from it." But
Cavendish did more than just detect the difference, he measured it. He
claims to have measured the deflection of the smaller masses within 0.25mm.
nor can anybody else because that would take infinite precision, but
lets calculate what he did do. The gravitational time dilation factor
is sqrt[1 - 2GM/rc^2] so if G is 6.67 × 10^-11 m3 kg^-1 s^-2 and c is
1*10^8 meters/sec then the gravitational time dilation factor caused
by being 9 inches (.0002 kilometers) away from the center of a 248
pound (112 kilogram) lead ball is, if I did my arithmetic correctly,
1-square root of [1- 112 *(13.3 *10^-11) / .0002 *(3*10^8)^2)] =
1-(2.6*10^-6) = .999997.
?? Not even close. I'm not sure why you converted 9" to kilometers
instead meters, but:
2(6.67e-11)112/0.2(9e16) = 8.3e-25
So using sqrt(1-x) ~ 1 - x/2 The time dilation factor is 1-(4.1e-25)
Waaay closer to 1.0 than the time dilation due to raising a clock 0.01m
above the Earth's surface, which is given by
[2(6.67e-11)5.97e24/9e16][1/6.36e6 - 1/(6.36e6 + 0.01)] ~
1.39e-9[1 - 1 + 0.01/6.36e6] = 2.19e-18
which gives a time dilation 1-(1.09e-18).
So Cavendish was able to detect a gravitational field that would make
a time dilation of about 3 parts in a million, but this new clock
could detect the change in a gravitational field that would make a
time dilation of about 3 parts in a billion, a thousand fold improvement.
Which is not only wrong arithmetically, it is comparing the time
dilation factor which depends on the potential (goes as 1/r) to the
acceleration (which goes as 1/r^2).
Brent
John K Clark
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