Re: A new computation of G from the Cavendish experiment

On Sep 15, 2:16 pm, pioneer1 <1pione...@xxxxxxxxx> wrote:

x = theta d << a
.... The change in a is relevant to this experiment.

Let's clear this up first. There is no change in a. On my page 11, a
is defined as the distance from the center of the big ball to the
center of the box holding the pendulum. This approximation lets x
represent theta directly without trig functions (aka the small angle
approximation), and it lets us approximate the gravitational field's x-
dependence by a straight line tangent to the real field at x=0 (good
for small x)

GMmd/(a-theta d)^2

Yes, multiplied by two, this is an accurate estimate of the torque on
the wire due to gravity. I approximate this by A + B*x, where
x=theta*d, A is the torque at x=0, and B is the torque's derivative
w.r.t. x at x=0 This is a first order Taylor expansion.

If we remove theta d the force term no longer acts on the arm:

GMmd/a^2

does not act on the pendulum arm because it is independent of theta.

No. Remember, this is half the torque due to gravity. This
contribution to not having a theta dependence simply means it is
constant. It's still acting to shift the equilibrium theta. When the
big balls are rotated to the opposite side, the *sign* of the torque
changes, causing the equilibrium theta to shift in the other
direction. The goal of the analysis is to determine that shift
accurately. Note, however, as explained, I do not make such a crude
approximation. The Taylor expansion accounts for its dependence on
theta, but only to first order. This is not absolutely necessary, but
it improves the accuracy of the analysis to better than one percent.


.

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